Exercises in Computer Systems Analysis (Lecture Notes in Computer Science)

Lecture Notes in Computer Science Edited by G. Goos and J. Hartmanis

35 W. Everling

Exercises in

Computer SystemsAnalysis Corrected Reprint of the First Edition

Springer-Verlag Berlin.Heidelberg. New York 1975

Editorial Board" P. Brinch Hansen • D. Gries C. Moler • G. SeegmLiller • N. Wirth Author Prof. Dr, W. Everling institut fiJr Angewandte Mathematik und Informatik der Universit~it Bonn Wegelerstral]e 10 53 Bonn/BRD

Formerly published 1972 as Lecture Notes in Economies and Mathematical Systems, VoL 65 ISBN 3-540-05795-1 1. Auflage Springer-Verlag Berlin Heidelberg New York ISBN 0-387-05795-1 1st edition Springer-Verlag New York Heidelberg Berlin

Library 0£ Congress Cataloging in Publication Data

Everling, Wolfgang. Exercises in computer systems analysis. (Lecture notes in computer science ; 55) Bibliography: p. Includes index. 1. Electronic data processing. 2. System analysis. 3. Queving theo?y. I. Title. II. ties. T57.5.E84 1975 001.6'4 75-29431

So-

AMS Subject Classifications (1970): 6 8 A 0 5 CR Subject Classifications (1974): 4,0, 4.6, 4.9

ISBN 3-540-07401-5 Korrigierter Nachdruck der 1. Auflage Sprioger-Verlag Berlin Heidelberg New York ISBN 0-387-07401-5 Corrected Reprint of the 1 st edition Springer-Verlag New York Heidelberg Berlin This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, specifically those of translation, reprinting, re-use of illustrations, broadcasting, reproduction by photocopying machine or similar means, and storage in data banks. Under § 54 of the German Copyright Law where copies are made for other than private use, a fee is payable to the publisher, the amount of the fee to be determined by agreement with the publisher. © by Springer-Verlag BerLin • Heidelberg 1972, 1975 Printed in Germany Qffsetdruck: Julius Beltz, Hemsbach/Bergstr.

INTRODUCTION

Planning for, and installation of, a Data Processing System is a complex process. At a certain stage, decisions are required concerning the type and number of components which will form the Computer System. A certain variety of components is offered for choice, such as terminals for data entry and display, long distance communication lines, data storage units, data processing units and their interconnections. Also, operating procedures have to be choserL Let this set of decisions be called -

Sy st e m s

D e s i g n.

Its major criteria are

whether the system can adequately handle the expected workload, whether it can be expected to have adequate reliability and availability,

-

whether its cost is optimal and in an appropriate relation to the service produced.

An essential part of the design process is the

Systems

Analy

sis,

- the subject of

this course. It can be understood as the set of considerations and computations which establish whether a supposed system satisfies the design criteria. The results of Systems Analysis serve within the design process either by immediate backward computation, or in some form of triM-and-error procedure until the criteria are satisfied. The methods of Systems Analysis are largely determined by the random nature of many design d a t a : Volume and transmission times of 'long-distance data', record lengths, and locations in direct access storage, execution times of programs depending on conditions determined by the data,

- all these are variable and can at best be predicted statistically. Hence

probability theory plays a key role in Systems Analysis. Asynchronous operation, with the consequences of possible interference and queuing, is an essential technique in modern computer systems, - either enforced by the fact that data are presented to the system at unpredictable times, or as a result of hardware and program structures. Hence the considerable share which queuing theory takes within this course. The methods discussed in this course are briefly presented and referenced in the text. More space, however, is devoted to exercises and their solutions, - stressing the application of the methods. A Case Study has been chosen as a starting point. It provides numerical values for

IV the exercises ; it gives a motivation for the particular selection of methods for this course ; and, by the order in which the analysis could proceed for this case study, it also determined the order in which the methods are presented. The course can be studied with two different aims : The reader who is mainly interested in the applications of Systems Analysis will concentrate his attention on the exercises related to the Case Study (which are quoted by topic and number in the table of contents ) and will find some convenience in the numerical tables collected as appendix C med computations, he will find formulae and procedures in appendices

. Also, for programB and

E.

A reader with this aim should be familiar with the basic notions of calculus, probabilities

and statistics. Also, some familiarity with the available components and programming concepts is assumed. The other view would be at the theoretical reasoning that leads to the formulae and tables. Here, the reader may gain a deeper understanding of their applicability and a basis for further analysis of problems not discussed in this course. The mathematical notions referenced in this context are surveyed in appendix D

The notes are based on lectures given by the author at the

IBM

European Systems Research

Institute in Geneva, Switzerland. My sincerest thanks go to the Institute, its direction, faculty and students, for their encouragement and critical discussions.

Geneva December 1971

Wolfgang Everling

TABLE

OF

CONTENTS

The Case Study

Section 1 : Communications network design

6

Chapter 1 : Some statistical analysis of design data

7

Some estimates from random samples (8). Worst ease estimates of worldoad (10). A probability m o d e l for the arrival of requests (16). Some properties of the POISSON process (18). Chapter 2 : Line service times and utilization

24

Line service times (25). Expected line utilization (26). The effect of errors on service times and utilization (30). A message length for m i n i m a l overhead (82). Chapter 3 : First concepts and relations of Queuing Theory

35

Relations of first order for the single-server queue (36). A sufficient condition for convergence (38). Simulation of a single-server queue (39). Confidence intervals for the averages of queuing variables (41). Distributions of queuing variables (43). Chapter 4 : A first result on the single-server queue

46

The expected remaining service t i m e (47). The POLLACZEK-KHINTCHINE formula (47). An approximation for the distribution of wait times (50). Further results on the remaining service times (82). Chapter 6 : A general result on the single-server queue

56

Chapter 6 : T e r m i n a l systems with polling

62

C y c l i c polling of terminals (62). The probability assumptions (64). The expected cycle t i m e (65). An approximation for the variance of cycle times (66). The expected wait t i m e (68). An improved approximation for the conditional c y c l e t i m e (70). The arrival pattern for conversational use of terminals (73). Generaltzed terminal load and polling lists (74). A rough approximation for T w (77). Chapter 7 : Some results on queues with priority disciplines General notions of priority disciplines (78). The probability assumptions (79). Non-preemptive discipline (80). The p r e e m p t - r e s u m e discipline (85).

78

Table of contents, cont'd. Chapter

8 : Some applications of Markov processes

88

Queuing processes with Markov property (89). Markov queuing processes with constant intensities (91). The limit probabilities (96). The wait time distribution (99), Section 2 : Computer center analysis Chapter

9 : C o r e buffer requirements

102 104

The probability assumptions (105). Expectation and moments of the buffer allocation (106), Probabilities to exhaust the buffer pool (107). Chapter 10 : Service times of direct-access storage devices

112

Probability assumptions (lla). Expectation and moments of the seek time distribution (114). Alternating access to an index and data (117). Chapter 11 : Q u e u i n g for data accesses

118

Queues with finite population (119). The probability assumptions (120). Relations between utilization and queuing variables (121). The relation between utilization and return rate (129), The throughput ratio for given utilization (131). Chapter 12 : F i n a l remarks on the computer center analysis

133

Waiting for the CPU as server (135). Waiting for a task as server (t39). Message response times (142). Conclusions (143). Section 3 : Further results of Queuing Theory Chapter 13 : Imbedded Markov chains of some queuing processes

145 146

The transition probabilities (149). Limit probabilities of the imbedded chain(150) Examples for the rate of convergence (153), Lag correlations in the chain (157), Chapter 14 : L i m i t utilization and wait time distributions

159

Limit probabilities at a random time (159). Some evidence about wait time distributions (160). Wait time distributions (163). Appendices : A - Remarks concerning the notation (167). B - Collected formulae of Queuing Theory (168). C - Tables for some formulae of Queuing Theory (170). D - Mathematical reference material (176). E - Computing procedures (182).

T a b l e of contents,

cont'd.

Application-oriented exercises, Subject

Exercise nr.

Distribution of message lengths

1 , 2

Worst case arrival rates

3

Distribution of line service times

8 , 9

28

Line service times affected by errors

13

, 14

33

Times waited for a line, first result

19

, 21

48,50

Times waited for a line, with cyclic polling

27

, 30

Times waited for a line, with output priority

37

Times waited for a terminal

47

Core buffer requirements for teleprocessing

49,

50 , 51 , 52

108

Seek times of direct access storage devices

54

55

114-115

Times waited for a data channel

56

57

122-124

Times waited for a direct access storage device

58

59

124

Utilization of the Central processing unit

60

61 , 62

134

Times waited for the CPU

63

64

137

Times waited for a program

65

66

140

CPU

page II

, 4

, 6

12,20

, 31

67,70,73 81

, 48

96

REFERENCES Books : Bodewig, E.

Matrix Calculus

Brunk,

An Introduction to M a t h e m a t i c a l Statistics,

H.D~

Cox, D . R . ,

Lewis, P.A.

North Holland Publishing Co.

The statistical analysis of series of events,

Martin, J.

Design of Real T i m e Computer Systems,

Parzen, Saaty, Takacz,

E. Th. L. L.

Widder, D.V.

Blaisdell Publ.

Co,

Methuen

1959 1963 1968

Prentice Hall

1967

Stochastic Processes,

Holden Day Inc.

1962

Elements of Queuing Theory

McGraw - Hill

1961

Introduction to the Theory of Queues,

Oxford Univ. Press

1962

The Laplace Transform

Princeton Univ. Press

1946

(several other equivalent references might replace

Bodewig, Brunk,

Parzen,

and Widder).

R E F E R E N C E S , cont'd.

Other publications : Chang, W.

Single server queuing processes in computing systems, IBM Systems Journal no. 1

Cooper, R.B.

and

Murray, G.

1970

Queues served in c y c l i c order

The Bet1 System Technical Journal no. 3 Kendall, D.G.

1969

Stochastic processes occurring in the theory of queues and their analysis by the method of the imbedded Markov chain Annals of Math. Statist.

Khintchine, A.Y.

24

1953

M a t h e m a t i c a l theory of a stationary queue (in Russian) Mat. Sbornik

no. 4

1932

Konheim, A.L.

Service in a loop system

(to appear in

Journal of ACM)

Leibowitz, A.

An approximate method for treating a cIass of multiqueue problems IBM Journal of Research and Development

Lindley, D.V.

1961

The theory of queues with a single server Proc. Cambridge Philos. Soc.

Mack, C.

1970

and

Murphy, T . ,

Webb, N.L.

48

1952

The efficiency of N machines unidirection-

ally patrolled by one operative when walking t i m e and repair times are constark "

2

Jour. Royal Statist. Soc. B19 no. 1

1957

A further paper adressing variable repair times continues the previous reference in the same issue.

Mises, R.v.

and

Pollaczek, F.

Geiringer, H.

Zeitsehr. f. angew. Math. und Mechanik

Uber eine Aufgabe der Wahrscheinlichkeitsrechnung

I-II

Mathematisehe Zeitschrift Spitzer, F.

1930

The Wiener-Hopf equation whose kernel is a probability density Duke M a t h e m a t i c a l Journal

Syski, R.

1967

Determination of waiting times in the simple delay system A T E Journal

Takacz, L.

1929

2

13

1957

On certain sojourn t i m e problems in the theory of stochastic processes Acta Math. Acad. Sci. Hungar.

1957

THE

The of

DUMMY

CASE

STUDY

Company, with headquarters at Geneva, is running five offices in the cities

FRA, KOP, LON, MIL, PAR; an extension to at least three more, viz. AMS, MAD, VIE,

is planned for the next year. The company's activities require fast response to customer calls and therefore frequent and fast reference to a common data base. The currently used technique of communicating with headquarter personnel through the public telex and telephone starts to create unbearable waittimes and errors. The"president, Mr. L. O. Okahead,

eagerly grasped the idea of a Communications Based

Computer System to which terminals in all offices are connected. He established a study group in order to define the functions of the system, and to specify the expected amount of data to be transmitted and processed. As result of their work, the study group presented the following descriptions and statistics :

Data base : T w o types of data have to be referenced, viz.

CUSTOMER

and

ITEM

descriptions. Cur-

rently there exist written files which contain 47,000

CUSTOMER

records consisting of

20

characters identifying key and vary-

ing data length ( 150 characters m i n i m u m , 380 characters average ) 18,000

ITEM

records consisting of

10

characters identifying key and

racters data. This file has a yearIy growth of

180

cha-

8 %.

Terminal operation in the offices : References to the data base are caused by customers arriving at, or calling, the office. These are first routed to the next free terminal operator. While answering the customer, the operator keys in and transmits some data base references as the customer's request requires. Also, he receives some response from the computer center. There are four types of customer's requests, with different requirements for data base references and response time. They are listed in the following table.

Types of customer's requests CR1

ITEM inquiry

asking for information about a particular item, terminal input : identification of response via terminal : part of

CR2

ORDER

ITEM and question asked, ITEM description.

ordering some operations with several terminal input : identification of

ITEMs,

CUSTOMER, several

ITEMs

and operations, response via terminal : one short acknowledgement. CR3

NEW customer

presenting a new customer's description, terminal input : a

CUSTOMER record of

150 characters,

response via terminal : one short acknowledgement. CR4

D E L E T Ecustomer

causing deletion of a

CUSTOMER record,

terminal input : identification of

CUSTOMER,

response via terminal : one short acknowledgement.

Message formats : Some experience with the current telex operation suggests the formats of 24 characters for

identification of

CUSTOMER,

36 characters for

identification of

ITEM and question or operation,

48 - 72 characters for 24 characters for

the response with part of an

ITEM description,

one short acknowledgement.

Duration of terminal use : The time it takes to enter a message at a terminal is determined by the message format and by the average key-in rate of

3

characters per second. An experiment which simulated the

terminal keyboard by a typewriter showed that the total time for a customer's request is 3

times the key-in time for inquiries,

1.6 times the key-in time for the other three request types. In this experiment, no delay by telecommunications and processing was considered.

3 Workload of the offices : During the month of June 19xx which had 20 working days, the number of customer's requests was observed at each office, viz. GE

25,000

FRA

32,800

KOP

15,800

LON

31,700

MIL

21,600

PAR

28,200

A year earlier,

requests

requests

all these values were about

have the same load as

MIL , while

Some more detail was observed at least active day had a load of hour of a day,

20 %

MAD

GE

1490

1055

ITEMS

470

NEW

210

DELETE

are estimated equivalent to

6%

4 - 6

in

24 %

7 - 10

in

50 %

Ii

15

in

15

16 - 20

in although

KOP.

The most active and the

requests, respectively. During the peak The distribution of the request

in an order was

in

tions in proportion

VIE

is estimated to

orders

1 -3

distributions,

and

AMS

inquiries

7,500

These

less. The location

during the same month. and

16,820

Finally, the number of

9 %

of the day's workload was handled.

types was

-

:

of the orders

%

5 %

observed

of the orders in

GE

only,

are supposed to hold for all other l o c a -

with their request numbers.

Message processing in the computer center : The different types of customer's requests require different processing by the computer center, which can be described roughly by five types of processing, PRO

PR4 :

PRO

applies to atl incoming requests and consists in logging the request as entered

and in a first anatysis of request type and format. PR1

locates and reads a record from the

ITEM

file as identified by the request

then sends part of this record as a response to the terminal. PR2

locates and reads a record from the

CUSTOMER

file as identified by the

request, then locates and reads several records from the by the sequel of the request. Each its place. Finally, the

ITEM

CUSTOMER

ITEM

file as specified

record is updated and written back into

record is also updated and stored, and a short

acknowledgement is sent to the terminal. PR3

writes a new record into the

CUSTOMER

file and sends a short acknowled-

g e m e n t to the terminal. PR4

locates and reads a record from the

CUSTOMER

file as identified by the

request, flags it for later detetion and writes it back into its place. A short a c knowledgement is sent to the terminal.

A p p l i c a t i o n - d e p e n d e n t processing times : The capacity of the computer center will be devoted partially to the control of the c o m m u nications network and the data base. The time which the central processing unit will spend on these functions depends mainly on the choice of general programs and access methods. It is therefore only the processing times dependent on the particular application which have been estimated with reference to one certain type of central processing unit. The estimates are, in milliseconds, for the types of processing PRO

30, of which the last

PR1

10 before reading, 10 overlapping the read, 20 after the read

PR2

15 before the 10 before each

10

CUSTOMER ITEM

ITEM

80 before writing back the

PR2, PR3 and PR4

read, 28 overlapping it

read, 10 overlapping it, 10 after it and

before writing back the

PR8 and PR4

may overlap the logging

record

CUSTOMER

record,

t 0 before the first data base reference 5 to finally form the short acknowledgement.

All times quoted are subject to a random variation of

+_ 40 %.

This completes the results c o l l e c t e d by the study group. As a minor item, they noted that of the currently rences. Further remaining

18,000 5,400

10,800

ITEM records a certain subset of

1,800

ITEM records are adressed by a further

ITEM records are adressed by only

observations were y e t made for the

CUSTOMER

absorbs 40 %

50 %

of ali r e f e -

of all references. The

10 % of all references. No similar

file.

After a presentation of the above descriptions and statistics by the study group, the president decided that a Systems Design should be based on these data, and set forth the further requirements that -

the offices planned to date and the growth' p r e d i c t a b l e for the next three years should be accounted for

-

a calling customer should g e t switched to an operator within at most one out of

-

10

20 seconds except for

calls

also, the t i m e from the last key stroke for a message to printing the first c h a r a c ter of a response, if any, should not exceed the preceding k e y - i n t i m e except for the same small fraction as above.

He understood that every effort would be made during the Systems Design to o p t i m i z e the cost of the system and to assure a satisfactory a v a i l a b i l i t y .

Section COMMUNICATIONS

1 NETWORK

DESIGN

As a guidance through this section the phases of Communications Network Design are stated below in the form of an operational instruction, with references to the chapters discussing the a p p l i c a b l e methods of analysis. Note that chapters 3 and 5 are of more general nature.

GET DATA

about

Locations, i . e .

Terminal use, i . e . Line use,

i.e.

their geographical situation

frequency and duration, per location

message frequency and length, per location

ESTIMATE

worst-ease frequencies, distributions of duration and length

SELECT

Terminal type, implying time characteristics, e . g .

COMPUTE

Terminal workload per location

chapter

1

transmission rates

required number of terminals per location for l i m i t e d wait t i m e

chapter

8

chapter

2

If necessary, review the selection and repeat the analysis. Else SELECT

Line type, implying time characteristics, error rates, line control procedures

COMPUTE

Line workload including error overhead, per terminal Bounds on terminal number and workload per line, for l i m i t e d wait t i m e

chapters

4, 6, 7

association of all terminals with lines, observing the above bounds and minimizing the total line cost. If necessary, review the selection and repeat the analysis. END

Methods for minimizing the total line cost are not discussed in this course. They have to solve a finite combinatorial problem, the data for which are the geographical description of the locations, the a p p l i c a b l e line cost tariff, and the restrictions from the wait t i m e l i m i t . T r i a l - a n d - e r r o r methods of stepwise changes in the network have been used successfully for programmed computation.

Chapter

1 :

Some statistical analysis of design dat a

The majority of the Systems Design Data as e x e m p l i f i e d by the Case Study, are statistics of variables, the individual values of which are unpredictable. An assumption basic to all systems analysis is that these variables can at least be described by probability laws, and hence are random variables in the sense of Probability Theory. As a consequence,

those design data are themselves random variables and can only give

estimates for the assumed probability taws. The problems arise, which are the best estimates derivable from the design data,

and

which confidence one may have in them. The systems designer has a further typical problem. He usually has to plan for the worst case or rather for a case such that a worse case will not occur with more than a specified few percent probability. This poses the problem of e s t i m a t i n g those values of certain design data which have a specified - usually high - probability of not being exceeded. Satisfying solutions to these problems exist only if certain general properties of the probability laws under study are known. A survey of the case study data shows in fact onIy a smati n u m ber of typical statistical situations, with simple general properties. The first situation, which is the simplest in its statistical consequences, generates a sequence of random variables independent of each other, viz. the type of request presented to some terminal operator, the number of ITEM

references in an order

Of a similar nature are the random variables which are a function of one or more independent random variables, viz. the duration of a terminal use or of an application processing, both functions of r e quest type and possibly the number of items. The second situation is that of a sequence of random variables which might have some correlation to the neighbours in the sequence, viz. the particular

ITEM

record to be accessed, correlated if a customer states them

in, say, alphabetic order the length of a message to be transmitted, correlated if a long message has an i n creased probability to be followed by another long one. Such situations either require a more detailed analysis than presented in the case study; or the possible correlation is neglected, reducing the situation to the first type. A third situation is the t i m e - d e p e n d e n t trend, with random deviations from it. Examples are the data file sizes, growing linearly or exponentially with time, and showing unimportant deviations from this trend the daily numbers of requests, with a similar trend, and with significant deviations which appear as a m u l t i p l i c a d v e factor to the trend value. Correlation between neighbouring deviations may exist but is neglected unless analysed closer. Thus, the deviations from the trend create another situation of the first type.

Some estimates from random samples. Observed values

v

= ( ~?1' ~2 . . . . . ~)N )

of

N

random variables

v 1, v 2

. . . . .

vN ,

all with the same probability distribution Fv(X)

=

Prob ( vn = x ),

and mutually independent, - constitute a random sample. ( For this and some of the following concepts, any textbook on Probabilities and Statistics, e.g.

Brunk1, may serve as refe-

rence. ) A random sample can be described by an empiric distribution function F~?(x) which, for each

x, equals the fraction of observed values

pression

(9 n ~ x ) is understood to have the value

9n 1

relation does not hold, then a formal expression for 1 Fg(x)

-

N

not exceeding

x. If the relation ex-

if the relation is true, and F~7(x)

0

if the

is

N 7 n = 1 (~n ~= x)

The empiric distribution gives the same information as the list

9

of observed values. It is

helpful only because it permits a general statement about estimates from random samples :

The v a l u e of

F9(x)

at any particular

x, and all m o m e n t s about constant c,

O3

E( (9 - c) k )

=

, p (x - c) k dFg(x )

;

k = 0,1,2 ....

-(30

are 'best a v a i l a b l e ' estimates for the corresponding values and m o m e n t s of the u n known distribution Hence,

Fv(X).

the e m p i r i c distribution gives the closest picture of

Fv(X) that can be derived from

the observed values. T h e closeness increases with increasing s a m p l e size

N.

the e m p i r i c m o m e n t s are of course written as the sums which in fact they

For computations, are, viz.

1

E( Especially, foi

(7

k = 1

- c) k )

=

E(¢) E(v)

t i m a t e the v a r i a n c e

V(v)

N =

~

n ~ 1 (vn

N

-

c)k

c = 0 , the ~ample average

and

estimates the m e a n

-

1

N

N

n=l

n

k = 2

and

Fv(X). For

of of

Fv(X). However,

g(v)

c = E(v),

E( (g -E(v))

2

)

would es-

is usually not a v a i l a b l e to c o m p u t e

this estimate. The sample v a r i a n c e E( (¢ - v--)2 )

has the e x p e c t a t i o n s a m p l e average

V

=

V(¢)

(1 - l/N) V(v) ,

1 N

-

N n~l

- slightly less than

(on - v)2 V(v)

because the use of the

somewhat outbalances the observed variations. Therefore, t v(v)

/(1-*/N)

is preferable as e s t i m a t e of

-

N

N - 1 n~l= (°n" v)2

V(v).

The term 'best a v a i l a b l e ' estimates is used in ~ i s context for the more precise term ' c o n sistent unbiased' estimates of e . g .

Brunk1. In fact,

the r a n d o m variables of interest in sys-

tems analysis h a v e by necessity f i n i t e values and variance, The case study data are,

as happens frequently in practice,

as required for ' c o n s i s t e n c y ' . an i n c o m p l e t e presentation of the

underlying observations. The e m p i r i c distribution or e q u i v a l e n t i n f o r m a t i o n is given only for

10 the request type, the number of items in an order, and the access probabilities of certain ITEM

subsets.

The average and a range are given for the daily request number and for the a p p l i c a t i o n - d e pendent processing times. Averages only, some of them based on not more than one or two observations, are stated for file sizes, the ratio the ratio

terminal use duration / key-in t i m e ,

and

peak hour load / daily load.

Confidence intervals for estimates. Any estimate is a random result and a c c e p t a b l e only with l i m i t e d confidence. For a general discussion of this confidence concept, the reader is referred to, say, Brunk 1. In the context of this case study, it is assumed that the study group took care to assure a sufficient confidence in their results. Only the confidence in estimates of arrival rates will be discussed l a ter in this chapter.

Worst case estimates of workload. A computer system usually requires more or more expensive components as the workload i n creases. The designer will therefore plan for the lowest assumed workload which still gives him a probability of at least

p

that it is not exceeded. If the workload has a distribution

Fv(X), the worst case value to some given Fvl(p)

=

rain

~ x

p

can be denoted as ;

Fv(X) = p it is a l s o called the lowest

100 p - t h percentile point of

Fv(X).

The inverse function

is defined by the above equation in a form adapted to the fact that

Fv(X)

F;l(p)

is a probability.

distribution. If an empirical distribution

F~(x)

estimates for the percentiles of

is given as an estimate of

N

p ~

0

is one of the observed values

observation estimates the

F~?(x) , - i t is a stepwise in-1 its percentile F~ (p) for x = ~?n'

Fv(x ). Due to the nature of

creasing function, with steps at the observed values any

Fv(X), its percentiles serve as

~n" Specifically, the m - t h largest vaiue out of

100 p - t h percentile if

11

m

z--N If the largest v a I u e

9ma x

such that

m-I

{

p

~

IN

was observed only once,

it is a p e r c e n t i l e e s t i m a t e for

all p

I 1 - _

p

•

~

1

;

N

in other words,

the probability that a further observation

observed so far, is e s t i m a t e d as

w i l l r e a c h or e x c e e d the m a x i m u m

t/N.

Note that for i n d e p e n d e n t random variabies

v, v 1 . . . . .

vn

with continuous distribution

of any shape,

Fv(X)

1 Prob(v

~

max(v 1. . . . .

vN )

)

= N+I

Then,

a m a x i m a l observed v a l u e would e s t i m a t e the p e r c e n t i l e with p = 1 - i / ( N + 1) .

Another situation frequently arises in which only the m e a n and the v a r i a n c e of known or e s t i m a t e d .

E f f i c i e n t p e r c e n t i l e e s t i m a t i o n then requires an assumption about the g e -

neral functional form of

Fv(X). Appendix

D

terial to distributions useful for this purpose. show the p a r t i c u l a r i m p o r t a n c e of

Exercises 1

contain s o m e r e f e r e n c e m a -

Examples arise in l a t e r chapters,

which w i l l

N o r m a l and G a m m a distributions.

Plot the e m p i r i c distributions of the request type and of the n u m b e r of T h e s e Case Study D a t a appear on p a g e

F r o m the two distributions referred t o

of the input message l e n g t h in characters, request.

, pages 176 ff.,

.

rences in an order. 2

Fv(X) are

3 .

in e x e r c i s e

1

A - every

CUSTOMER

B - within e a c h request, as fit c o m p l e t e l y into a buffer of

120

find the e m p i r i c distribution

and the a v e r a g e n u m b e r of messages g e n e r a t e d per

S o m e message formats are Case Study D a t a on p a g e

assumptions :

ITEM r e f e -

and

ITEM

as m a n y

2

. Consider the f o l l o w i n g

r e f e r e n c e is a s e p a r a t e message ; CUSTOMER

and

ITEM

references

characters are c o l l e c t e d into one message ;

C - all data for e a c h request are c o l l e c t e d into one m e s s a g e . Also find the e m p i r i c l e n g t h distribution for output messages,

and for B and output together.

For e a c h distribution c o m p u t e the first three m o m e n t s ( see p a g e

9

, with c = 0 ), and V(9).

12 3

Assume a trend and deviations ha(t)

=

e

ct

v(t)

for the number of requests arriving at the computer center on day

t.

The deviations

on different days are considered as independent, but all with the same distribution varying with

t . Estimate, for a date

t

which is

3

v(t)

Fv(X) not

years later than the Case Study, the

expected value and the 95-th percentile of na(t ). 4

Which worst case daily number of input and output messages results from exercise 3

combined with the cases

A - C

of exercise

2 ?

How many of these arrive in the peak

hour ?

Solutions. In the first example, a numbering of the request types, say, 1 through

4 is used.

A table is formed which lists the type values, the number of observaHons with at most this type, and this number divided by the total N. 1

16,820

0.673

2

24,320

0.973

3

24,790

0.992

4

25,000

1

This table describes the empiric distribution completely. In the second example, only the values of the empiric distribution for some ITEM numbers are reported, viz. 3

0.06

6

0.30

10

0.80

15

0.95

2O

1

Values for the other approximated by linear interpolation.

ITEM

numbers could be

13

2

The computations which lead to the empiric distributions and the average numbers

of messages per request are presented on the next page in a tabular form. They are carried out in relative frequencies per request, i.e.

all numbers of observations are divided by the

total number of requests. Thus, all results remain in a convenient range. A first step computes the frequencies with which the customer request

( C R ) types 1, 3, and 4

and the different possible item numbers within type 2 occurred. Then, for the different cases input, A - C

and

Output, the multiplicity with which each occurrence generates messages

of a certain length is noted• E.g. ces generates i.e.

108

1

message of

in case

24 + 36 + 36

characters, and finally

only the lengths

48, 72

B, a request of type

1

=

96

message of

2

with

2

messages of

characters,

36

9

item referen3 x 36,

characters. For output, occurrence of

with equal frequency is assumed as a simple approximation.

Total frequencies ( p e r request ) are found for each possible message length by collecting occurrence frequencies multiplied by the multiplicity applicable for this length. E.g. for Input case B, the length of

108

characters has the total frequency

.024 ( 1 + 1 ) + . 0 3 7 5 ( 1 + 2 + 2 + 2 ) + . 0 0 9 ( 3 + 3 + 3 + 4 + 4 ) + . 0 0 3 (4-+ 5 + 5+ 5+ 6 ) The multiplicities used here are not all shown in the table, nor are all computations. The sum of the total frequencies taken over all possible message lengths

gives the total

number of messages divided by the number of requests, in other words the average number of messages generated per request. Finally, a division of all total frequencies by their sum leads to frequencies relative to the number of messages• These values in the last column of the table determine the empiric distributions of message lengths. For the computation of moments xm

which was observed

the factor

nm / N

nm

E( 07 ~ c) k )

times contributes a summand

•

243

+

.907

" 363

Input case A +

.005

• 1503

9

, note that a value

n m (x m - c) k / N , and that

is the relative frequency of occurrence of

moment of the message length for • 088

as defined on page

x m.

Thus, e.g.

the

3 rd

is computed (with c = 0, k = 3 ) as :

61. 727

.

A further table on the overnext page gives the complete results of a programmed computation. The sample standard deviations, defined as the square root of the sample variance, are added for comparison with the averages (with which they have common dimension) .

14 T y p e CR

1

f r e q u e n c y . 672

3

4

.019

.008

.3

items

1

frequency

• 006

2

3

4

each

5

.024

6

I

each

FormatIm u l t i p l i c i t y Input,

8

17

9

• 0375 e a c h

total

case A 1

24 36

1

for e a c h request

equals the a b o v e i t e m s n u m b e r

150 a v e r a g e number of m e s s a g e s / r e q u e s t : Input,

.308

• 088

3.198

• 907

.019

• 005

3.525

case B

24 36

008

.005

761

437

006

003

ere

114

065

1

60 72

1

96

1

1

1

108

1 1

1

1

1

1

294

•

1

1

1

2

2

539

.310

019

.011

150 a v e r a g e number of m e s s a g e s / r e q u e s t : Input,

frequency relative

169

1. 741

case C • 008

24 36

1

150 24 + 3 6 m

occurs

for

m

items,

• 672

same as

.019

total

with above frequencies 1,000

Output .328

24 48

.5

• 336

72

.5

• 336

same as

total

I. 000

15 Moments of empiric message length distributions

Average

Case

2ndmoment

3rd moment

Sample variance,

standard deviation

Input, A

35.6

1347

61,727

B

72.1

6334

621,875

1142

33.8

C

125.4

38,943

16,558,246

23,226

152.4

48.2

2705

167,105

Output

Now, consider the combined messages for

82.4

9.1

382.4

Input case B

19.6

and

Output. The total frequencies

for each possible message length are found by addition, leading to for the length 24,

to

.336

other values remain as for

simply for the length 48, and to

Input

.008 + .328

=

.336

.450 for the length 72. The

B. The new sum of total values is

1. 741 + 1. 000 = 2.741.

Frequencies relative to the combined number of messages can again be found by a division, and moments can be computed from them. However, it is simpler to use the fact that each of these values i s . a linear combination of the corresponding values from

Input B, with the

weight factor

1 / 2.741.

1. 741 / 2. 741,

and from Output, with the weight factor

the frequencies are for length

The moments are

24

36

48

60

72

96

108

150

• 123

.278

.123

.002

.164

.107

.197

.007

Average I 2rid moment

13 rd m o m e n t

i

I

I

63.3

Thus

5010

:

:

455,960

The sample variance is not a linear combination of the above values. It is found from the general relation v(¢) and has the value

3 equals tions of

Let

t

=

E(¢2)

_ -v2

997,1 . Its square root, the sample standard deviation,

=

0

averaged over

v

can hence be estimated from observa-

The total daily request number, including 20

days. This estimates

ed from the Iargest of

31.6 .

denote the date of the case study observations. For this date, na(0 )

v(0). Expectation and 95-th percentile of na(0 ).

is

20

E(v).

The

daily values observed at

1490 / 1280 -fold, of the mean value. For the mean

AMS,MAD and VIE

95-th percentile GE

F j I ( . 95)

was

10410

is estimat-

as a certain multiple, viz. the

10410

this multiple is

12409 .

16 The constant a value of

c

is estimated from the observation that one year earlier,

9 %

less, i . e .

of

.91 . Then,

the factor

three years later its value will be

e

ct

• 91-3

had

= 1.33

A worst case assumption for the total daily request number is therefore 12409 • 1.33 a value which is 4

58 %

=

16467

,

higher than the observed average.

Average numbers of messages per request were found in exercise 2 . Multiplied

with the worst case request number of exercise 3 , they produce the following numbers : From :

arise messages daily :

in the peak hour :

Input, A

58,046

11,609

B

28,700

5740

C

16,467

3293

Output

16,467

3293

Input B and Output

45,164

9033

It is on this peak hour, worst case workload that later exercises will be based. It should be noted that this approach neglects another effect which possibly increases the future workload significantly, and is sometimes referenced as the 'turnpike effect' :

A good highway attracts

traffic. Similarly, an efficient Data Processing System may attract an unforeseen workload. The case study gives no statistical basis for estimating such an effect.

A probability model for the arrival of requests. The hourly number of requests is still a very summarizing description of how the requests are placed in time. It only states the number tively long t i m e interval from

0

to

Na(0, T)

of requests expected m arrive in a

rela-

T, say.

For an analysis of the expected utilization of system components,

this information is suffici-

ent. However, as the discussion in later chapters will show, for an analysis of wait times and response times,

a more detailed description is required. In the case study, where no further

information is reported, assumptions.

this detailed description must take the form of some plausible

17 A plausible probability model for the arrivals of customer requests is provided by the famous POISSON process : Let arrivals ~ [ 0

!

/

!

X | to

denote the n u m b e r of arri-

vals in a t i m e interval ~

I

na(t 0 , t l )

I

For every i n t e r v a l v t1

I

t'- t "F

to <

(t0, t l ]

t

~<

t 1.

this is a r a n -

dom v a r i a b l e with integer values

j = 0,1,.

and with probabilities na(t O, tt)

=

2 Pa, j(t0' tl)

Since in

=

Prob(

na(t 0, tl) = j

) .

(0, T) , say, one can choose an infinity of intervals, accordingly an infinity of ran-

dom variables is considered here. Now assume that the

na

in any n o n - o v e r l a p p i n g intervals are i n d e p e n d e n t r a n d o m variables;

this is a plausible assumption if the arrivals are caused by i n d e p e n d e n t customers. -

the probabilities

Pa, j

depend only on the duration

t I - t O of the i n t e r -

val considered; this is a plausible assumption for a t i m e period

(0, T) , say, over which the

customers as a whole have a constant interest to p l a c e requests,

e.g.

-

the probabilities

babilities, viz. Pa, j (t0' tl)

Pa, j

have more specifically the values of POISSON pro-

[ Ra (t l - t 0 ) ] j j ,

=

e

this is a plausible assumption if the requests during large n u m b e r

N

of customers,

for a peak hour.

- Ra ( t l - t 0 )

(to, t l )

with some constant

Ra;

are p l a c e d by several out of a

each customer h a v i n g i n d e p e n d e n t l y the same probability to

p l a c e or not to place his request. p

If each of the -

N

customers has the probability

Ra (t 1 -t 0) N

to arrive, the probabilities that e x a c t l y ,~ Pa, n(t0 ' t 1)

=

n

arrive are b i n o m i a l , viz.

N ) pn (1 - p)N-n (n

;

n = 0,1

These probabilities have the above POISSON p r o b a b i l i t i e s as l i m i t s for ed in exercise

.....

N

N~

5 , and are well approximated by these l i m i t s even for small

.

oo , as discussN.

Arrivals satisfying these three assumptions are said to form a POISSON process.(E.g.

Parzen 1)

18

S o m e properties of the POISSON processes T h r e e g e n e r a l properties of the POISSON processes w i l l find a p p l i c a t i o n in this course.

The

first of these concerns the statistical b e h a v i o u r for long intervals. POISSON probabilities with s o m e p a r a m e t e r are known (see appendix h a v e m e a n and v a r i a n c e e q u a l to this p a r a m e t e r . has POISSON probabilities with the p a r a m e t e r Na(0, T) Therefore,

the number

The n u m b e r

Ra T

D

, p a g e 178 )

na(0, T)

of arrivals

V( na(0, T) )

=

=

ha(0 , T) /

of arrivals per t i m e unit,

T

in (0, 7)

and h e n c e the e x p e c t a t i o n and v a r i a n c e

= v

to

Ra T . has the e x p e c t a -

tion E(v)

=

Na(0, T) / T

V(v)

=

V( na(0, T)

=

Ra

and the v a r i a n c e Ra /

T .

T

a b l e ' e s t i m a t e for the constant

Ra which is therefore c a t t e d the m e a n arrival rate or the Exercise

Hence,

=

This v a r i a n c e tends to zero as

intensity of the POISSON process.

increases.

) / T2

6

an observation of

v

is a 'best a v a i l -

discusses c o n f i d e n c e intervals for its e s t i m a -

tion. The second property concerns the t i m e i n i n t e r - a r r i v a l times

tervals b e t w e e n c o n s e c u t i v e arrivals, c a l l e d inter-arrival times

ta, k ;

k = 1,2 ....

These are easy to observe and lend t h e m 11

I

i

I

o

selves better to a statistical analysis than

I

T the numbers of arrivals in the infinity of possible t i m e intervals

Let

to

be the t i m e of an arrival.

greater than s o m e fixed v a l u e i.e.

The probability that the t i m e to the n e x t arrival is not

x = t1 -t o

equals the probability that

na(t0, tl)

is no_.!t 0,

it is i

Thus,

(t 0, tl) .

Pa, 0 (to' tl)

=

1

e" Ra x

the t i m e to the n e x t arrival is a r a n d o m v a r i a b l e Fa(X )

=

Prob(

ta ~

x

)

=

1

ta

with the probability distribution

-

e- R a x

,

19

an exponential distribution ( s e e also

appendix

D

, page 176 .) Furthermore, all inter-

arrival times are mutually independent since, due to the assumed" independence of the arrivals in non-overlapping intervals, the distribution of any

ta, k

is not influenced by where

the preceding inter-arrival times might happen to have piaeed its reference point AU interarrival times have the same expectation, exponential distribution these are found to be Ta

=

;

1/R a

Va

=

T a , and variance, (see

appendix

Ta2

D

t 0.

Va . From the above , page 177 )

1/ Ra2

=

Conversely, an arrival process which produces independent inter-arrival times with the same exponential distribution is always a POISSON process as defined before. ( The proof of this fact is given in a later chapter on Markov processes.) The most efficient method to test whether some observed process is a POISSON process is therefore to test whether its inter-arrival times have exponential distribution, and whether they are independent. Exercise

7

suggests some

applicable techniques. Another important application of this second property arises with the simulation of POISSON arrival processes. The simulation of systems in which POISSON arrivals occur requires a procedure to find consecutive arrival times by computation. There exist algorithms, known as 'random number generators', which produce number sequences tribution over the range

91 , 02 . . . .

with uniform dis-

(0,1), say, and mutual independence of the individual

in a statistical sense. The values

T a. ( - l o g ( g n ) )

Vn at least

then have the above exponential distri-

bution and can be added one after the other to compute arrival times of a POISSON process. This technique is used in an exercise of chapter

3.

A third property of the POISSON process will be used in some reasonings of queuing theory as discussed in later chapters of this course. It explains why ly

'random arrivals'

is often used as a

uniformly distributed arrival times

I I

I

I

I

'random arrival process'

synonym for the POISSON arrival process. Given that some fixed

|,t T

or short-

the

M-th

the

M -1

T

is the time of

arrival of a POISSON process, preceding arrivals in (0,T)

have the same probabilities (for numbers in

20

intervals,

and for i n t e r - a r r i v a l times) as if e a c h of the

M - 1

arrival t i m e s is an i n d e p e n -

dent r a n d o m v a r i a b l e with a probability distribution uniform o v e r the i n t e r v a l (0, T). A proof of this s t a t e m e n t is found in Parzen 1. This property permits to consider e a c h i n d i v i d u a l arrival t i m e as a ' r a n d o m l y ' chosen s a m p l ing point in the i n t e r v a l

(0, T).

Thus,

POISSON arrivals 'probe' the properties of other a c -

c o m p a n y i n g processes in a p a r t i c u l a r ' r a n d o m ' way. A first a p p l i c a t i o n of this reasoning arises in chapter

4.

As an e x a m p l e g i v i n g s o m e n u m e r i c a l e v i d e n c e , third property,

the s i m u l a t i o n of random arrivals with this

and their statistical analysis with respect to i n d e p e n d e n c e and distribution of

the i n t e r - a r r i v a l times,

is discussed in e x e r c i s e

7.

Exercises. 8

Show that the b i n o m i a l probabilities

Pa, n(t0, tl)

of page

POISSON probabilities as

N ~ m , if

Ra(tl-t0)

is fixed.

some of the

N = 10 and

N = 100,

and of the

Pa, n

with

e.g.

read t h e m from appropriate tables in 8

Given an arrival n u m b e r estimates the intensity

~a/T ber

fia

na(0, T).

the 95-th and the

For n u m e r i c a l e v i d e n c e , Pa, j , if

Ra(tl-t0)

compute = 2 , -or

Brunk t .

observed in the t i m e i n t e r v a l (0, T),

Ra . Find the values of 5-th p e r c e n t i l e ,

Ra

respectively,

The i n t e r v a l b e t w e e n these values can serve as a

e s t i m a t i o n of

the v a l u e

which m a k e the observed h u m of the

POISSON distribution for

90 %- c o n f i d e n c e i n t e r v a l for the

Ra.

Which i n t e r v a l applies if page

r~a

15 c o n v e r g e to the

5740

input messages were observed in a p e a k hour,

as quoted on

16 ?

7

(This exercise supposes the use of a computer. )

of a random v a r i a b l e uniformly distributed over t e r - a r r i v a l times, Obviously,

using

T = M

(0,M).

Compute

M - I

s a m p l e values

Sort t h e m in order to find

M

in-

as a further arrival t i m e .

the a v e r a g e i n t e r - a r r i v a l t i m e is

t-a

=

1. C o m p u t e the s a m p l e v a r i a n c e of the

^

t a.

As a test for i n d e p e n d e n c e ,

c o m p u t e s o m e of the s a m p l e covariances with lag

They should be small if c o m p a r e d with the s a m p l e v a r i a n c e .

n --1,2,..

As a test for e x p o n e n t i a l distri-

21 bution compute a chi-square statistic for the goodness of fit as described in Brunk 1. The probability to exceed this computed value should not be small.

Solutions. Write the binomial probabilities of page

5

N ( N - l ) - . • (N-n+l)

( n: If now

Np

the l i m i t

Np

N-n

(Np) n

)

=

-

N

Then

=

.2

=

for

2, i . e . N

=

N

N ( N - l ) - " (N-n+1)

N

Ra(tl-t0), and

(N -Np) n

N--~ m, the second factor above has

i;

e -Np, whereas the last factor has the l i m i t Ra(tl-t0)

Np

( 1 - --)

-

n'

is held fixed at the value

Consider p

Np n ) (1N

17 in the equivalent form

this holds for any n.

a t i m e interval with the expected arrival number of

10, and

p

=

.02

for

N

=

2.

100. Values of the binomial

and POISSON probabilities are = =

0 0.107

1

2

3

4

5

6

0.268

0.302

0.201

0.088

0. 026

0.006

, N = 100 1(1-.02) 100=

0.133

0.271

0.273

0.182

0.090

0. 035

0.011

0. 135

0.271

0.271

0.180

0.090

0. 036

0.012

=1 ~ 0 ~a,n ' N

for n , j ( t - . 2 ) 10

/

Pa, j

e-.2

=

The probability that 1-

Prob(

na(0, T)

is at least equal to

na(0'T)~< f i a -

1

)

=

1-

=

1-

fia

is

fia-1 ~ pa, j(0, T) j=0 ~a-1 (RaT)J e- RaT j__~ J'

and can also be read as the value of a G a m m a distribution as defined in appendix D page 1 7 7 ,

viz. as

Gammal~ a(RaT) . The

100 (1-p)-th percentile of

,

ha(0, T) is then d e -

fined by Gammafia (RAT)

=

p ,

or as the 100 p - t h percentile of the G a m m a distribution. Given

~a

and

and the variance large

fia

T , the G a m m a distribution fia/T 2 . Furthermore,

Gammafia (x T)

has the expectation

the table of page 177 in appendix D

the distribution is approximately normal and has its

ffa / T

shows that for

5-th (95-th) percentile at

22

ff 1. 645 - g ( 1 - (+) - ~ ) . T

This l o c a t e s the

Its upper bound is a

95 °~o w o r s t - c a s e l i m i t for the s a m e e s t i m a t e .

V"a

i n t e r v a l has the r e l a t i v e width of would be

5865

instead of

s m a l l e r counts

?

ffa

90-%

c o n f i d e n c e i n t e r v a l for the e s t i m a t e d

+ 1. 648 / ~ / 5 7 4 0 "

=

fia

=

5740 . But this difference is not very important.

ARRPROC

ed on page 182 of appendix

5740,

the

+ 2 . 2 % . Thus, a worst case rate

the r e l a t i v e i n t e r v a l width increases,

A procedure

With

Ra .

e.g.

g e n e r a t i n g and sorting

to

1000

Obviously, for

+ 16.45 %

for

fia = 100,

random arrival times is list-

E . S o m e values of their e m p i r i c distribution are i n d i c a t e d in

the figure on the next page. T h e unbiased e s t i m a t e of v a l u e is close to

~-2

V(ta)

viz. C°Vn -

where

Pa

( Pa )

n

1 M - n

Coy n e s t i m a t e s the c o v a r i a n c e

M-n m=l~ ( t a ' m - Pa) (ta, m+n - ta ) ^

Thus,

This

ta .

A

(last)

C o v ( t a , m ' ta, m+n)

with an index distance or ' l a g ' be assumed to depend only on

1. 039.

has the v a l u e

are defined by a sum of products of deviations from s a m p l e

are the averages of the first

v a r i a n c e s are zero.

9

as should be for an e x p o n e n t i a l distribution of the

S a m p l e c o v a r i a n c e s with lag averages,

as r e c o m m e n d e d on p a g e

n

M - n

,

=

1,2,...,

observations. (see Cox, Lewis 1.)

of two i n t e r - a r r i v a l t i m e s occurring

in the s e q u e n c e of observations,

n , but not on

n

m.

- i f this c o v a r i a n c e can

For m u t u a l l y i n d e p e n d e n t

ta

all c o -

considerable s a m p l e c o v a r i a n c e would m a k e the i n d e p e n d e n c e as-

sumption doubtful. N u m e r i c a l results w e r e found using the procedure For a lag of

n

=

t

the s a m p l e c o v a r i a n c e was the s a m p l e v a r i a n c e .

2

-.'029

3

.061

Fa(X )

-. 048

shown in appendix E, p a g e 182. 4

10

.035

20

-. 027

-. 035 t i m e s

These values do not i n d i c a t e significant dependencies.

For a c h i - s q u a r e test of the goodness of fit, distribution

STANAL2

=

1 - e -x

the

w e r e considered,

10 m - t h p e r c e n t i l e intervals of an assumed viz.

m-1 log ( I

-

) ~ I0

x

< --

iog(l

m - --) I0

,

m:

1 ..... i 0

.

23 The counts of observations failing into t[lese 100, 117, les

98,

87,

90, 107 ) .

10

.1" M

10 =

=

( 96,101, 99,105,

'chi-square distribution with

9

=

100.

The

chi-square statistic

100 Fv0(X)

=

G a m m a 4 . 5 < x / 2 ) , also

degrees of freedom'

(Brunkl). This distribution

has according to page 177 of appendix D

the expectation

V(v0)

6.54 which results from the above

=

located at

18. The observed value E(v0)

an even greater

v0

.58 ~

,

¢0

:

random variab-

(v m - 100 )2

m:l

is known to have approximately the distribution called the

g

They are themselves observations of 10

v m , each of which has the expectation

v0

intervals were

E(v0)

=

9

and the variance

and is not contradicting the assumption of

could arise with probability . 69

(see appendix D

9

is hence Fa(x )

since

, page 177 ) Fa(X) and its 10m-th percentiles are shown in a figure, together with values of the empiric distribution at the percentiles of Fa(X) .

There are in fact several other tests which could have been applied here (Brunkl). The particular choice of the covariances and the chi-square statistic was made because they have some relation to questions discussed later in this course. In fact, the covariances of mutually dependent random variables play a key role for the confidence in simulation results. For the Gamma distributions, there will be a variety of further applications.

24

Chapter

2 :

Line service times and utilization

Terminals and lines are the two major components of a communications network, The need for a n analysis to determine

the required number of these components was stated in the sur-

vey of Section

6

t

on page

For both components, the decision criteria are to utilize

their capacity as well as possible, at the same time avoiding undesirably long wait times. It is the line analysis which is discussed in this and some following chapters. The terminal analysis is deferred to chapter

8

because it requires a more complex mathematical model.

In this chapter, the capacity of a line and its utilization by transmission, line control, and error checking are considered. Wait situations form the topic of the following chapters. T e l e c o m m u n i c a t i o n lines have a transmission capacity which is determined by several factors, viz.

a line speed, usualty stated in bits per second, which is an upper l i m i t for the

transmission capacity; a time per character,

Tch , which is required to transmit the bits that constitute

a character, including some redundancy. The line speed gives a lower l i m i t for ever, it is frequently the hardware of a terminal which determines delays

Td

Tch ; how-

Tch;

caused by electrical signal propagation on the line and in switching

circuitry; control characters,

Nb

in number, which have to be transmitted with every con-

tiguous block of data. They serve the purposes of selecting an individual device attached to the line as receiver ('adressing') or sender ('polling'). providing a longitudinal redundancy check on a block of data. The relative effect of these extra characters on the line capacity depends on the possible size of data blocks, and thus on buffer sizes; error probabilities, usually stated as a rate

Re

of errors per character. Their ef-

fect on the line capacity depends on the amount of re-transmission required after an error. S o m e typical combinations of the above factors are given in the following table.

25

Factors of line capacity, combination

II

III

IV

bits/sec

200

600

2400

40,800

Tch

msec

67.5

15

4.0

0.2

Delay by propagation

msec

Line speed Character time

by line turnaround *)

0.01 per kilometer

msec

Control characters per block Buffer size

150

30

8

8

?

7

14

14

120-480

480-1920

any

3.10-4orless

3.10 -5

in characters

Error rate

per character

*) Line turnaround, i . e .

3'10

-5

switching the line from transmission in one direction to the other

direction, occurs for lines that are alternately used for transmissions in both directions, called 'half-duplex lines'. The delay stated arises twice for each of the actions polling, requesting, receive after requesting, adressing and send after adressing. Only the receive after polling follows the polling action immediately without further delay.

Line service times. Consider the time interval for which a line is devoted to the transmission of a contiguous block of data,

- viz. the messages of the previous chapter.

a service time

The interval's duration is called

t s . For this time, the line gives service to, or acts as a server for, the

message. Different messages represent units of work, or are items that require service. The line service time depends linearly on the number ts where

nch

=

nchTch

+

rich

Nb Tch

may vary from message to message while

constant at least for certain groups of messages, e.g.

of characters in the message, by +

Tch, Nb

Td , are constant, and

Td

is

all input messages from a particular

location. If, for a first discussion, "polling is not considered, and propagation delays are neglected,

Td

equals four line turnaround times for any message.

The simple relation between line service time and message length entails an equally simple relation between their respective probability distributions. With the abbreviation c

=

Nb Tch + T d

,

26

the distribution

Fs(X)

of the l i n e service t i m e follows from the distribution

Fch(X)

of the

message length as Fs(X)

=

Prob(

t s <.~ x

)

=

Prob(

nch Tch + c ~ x

)

=

Prob(

nch ~ (x - c ) / T c h

)

=

Fch( (x - c ) / T c h )

The m o m e n t s of this distribution are l i n e a r c o m b i n a t i o n s of the m o m e n t s of the definition

E( t k

s )

eo

=

xk

f

dVs(X)

Fch(X ). From

'

-(13

substitution of

Fch

and an appropriate v a r i a b l e of integration leads to

oo E(t?)

=

)k

,,/" ( x Tch + c

dFch(X)

-00

Therefore,

k E(ts )

Especially, the e x p e c t a t i o n length

Nch

Ts

k k em k-m ,k-m ~ (m) Tch E( nch ) . m=0 of the l i n e service t i m e is found from the expected message =

as Ts

The v a r i a n c e

Vs

=

Nch Tch + c .

of the l i n e service t i m e is found from the v a r i a n c e

V(nch)

of the m e s -

sage length as Vs

=

V(nch ) T 2

, i n d e p e n d e n t of c.

Some further examples for the use of the above general r e l a t i o n arise in exercises of this chapter. Expected Line Utilization. An i m p o r t a n t p a r a m e t e r of any system giving service to arriving items,

- as discussed on the

previous page - , is the effective use m a d e of the System's service capacity.

For a single

t e l e c o m m u n i c a t i o n line, an appropriate measure of use is the fraction of t i m e in some i n t e r val for which the l i n e is a c t u a l l y giving service.

This fraction is called the u t i l i z a t i o n of the

l i n e in (0, T). It is computed as the sum of all service times occurring in (0, T), divided by ~,

T:

~(0, T)

=

T)

m=_l

ts,

m

/

T.

If the service times and their n u m b e r are random variables, so is the u t i l i z a t i o n computed from them. Its probability distribution for finite

T

m a y be difficult to derive. However, as

ns---~ eo , the distribution permits a simple description under fairly general conditions.

2?

In fact

it can often be assumed that all service times

h a v e a c o m m o n probability distribution

Fs(X) ,

ts, m

are m u t u a l l y i n d e p e n d e n t

with the possible e x c e p t i o n of the first

and the last s e r v i c e t i m e s w h i c h m a y be truncated by the endpoints of the interval This e x c e p t i o n b e c o m e s i n s i g n i f i c a n t for large

T

T h e u t i l i z a t i o n m a y be e x t e n d e d by the factor

5-(0,

ns(0, T) T)

=

T

where

ts

~-S

and

8

ns/n s , and thus rewritten as

'

wording for this f a c t is that

other i m m e d i a t e l y . a b o v e expression,

with the properties

if. T h e consistency m e n t i o n e d there i m p l i e s that for any

I-t s

the probability of deviations

As a first c o n s e q u e n c e ,

(0, T).

n s.

is the a v e r a g e service t i m e c o m p u t e d from a random sample,

as discussed on pages

' -is

-

and

I

Ts > c

converges to

tends to z e r o as T s (in probabilities)

c •

0 ,

n s-4~ co. Another as

n s --~ co '.

consider a system which is fully u t i l i z e d by services f o l l o w i n g e a c h

Its u t i l i z a t i o n ,

for any interval,

is

E

=

1. T h e first factor in the

ns(0, T) / T , which can be interpreted as a rate of service,

ways r e c i p r o c a l to

~s . H e n c e ,

The v a l u e

Rs

=

it converges to

1/T s

1/T s (in probabilities)

as

ns ~

is then a l co.

is the largest service rate that can be a c h i e v e d in

the long run by a single server with the e x p e c t e d individual service t i m e A second c o n s e q u e n c e arises for systems in which the number p e n d e n t of the i n d i v i d u a l service times.

As an e x a m p l e ,

ns(0, T)

T

s

of services is i n d e -

a t e l e c o m m u n i c a t i o n l i n e is often

used such that the number of services is d e t e r m i n e d by an arrival process (see pages

16 if.)

i n d e p e n d e n t of the service times. With the a d d i t i o n a l assumption that the arrival rate

na(0, T) / T , and h e n c e the s e r v i c e

rate,

T-.-~ co, the u t i l i z a t i o n converges (in

c o n v e r g e to a v a l u e

Ra (in probabilities)

as

probabilities) to the v a l u e U Apparently, wise,

since

U

=

R

a

T

cannot e x c e e d

s

1, the arrival rate

Ra

must not e x c e e d

Rs. O t h e r -

it cannot be the arrival process which d e t e r m i n e s the service rate in the long run.

Exercise

9

shows a first a p p l i c a t i o n of this result.

The e s t i m a t i o n of

U

from the

K(0, T)

28

o b s e r v e d in an i n t e r v a l of f i n i t e l e n g t h

T

is t h e s u b j e c t of two further e x e r c i s e s

10,11.

Exercises. 8

For t h e l i n e / t e r m i n a l

l e n g t h distributions a p p l y i n g to

combination

II

(page

I n p u t B, Output,

26 ) and for e a c h of t h e m e s s a g e

and the c o m b i n a t i o n of b o t h

(page

15

),

c o m p u t e t h e v a l u e s of T s , Vs , E(t?), using t h e f o r m u l a e of p a g e 9

26

traffic of p a g e 10 page

8

find t h e m a x i m a l s e r v i c e r a t e

=

1/T s

for

16 ?

180 to a p p r o x i m a t e t h e v a r i a n c e of U

Rs

How m a n y l i n e s are at l e a s t r e q u i r e d to h a n d l e t h e p e a k hour

A s s u m e POISSON a r r i v a l s w i t h a r a t e

i n t e r v a l for

,

.

F r o m t h e results of e x e r c i s e

c o m b i n e d i n p u t and output.

Etts3)

w h e n e s t i m a t e d by

Ra.

u(0, T)

Use t h e f o r m u l a e of a p p e n d i x for l a r g e

T.

D

,

S p e c i f y a 90% c o n f i d e n c e

K .

11

[f on a l i n e as discussed in e x e r c i s e

9 , a u t i l i z a t i o n of

.64

is o b s e r v e d w i t h i n

o n e hour,

w h i c h is t h e e s t i m a t e d a r r i v a l r a t e and the 90% c o n f i d e n c e i n t e r v a l for

U ?

Solutions• F r o m the t a b l e on top of p a g e

T h e s e v a l u e s l e a d to

Tch

=

15 m s e c

Nb

=

7

Td

= c

25

. 0 1 5 sec

=

,

• 12

=

sec ,

T s, V s

of p a g e

are e x p l i c i t l y :

26 is s t r a i g h t f o r w a r d . 2

E(t s )

characters/sec.

four t u r n a r o u n d t i m e s .

15

; t h e i r use in t h e f o r m u l a e

T h e s e c o n d and third m o m e n t

2

=

Tc2h E(nch )

=

T c h E(nch)

3

E(t s )

67

. 2 2 5 sec.

M o m e n t s of m e s s a g e l e n g t h distributions a r e s t a t e d on p a g e for

some

II

characters

120 m s e c

=

r e a d for c o m b i n a t i o n

3

3

of s e r v i c e t i m e s

2

+

2 c T c h Neh 2 3 c Tch

T h e n u m e r i c a l results are c o l l e c t e d in t h e f o l l o w i n g t a b l e :

+

c

2 c2 E(nch) + 3 T c h Nch +

c

3

29

T a b l e of results for e x e r c i s e

8.

Ts

Nch characters Input

B

sec 2

sec

I, 31

• 256

Output

48.2

0.95

• 086

combined

63.3

1.17

• 225

The r e c i p r o c a l of the e x p e c t e d

above results),

is

Rs

=

1/T s

E(t~)

sec 2

sec 8

51

1.96

3.24

•29

0.98

1•10

48

1.61

2,46

see

72.1

9

E(ts2)

Vs

=

•

•

service t i m e

.85

Ts

services/sec,

or

=

1.17 sec , (the third of the

8068 services/hour•

largest service rate which a single l i n e can a c h i e v e in the long run. a c h i e v e the corresponding m u l t i p l e of this rate. hour (from page

16 ) requires at feast

T h e r e are several reasons,

3

Thus,

M u l t i p l e lines can

the p e a k hour rate of

!ines with a c a p a c i t y of

to be discussed in this and l a t e r chapters,

This is the

9033

services/

9190 services/hour. why even m o r e c a p a c i -

ty should be planned for.

10

The f o r m u l a e of appendix

D

, p a g e 180 are c o n c e r n e d with a sum of random

variables with i d e n t i c a l probability distribution. They m a y be applied to the n u m e r a t o r of ns~T) g(0, T ) , viz• m=l ts, m , if T is large because then the possible truncation of two service times m e n t i o n e d on p a g e

27 b e c o m e s insignificant•

When a POISSON arrival process d e t e r m i n e s

ns(0, T),

pectation

Ra T

N

and v a r i a n c e

V(n)

equal to =

N Ts / T

V ( g ( 0 , T) )

:

2 ( N V s + V(n) T s ) /

:

E(t 2) U ~ Ts

N o t e that the v a r i a n c e tends to v e r g e n c e (in probabilities) of g(0, T)

(see p a g e

E ( g ( 0 , T) ) and

bution of

this number of summands has the e x -

0

g(0, T)

as

T ~ to

=

/

18 ). Thus,

Ra T s T2

for l a r g e

=

U ,

:

2 Ra E(t ) / T

T,

T•

oo. Thus,

the a b o v e results confirm the c o n -

U, stated on p a g e

is a p p r o x i m a t e l y normal for large

T.

The

27 . Furthermore,

the distri-

90°70 c o n f i d e n c e interval is

30

therefore approximately

g(0, T)

+

1. 646 V V (

a t a b l e of the s t a n d a r d n o r m a l d i s t r i b u t i o n ,

11

From

Ts

=

e.g.

g(0, T) )

appendix

D

, as can b e found f r o m , page 177.

1 . 1 7 sec , and an o b s e r v e d u t i l i z a t i o n

g(0,3600)

=

.64 , the

a r r i v a l r a t e is e s t i m a t e d as Ra a r r i v a l s in an hour.

=

~ / Ts

or

1962

g,

- e n o u g h to m a k e t h e result of e x e r c i s e

The variance

V( 5(0, 3600) )

=

.56 arrivals/sec

About that many service times must have entered the observed 10

is e s t i m a t e d by

applicable.

(see e x e r c i s e s

8

and

10 )

1.61 • 64

/

3600

.000243

=

1.17

The corresponding

90% c o n f i d e n c e i n t e r v a l is

.614

~

U

<

• 64

+

. 0 2 6 , or else

.666

This w o u l d b e an a p p r o p r i a t e l y c l o s e r a n g e for m o s t p r a c t i c a l purposes.

T h e a b o v e c o n s i d e r a t i o n s apply to u t i l i z a t i o n s o b s e r v e d in a c t u a l systems as w e l l as in a simulation.

For t h e l a t t e r ,

an e x a m p l e follows in c h a p t e r

3.

T h e e f f e c t of errors on s e r v i c e t i m e s and u t i l i z a t i o n .

T h e error r a t e

Re , i n t r o d u c e d on p a g e

24

, c a n b e i n t e r p r e t e d as t h e p r o b a b i l i t i y t h a t an

i n d i v i d u a l c h a r a c t e r is n o t t r a n s m i t t e d c o r r e c t l y . ne /n Then As a

from t h e n u m b e r n /n e

=

1/Re

ne

Obviously,

of errors in a l a r g e n u m b e r

Re n

w o u l d b e e s t i m a t e d as of t r a n s m i t t e d c h a r a c t e r s .

is the a v e r a g e d i s t a n c e (in c h a r a c t e r s ) b e t w e e n c o n s e c u t i v e errors.

simplifyinga p p r o x i m a t i o n ,

r a c t e r s are m u t u a l l y i n d e p e n d e n t .

it is also a s s u m e d t h a t t h e errors o c c u r r i n g on d i f f e r e n t c h a Then,

c o r r e c t t r a n s m i s s i o n is t h e p r o d u c t of

for a string of

m

m

characters,

t h e p r o b a b i l i t y of a

p r o b a b i l i t i e s to c o r r e c t l y t r a n s m i t o n e c h a r a c t e r ,

i.e. p(m) For the s m a l l v a l u e s

Re p(m)

=

( 1 - Re ) m

t h a t arise in p r a c t i c e , =

e -Rem

is a good a p p r o x i m a t i o n .

31

Consider a s c h e m e of error c h e c k i n g which detects errors at the end of e a c h message,

and

causes the l i n e to r e m a i n d e v o t e d to the particular message until its transmission was successfully repeated.

T h e transmission of a message of g i v e n length

characters that m i g h t be erroneous.

T h e service t i m e is

nch

ts

=

involves

m = rich + N b

m Tch + T d

if no

errors occur; the probability for this is p(m)

ts - T d p(.---) Tch

=

With probability

(1 - p(m) )J p(m) j

;

j

=

0,1,2...

transmissions are erroneous before one is successful. t's

Thus,

g i v e n the v a l u e

=

(j

=

I

i.e.

and its powers are b

~E ( 1 - p ( m ) ) J j=0

consider the e x p e c t e d service t i m e

the e x p e c t e d 'error free' length.

p(m)

(j+l)kts

E( t's I i s )

With this assumption,

transmission is as s m a l l as

; k = 1,2 ....

for the length

m = 1/Re ,

the probability for a successful

-1 p(m)

Consequently,

t[

GO

L

the service t i m e is

ts

+ t)

t s, the c o n d i t i o n a i e x p e c t a t i o n s of

E( t'sk Its ) As an e x a m p l e ,

In this case,

=

e

=

0. 368 .

the e x p e c t e d s e r v i c e t i m e as found in e x e r c i s e

12

to be

e ts

=

2. 718 t s

is m u c h higher than without errors. If

m

is restricted to values which hold .pim)

constant or linear in

1 -p(m)

r e m a i n significant.

E( tsk [ is) Furthermore, Finally,

1 - p(m)

close to

=

1 , only the summands that are

Then,

approxt..matety,

tsk ( 1 + (2k-1) ( 1 - p ( m ) )

is a p p r o x i m a t e l y i i n e a r in

the known probability distribution

Fs(X )

Re , viz.

)

equal to

of the service times

Re (t s - Td) / Tch. ts

is used to find

the (unconditional) e x p e c t a t i o n s O0

E(@)

:

,/" E(t[k[ x)

dFs(X ) .

-GO

S i n c e the conditional e x p e c t a t i o n s contain two summands with the factors k - t h and k+l -th m o m e n t of E( t'sk)

k+l t k , ts , the

enter the result

Fs(X ) =

E( t k s )

+

(2k-l)

Re Tch

t k+l ) - T dE( tsk ) ) . ( E( s

32

T h e e x p e c t e d service t i m e

T~

T's

results for

=

Ts ( 1

the s a m e

c'

=

Ra T s

U'

1

as

Re E(t~) - ( - Tch Ts

Td )

c').

can be interpreted as the o v e r h e a d

a f f e c t e d by the errors,

=

+

Ts(l+ C'

k

caused by errors. If the arrival rate

Ra

is not

applies to the overhead in u t i l i z a t i o n : =

U(l+c').

Exercise 13 provides s o m e n u m e r i c a l values for c' , as w e l l as for the second m o m e n t 2 E(t~ ) which is of m a j o r i m p o r t a n c e in w a i t situations (see later chapters.)

A message .length for m i n i m a l overhead. The o v e r h e a d

c'

b e c o m e s e x c e s s i v e (see p a g e

On the other hand,

31

) if the messages are m a d e too long.

there is an overhead caused by error checking,

viz.

C C"

-N e l l Tch

with the

c

=

of page

N b Tch + T d

25.

This overhead b e c o m e s e x c e s s i v e if the messa-

ges are m a d e too short. The product ably.

( 1 + c' ) ( 1 + c" )

w i l l therefore h a v e a m i n i m u m if

In the s i m p l e case w h e r e all messages h a v e the s a m e length,

d e t e r m i n e d by

2 Nch

Nch

is chosen suit-

the o p t i m a l

Nch

is

c =

(t

+ NbRe)-

1/Re

,

Tch as derived in detail by e x e r c i s e 1 . Therefore,

14.

The first of the above factors will usually be close to

the o p t i m a l message length is found as the g e o m e t r i c m e a n of the e x p e c t e d

error free l e n g t h

1/Re , and the l e n g t h

C/Tch

which could be transmitted in the t i m e

required for one checkpoint. Exercise

14

also shows n u m e r i c a l values for

insensitivity of the m i n i m u m . the overhead.

In fact,

values

Nch , c' , c" Nch

and gives e v i d e n c e for the

off the o p t i m a l do no__~tstrongly affect

c

33

Exercises.

12

F r o m the

13

Use the

E( t[ I ts) E( t'sk)

of p a g e

of p a g e

m i n a l c o m b i n a t i o n II, p a g e

31

31, with

25 , to find the values

8.

14

For messages all of the s a m e l e n g t h T 2 . Then

' o v e r h e a d factor'

c'

:

Re (Nch + Nb).

for

m

:

1/Re.

T~ , E( t's2 ) , V~

Nch , T s

from the results of

has no v a r i a n c e ,

Find the v a l u e

Nch

and h e n c e

g E(ts)

which m i n i m i z e s the

(1 + c') (1 + c").

With the s a m e data as in e x e r c i s e Also,

T's

k = 1, 2 , and the characteristics of l i n e / t e r -

exercise

equals

find

13

c o m p u t e the o p t i m a l

c o m p u t e the o v e r h e a d factor for the

Nch

Neh

and o v e r h e a d factor.

of the c o m b i n e d message stream.

Solutions.

o3 12

The sum

E(t's I t s )

=

~

)j (1-p

(1

p ( j + l ) ts

-p

1 - ( 1 -p)

(j+l)

ts

j=O o3

=

~ (1 -p)J j=0

t

+

c a n c e l l i n g terms

s

ts reduces to With the

e(c s Its) p(m)

:

e

-1

:

:

1 - (l-p)

stated in the text, and constant

T'

=

ts

/

P

t s , the desired result

e tS

s

is proven. The results,

13

c o m p u t e d with

T~

C'

sec i

Re

:

3 . 1 0 -4

E(@)

and data from e x e r c i s e overhead

v~

8 , are : overhead

sec 2

%

sec 2

o7o

i

1, 85

2.8

2.14

9.2

.328

22

Output

0.9'/

1.8

i. 04

6.0

• 103

2O

combined

1.19

2.5

1. '/5

8.4

.307

36

Input

B

34

14 while

D i f f e r e n t i a t e d with respect to c"

=

c/ NehTch

Nch ,

c'

has the d e r i v a t i v e

=

Re( Nch + N b )

- c " / N c h . Thus,

has the d e r i v a t i v e Re

the d e r i v a t i v e of the o v e r -

head factor is Re(l+c"

)

(l+c')

c" / N e h

=

Re - ( 1 + ReN b )

c" / N c h

The zero of this d e r i v a t i v e is found from C

R e With the data

Re

=

3 . 1 0 -4 ,

=

Nb

the o p t i m a l length is such that

=

7 ,

C/Tch

=

2

Nch

15

-3

=

( I + 2.1.10

Nch

for a total o v e r h e a d of

10

c'

=

c"

=

for a total overhead of tributions

c',

c"

11

),

=-4-

100 V 5 . 0 1 224

characters.

3 . 1 0 -4 (224 + 7) 7 / 224

Nch

c'

=

c"

=

=

6.9%,

=

3.1

=

63

%,

characters , (see p a g e

3" 1 0 - 4 (63 + 7) 7/63

=

2.1

=

9.1%,

% . This is still close to the o p t i m a l value,

show another relation.

of the service times,

28

% .

The o v e r a l l v a l u e from the case study, result in

25,

f/----.-

= =

It results in

(see pages

15

)-3" I0

i.e.

, as stated in the text.

( 1 + ReN b ) ---'2---Nc h T c h

Note that this last

c'

15 ) would

% ,

although the c o n -

n e g l e c t s the v a r i a n c e

and therefore is s m a l l e r than the result of e x e r c i s e

13.

35 Chapter

3 :

First concepts and relations of Queuing Theory.

Many problems in computer systems' timing and resource allocation arise from queuing situations. At many points of a system, units of work (see page 25 , e.g.

messages to be transmitted,

data to be stored or processed, requests to retrieve data etc. ) arrive at random times while the required server may have to work on other items. Therefore, they cannot always get i m m e d i a t e service. In this case, the items have to enter some buffer, i . e .

a space provided for waiting items,

and to wait there until selected for service. A linear order, e.g.

the orde~ by arrival times,

is often natural for the set of waiting items. Therefore, wait line or queue are other names used for this set. As an example, the messages requiring a certain t e l e c o m m u n i c a t i o n line for transmission may have to wait at offices, in terminal buffers, in a message concentrator, or in the computer center. Thus the queue associated with the line may even consist of several subsets. A definite queuing discipline must be established which at any time determines which item has to get service. A t e l e c o m m u n i c a t i o n line, once it started a transmission, remains devoted to a message until it is successfully transmitted; therefore, the discipline is apptied only at the times when a service ends. The discipline which always selects the item with the earliest arrival time is called the 'first-in-first-out' or FIFO discipline. Problems which arise from queuing situations are : Which wait times tq

tw

will the items spend in the wait line, which queuing times

will be taken for their complete execution, spent in the wait line or the server. Which space must be provided for the waiting items. Here it is essential to know

the number

nw

of items which will be in the wait line at different times.

The 'queue variables'

tw , tq

for a particular item, as well as

nw

for a certain time,

36 use to be random variables. It is their probability laws which the Queuing Theory aims to determine from certain basic assumptions about the queuing system. The possible assumptions are manyfold as reference books like

8aaty 1, Takacz I

show. Some of these assumptions

and results will be used in this course.

Relations of first order for the single server queue. A basic model of a queuing situation is described by the following assumptions : There is a single server, giving the same type of service to all items. The individual service times are mutually independent, with the common distribution

Fs(X). The arrival times of items, i.e.

the earliest times at which the service could start

for each item, form a POISSON process with rate

Ra.

The wait set may contain any necessary number of items. A graphical symbol for this

"7,"

IJI

arrivals

1-- l Fsx, 1--

queue

server

situation might be as suggested in the figure.

The actual behaviour of such a queuing system, as time proceeds, may be described by the number

nq(t)

of items in the system, given as a function of time

t . Its points of discon-

tinuity mark the times of arrival by an increase, the times of service completion by a decrease. Individual queuing times, wait times and service times can also be read from nq(t) if the queuing discipline is known. The following figure assumes

FIFO discipline :

nq(t) *'ts,1

~

ts,2 ~ t s , 3

-,tl,

•

f

4---- t q, 3 -------~

arrivals

b

37

Irrespective of the queuing discipline,

the averages of the queuing variables are r e l a t e d

simply. Consider any set of items, line,

viz.

or of i t e m s in service.

the set of i t e m s in the queuing system,

of i t e m s in the w a i t

T h e n u m b e r of i t e m s in the set is a function

Its t i m e a v e r a g e over s o m e i n t e r v a l

(0, T)

n(t)

of t i m e .

is defined as

1 g ( 0 , T)

During the s a m e interval, tain t i m e in it.

-{

Then,

n(0)

assuming

T , j " n(t) dr. T 0

-

na(0, T)

a number

of i t e m s appears in the set and spends a c e r -

denotes the averages of such t i m e s as introduced in chapter =

n(T)

=

1.

0 , inspection of the previous figure shows that na(0, T)

~(0, T)

=

i- . T 2.

One result of this type was already found in chapter n(t)

i n d i c a t e s the use

Then,

u(t)

of the server,

and is

0

For the set of i t e m s in service, for the i d l e server,

1

if it is busy.

the above result reads na(0, T) ~(0, T) T

as on p a g e

27 .

The e f f e c t of truncation by the i n t e r v a l ends, by the assumption that only

n(T)

=

0

n(0)

=

n(T)

=

also m e n t i o n n e d in chapter

2, is e x c l u d e d

0 . It b e c o m e s insignificant for any l a r g e

happens often enough w h i l e

T

increases,

T

if

thus l i m i t i n g the values of the

possibly truncated times. Now,

the relation

U

If e i t h e r the a v e r a g e n u m b e r as

T~

tion w i t h

=

Ra Ts

K(0, T)

of chapter

or the a v e r a g e t i m e

oo , so does the other a v e r a g e ,

2 i-

can also be g e n e r a l i z e d :

converges (in probabilities) ,

and their l i m i t values are r e l a t e d by a m u l t i p l i c a -

R a . Especially, Nq

=

Ra T q

for the i t e m s in the queuing system,

NW

=

Ra T w

for the i t e m s in the w a i t line.

S i n c e i t e m s in the system are either w a i t i n g or in service, at any t i m e w h i l e for e a c h i t e m

tq, m

=

tw, m

*

ts, m "

nq(t) = nw(t ) + u(t),

This i m p l i e s corresponding relations for

88

the averages and their l i m i t values, *eiz. Nq and

=

Tq

Nw

+

U

Tw

+

Ts

The expected service time is known. Thus, from any one of the l i m i t values

Nq, Tq, Nw,

Tw , the other three can be computed for any meaningful arrival rate. Note that the above relations between expectations (i. e. between moments of first order ) hold for any queuing discipline, even including some situations with priority schemes chapter

(see

? ). However, relations between higher moments of the queuing variables, e.g.

the variances, d__oodepend on the queuing discipline (see chapter 14. )

A sufficient condition for convergence. For the single server queue, a very plausible condition is sufficient to make the preceding resuIts applicable. As discussed in chapter U

=

if this value is less than

Ra T s

1 , i.e.

Ra

2, the l i m i t of the utilization is

if

I/T s

=

Rs

T h e s a m e condition assures that

the server continues to have idle times as

T

increases. Thus, truncations tend to

become insignificant. the average numbers

nq'

nw

converge (in probabilities). Plausibly, with every

idle period the queuing process starts afresh with the same probabilities for its future devellopment. Of course, the independence of individual

ts

and of arrivals from a POISSON

process are essential for this reasoning. -

the averages of wait time and time in the queuing system converge (in probabili-

ties). This follows from their reIation with the average numbers. The convergence of the averages can be interpreted as a 'statistical stability' in the long rum The above stability condition is proven in later chapters. Note that instable queuing processes with

Ra >

Rs , i . e .

an 'overloaded server', may still be of practical interest, although

only for a finite duration.

39

Exercise. 15

Assume that

T s, U ,

and

Tw

are known.

G i v e expressions for

N w, Nq, T q .

Solution. 15

From

U

and

Ts,

the arrival rate is found as

Ra

=

U / T s . Thus,

the

desired expressions are Nw

:

Tw U / T s ,

Nq

:

( Tw / T s

Tq

:

Tw + T s .

+

1)U,

S i m u l a t i o n of a single server queue. In order to get s o m e n u m e r i c a l e v i d e n c e ,

a p a r t i c u l a r queuing process is s i m u l a t e d by c o m -

putation. A l a r g e n u m b e r of arrival times,

with e x p o n e n t i a l distribution of their intervals

puted successively as e x p l a i n e d on p a g e unit.

This m a k e s

Ra

=

19 . The e x p e c t e d i n t e r v a l

The

FIFO

Ts

=

Then,

.4

.7 , and the l o n g - r u n u t i l i z a t i o n

d i s c i p l i n e is considered,

no i t e m in the system.

is used as t i m e

1 . To e a c h arrival there corresponds a service t i m e ,

a random n u m b e r with uniform distribution b e t w e e n service t i m e is

Ta

and

t

Arrival t i m e s up m

=

0

and U

1

t a , is c o m -

t i m e units.

c o m p u t e d as

The e x p e c t e d

has the s a m e n u m e r i c a l v a l u a

is assumed to b e a t i m e when there is

1000 t i m e units are taken into account.

e a c h i t e m starts its service at the J.ater of its arrival t i m e and the departure t i m e of

the p r e c e d i n g i t e m .

It departs after its service t i m e has eiapsed.

can be c o m p u t e d one after the other. the queuing t i m e s wait times

Thus,

the departure t i m e s

Subtraction of the corresponding arrival times results in

tq ; a further subtraction of the corresponding service times produces the

tw .

Ideparture I

I

larrival m

ts, m

m-1

T h e figure shows a short section

i departure m

II

!I

ts, m + l

iarrival m + l

of the s i m u l a t e d process in order to e x p l a i n the c o m p u t a t i o n .

40 A procedure for this simulation, written in It produces

10

interval

=

T

PL/I,

is shown in appendix

E

, page 183.

different sequences of random arrival and service times, each for the same 1000 time units. The

10

sets of resulting statistics are reported in a table:

ha(0, T)

~(o, r)

~q

1

1050

.742

1.85

I. 14

1.54

2

1027

•

721

1.65

O. 95

1.46

3

1003

.703

I. 83

I. 12

2.18

4

1003

700

1.37

0.67

O. 65

5

1002

710

1.68

0.97

i. 56

6

1056

736

I, 72

1.02

1.51

7

1037

741

1.84

i, 13

2.15

8

963

683

1.36

O. 65

O. 62

9

992

700

1.49

0.79

O. 90

i0

955

662

1.52

O. 81

1.15

Simulation hr.

Average of

10

sample variance of the tq

resuIts

0.925 Sample variance of

10

results

j

Sample standard deviation of

[ Interval of

10

!

results

33

r

.020

.1781.176

1.65 standard deviations to both sides of average of ., I 955 - I0631.677-.74311.34-1.921.64-1.22

The convergence (in probabilities) of closely with the results of chapters variance of

na

ha(0, T) 1

and

lation is based, the values

and

is apparent. Its 'rate' checks

2 . In good agreement with page

Ra

Ra E(t?) / T . =

1 , T

follow. The approximation value is then i'q

~(0, T)

results

I

almost equals its expectation. For the utilization

approximation of the variance as

With

and

10

=

g,

page

18 , the 29 states the

From the assumption on which the simu-

1000 , and

E(t?)

=

.52

(see page 177).

.00052, and the observed value is close to it.

~w ' however, the convergence (in probabilities) appears to be 'slower'.

41 In fact, the observed values differ so widely that the confidence in observed averages of queuing variables appears as a serious problem. This problem arises with simulations as well as with observations of an existing queuing system '

Confidence intervals for the averages of queuing variables. Some theoretical considerations will be discussed here, on which two rules for practical application can be based. Consider an average of

M V(V)

random variables =

M ~" m=l

v 1 . . . . . vM

M ~ n=l

Its variance is in general

.

cov(v m, Vn)

(Brunkl). cov(v m, Vm) is meant to denote the variance of

/

Ms

vm .

It is only for independent random variab!es that this sum reduces to a simple sum over variances. Positive covariances of the random variables increase the variance

V(V) beyond

this simple sum. This effect can be very strong. The above simulation gives some evidence. In simulation hr. dividual

tq

is estimated by the sample variance

simple sum of variances, divided by ever, from the sample variance of 10..0816 a value which is about

24

1.54

(see last column of the table). The

M 2, would result in 10

/ 9

results, =

1 , the variance of each i n -

1.54 / 1050

V(i-q) is estimated as

.0351

= . 00147 . How(see page

9

)

,

times as high. Most of this increase goes onto the account of

covariances. In fact, queuing times show a positive covariance if they arise at not too long a time distance. E . g . ,

a long (short) individual queuing time makes a long (short) next

queuing time more probable, especially under the

FIFO

discipline.

One method to estimate the variance of averages, taken over a single sample process, consists in estimating the covariances

cov(v m,vn), summing them up to

V(7). This is applic-

able to statistically stable processes in which the covariances depend only on the lag as discussed on page

22 . By collecting all terms with the same lag,

m-n,

V(V) can again be

simplified to a simple sum. Only a l i m i t e d number of terms in this sum is significant, since

42 the covariance tends to fade away for longer lags. A further term

T2/(RaT)

should be

added in analogy to page 29 , in order to approximate the effect of varying arrival n u m bers. The procedure

STANAL2

of appendix

E

, page 182 , was applied to the queuing times

of simulation nr. 5 , and produced estimated covariances with a smooth pattern, characterized by the values below,

- which are given as multiples of

lag

0

1

2

3

covariance

1

.88

.79

.71

6 • 55

Vq - •

10

20

30

.44

.25

•1

times Vq.

The same procedure also evaluates the expression V q na

(

+

1

which for limited lags and large

na

For simulation nr. 5 , the result is Vq/n a = .00156.

The further term

2 - ~" covariance/Vq ) lag/0 is a simple approximation of the double sum in .0361

or

• 00156

y2 / (RaT)

• 23.3

1. 6832 / i000

for comparison =

.0028

v(~).

with is rela-

tively insignificant• A second method to estimate

V(Y) is to observe a certain number of averages

i" , i n d e -

pendent of each other. Their sample variance yields an estimate with very little statistical computation. This was shown for the simulation results. A weakness of this second approach is that it takes much more time to observe several averages. A more practical suggestion is therefore to record the averages over some subintervals of one single observation period.

The variance of sub-averages can then be estimated as sug-

gested above• Now, the sub-averages can be made almost independent random variables by assuring that each subinterval contains at least so many individual times that the first and last of these are no longer dependent. The covariance values stated above give some feeling for the required length. Finally, the variance of the average over the whole observation period is estimated from that of the subaverages by a simple division with the number of sub-averages, as is appropriate for independent random variables. As an example, collecting ali the reported simulations into a single larger one, the average queuing time with the observed value

1. 631 , has the estimated variance

.00351 .

43 Confidence intervals should in all the above cases be based on an approximately normal distribution. Then they are determined by the estimates of expectation and variance.

Distributions of the queuing variables. A further evidence of practical interest can be drawn from the above simulations: The e m piric distributions of queuing times and wait times. A computer procedure was used to print plots of these empiric distributions. The results for simulation nr. ~

are shown on the next page. The averages

gq

and

gw

are marked as

points of reference. Apparently, the individual times are widely spread around the averages. The largest observed queuing time was time

0

8.44 time units, the largest wait time

occurred frequently, viz. for all arrivals that found the server idle.

The corresponding distributions for all shown here, and

tw

7.77 time units. Also, a wait

10

simulations, - which were plotted but are not

- are close in shape to each other. Thus, one single distribution each for

t

q

appears to describe the queuing system under study, at least in some sense of sta-

tistical convergence. This fact is of great theoretical importance, and is therefore studied in several later chapters. It is as well of practical importance. The systems designer is often asked to assure that wait times or response times (for the user of a terminal, say) will not exceed a given limit. A definite limitation is rarely possible. However, knowledge of the probability distribution of the l i m i t e d quantity permits to state how often,

- or rather how seldom,

a violation of the l i m i t is to be expected in a large

number of observed cases. The designer must then choose a system such that the given l i m i t becomes some high percentile point of the distribution in question.

Exercises. 16

From the plot of the next page, read approximately the relative number of items that did not wait ,

-

that waited longer than twice the average

that spent less than

. 4 time units in the queuing system. Also, read the

Fw

,

90-th

44 ,I,

~

m

0 -0

+-.9

.0.

I I I I I I I

Empiric distribution of

wait

F~(x)

0

times o

"t".8

0

average ~r

wait time i-

=

w

q-.7

.97

0

"

w

4-.6 t I I I I t I

average 0

queuing time

Q

~q

=

1.68

"I'.5 Empiric distribution F~(x) 3t-

of

queuing

times

0

-t'-.4

,,.2

~--.1

~-

O.5

1

1.5

2

2.5

3

: X

45

p e r c e n t i l e of b o t h e m p i r i c d i s t r i b u t i o n s ,

i.e.

t h e t i m e s e x c e e d e d in less t h a n

10 %

of all

cases. I?

Choose the time unit for the above simulation, and hence

a response time of

2 seconds is not exceeded in more than

then the average response time

I0 %

Ra , T s , such t h a t of t h e cases.

W h i c h is

i-q ?

Solutions. 16

Up to t h e p r e c i s i o n of t h e plot, • 275

f r a c t i o n of t i m e .85

, i.e.

2 7 . 5 °~0 of t h e i t e m s did not w a i t .

1 - U , i.e.

This is c l o s e to t h e e x p e c t e d

for w h i c h t h e server is i d l e . 85 %

of the i t e m s w a i t e d less t h a n

No i t e m s p e n t less than

2 • .97

. 4 t i m e units in the system,

=

1 . 9 4 t i m e units.

s i n c e this is the m i n i m a l

s e r v i c e t i m e assumed. The

17 Then

90-~

p e r c e n t i l e is at a b o u t 2 . 4 t i m e units

for t h e w a i t t i m e s ,

8 t i m e units

for t h e q u e u i n g t i m e s ,

A t i m e u n i t of Ra

=

2 / 3 seconds p l a c e s t h e

a / 2 s e c o n d s - ,1 T s

=

.7.2/8

=

90-th

or

2 . 5 t i m e s the a v e r a g e ; or 1 . 8 t i m e s t h e a v e r a g e .

percentile at

. 4 6 7 seconds , a n d

3' 2 / 3 = 2 seconds. Fq = 1 . 1 2

seconds.

46 Chapter

4

:

A first result on the single server queue.

A basic m o d e l of the single server queue was defined in chapter queuing system,

the l i m i t

Tw

3, page

86

. For this

of the a v e r a g e w a i t t i m e can be derived by a reasoning

( P o l l a c z e k 1 Khintchine 1) which assumes as g i v e n the statistical stability of the queuing p r o cess,

as considered on p a g e

38 .

Such reasoning,

even though i n c o m p l e t e in its m a t h e m a t i c a l analysis,

possible results.

Also, the s a m e reasoning can be g e n e r a l i z e d to a variety of other queuing

situations,

is useful as a guide to

as l a t e r chapters w i l l show.

T h e w a i t t i m e of a particular i t e m which arrives at t i m e

t

is the sum of

a possibly r e m a i n i n g part of the service t i m e underway,

if any,

and

the service t i m e s of aI1 further items which p r e c e d e the n e w c o m e r in the server, thus tsl t I

=

+

ts, m"

over p r e c e d i n g i t e m s m

Now, form the a v e r a g e of m a n y such w a i t times,

and consider its l i m i t (in probabilities) as

T - - ~ co. T h e n Tw where

N

=

Tsl t

+

N Ts ,

is the l i m i t for the a v e r a g e number of i t e m s that p r e c e d e a n e w c o m e r w h i l e it

waits. For the

FIFO discipline,

N

is e v i d e n t l y the v a l u e

a v e r a g e number of waiting items,

Nw

=

Ra T w

, the l i m i t of the

because just these p r e c e d e the n e w c o m e r .

s a m e e q u i v a l e n c e holds for any discipline.

However,

In any intervals of an a v e r a g e l e n g t h

the

Tw,

the

server must on the a v e r a g e start as many services as arrivals are e x p e c t e d in this t i m e , R a T w ; otherwise,

statistical stability would be impossible.

is i m p l i e d in the results of chapter Substitution for

N

i.e.

A f o r m a l proof of this s t a t e m e n t

14.

g e n e r a t e s an equation which can be solved for the unknown Tw

=

Ts[t

+

R a T s Tw

Tw

=

Tsl t / ( 1 - U )

=

Ts It

+

T w, viz.

U Tw ,

and Apparently,

this result is m e a n i n g f u l only for

. U < 1, i . e .

for a stable queue (page

38 .)

47 The e x p e c t e d r e m a i n i n g service time._:_ The r e m a i n i n g service t i m e considered.

tsl t

is c o n d i t i a n n e d by the t i m e

This is i n d i c a t e d by the subscript

POISSON arrivals h a v e the property,

s[~, viz.

discussed on page

of arrival at which it is

'service t i m e , given t ' . !9

dependently 'probes' at a t i m e uniformly distributed in sult for

t

, that each individual arrival i n -

(0, T).

This implies a p a r t i c u l a r r e -

Ts[ t .

For a given pattern of service times,

the r e m a i n i n g t i m e

tsl t

is a function of

t , the

t i m e of a possible further arrival.

The figure shows an

example. L

t

0

ts, 1

If now tation

t

ts, 2

ts, 3

T

is a random v a r i a b l e with uniform distribution over

(0, T) ,

ts[ t

has the e x p e c -

(Brunk 1) T ~0

dt T

tslt

=

niT) m=l

2 ts, m / ( 2 T )

Inspection of the above figure makes evident that the v a l u e of the integral is in fact a sum of areas of triangles with the bases As

ts, m , divided by

T.

T , . ~ m , this expression converges (in probabilities) to Tsl t

=

Ra E ( t s 2) /

2 .

Note the close analogy to the reasoning of p a g e

26 which concerned the u t i l i z a t i o n .

The POLLACZEK-KHINTCHINE formula. A substitution of the above result into that of the previous page leads to the expression Tw

in terms of data assumed as known : Ra E(t?) T

= w

Other

E(ts2)

arrangements

=

TS2

+

of the same

( P o l l a c z e k 1, K h i n t c h i n e 1) 2(1 formula

V s , one arrives at

-U) can

be useful.

With

Ra

=

U / T s

and

for

48

Tw where Tw

T

w

Vs T s ( 1 + Ts2 )

=

a p p e a r s as a m u l t i p l e of

U 2(1

-U)

T s . T h e second factor,

i n c r e a s e s w i t h the i r r e g u l a r i t y of s e r v i c e t i m e s ,

in p a r e n t h e s e s ,

shows t h a t

expressed by t h e i r v a r i a n c e

V s . This

f a c t o r has its s m a l l e s t possible v a l u e ,

1 , if all s e r v i c e t i m e s are e q u a l to t h e c o n s t a n t

and h e n c e h a v e t h e v a r i a n c e

0.

V

=

S

T h e third f a c t o r r e f l e c t s the i n f l u e n c e of the u t i l i z a t i o n tor small; u t i l i z a t i o n s close to

U . Low u t i l i z a t i o n m a k e s this f a c -

can m a k e i t very l a r g e ,

1

Ra.

f a c t o r depends on the a r r i v a l r a t e

(exercise

18).

Only this third

Thus, w i t h g i v e n d i s t r i b u t i o n of the s e r v i c e t i m e s ,

Tw

a c h o i c e of t h e a r r i v a l r a t e a f f e c t s

Ts

only t h r o u g h this t h i r d f a c t o r

(exercise

19).

Exercises. 18

Plot

Tw/T s

exponential distribution 19

over

U

Fs(X)

for c o n s t a n t s e r v i c e t i m e s , =

1 - e -x/Ts .

A s s u m e t h e s e r v i c e t i m e s of a t e l e c o m m u n i c a t i o n

page

38

for c o m b i n e d i n p u t and output.

Tq,

Also,

N w,

find t h e

Nq Ra

for t h e a r r i v a l r a t e w h i c h is which makes

Tw

t i n e as f o u n d in e x e r c i s e

18,

S t a t e further t e c h n i c a l p r o p e r t i e s of t h e l i n e w h i c h

w o u l d m a k e t h e m o d e l of a q u e u i n g s i t u a t i o n a p p l i c a b l e . Tw,

and for s e r v i c e t i m e s w i t h an

8/4

e q u a l to

Assuming the applicability,

of t h e s e r v i c e r a t e

Rs

=

find

t/T s .

3 seconds.

Solutions. 18

For c o n s t a n t s e r v i c e t i m e s , Tw/T

s

=

the variance .5U/(1

T h e e x p o n e n t i a l d i s t r i b u t i o n (see a p p e n d i x Hence

Tw /T s

=

-U) D

U /(1

Vs

is zero.

. has the v a r i a n c e

, p a g e 177 -U)

Then,

Vs

=

T

2 S

,

w h i c h is t w i c e as high.

•!

T h e figure was p l o t t e d ,

~

using

values read from appendix page

170.

C

49 19

C h e c k i n g through the assumptions of p a g e

single server queue,

36 , which defined the m o d e l of a

one notes that

the i n d i v i d u a l t e l e c o m m u n i c a t i o n l i n e is a single server, g i v i n g the s a m e type of service to all messages. the transmission times for messages, rated by one request, -

e v e n if not e x a c t l y i n d e p e n d e n t when g e n e -

are not strongly dependent.

the arrivals of requests form a POISSON process.

request,

and the responses to a message,

Although the messages within a

m a y follow a s o m e w h a t different pattern,

domness of the t i m e s b e t w e e n t h e m (viz.

for reactions of the keyboard operator,

ing in the C o m p u t e r Center) weakens the possible dependencies,

the r a n -

for process-

and l e a v e s the POISSON

process as a good a p p r o x i m a t i o n ' for the arrivals of messages. an u n l i m i t e d w a i t set is a good working hypothesis. T h e a r g u m e n t a t i o n of chapter

3

i m p l i e s however,

as part of the queuing discipline,

that

the server starts a new service i m m e d i a t e l y when it ends a previous service and finds an i t e m waiting,

or when it is idle and an i t e m arrives.

This part of the discipline

holds all queuing variables to a m i n i m u m . For a t e l e c o m m u n i c a t i o n l i n e ,

this f a v o r a b l e assumption is not quite realistic.

It can be

m a d e a p p r o x i m a t e l y true by buffering any w a i t i n g message in s o m e d e v i c e (a processor storage,

a c o n c e n t r a t o r e t c . ) where the l i n e can fetch it i m m e d i a t e l y for transmission,

and by

a fast l i n e control procedure w h i c h notices arrivals and the end of s e r v i c e i m m e d i a t e l y ,

and

a c c o r d i n g l y starts the required actions. For p r a c t i c a l purposes, From e x e r c i s e They result in equals

the f o l l o w i n g results may serve as o p t i m i s t i c bounds.

13 , the e x a m p l e values Rs

• 63 sec -1

=

1/T s

=

Ts

1 . 1 9 sec,

Vs

=

. 8 4 sec -1 ; the suggested arrival rate

and results in a u t i l i z a t i o n

Then

U

=

.75 . Also,

.307 sec 2 Ra

=

V s / T 2~

are taken. 3 Rs / 4

=

.216 .

.75

T

=

1.19 .1.216 .

w

and

=

-

2 . 1 7 sec ,

=

3.36

=

1.37

items

waiting

=

2.12

items

in

2 • (i -. 75)

Tq

=

T w

+

N w

=

Ra T w

Nq

=

N w

+

T s

U

sec

,

the

, system.

50 The general relation T

=

w can be solved for

and to

U

.72 U

2.(1 -U)

(I - U )

1.19.1.216.

U

=

Tw / ( T w + . 7 2 )

U

=

3 / 3.72

Ra

=

U / Ts

.

The example 'values lead to =

. 8 0 5 / 1 . 1 9 sec

.805

,

.68 sec "1

An approximation for the distribution of wait times. Distributions of the queuing variables were considered in chapter

3 , page

wait time distribution Fw(X) , the knowledge of the expectation a reasonable approximation. In fact,

Fw(0 )

Tw

43 ff. For the

is sufficient to find

is also known. It is the probability that an

item does not wait at all. For POISSON arrivals, this probability is equal to the fraction of time for which the server is idle, due to the 'uniformly probing property' of random arrivals. Thus, for

T.-bco,

Fw(0 ) = 1 - U.

The plot of page 44 then suggests an approximately exponential distribution of the non-zero wait times, i . e . Fw(X)

~

1 - U e -cx

Since this distribution has the expectation taking Once

c Tw

=

U / Tw

for

x ~

0,

zero for

U / c, the knowledge of .(see exercise

Tw

x < 0

is included by

20)

has been found from the POLLACZEK-KHINTCHINE formula, percentiles of Fw(X)

can be approximated by -1 Fw (p)

Exercise for

21

~

Tw ( l o g U - l o g ( I - p ) ) / U

for

p~

1 -U

0

for

p~

1 -U

,

uses this approximation for a typical situation of Line design. An exact result

Fw(X), and other approximations, are discussed in later chapters.

Exercises. 20

Show that the above suggested approximation for

Fw(X) has the expectation

u/¢

Find its inverse function. 21 makes the

With the above approximation, and the data of exercise 90-th percentile

Fwl(. 9)

equal to

19,

find the

3 seconds. Compare with exercise

Ra

which 19.

51

Solutions. 20

F(x)

=

1 - U e -ex

b u t i o n for any p o s i t i v e

c;

for

x ~ U

note that

0 ,

z e r o for

x <

0

is a p r o b a b i l i t y d i s t r i -

by its d e f i n i t i o n l i e s b e t w e e n z e r o and 1. T h e e x -

p e c t a t i o n of this d i s t r i b u t i o n is co ./'xF'(x) 0

O0

,ff x d F(x)

0' (l-u)

=

+

-(13

U dx

co J~ cx e c 0

-

-CX

c dx

U --" c

=

1:

as used in t h e text. -1 T h e i n v e r s e d i s t r i b u t i o n is found by solving This is possible for all

p F

between

-1 (p)

p

1 -U T

and

1 - U e -U

=

F(0)

which 21 for

~

F(x)

p , while

>= p

(see page

< 10 )

p

for all n e g a t i v e is

T h e POLLACZEK - KHINTCHINE Tw

in order to express

Fwl(p)

0.

x.

between Thus,

0

and

-1

1 -U

the s m a l l e s t

x

This v a l u e c o m p l e t e s t h e f u n c t i o n

(p) .

48

note

for

F-l(p).

is s u b s t i t u t e d

by design d a t a : vs

l o g ( u / (1-p)

Ts(l+7~) Ts

w h i c h for t h e e x a m p l e v a l u e s is

p

f o r m u l a in t h e version of p a g e

-1

Fw (P)

F

p - 1 ( _log__ -U

w U

F(x)

for

1 , with the result

which is e q u i v a l e n t to t h e o n e s t a t e d in t h e t e x t . For t h e that

(p) /ww

F

t . 445

log(10u

2(1

-U) / (2(1-u))

=

In t h e figure, f(U)

=

,72 • f(U) . the function

log(10U)

p l o t t e d over

/(l-U)

is

U . The utiliza-

tion

U

for w h i c h

f(U)

=

3 / .72

= 4.16

is

r e a d f r o m t h e f i g u r e as a p p r o x i mately equal

to

.58 . The

c o r r e s p o n d i n g a r r i v a l r a t e is Ra

=

.58 /

1.19

=

. 4 9 s e e -1.

This is only

72 %

of the load

52 which was found as permissible under the assumptions of exercise all design specifications which l i m i t queuing variables

19 . Generally speaking,

(such as wait times, response times,

space for wait lines ) tend to also l i m i t the usabIe capacity of the servers concerned. Note that for an exponential distribution of the service times,

Fwl(. 9)

=

T s flU)

exact theoretical result, as shown in later chapters. Then, the above ptot of to read directly the

90-th percentile of the wait times in units of

is an

f(U) permits

T s.

Further results on the remaining service times. In preparation of later chapters, some generalizations of the expression for are discussed here. Throughout this discussion,

t

Ts[ t

(page 4?

continues to denote a 'random arrival'

time. The first generalization arises in queuing systems with priority disciplines. Here it is necessary to distinguish several sub-streams of arriving items, the streams numbered from

1 to N,

say. The items in each sub-stream have different distributions of the service times, viz. Fs, n(X)

for

n = 1 . . . . . N. As a simple example, consider the two streams of Input messages

and Output messages mentionned in exercise

8, page 28 if.

The pattern of service times, and the remaining service time as a function of have as shown by the figure of page 47 n~,

T)

. Hence, the expectation of

ts[ t

t , still be-

is still

2 ts, m / (2 T) .

m=l However, in this sum the contributions of the different streams of items can be distinguished. The number of items from sub-stream the arrival rate stream

n

Ra, n

n , divided by

T , converges (in probabilities) to 2 of that stream. The average of terms ts, m contributed by the sub-

converges (in probabilities) to the second moment of

Therefore, Ts[ t

=

N ~ n=l

Fs, n(X), i . e .

to

En(ts2).

Ra, n En(ts2) 72 :

The contributions of all streams, computed individually with the formula derived earlier, add into the total

Ts[ t .

53 The second g e n e r a l i z a t i o n is to consider the higher m o m e n t s of

E(t note that the case

k

=

s~ t) 1

=

T k lira ~tsl t T,,,lpco 0 ~

dt

tsl t , viz. for

k = 2,3 ....

;

T

was considered on page 47

In order to e v a l u a t e the integral, again the fact is used that tsl t is a l i n e a r function of t k over each service period. Thus, the integral of ts| t over a service period of duration ts, m equals t The integral from na(0, T) / T

0

to

times the

T

Fs(X) , and

/(k+l)

is then a sum of such terms, divided by

9

k E( ts| t )

and g(t

=

s~ t)

By the sequence of m o m e n t s ,

=

Ra E ( t s k + l )

/ (k+

Fs(X)

of the service

=

1)

for

k = 1,2 .....

these higher m o m e n t s also add , as discussed

N k+l ~ Ra, n E n ( t s ) / (k+ n=l

also the distribution of

LFslt(s)

Go (-s) k ~ k=0 k :

ts[ t

1)

for

is d e t e r m i n e d .

k = 1,2 .....

The series

E( ts] t )

is known as a m o m e n t - g e n e r a t i n g function of the distribution D

T. This is now

, the l i m i t (in probabilities) is then the corresponding m o m e n t

Note that for several sub-streams of items, above,

.

k+l -th m o m e n t of the e m p i r i c distribution

times. As quoted on page of

k+l s,m

Fslt(x ) . Page 179 of appendix

discusses some of its properties. With the m o m e n t s as found above, it can be related to

the m o m e n t - g e n e r a t i n g function LFs(S)

=

of the given service t i m e distribution LFs~t(s )

=

co ~ k=0

(-s) k

b

E( t s" ) k '

Fs(x ). In fact,

exercise 22

shows that

Ra 1 - U + - 7 - ( 1 -LFs(s ) ) .

General rules for the m o m e n t - g e n e r a t i n g functions p e r m i t to c o n c l u d e that therefore X

Fs[t(x)

:

1 - U + Ra ¢ ( 1

- F s ( y ) ) dy .

Some consequences of this result are discussed in exercises

23, 24 .

54

A last g e n e r a l i z a t i o n is useful for the analysis of buffers which are d y n a m i c a l l y a l l o c a t e d for message transmission (see chapter

9 ). The size of buffer space required at a random t i m e

follows a pattern similar to that of the figure of p a g e t i m e s of a t e l e c o m m u n i c a t i o n line.

However,

47

. The base intervals are service

the v e r t i c a l scale is a buffer size.

Therefore,

an additional scaling factor appears in the a b o v e results.

Exercises. 22

k E( is-it )

Using the expressions for LFsIt(s )

23

=

stated in the text, prove that

1 - U + R a (1-LFs(S) ) / s

Assume a server in continued use.

and the u t i l i z a t i o n is constant

Ra

is d e t e r m i n e d as the service rate

Rs

1. Find the distribution of the r e m a i n i n g service times for the cases of

service times,

24

Then,

and of s e r v i c e times with the e x p o n e n t i a l distribution

1 - e -Rsx

In order to v i s u a l i z e the s i g n i f i c a n c e of a ' r a n d o m a r r i v a l ' t i m e for the above

results,

find as c o u n t e r - e x a m p l e s the distribution

l a t e r than the begin of a service period.

Fs] t, g i v e n that

t

is by a constant

tO

Consider the service t i m e distributions of e x e r c i s e

23.

Solutions.

22

for

co ~ k=0

(-s) k k'

k E(tslt)

s ~ 0 equals

=

1

+

co (-s) k ~" ~ k=l k'

=

1

+

03 Ra ~,. j=2 -s

E( tsk+l ) Ra

k + 1

(-s) j E(t])

with

j

=

k+l,

j :

Ra

=

1

+

--(LFs(S)

- I + sT s )

-S

= 23

1

-

U

+

Constant s e r v i c e times h a v e the distribution

function as introduced on p a g e z e r o otherwise. T h e n

Fs[ t(x)

8

=

Thus,

1 - Fs(X)

Rs >n(x, T s) 0

dy

Ra( 1-LFs(S) ) /s Fs(X ) equals =

=

(x~ 1

q.e.d.

T s) , a 'truth-value'

for all

min(x, Ts) / T s

x ~ T s , and is for

x ~_ 0,

55

which is the uniform distribution b e t w e e n

0

and

T s.

The e x p o n e n t i a l distribution of s e r v i c e times makes X

Rs j ' ( 1 0

Fslt(x)

~ 0 =

-(1

- e -Rsy) ) d y

e -Rsy Rs dy

=

1 - e -Rsx

Fs(X )

For an e x p o n e n t i a l distribution of the interval length,

the t i m e to the n e x t end of an i n t e r -

val has the s a m e distribution as the interval l e n g t h itself.

This result corresponds to the p r o -

perry of the POISSON arrival process 'to p r o c e e d i n d e p e n d e n t of history from any point in t i m e ' , already used on p a g e 24 than

ts[ t ~=

x

19 .

occurs if a service t i m e is greater than

t o + x . T h e probability

probability,

Fslt(X )

viz.

=

Prob( ts/t

Prob( t O ,~ Fsit(x)

ts ~

~

t o , and also is not greater x )

is therefore a conditional

tO + x )

= Prob( t s > t O ) Fs(t 0 + x) - Fs(t0) =

if 1

For constant s e r v i c e times,

only

Fs/t(x)

tO ~ T s

=

Fs(t0

w h i c h is again a 'single step' distribution, derived in e x e r c i s e

Fs(t0)

/

m a y be considered, + x)

so that

Fs(t0) = 0 . T h e n

,

and is different from the uniform distribution

23.

For the e x p o n e n t i a l distribution, -

Fslt(x)

:

again l e a d i n g b a c k to the distribution

1 .

Fs(t0)

e-Rs(t0 + x) + e-Rst0

of intervals.

e -Rst0 =

1 - e -Rsx

62 Chapter

6

:

Terminal systems with polling.

A telecommunication line as discussed in chapter

2

gives its service to the messages which

arise in geographically different places, the computer center and at least a terminal. In fact, economic reasons often suggest to share a single line of high capacity among several t e r m i nals, even if they are at different locations. Such a situation has two evident consequences, Firstly, the messages rarely arise at scheduled times. Thus, a queuing situation is unavoidable. The set of waiting items consists of several sub-queues at the different locations, as discussed in chapter

3 , page 35

Secondly, it takes time to maintain the queuing discipline. The line itself is used to collect the information on which each decision of the discipline is based, tt carries signals which announce the presence of messages ready for transmission, as well as other signals which test for such presence. This additional use of the line contributes considerably to the total line utilization. It also tends to delay the required transmissions and causes longer wait times. Formulae to evaluate this effect are the subject of this chapter. A particular model of the queuing situation will be used which does not cover all techniques known today but rather the one in most frequent use. The model is described below. Some remarks about other techniques, and references to their queuing theory, are added to this description.

Cyclic polling of terminals. Consider a telecommunication line which is shared among

M

terminals. Assume that it is

devoted to one terminal and message for the whole time of message transmission. In fact, other techniques are known which devote a line on a character-by-character basis. Their queuing theory was considered by Konheim 1. It has proven most efficient to make the computer center responsible for maintaining the queuing discipline. Under such a scheme, the center tests or polls the terminals for messages that are present there, whenever the discipline is to be applied. Of course, output messages

68 which are present in the computer center are considered at the same times. The center polls a terminal by sending it a special message of a few characters length, just enough to identify the special nature of the, message and the terminal concerned. If the terminal holds a message, the line is now devoted to it; if the terminal does not hold a message, it passes its mrn to another terminal. In fact, there is a technique to pass this turn i m m e d i a t e l y from terminal to terminal in an order prescribed by the physical arrangement of the

M

terminals along a looped line. If

passing loops around m the computer center, it indicates that there was no input message at all. Note that this technique always starts the loop with the same terminal, thus crea ~ ting a priority scheme as discussed in the next chapter. The model considered here assumes, however, that a terminal which is polled and does not hold a message, responds with a negative answer of one character length sent to the computer center. The center decides which terminal to poll next. A cyclic order of the polled terminals can be m a i n t a i n e d by using a iist of all terminals which is referenced from top to bottom repeatedly. As an alternative, certain terminals may appear several times on the same list. This would give them some service preference, (exercise 82 ). In order to serve each arriving message as soon as possible, the polling action must occur whenever there is no productive data transmission going on. The line is then fully utilized, although only part of this utilization is productive, i . e .

caused by the external workload.

If continued polling causes too high a load of the computer center, polling may be suspended for fixed time intervals which are interspersed whenever, say, the bottom end of the polling list was referenced. The effect is then a certain increase of the wait t i m e s . Output requirements also contribute to the wait times, tt is often advisable to give them service by the line with a priority over the input. This reduces the space for storing output data in the center, and also tends to speed up the response to each individual input message. Therefore, the contribution of output to the wait times is discussed in the next chapter. A graphical symbol for the queuing situation created by a line with several terminals is suggested in the following figure.

64

Ra,1

Ilk-

v

illl

queues

arrivals

:

"

!

~

b-

Fs(X)

a,M

The probability assumptions. A result which shows some analogy to the POLLACZEK-KHINTCHINE formula can be derived from the following assumptions : Arrivals to each terminal form a POISSON process, with rates general are different for each terminal, i . e .

for

Ra, m

which in

m = t . . . . . M . The sum of these rates is

Ra , a total arrival rate for the line. The case where all

Ra, m

equal

Ra/M

is discussed

first. Also for the first discussion, continued polling according to a list which shows each terminal once is assumed. Extensions of this assumption follow later. Each polling of one terminal takes the same t i m e combination times and

II 7

Tp . For the l i n e - t e r m i n a l

of page 28 , an unsuccessful or negative poll takes two line turnaround

character times. Thus,

Tp

equals

.168 see.

The distribution of transmission times is the same for all terminals, In the following derivations, the t i m e for a polling which precedes a transmission, i . e . is considered separately, and is assumed equal to with the distribution exercise

Fs(X), are by

Tp

a positive poll,

Tp . Then, the transmission times

ts ,

and two additionally assumed turnaround times (see

8 , page 28 ) smaller than the service times considered in exercises

8

and

18.

In fact, they are thus reduced to the times for transmission of externally given data. E . g . , for the l i n e - t e r m i n a l combination II the reduction is by the average

Ts

for input B

ances remain unaffected, e . g .

is now Vs

1.12, =

.22G sec

(see exercise

.328 for input B.

(see page

13, page

28 ) .

Thus,

38 ) etc. The v a r i -

65 A last assumption m a d e here is

that a positive poll causes the transmission of just

one message from the terminal concerned. This transmission being completed, the polling scheme proceeds to the next terminal on the list. Other possibilities which also have been suggested for p r a c t i c a l applications, and may be enforced by technical properties of the terminals, are to

flag all messages currently at the terminal, then transmit them all ; or

to

transmit all messages from the positively polled terminal, including tkose

which arrive during the transmission. The queuing theory for these cases was studied by L e i bowitz I and

Cooper et al. 1

The expected c y c l e time. A polling line gives attention to a particular terminal cycle t i m e s ,

denoted by the random variable

m

in intervals which will be called

t c.

In a time interval of duration

T

terminals was polled

times. This polling held the line busy for a t i m e computed

nc(O, T)

which covers

nc(0, T)

c o m p l e t e cycles, each of the

M

as

T' The remainder

T"

=

nc(0, T) M Tp .

of the interval considered was used for a number

transmissions. This remainder is the sum of

ns(0, T)

T"

=

ns(O, T) r s

T

=

nc(0, T) M T p

ns(0, T)

of productive

service times, hence equal to

In total, +

ns(0, T) Ts

According to the above assumption that one transmission follows each positive poll, ns

cannot be greater than

n c M.

In analogy to the reasoning of chapter

2 , consider as a first consequence the situation

where each poll is positive. This results in the highest achievable productive utilization of the line. The numbers of polls and services b e c o m e equal, and as a result the number of services per time unit is ns(0, T)

1

T

T p + ~s

Its l i m i t (in probabilities) as

T-4~eo

is apparently

1/( Tp + T s ), a value which is 1..ess

66

than the maximal service rate

Rs

= 1/Ts

achievable in the long run without polling. Cor-

respondingly, the productive utilization ~(O,W)

=

T" / T

cannot have a l i m i t (in probabilities) greater than T Tp

1

S

+

1 +R

T s

This bound depends on the poll time ratio

R

=

T p / T s , and becomes lower as the poll

time increases relative to the service time. A second consequence arises, again in analogy to chapter number of services is determined by for

T T

na(0, T), the number of arrivals. Then the expression

is T

But

2 , for systems in which the

=

is also the sum of T

=

1

---

A division leads to

n c ( 0 , T ) M Tp nc

+

na(0, T) gs

cycle times, i . e .

nc(0, T) ~c

/

M Tp

0

na(0, T) +

~S

rc

,

T

Thus, for the assumed model of the polling line, the average cycle time probabilities) as

T ~

m

to a value

T

t-c

converges (in

which can be found from C

M Tp 1

=

+

Ra T s

Tc M Tp

as

=

T c

=

M Tp / ( 1 - U )

,

1 - Ra T s where

U

is the productive utilization in the long run. As indicated above, this result is

applicable only if

U <

I/ ( I + R )

.

An approximation for the variance of cycle times. In order to evaluate the remaining time of a cycle during which an item arrives, the variance of the cycle times needs to be known (see chapter 4 ). Among the

nc(0, T) M

polls executed,

ha(0, T)

were positive. The relative number of

67

p o s i t i v e polls, na~0, T)

n a ( 0 , T)

gc

z

,.,

no(0, T) M

T

M

has t h e l i m i t (in p r o b a b i l i t i e s ) p

Ra T e / M ,

=

w h i c h c a n b e i n t e r p r e t e d as a p r o b a b i l i t y t h a t an i n d i v i d u a l poll of any t e r m i n a l is p o s i t i v e .

The number fact,

ns

of s e r v i c e s in o n e c y c l e t i m e has a p p r o x i m a t e l y b i n o m i a l p r o b a b i l i t i e s .

if e a c h t e r m i n a l has i n d e p e n d e n t l y t h e s a m e p r o b a b i l i t y

number

ns

of p o s i t i v e polls a m o n g the

M

p

of a p o s i t i v e p o l l ,

In

the

polls of a c y c l e has e x a c t l y b i n o m i a l p r o b a b i -

lities, with the expectation

and t h e v a r i a n c e (see a p p e n d i x

D

E(n s)

=

M p

=

Ra T c ,

V(ns)

=

M p (1 - p)

,

, p a g e 178 ). M a c k 2 i n d i c a t e d t h a t this i n d e p e n d e n c e does n o t a p p l y

exactly under general assumptions.

In t h e s a m e sense a p p r o x i m a t e l y ,

the cycle times n s

have the variance

tc

=

M Tp

+

Vc

=

E(ns) V s

=

Mp

m'~=l ts, m +

2 V(ns) T s

(see a p p e n d i x

D,

page

180)

( Z ( t fi) - p Ts2)

Exercise

27

Assume

M

m o d i f i e d on p a g e

64

=

6

t e r m i n a l s on a l i n e w i t h t h e s e r v i c e t i m e s of e x e r c i s e

. C o n s i d e r t h e a r r i v a l rates

resulted from exercises

19

and

Ra

=

21. F i n d t h e v a l u e s of

. 6 8 sec -1

and

13

as

. 4 9 sec -1 w h i c h

T c , p , and t h e a p p r o x i m a t e V c.

Solution. 27

Line d a t a are

poll t i m e r a t i o is results d e p e n d i n g on

R

Ts =

Ra

=

1 . 1 2 sec , V s

.147 .

=

. 3 2 8 sec 2

and

Tp

=

. 1 6 5 sec. T h e

T h e b o u n d for the p r o d u c t i v e u t i l i z a t i o n is

are a r r a n g e d in t h e f o l l o w i n g t a b l e :

.872.

The

68

e a

U

=

p

T C =

see-1

=

VC

sec 2

SeC

.63

• 706

3.36

• 353

2.41

.49

.549

2.19

.179

1.46

The expected wait time. A reasoning analogous to that of chapter time

Tw

4

now leads to a formula for the expected wait

• It assumes as given that there is statistical stability•

The wait time of a particular message which arrives at time of

t

to any terminal is the sum

a remaining part of the current cycle time for that terminal, and one cycle time for each further item which precedes the newcomer in the server,

thus t

w

=

tel t

~"

+

over preceding items m

tcll, m

The cycle times per preceding item are conditioned by that each of them starts with a transmission• The l i m i t (in probabilities) for Tw where

Ra / M

=

T-)-oo Tc[ t

is then, in analogy to page

+

Ra T w Tc~ 1 / M

Tcl t Tw

= 1 - Ra T 4 t

/ M

denotes the expected remaining cycle time, and equals, according to exercise Tc] t

Tc! 1

,

enters as the rate of arrivals to the particular terminal considered• This

equation solves for

Tcl t

46 ,

=

E( tc2 )

/ (2 Tc) .

denotes the expected cycle time, given that its starts with

1

terminal considered. If the probability of a positive poll for the other were not affected by this condition, Tc[ 1

=

M Tp

23,

Tci 1 +

transmission from the M - 1

terminals

would simply equal Ts

+

(M - 1 ) p T s .

However, this argument is approximately valid only for high poll time ratios or low productive utilization• In the other cases, the conditional cycle time is further increased by the influence of additionally expected arrivals at the other terminals.

69 Exercises. 28

Find the l i m i t s of

29

Write

Tw

T c, p,

Vc, Tc/t

and

Tw

as the u t i l i z a t i o n

U

tends to 0.

as a product of the POLLACZEK-KHINTCHINE result of page

a factor of the form

(1+

clp)

/(1+

c2P)

with the probability

the fact that this factor must be greater than

p

of page

1 , derive a lower bound for

47

, and

67 . From

Tc[ 1 .

Solutions. 28

The u t i l i z a t i o n

tends to zero.

tends to zero, with the other design data unchanged, if

Ra

T h e f o r m u l a e of the three preceding pages show that then

Tc

tends to

M Tp , i . e .

the t i m e for

M

n e g a t i v e polls,

the probability

p

Tc[ t

M Tp / 2 , half a c y c l e of n e g a t i v e polls. F i n a l l y ,

Tw M(Tp+T

U

tends to

of a positive poll, and with it

tends to the same l i m i t

s ) , h e n c e its product with

M Tp / 2 , since Ra

V c, tend to zero.

Tc[ 1

Thus

c a n n o t be greater than

tends to zero.

There is a positive m i n i m a l w a i t r i m e expected e v e n for low u t i l i z a t i o n . 29

Into

Tw

= Tel t

Tel t / ( 1 - Ra T c i l / M ) substitute the relations 2 = E(tc2) / ( 2 T c ) = ( T c + V c ) / ( 2 Tc)

Tc

=

Mp/R

Vc

=

Mp(E(t?)-PTs2

T

=

a

=

MRTs/(1

-U)

,

) ,

in order to find

( M PAa ? + M p ( E @ w

2(Mp/R

a) (1 -R a(

- p

Ts + Tell/M-

) T s)

m

2 ( 1 -RaTs)

1 + p(MT

s - T e l l ) / ( M T sR)

which is of the form suggested in the exercise, T h e polling scheme must result in a g~eater Tw

than expected for i m m e d i a t e service; therefore

In terms of

Tc[ 1 , this a m o u m Tcl 1

with

c1

~ =

e2

is necessarily smaller than

to M T s ( 1 - c 1 R) ( M - U 2 ) / ( R a2 E(ts2) ) .

c 1,

70 _An improved approximation for the conditional cycle time. If the poll time ratio

R

is small relative to

-I cI ,

Tc~ 1

is bound to be close to

M Ts ,

indicating that almost all terminals are expected to contribute a transmission time to A simple iational approximation for and the approximation of page Tcl 1 so that M-1

P[1

Tc[ !

Tel 1.

which is a co mp'on:ise between this behaviour

68, is found by writing

=

M Tp

+

Ts

+

p~l(M-1)T s

,

can be interpreted as a conditional probability of a positive poll for any of the

remaining terminals. Then

pt 1

is bounded by M

p lI

>

1

-

c3 R

with

c3

-

( 1 + c I) M-I

The compromise between this bound and the fact that P]I ~ P approximation

I -p P~I

~

P

+

1+

c3R

This expression is always greater than either bound; it approaches 1 - c3 R

is then expressed by the

for small

p

for large

R , and

R.

Exercise. 30

With the data of exercise

resulting

T w.

27,

compute

Tcl I

as described above, and the

Solution. 30

The intermediate and final results are stated in the following table.

Ra see

Cl

c3

Pll

-1

Tcll

Tw

sec

sec

.63

8.76

11.7

.59

5.42

4. 73

.49

15.0

19.2

.39

4.31

2. 21

Note the increase of found in exercises

Tw

19,21 .

over the comparable values

3 see and I sec

without polling as

71

The large number of parameters entering the computation of

Tw

makes it difficult to give

a comprehensive graphical survey of their effects. The following figures show somewhat the nature of the dependence of variance

Vs

Tw

on varied terminal number

of the transmission times.

more of which are given in appendix

::~

.222 2_~--~; ~ [ 2 + H Ti

__~__

I;

''

l

~

I[i

The figures as based on computer remits,

some

C , page 173 .

;;:'

....

The first figure shows

Tw

its dependence on

for the

U

in

i

poll t i m e ratio :

R, and

M, poll t i m e ratio

R

=

.15 (see

_ ~ ~ a i ~ ----4-:

exercises), and constant service time. 20.

M varies between

,,.i,]],-J.

I

'i

;

,

i iliii!l

!

: !2~

?,'T; ; i ! [

~;l:~z:~::t 1. t ii!~-:l_-._i:.--

i-~:~!~¢-!ZiqP:i .... ~llmr-

2 2 : I--~:~."~:_-: 1 ! ~

#

: '

-

~

'f

'-

~I""

_1. . . .

and

The figure gives e v i d e n c e

that the dependence on . ~H:,TI~ [ ]:i-~- x~i[i ~-. k.,[[idil [ /

2

M

is

almost linear.

7-

~

The systems designer is usually

"

-!2":r=~iK_" ,t ; ~,~J~& :.: :it' 7 r ~ . ~ _ + L ~ : . L A ; 4 i : ' : : ~ ram:

faced with the problem to decide which terminals out of those l o c a t e d in some geogra-

T--.XI ~m.47 ; [ TTi - T . 2 ~ T

2 1 ~ T+ i - ' T ' - I - ~

~

phical area may share a line.

........................

The workload of each terminal is then given as a fixed contribution

Um

=

Ra, m Ts

to the productive utilization of the line to which it will be a t -

tached.

The selection of certain terminals now influences both

M

and

U.

A useful aid in this situation can be a table of highest permissible utilizations different

M

considered, assuming a certain l i m i t for

the above figure, assuming

Tw

=

U

for all

T w. One such table was read from

2.68 T s , corresponding to

3 sec with the data of the

preceding exercises: Number Bound for

M U

4

5

6

7

8

9

i0

I!

12

13

14

18

16

.66

.64

.62

.60

,57

.85

.53

.51

.49

.47

.45

.48

.41

An application of this table is shown in the next exercise.

72 A second figure shows the dependence of t i m e ratio minals, R

Tw

=

Tw

on the poll

R. For

M = 8

ter-

and the ratios .3,

.16,

.075, .025, .01,

is plotted over

bound

1/(I+R)

U. The

for the u t i l i z a -

tion is indicated by a vertical dotted l i n e with each curve. Vs

is assumed as in the pre-

vious figure. Within the precision of the plot, the last curve coincides with the lower bound of

Tw

given by

the POLLACZEK -KHINTCHINE formula (see page

47

.)

Finally, the effect of service t i m e variance and arrival p a t tern is shown by plotting for

M = 8, R = .15

ral

Vs

Tw

and seve-

which arise for e x p o -

nential distribution F s, for the exercises, and for constant serv i c e times. For comparison, the lowest curve gives

Tw

assuming con-

stant service times and ' c o n v e r sational' arrivals as discussed under the next heading.

73

Exercise.

31

Assume that each terminal produces messages with an average interval of

equal to

40 sec . This is roughly the shortest possible interval with which the customer

requests

CR1

of the case study can be entered at a terminal,

Based on the data of exercise contributed by each terminal,

27 , page

67

Ta, m

as described on page

, find the productive iine utilization

2 U

m

and the largest possible number of terminals that could share

a single line, with reference to the table of page

71

Solution. 31

The contributed utilization per terminaI is computed as Um

The total contribution of table for all

M

=

Ts/Ta, m

M

terminals,

=

1.12/40

.028,

or

2.8 % .

M U m , remains below the bound of the quoted

15, where it reaches the value

up to

=

.42 .

The arrival pattern for conversational use of terminals. Keyboard terminals are often used in a sort of conversation where after each input message the operator waits for a response before he prepares his next input. It is evident that the arrival times in such a case cannot form a POISSON process with mutually independent arrivals After each arrival, at least the wait time,

then the transmission t i m e have to elapse before

another arrival becomes possible.

Assume that the t i m e

tr

from the end of input transmission to the arrival of the next m e s -

sage has an exponential distribution with the expectation mutually independent.

The time

tr

t i m e for output of a message if any,

T r , and that all such times are

includes message processing in the computer center, the and the t i m e to prepare the next input message.

Then the queuing situation for each terminal is equivalent to that of a m a c h i n e waiting for repair by a cyclicly patrolling repairman. for the repairman,

and

tr

There,

ts

is a repair time,

a t i m e to the next failure after a repair.

Tp

a walking time

Mack et al. 1 studied

this problem for constant servioe times; Mack 2 also considered variable service times.

74

T h e i r essential result is a recursion formula for the probabilities that e x a c t l y

m

services are g i v e n in one c y c l e ,

Pm

=

Prob( n s

m )

for

m = 1, 2, . . . . M,

viz.

M -m+l Pm

=

( cm-lc'

- 1 )

Pm-1

m where

e Ts/Tr C

aIld

The e x p e c t e d number of transmissions per c y c l e , M g(ns) = ~" m Pm ' m=l can be c o m p u t e d from the Pro"

bution of

tr

97

, this determines

assures that the

Tcl t

=

and the v a r i a n c e of n s , viz. M V(ns) = ~ m2 Pm m=l

T c , E(tc2)

and

=

Tc~ t

E2(ns )

T c l t . The e x p o n e n t i a l distri-

is still a p p l i c a b l e , s i n c e it m a k e s

t i m e to a n e x t arrival in a POISSON process. Finally, Tw

e

1 , they are c o m p l e t e l y d e t e r m i n e d by the reeursion.

S i n c e the sum of these probabilities is

In analogy to p a g e

M Tp/T r

c'

=

tr

b e h a v e l i k e the

the e x p e c t e d w a i t t i m e is

,

since there are no other messages at the s a m e t e r m i n a l which p r e c e d e the n e w c o m e r . T h e procedure MACK of appendix

E

, p a g e 183 , shows the required computations.

r i c a l results of this procedure are i n c l u d e d in the tables of appendix previous figure m a k e s e v i d e n t that the resulting rivals.

Tw

This is due to a similar e f f e c t as discussed in chapter 8, p a g e

for g i v e n

T r. In design situations,

for g i v e n

U

however,

U

, p a g e 173 . The

is considerably less than for random a r -

Note that the procedure MACK c o m p u t e s the u t i l i z a t i o n

Tw

C

Nume-

U

=

61

, already.

E(ns) T s / T c

is a design d a t u m rather than

as a result T r . The

can be found by r e p e a t e d use of the procedure in a t r i a l - a n d - e r r o r

s c h e m e l i k e the r e g u l a falsi. A p l o t of the result for s o m e c o n v e n i e n t values

Tw

o v e r the result

U , both c o m p u t e d

T r . is another possible t e c h n i q u e .

G e n e r a l i z e d t e r m i n a l l o a d and polling lists. If t e r m i n a l s with different arrival rates p e r m i t that some terminals, list than others.

viz.

Ra, m

are considered,

those with a higher load,

it also b e c o m e s plausible to

appear m o r e often in the polling

75 Some results for this case are stated here. However, their difference from those given above is significant only in certain extreme situations. Let terminal

entries.

m

Lm

times in the polling list, which then has a total length of

=

M ~2 m=l

occur

Terminal

m

b m

is polled whenever an entry

over the list. In analogy to page 65 interval

(0,

T)

is denoted by T

but also where the

tc, m

'm'

is encountered during the scan

, the number of polls for terminal

m

during a time

nc, m(0, T). Then

=

L nc, m(0, T) ~m Tp

=

nc, m(0, T) tc, m

+ na(O,T) ~s

'

,

are the intervals between consecutive potls of terminal

m. Their average

has a l i m i t (in probabilities) found from L 1

=

T

Lm Tc, m i.e.

L T

Among the logy to page

Ra

T s

,

Tp

=

c,m

nc, m(0, T)

+

P

Lm

( I -U)

polls of terminal

m, na, m(0, T)

were positive. Therefore, in aria-

67 , the relative number of positive polls has the l i m i t Pm

=

Ra, m Tc, m

'

which can be interpreted as the individual terminal's probability for a positive poll. The expectation and variance of the number

ns

of services between two polls are then approxi-

mately

E(ns)

=

L Pm/L m

and

V(ns)

=

E(ns) (1 - Pm) •

Thus, the values

Vc, m '

Tclt, m ' T c l l , m and finally

dually for each terminal with the formulae of pages numerical results, comparable to those of exercise Ra, m

Tw, m

can be computed indivi

67 ft. The next exercise gives some 30. In fact, the influence of varying

on the average wait time expectation appears not very strong.

76

Exercise. 32

C o m p a r e the results of e x e r c i s e

nals,

3

with the following situation :

h a v e an arrival rate t w i c e as high as the

into the polling list. minals,

30

C o m p a r e the i n d i v i d u a l

3

others,

Of

6

and are p l a c e d t w i c e as often

Tw, m , and their averages over a l l

for the same total l i n e utilizations as in e x e r c i s e

termi-

6

ter-

30.

Solution. 32 Lm

The suggested polling list could h a v e =

1

for t e r m i n a l s

could be

4 to 6. Thus

L

=

Lm

=

2

for t e r m i n a l s

1 to 3 , and

9. A possible a r r a n g e m e n t of the entries

I, 2, 4, 3, I, 5, 2, 3, 6

T h e f o r m u l a e to be used are Ra, m

=

Lm R a / 9

T c , in,

=

9 Tp/(L m (l-U))

Pm

=

Ra T p / ( l - U )

Vc, m

=

9 p ( E(t s2 ) - p Ts 2) / Lm

Tc{t, m

=

. 5 ( T c , m + Vc, m / T c , m )

Tel1, m

=

--

=

p

9

and a p p r o x i m a t e l y

(Tp+p

T s)

+ (1 - p )

Ts ,

Lm from which

T

can be found• Lm

is either

1

Ra

Ra, m

Tc, m

Vc, m

Tclt,m

Tel1, m

Tw, m

weighted average Tw

see -1

sec

sec 2

sec

sec

see

see

I

.054

3.28

2.19

1.98

4.21

2.56

2

.109

1.64

1.09

1.15

2.56

1.60

1

• 070

5.03

3.62

2.88

5.76

4. 82

2

. 140

2.51

I. 81

I. 62

3.24

2.96

Lm

sec-1

• 49

• 63

w,m

=

Tc[t, m / (1 - Ra, m Tc[1, m ) or

2. With the data of e x e r c i s e

30, the results are :

1.92

3.58

T h e results for e q u a l t e r m i n a l load,

c o m p u t e d with the a b o v e a p p r o x i m a t i o n of

Tcl 1 , are

77 1.91

and

3.56 sec , respectively, i.e.

the same as the average

the difference between individual terminals. nevertheless the lower values of

T w. More noticeable is

The terminals with the higher load

Ra, m

have

Tw, m " The preference given to these terminals by placing

them twice onto the polling list is stronger than would be required to balance out the wait times caused by different load.

A rough approximation for

Tw .

Inspection of the figure on page 71 suggests an approximation

Tw(U, R)

=

Tw(0,R) Uma x

+

Uma x where the point of infinity, tion,

Tw(0,R )

=

Urea x

M Tp / 2 Tw(U, 0)

, =

Tw(U, 0) U

= I/(I+R) , the wait time expectation for low utilizaand the bound for low poll time ratio

R,

E(~) U / 2(1-U) ,

are known. With some rearrangement, MR

rw(U, R)

+

U (I+R) (l+Vs/Ts2)

= 2

[I - U (I+R)]

Ts

results. Values computed with this expression are greater, i.e. earlier expressions, but by not more than practice.

t0 %

more pessimistic, than those of the

for the range of arguments which arise in

78 Chapter

7

:

Some results on queues with priority disciplines.

Queuing disciplines which consider priorities of some items are an important technique for improved service. They permit to reduce the response t i m e for services of high importance. In many cases they also reduce the overall response t i m e of the system. As a first example, chapter

6

mentionned that it is often advisable to give the output a

priority over the input of messages. The effect of such a discipline is considered in this chapter. To begin with, some general notions of priority disciplines are discussed. Then some results are reported which are,

- similar to the approach in chapters

4

and

6 -, concerned with

the l i m i t expectations of averages, assuming as given the statistical stability of the queuing system. Exercises for their application follow here, but also in later chapters. Complete results of this part of queuing theory can be found in Chang's 1 paper, which also gives further applications to computer systems analysis.

General notions of priority disciplines. Priority disciplines assume that each i t e m entering the system belongs to one of

N

classes

which have a well defined order. The classes are identified by elements of an ordered set like the first

N

letters of the alphabet, or the integers

n = 1 .....

N

.

In a n o n - p r e e m p t i v e priority discipline, an item may enter service only if there is no i t e m of a preceding priority class waiting. Among items of the same priority, FIFO discipline is assumed. Furthermore, the server always completes a current service before a new item is considered. In a preemptive priority discipline, the priorities are also considered at each arrival time. Thus, the service being given to an i t e m may be preempted by the arrival of items of a preceding, i . e .

'stronger', priority class. The preempted service will later have to be

resumed at the point where it was interrupted, at some earlier checkpoint, or even restarted completely. Only the first of these possibilities will be discussed here.

79 A further assumption for the following discussion is that each item maintains its priority class throughout its presence in the system. Thus, the priority is defined for each arriving item. The type of time sharing disciplines which modifies priorities according to, say, the time already spent in the system is not covered here. The probability assumptions. The items of a class

n

can be distinguished upon arrival. Their arrivals are assumed to

form a POISSON process with rate

Ra, n

(n = 1 . . . . . N) . The arrivals for all classes toge-

ther wiI1 then form another POISSON process with rate Ra as discussed in exercise

N ~n=l

=

Ra, n

'

88.

Mutual independence of the service times, and a common distribution Fs(X), are assumed as before. However, each priority Class may have a different conditional service time distribution

Fs, n(X) , given that an item is in class

n. This condition has the probability

Ra, n / R a , so that the total service t i m e distribution can be expressed as Fs(X)

=

N Z n=l

Ra, n Fs, n (x) / Ra "

Its expectation and moments are similar sums over the conditional moments, respectively, e.g. Ts

=

N ~ n=l

Ra, n Ts, n / Ra

As results, the conditional probabilities for the queuing variables, given the priority class of an item, are of interest. They are all denoted with an extra index

, n . Total probabilities

and moments are found as sums analogous to those above. Statistical stability is assumed for each class. In particular, the utilization of the server by items of class

n

wilt converge (in probabilities) to Un

=

Ra, n Ts, n n

An abbreviation useful in the following formulae is for the utilization by all classes up to

n ~ with

Sn SO

O.

=

~ m=l

Um

80 Under a non-preemptive discipline, the wait time for a particular item of class arrives at time

t

n

which

is the sum of

a possibly remaining part of the service time underway, if any, the service times of all not 'weaker' items already waiting, and the service times of all 'stronger' items arriving during this wait time. Proceeding i m m e d i a t e l y to the limits (in probabilities) of averages, the equations

Tw, n = Tslt, n arise for

+

n ~ Ra, m Tw, m Ts, m m=l

n-1 + ~ Ra, m Tw, n Ts, m m=l

n = 1 . . . . . N , is close analogy to the reasoning of page 46 • They can be rear-

ranged into the recursive form T w , n ( 1 - Sn )

=

Ts[t,n

n-1 ~-" m=l

+

Um T w , m

and can therefore be solved by repeated substitution. E . g . ,

Tw, I

=

Tslt,i/(

Tw,2

:

(Tslt, 2

I - v I)

+

for

,

n = 1

and

2

the result is

,

Ts~t,l UI/(I

-UI)

) /(I

-S 2)

The solution is especially simple if the expected remaining service time is the same

Ts[t

for all items. Then =

TW, n

Ts[t

for n = t , . . . , N

,

(1 -Sn_ 1)(1 -S n) 34. In fact this is rather the usual situation with non-preemptive prior-

as shown in exercise ities.

Values for the other queuing variables follow from the general results of page 37

and

Tq, n

=

Tw, n

+

Ts, n

Nq, n

=

Ra, n Tq, n

Nw, n

=

Ra, n Tw, n

, viz.

, ,

This last equation did already enter the above reasoning. Overall values for the stream of all items are found as sums with the weight

Ra, n/Ra, e.g.

N T

W

Tq

=

n=l =

Ra,n Tw, n / R a

Tw

+

Ts

=

Nw / Ra

,

81 Exercises. 33

Show that the time to the next arrival from any of N

with the rates

Ra, n

has the same probability distribution as the time to the next arrival

from one POISSON arrival process with a rate 84

PO[SSON arrival processes

Ra

which is the sum of all

Ra, n,

Show by induction that the equations

Tw,n(1-Sn)

=

T

=

Ts] t

are solved by

Um T w , m

Tsl t w,n

(1 -8n_ t)(1

$5

Assume that there are three classes

val rates

.1,

.25 aad

seconds. The priorities

merits m i n i m i z e

-S n)

I, II and III of items with the respective arri-

1 see -1 , and the respective constant service times .2, 1,2, and 8

can be assigned to these classes in

For each assignment, compute the three values

36

n-1 ~ m=l

+

Tw, t , which m i n i m i z e s

6

Tw, n , and the overall

different ways. T w.

expression for 37

1

in order to m i n i m i z e

Fs(x)

of service

Tw,1 ? Also, find the

Tw.

Assume the data of exercise

resulting from

Which assign-

Tw ?

Consider the case in which all items have the same distribution

times. Which class should have priority

1 and . 4

Input B and

27 , page

Output (see exercise

over the input. Find the approximate

Tw, n

67 , and consider the Ra 4,

used there as

page IG ). Give the output priority

for both output and input messages.

Solutions. 33

The time to the next arrival from any of

m i n i m u m of to

t

-

N

independent random variables

ta, n

N

POISSON arrival processes is the with the distributions

Fa, n(X) equal

e - R a , n x . The distribution of the m i n i m u m is then Fa(X)

=

1

=

1

Prob( all ta, n > x ) N -

~

(1-Fa,n(X ) )

n=l

= with the

Ra

as stated above.

1

-

e-RaX

(see appendix D, page 180)

82 34

n = 1

For

rect for a l l

the c l a i m e d solution is evidently correct.


m

Now assume that it is c o t -

Then Um Tw, m

=

(Sm

- Sin-l)

(I

Tw, m

- Sm_ 1 - I + Sin)

(1 -S m_l)(1 =

-Sm)

Tslt 1

Tslt

Sm

-

Tsl t

I - Sm_ 1

If this expression is substituted in the sum of the given equations, leaving

all tema~ but one cancel,

Ts[ t Tw, n( 1 -S n )

= ( i - Sn_ 1 )

as had to be proven. 35

The total rate of arrivals is Ra Ts

=

(.1".2

The dividend in S3

Ts

=

+ .25+

.4)

U 1 + U 2 + U 3

=

According to page

1,35 sec -1. The overall

/R a

=

Ts

is found as

,496 sec .

gives at the same t i m e the total u t i l i z a t i o n

The expected r e m a i n i n g service t i m e Ts[ t

=

(.1..04

=

.67

Ts~ t

+ .25 + .16)

.

is c o m p u t e d as /2

=

Ra E ( t : ) /2,

as

i.e.

.207 sec.

52, each s u m m a n d in this c o m p u t a t i o n is a r e m a i n i n g service t i m e e x -

pected by items of one class if they were alone in the system. Further results are arranged in the following table : Class assigned to priority

S1

S2

Tw,l

Tw,2

Tw,3

Tw

see

see

see

sec

1

2

3

I

II

III

.02

27

.211

289

.859

• 706

II

I

III

.25

27

.276

378

.859

•

II

III

I

.25

65

• 276

789

1.79

.768

I

III

II

.02

42

•

211

364

1.08

•486

III

I

II

.4

42

• 345

• 595

1.08

• 500

III

II

I

.4

65

• 345

• 986

1•79

•

The v a l u e

Tw, 1

716

571

is m i n i m i z e d by any priority assignment giving strongest preference to

83 class 1/(1

-

I U1)

which has the lowest u t i l i z a t i o n . In fact, such a choice m i n i m i z e s the factor in

Tw,

1

•

T h e smallest overall T w

is found by choosing class

class has the second lowest service time.

III

The m i n i m a l

for the second priority l e v e l . This Tw

• 520 sec , while without

is

priorities the expected w a i t t i m e would be Tw

36

If all

Ts, n

:

Ts~ t / ( 1 - S 8 )

:

• 628 sec

are equal to

T s, the utilizations

Tw, 1

Ts] t / ( 1 - U 1 )

Un

=

Ra, n Ts

are proportional

to the arrival rates• In order to m i n i m i z e

=

, priority

1

should be given to

the class of items with the lowest u t i l i z a t i o n , h e n c e the lowest arrival rate. The overall expected wait t i m e Tw

=

Um Tw, m

Tw

is in this case

N ~ Ra,n Tw,n /Ra n=l N ~... Un T w , n / ( Ra T s ) n=l

:

Using the expression for

Tw

derived in exercise

=

rs It ( - 1 - SN

=

Ts] t / ( !

•

84 , one finds

Ts{ t ) /

SN

-S N )

This is the same as the POLLACZEK-KHINTCHINE result for an equally loaded queuing syst e m without priorities:

Without different conditional service times,

a n o n - p r e e m p t i v e priority

discipline can no longer i m p r o v e the overall wait situation . 37

Since priority

1

is g i v e n to output, priority

=

.103 sec 2,

Ts, 1

=

,9'7 sec, Vs, 1

Ra, t

=

Ra /

Ts, 2

= 1.35

Ra,2

=

2. 741

(from page 15

sec, Vs, 2

1. 7 4 1 R a / 2. 741

=

El(ts2)

2

to input, the p r o b l e m data are:

= 1 . 0 4 sec 2

(from page

33 ),

),

. 3 2 8 sec 2, (from page

E2(ts2) 15 ).

= 2 . 1 4 sec 2

(from page 33

The polling t i m e is

Tp

=

),

• 165 sec.

Some values resulting from these data were computed for the following table. They i n c l u d e

84 the expected service time remaining upon arrival of an output message, according to page 52

Tstt, 1 , which

is found as the sum for three types of remaining service, viz. another

output transmission, an input transmission preceded by the positive poll, or a negative poll. Thus, Ts|t, 1 Note that

1 - S2

=

( Ra, 1 El(ts2) + Ra, 2 E~ts2) + (1-$2) Tp )

is the non-productive utilization, hence

(1 - $2) / Tp

/ 2 the rate of n e g a -

t i r e polls. Ra

Ra,1

Ra, 2

U1

U2

sec -1

sec -1

see -1

.63

.23

.40

.22

.54

.49

.18

.31

.17

.42

S2

Tslt, 1

Tw,1

sec

sec

.76

.57

.73

.69

.46

.56

The effect of the priority discipline on the input wait times is not expressed by the formulae of this chapter but may be approximated by results of the previous chapter. For messages waiting at a terminal, it is still the c y c l e t i m e which determines the wait times. The equation of page 66 T

=

which lead to

nc(0, T) M Tp

+

T c , viz.

ha(0, T) gs

must now be understood to account for both input and output messages arriving and serviced. Hence,

M Tp Tc

=

1 - Ra T s

still holds under the priority discipline, where

Ra

and

Ts

refer to the combined message

stream. Vc, Tcl t

and

Tc] 1

will be somewhat affected by the priority discipline. Anyhow, they

were considered only approximately in the previous chapter. Thus. for

Tw, 2

to use the results of that chapter, with a productive utilization

S2

U

=

it is suggested

from both input

and output, and with the service t i m e distribution for the combined message stream. The computations are then similar to those of exercises

27 if.

85 The preempt-resume discipline. Preempting priority disciplines are often used by programmed computers. Modern hardware and programming techniques tend to minimize the need to repeat interrupted service, and the time required to apply the discipline.

Thus, some results on preempt-resume priorities

(see page 78 ) are added here, in preparation for later chapters. An item which is interrupted while in service, returns into the queue where it will be reconsidered by the queuing discipline whenever applicable.

The time spent in the wait line may

therefore consist of several periods, all of which contribute to the time

tq

spent in the

queuing system. It is the total time in the wait line which is denoted here by which equals

t

w

=

t

q

- t

t w , and

s

The time which a particular item of class

n

spends in the system is the sum of

a possibly remaining part of the service time underway, if any, the service times of all not 'weaker' items already waiting, the service time of the item itself, and the service times of all 'stronger' items arriving during the time which the item spends in the system. The first two of these summands arose already with the non-preemptive priorities on page 80 Together, they can be interpreted as the wait time caused by the items present upon arrival of the newcomer. The results for non-preemptive priorities show already (see e . g . exercise that the expectation of this wait time can be found from the POLLACZEK-KHINTCHINE formula as T'w,n

=

Ts]t, n / (1 -Sn)

The expected remaining service time

Ts[t, n

is found by adding (see page

52 ) all the

contributions of the item classes which the newcomer cannot preempt, viz. as n

Tslt, n

=

~r"

Ra, m Em(ts2) / 2

m=i

Using these results, the expected time spent in the system satifies the equation n-1 Tq,

n

=

T'

w, n

+

Ts, n

+

m=~-'lRa' m

Tq, n Ts, m

I

38 )

86 which has the solution Tq, n By a subtraction of

:

Ts, n ' Tv,n

( T~c,n

+

Ts, n ) / ( 1 -Sn_ 1 )

the expected wait time =

( T~r,n

+

Tw, n

is found as

Sn-1 Ts, n ) / ( 1 - Sn_l )

Exercises. Show that the result for non-preemptive discipline can be arranged as

38

Tw,n

where

T'w , n

=

T(v,n

+

Sn-i Tw, n

follows from the POLLACZEK-KHINTCHINE

, formula stated on the previous

page. 39

Reconsider exercise

35

with preempt-resume priorities, and compare the results

36

with preempt-resume priorities.

of the two disciplines. 40

Reconsider exercise

Solutions• From

38

Ts]t, Tw, n

= (i -Sn_l) ( 1 -Sn)

follows

Tw, n ( I - Sn_ I)

=

Tw,n

+

T'w , n

and

:

Tw, n

Sn-1 Tw,n

as stated in the exercise. 39

The results are given in the fallowing table :

Class of priority

Tslt, 1

Tslt, 2

Tslt, 3

Tw, 1

Tw, 2

Tw, 3

Tw

1

2

3

see

see

see

see

see

see

see

I

II

I[I

• 002

.127

.207

.002

.198

1.01

• 783

II

I

Ill

.125

.127

.207

.167

.299

1.01

• 799

II

III

I

.125

.205

.207

.167

.914

2.16

• 868

I

III

II

.002

082

.207

.002

.152

1.81

.447

Ill

I

II

.08

.082

.207

.133

.369

1.81

.460

IH

II

I

.08

.205

.207

.133

2.16

.563

1.64

8q Comparison of these results with those of exercise duction of

Tw, 1

caused by p r e e m p t i o n .

which produces the s m a l l e s t

on page

Still i t is the class

82 I

shows the significant r e -

with the lowest u t i l i z a t i o n

Tw, 1 •

The overall e x p e c t e d w a i t t i m e assignments,

35

is further reduced by p r e e m p t i o n for some priority

Tw

but i n c r e a s e d for some others. The next exercise looks at the basic reason for

these effects. 40

If all

are equal,

Fs, n(X) Tslt, n

the expressions derived above simplify to

n Z m=l

= = =

is the sum of the

can be expressed as

+

Tw, n , w e i g h t e d with the factors U n / S N . tn the resulting sum,

( 1 - S n ) ( i - S n _ 1) SN T

s

Sn-i Ts ) / ( 1 - Sn_ 1 )

2 Ts

U n Sn

while

,

E(t 2)

(- -

1 - Sn Tw

2

Sn E(t 2) / ( 2 T s ) Sn

Tw,n

E(t 2) /

Ra, m

Ra, n / R a

E(ts2) / 2T s

which for equal

Fs, n

has the factors

Un

Un

(1-Sn)(1-Sn_ 1) S N

( 1 - S n _ l ) SN

has the factors Un

Un

SN

(I-Sn_l) S N

U n Sn- 1 / (I-Sn_I)S N =

The sum c

N

Un

n=l

(1-Sn_l) SN

= N 2Y" Un/S N = 1 . The sum of the positive n=l was a l r e a d y found in exercise 34 as 1 / ( 1 - S N ) . Thus finally

is term by term greater than the sum factors to

E(ts2) / 2 T s

E(t_2) Tw T h e first s n m m a n d c -1

=

~ 2 Ts

if the expected service time

Ts

E(ts2) +

1 - SN

is the P O L L A C Z E K - K H I N T C H I N E

is positive. T h e preempt-resume

E(ts2) / 2 T s

SN

(c - I) ( T s - - ) 2 Ts result as found in exercise

34

already.

discipline increases the overall expected wait time

is greater than the expected remaining service time

of the interrupted service, given that a service is underway (see exercise

In terms of the variance

V s , the condition for increased

Tw

is

Vs<

2 Ts .

28).

88 Chapter

8

:

Some applications of Markov processes.

One problem of Communications Network Analysis, though appearing early in the survey of page

6

, was delayed until now due to its mathematical complexity : The queuing situa-

tion in front of several identical servers. In the Case Study, this situation arises at all offices with more than one operator and terminal, all of which can give the same service to an arriving customer request. The queuing discipline directs an arriving or waiting item to any free server. This server is held for the total service time. The complexity of these situations, as compared to the system with a single server, lies in the fact that an individual wait time is no longer the fairly simple sum of service times as on pages 46 and 68

Instead, a certain number of remaining or full service times must be

completed before an arriving item can get service. The analysis of a queuing situation is very much simplified if there are times

t

such that

the future of the queuing system is conditionned only by its state at the time

t , but not

by the history of the process before this time. Processes with this property were first studied by MARKOV (see Parzen 1) . Within this course, the POISSON arrival process arose as the first example of a process which has the Markov property, in fact for any time 19

). Exercise

24

t (see page

indicated that it is the assumption of interval times with an exponential

probability distribution which gives the Markov property for any

t

to certain queuing pro-

cesses. Therefore, exponential distribution of service times is assumed throughout this chapter. The results of chapter

4

show that this tends to result in more pessimistic estimates of the

queuing variables than with the service time distributions actually occurring in most applications. The relatively high variance of the exponential distribution is the basic reason for this tendency. For the arrivals, a POISSON process will be considered. However, the same m a t h e m a t i c a l reasoning can be applied if only a finite number of items may request service, each with an exponential distribution for the time

tr

between the end of a service and the next arrival

of the item. The analysis of such cases is included here, for use in later chapters.

89 _Queuing processes with Markov property.

Consider the number of items in a queuing system at time

t, viz.

nq(t). It describes the

state of the queuing system. Assume that this state is a random variable,

with the probabili-

ties Pq, j(t) Then,

:

Prob(nq(t)

=

j )

for

j = 0,1,2 ....

the expected number of items in the system is CO

Nq(t)

=

~ j pq, j(t) j=O

,

and the expectations of other queuing variables in which the systems analyst may be interested will be related to by a convergence (as

Nq{t). Statistical stability of the queuing systems would be caused

t - - ~ co ) of the probabilities

pq, j(t)

to limit values

pq, j .

Assume further that conditional probabilities pq, jlk(tl, to) exist for any pair ing

t0,t 1

of times.

=

Prob(nq(tl)

= j ]nq(t0)

= k )

; j,k = 0,1,2...

They will be called transition probabilities since, suppos-

t I > t O, they refer to transitions of the system from state

k

to

state

j.

Also, the states at several preceding times m a y be given, so that conditional probabilities Pq, jlk, 1 . . . ( t l ' t0' t-1 . . . . ) arise. If a process has the Markov property for equal to

= P r o b ( n q ( t l ) = jlnq(t0) = k, nq(t_l) = 1. . . . ) t 0, these conditional probabilities are all

pq, jlk(tl, to) . Once these are known, they describe essentially the whole process.

In facL from the state probabilities state probabilities for a later

t1

Pq, j(t 1)

--

pq, k(t0)

and the above transition probabilities,

the

follow by the law of total probabilities as CO

i.e.

Z Pq, j{k (tl, to) Pq, k(t0 ) k=0

J = 0, 1 . . . . .

by a matrix multiplication pq(t 1)

=

Pq (tl, t 0) pq(t 0) •

Next, assume that the process has the Markov property also for lities for a later

t2

t 1. Then,

the state probabi-

are found from the matrix operations P4t2)

=

Pq(tg, tt)

pq(tl)

--

Pq(t2, tl) Pq(tl, to) pq(t0) ,

9O i.e.

a repeated matrix multiplication with the transition probabilities for adjacent t i m e inter-

vals. This argument can be extended to any number of intervals, supposed that the process has the Markov property in the points which separate the intervals. This approach has advantages for many applications. It is often easier to find transition probabilities for short t i m e intervals than for long ones. Then,

the matrix multiplication is a

simple computational procedure• Finally, in many cases the transitions probabilities are the same for all intervals considered. Then the procedure reduces to taking powers of a single matrix, and its l i m i t values can be found by even simpler computations.

Exercise. 41

h server with a single potential user is assumed idle at t i m e

its state probabilities at

ti

=

i time units, i = 1,2 . . . . .

=

t

when the server

if busy at t i m e

t i, is idle at t i m e

ti+ 1

with probability .1 ,

if idle at t i m e

t i,

ti+ 1

with probability

is busy at time

0. Compute

.4 .

Solution• 41

Only the transition probabilities

pq, 0}l(ti+l, ti)

=

.1

are stated. They are, however, sufficient to c o m p l e t e the matrix

and

i. The initial state probabilities are

Pq, 0(0)

= .4

since the sum

Pq(ti+ 1, t i)

over each of ira columns is a total probability to reach any state at

holds independent of

Pq, l | 0 ( t i + l , t i )

ti+ 1, hence equals

=

1 , Pq, l(0)

=

1.

0.

Some results of the repeated matrix multiplication, its general result and its l i m i t foUow: limit

i

1

2

3

4

general i

Pq, o(ti )

.6

.4

.3

• 25

. 2 + •8"2 -i

.2

Pq, l ( t i )

.4

,6

.7

.75

.8 - . 8 . 2 -i

.8

The l i m i t values can be found directly by solving the linear equations as components of an eigenvector of the matrix

Pq.

pq

=

Pq pq , i . e .

91

Markov queuing processes with constant intensities. Queuing processes in w h i c h the distributions of service times and times to the n e x t arrival are exponential, h a v e the Markov property at any t i m e be found easily for short intervals of length

t.

Their transition probabilities can

dt . Thus, their state probabilities satisfy simple

differential equations. Consider a system of system,

u(t)

=

with the rate tion

Fr(X)

servers, each with the s a m e

min(j,M)

Fs(X ). If there are

j

items in the

of the servers are busy. Also, consider either POISSON arrivals

Ra , or a f i n i t e population of =

At any t i m e

M

N

items,

each with the return t i m e distribu-

1 - e -Rrx , as assumed for the conversational t e r m i n a l use, page t, the t i m e to the next end of a service is the m i n i m u m of

service times. H e n c e its distribution is (see exercise 33 of page 81 Fslt(x)

--

u(t)

73 remaining

)

t - e - u ( t ) Rs x

The t i m e to the next arrival has the distribution Fair(X) for POISSON arrivals. If the m i n i m u m of bution

=

nq(t)

N -nq(t) Fair(X)

1 - e-Ra x of the

N

items from a f i n i t e population are in the system,

possible return times has to be considered, which has the distri=

1 - e - ( N - n q ( t ) ) Rr x

It is now characteristic for the queuing processes that the state

nq(t)

changes by just one on

any arrival or end of service. A transition from one state to a neighbouring state can be caused by one of these events, or by a larger, j, where

Ij - k] • i , requires at least

In a short t i m e i n t e r v a l

odd n u m b e r of events. A transition from

to

IJ - k] events, but can be caused by more.

dt, one arrival occurs with the probability

service with the probability

k

Fs~t( dt ). These probabilities are small,

Fa[t(dt),

one end of

since from the power

series of the e x p o n e n t i a l function it follows that Fa[t( d t )

= or

and

Fslt( d t )

=

Ra dt

+

O(dt 2)

(N - nq(t) ) Rr dt u(t) Rs d t

+

+

O(dt 2)

O(dt g) .

But they are nevertheless the only significant contributions to the transition probabilities. T h e

92 transitions caused by m o r e than one events occur within

dt

with a probability which is a

product of more than one of the a b o v e probabilities of order

dt , h e n c e is at least of the

order

O(dt2).

T h e transition probabilities

pq, jlk(t+dt, t)

are therefore

for

j

=

k-1 (end of service)

Pq, k_l~k(t+dt, t)

=

m i n ( k , M ) Rs dt

for

j

=

k+l (arrival)

Pq, k+l~k(t+dt, t)

=

Ra dt

or

for

j

=

(no change)

k

and of the order

=

O(dt 2)

O(dt 2)

(N - k) Rr dt

+

O(dt 2)

1 - sum of the above

for all other index combinations.

O(dt 2)

The equation

pq, klk(t+dt, t)

+

+

pq(t+dt)

=

Pq(t+dt, t) pq(t)

(see page

89, with t 1 = t+dt, t o = t)

can be rewritten as pq(t+dt) - pq(t) where

I

is the i d e n t i t y m a t r i x . p~(t)

Here,

18

k ;

k ;

Rai k

=

is

Pq pq(t)

or

=

c a g e d the transition intensities in analogy to

m i n ( k , M ) Rs

(N - k ) R r

in the p l a c e ,

in the p l a c e ,

in the diagonal p l a c e of column

- Rsl k - Ra~ k

dt --~ 0

!

Rsl k

Ra

,

After division by dr, the l i m i t for

=

Their values are

of column column

( Pq(t+dt, t) - I ) pq(t)

is a m a t r i x of constant coefficients,

P~

page

=

just above the diagonal,

just below the diagonal, k ;

of

and zero otherwise.

Some e x a m p l e s are written e x p l i c i t l y in the next exercises. Thus,

the state probabilities

constant coefficients.

pq, j(t)

satisfy a system of l i n e a r differential equations with

In the following exercises, some systems

are stated, solved and i n t e r -

preted.

Exercises. 42 and

43

Give the matrix

P$

for a single server, assuming P O I S S O N

N = 3. Solve the differential equations for

arrivals, then

N = 1

N = i.

By adding the first j + l differential equations (j = 0,1 . . . . ) find a system of e q u i -

v a l e n t equations with less n o n - z e r o coefficients. 44

By adding all differential equations,

differential equation for

Nq(t)

m u l t i p l i e d with appropriate integers, find a

and solve it as far as possible. When is

Nq(t)

bounded ?

93

45

Show t h a t t h e POISSON p r o c e s s c a n b e d e f i n e d by t h e p r o p e r t y to h a v e m u t u a l l y

independent

inter-arrival

times with a common

exponential

distribution.

Solutions. 42

With a single server,

a r r i v a l s r e s u l t in

P~

a l l s e r v i c e i n t e n s i t i e s for n o n - z e r o

states equal

R s. POISSON

=

-

Ra

-(Ra+R s)

0

Ra

0

i!

Rs

-(Ra+Rs) etc.

For

N

=

1 , only the states

0

and

1

are possible,

and

Rs

For

N

=

,IF

3 , the possible states are

0, 1, 2, 3, Rs

- 3R r p'

=

Apparently, is a l w a y s

p~,0(t) 1.

for

Rs

0

0

2 Rr

-(R r + Rs)

Rs

N

Rr

• 0

=

1

=

_ R r Pq, 0 (t)

P ; , 1(t)

=

g r Pq, 0 (t)

p~,l(t)

=

With the substitution

0

-

Rs

are

Pq, 0 (t)

+

0

-(2 R r + Rs)

0 equations

0

3 Rr

q

The differential

and

+

Rs Pq, 1(t)

for a l l

Pq, l ( t )

R s Pq, 1 (t)

t , s i n c e t h e s u m o f t h e two p r o b a b i l i t i e s

=

1

-

Pq, o(t) , a s i n g l e d i f f e r e n t i a l

equa-

tion Pq, 0(t)

+

(R r + R s) Pq, 0(t)

=

Rs

is f o u n d w h i c h h a s t h e s o l u t i o n Rs Pq, 0(t)

=

+ R r + Rs

Whatever run

the initial probability

Pq. 0(0)

is,

to

(

Rs Pq, 0 (0) - _ _ Rr + Rs

the probability RS

Pq, 0

=

lira t-~m

pq, 0(t) Rr + Rs

e -(Rr + Rs) t )

of s t a t e

0

tends in the long

94 If

ts

failure,

is interpreted as the t i m e to repair some device, while pq, 0(t)

tr

is the t i m e to the next

is the probability that the device is no___!in repair. Then Rs Pq, 0

Tr

-

= Rr + Rs

Ts + Tr

is a well known expression for the a v a i l a b i l i t y of the device, expected in the long run.

48

Addition of the first

cancels the factors of

j + 1

pq, 0(t) . . . . .

differential equations of the system

p~(t)

= Ph pq(t)

pq, j_1(0 , l e a v i n g

J p~, k(t)

=

- Ra[ j pq, j(t) + Rslj+ t pq, j+l(t)

j = 0, 1 . . . .

k=0 Thus, only two coefficients r e m a i n different from zero in each equation, viz.

the n e g a t i v e

of an arrival intensity and a service intensity. This result makes it especially simple to find l i m i t probabilities for

t - - ~ ca , as discussed under the next heading. GO

44

Nq(t)

=

O0

~

j pq, j(t)

has the derivative

N~(t)

:

5

j ph, j
sum results to the left of the equality sign i f alI differential equations are added after a m u l t i p l i c a t i o n with

j . To the right of the equality sign, ( k+ 1

k ) Ral k + ( k - 1

Thus, Nq(t)

=

oo X k=O

the service rate at t i m e

k ) Rs[k

Ral k Pq, k (t)

The two sums which arise here are expectations,

=

Ra

-

M

Rs[k

Rs[kPq, k (t)

viz. of the arrival rate at t i m e

t, and of

Nq(t)

=

Nq(0)

E(u(t) )

Ra

"

T h e expected service rate depends

E(u(t))R s

so that

Ra T s <

Ral k -

E(u(t) ) Rs . Then,

N~(t)

Nq(t)

:

will have the factor

t .

on the expected server use, and is

servers. Hence,

m X k=O

-

For POISSON arrivals, the arrival rate is constantly

The expected server use

pq, k(t)

+

Ra Rs t ( m Rs

1 t - --./~ t 0

E(u(t'))

at'

)

and its t i m e average c a n n o t exceed the n u m b e r

can only be bounded if

is a sufficient condition (see page

Ra/R s

=

Ra T s ~

38 for comparison). If

M

of

M . tn fact, Ra T s >

M , the

95 queuing system is no longer statistically stable. However, the expected number finite time approaches the expression Thus, a slight overload by an

Ra

Nq(0)

+

( Ra - M Rs ) t

not much greater than

Nq(t)

for a

as all servers become used.

M Rs

might not do much harm

in some finite time period. For a finite population, N~t(t)

=

(N-Nq(t))l~

-

E(u(t))R s .

The solution of this differential equation is always bounded. In fact, Nq(t) can never exceed the number 4~

N.

Consider the arrival process as a queuing process without service, i.e.

with

Rs = 0.

The assumed exponential distribution of interarrival times gives to the process the Markov property for all times applicable.

t. Thus, the above results, especially the result of exercise

Assume that no item is in the system at

the number of arrivals between and

pq, j(0)

=

0

for

0

j >

and

=

0. Then, the state

t . The initial probabilities are

0 . It need

probabilities with the parameter

t

only be shown that the

nq(t)

Pq, 0(0)

pq, j(t)

The probability to have no arrival satisfies the differential equation =

- Ra pq, 0(t)

Pq, 0 (t)

=

e-Ra t

with

pq, 0(0)

=

0

hence equals By induction, if then

pq, j(t)

pq, j_l(t)

=

( Ra t)J-1 e-Rat / (j-l) :

,

sitisfies the differential equation pq, j(t)

=

- Ra pq, j(t)

pq, j(t)

=

e -Rater a(R at') j-1/(j-1) :

f

+

Ra ( R a t ) j - t

e -Rat / (j-l) :

hence

(Rat)J e-Ra t / J : which completes the proof.

,

dt'

=

gives 1,

are POISSON

Rat.

p'q, 0(t)

42, are

96

The l i m i t probabilities. The l i m i t values

pq, j

=

l i m pq' j(t) t-~oo

of the state probabilities c a n be found without

e x p l i c i t solution of the differential equations. If the functions p'q,j(t)

their derivatives

values in a f i n i t e circle,

tend to zero.

pq, j(t)

The particular form of

P~

tend to constants,

places all its e i g e n -

and assures that no oscillations with high frequency are possible. 48 , page 94

Thus, the differential equations found in exercise

, tend to the a l g e b r a i c re-

iations 0

=

Rs[j+l

Pq, j+l

Ra{j Pq, j

J = 0,1 ....

By repeated substitution, Ra[j Pq, j+l

=

~

Rsij+ 1

The sum of all probabilities equals

Pq, 0

=

a/~

where the sum is extended from

Pq, j

J = ('IT

= "'"

I . Hence,

RaIk

k=0 Rs[k+1

Pq, 0

) Pq, O =

cj+l Pq, O"

can be found as

cJ

j

0 , with

co

=

1, to

or

co

N. Then, the other pro-

babilities can be found recursively from the previous equation.

Exercises. 46

For

M

servers and POISSON arrivals, find

the expected server use has the l i m i t the l i m i t s of the probability

P

-

U

=

Ra T s

1 - Fnq(M-1 )

pq,j

for

j -- 0 , 1 , 2 . . . .

if this v a l u e is less than

Show that M. Also find

that all servers are busy, and of the e x -

pected numbers of items in the system and in the wait line. 47

From the Case Study data,

find the expected hold t i m e

operator. With the worst case request rates according to exercise m i n a l use

U

Ts

of a t e r m i n a l and

4 , find the expected ter-

for the different locations.

Use the table of appendix

C

, page 172 , to find approximately the numbers of t e r m i n a l s

at least required in the different locations if the probability to find no free t e r m i n a l upon arrival has to b e less than

.2 .

97 Solutions. 46

With

Ra] j

=

Ra , and

e.l and The series of the

c: J

Rs] j

=

m i n ( j , M ) Rs , the coefficients

=

(Ra Ts)J / j :

for

=

(Ra Ts)J / ( M ' M J - M )

cj

are

j __< M , for

is g e o m e t r i c with the quotient

j

~

Ra T s / M

M . for

j

~

M.

If the

quotient is not less than

1, the sum over this series diverges. Then, the equations h a v e

only the trivial solution

pq, j

tend to zero as higher indices

t - ~ oo.

=

0

for all

j. This i n d i c a t e s that all probabilities

pq,j(t)

Consequently, the n o n - z e r o probabilities arise for higher and

j. The queuing system is unstable.

If the quotient is less than 1 , the sum of the cj oo M-1 pq~l = ~ ' - cj = ~ " cj + c M / ( 1

j=o

can be written as -R aT s/M)

j=o

.

The limit probabilities are pq, j For the states up to

M

=

cj pq, 0

they are in the same relation as POISSON probabilities; above

they form a g e o m e t r i c series. In particular, for the single server system with are g e o m e t r i c probabilities. One c o n s e q u e n c e is that in this case the state

M

M = 1, they

0 , with idle

• server, has the highest probability of a l l states. The e x p e c t e d server use has a l i m i t defined by probabilities are n o t all zero. m i n ( j , M ) pq, j so that

U

equals

=

Ra T s

U

=

0o Z j=l

m i n ( j , M ) pq, j , if the l i m i t

Now, m i n ( j , M ) cj Pq, 0

=

Ra T s Cj_l Pq, 0

=

R a T s Pq, j - 1

m u l t i p l i e d by the sum of all probabilities, which is

1.

The probability that all servers are busy usually is an i m p o r t a n t design criterion. Also, it arises within the expressions for other parameters in which the designer is interested. For the above case i t has the l i m i t

M-t P

=

1 j=O

which is

I

Pq, j

'

if the system is not stable, and (13

P

=

~ - Pq, j j=M

=

Pq, M / ( 1 - U / M )

if it is stable.

98 Two tables for this expression are given in appendix shows percent values of

P

for various

M

is increased to serve a given workload

and

C

U/M.

, page 172 . The first of them Note that

P

decays sharply if

M

U. The tab!e is then read in a Iower row and in

earlier columns. The second table gives

-logP

for various

U

and

M - U . It is used to find how many

servers more than those in use on the average are required to hold value. The logarithmic scale is more convenient if very small

P

at a certain small

P are required. The next

exercise makes use of this table. The expected number of items in the wait line of a stable system has the l i m i t O3

Nw since

j - M

j=~

items wait i f

(j-M)pq,

j ~

M

=

W

N

Finally,

q

,

are in the system.

sum form a g e o m e t r i c series, so that, N

j

The

pq, j

appearing in this 12, page

in a certain analogy to exercise

33

,

Pq, m (U/M) / (1 - U/M) 2

U/(M-U)

=

p

=

N

W

+U

47

The rimes for which a terminal and operator is occupied by a request are stated

on page

2

t i m e is

1 . 6 neh / 3

in relation to the k e y - i n times. For the input message length

the hold rime is 1.4 n c h / 3

=

sec

with the request types

3 nch / 3 sec, i . e . 1 . 4 - 12

=

CR2, CR3 and CR4. With the type CR1,

the value for the other types, plus an extra

16.8 sec.

The average amount of input data per request was found on page 15 Nch

=

nch , the hold

125.4 . The extra t i m e for

CR1

occurs in

16820/25000

, input case =

67.3%

C , as

of the r e -

quests. Thus the average hold t i m e is, not accounting for the response of the computer, T s

=

1.6"125,4

=

78.2 sec

/3

+

.673

• 16.8

sec

The computer response t i m e is bounded by the requirement of page 5 . The average key-in t i m e of an individual message is

12 sec, with the average length of input

A, page

15

99

In order to make this value the

90-th p e r c e n t i l e of the response t i m e distribution, its

expectation must be considerably less. Thus, Ts

=

a hold t i m e average of

90 sec

is a pessimistic assumption. 3293 per hour

The worst case request rate of

3

to the request numbers of page

is split into the values per location according

. The following table gives the resulting rates per hour,

per second, and the expected terminal use, assuming Location

GE

FRA

KOP

Ts

as above.

LON

i

MIL

PAR

|

Hourly rate

395

519

250

501

341

446

Rate in sec -1

.110

.144

.069

.139

• 095

.124

9.88

12.96

6.24

12.53

8, 54

11.15

Expected use

U

The second table of appendix

C , page 1 7 2 , shows a probability

value

-logP

U

M -U

=

3

= for

1,61

, if

P

=

.2,

i.e.

the

is integer, and the additional number of terminals is

U = 5or 6 ,

4

for

U = 7 through 11,

5

for

U = 12 through 19 .

Thus, the required numbers are roughly for location ml

M

i

=

With further

13

[

GE

FRA

KOP

LON

MIL

PAR

I

14

18

9

18

13

16

for

AMS, and

9

for

MAD

and

VIE

each,

terminals.

the total number is

119 terminals. Note that this total is still almost proportional to the average hold t i m e

T

. Any reduction S

of this value affects the terminal number to the same percentage. t i m e of

4

instead of the assumed

12 see

E.g.,

an average response

results in a total number of

112 terminals.

The case study states another criterion than considered in this exercise, viz. a bound for a percentile of the wait t i m e distribution.

This is the subject of the next exercise.

The wait t i m e distribution. The assumption of exponential service time distributions eliminates the difference between c o m p l e t e and remaining service times. Thus, the wait t i m e of an i t e m is

0

with the pro-

100

bability

1 - P

that not all servers are busy. It is not greater than s o m e

POISSON process of rate least

j -M

used on p a g e

M Rs

which is formed by the e n d - o f - s e r v i c e events,

events in the t i m e 21

already.

curs with the probability

x . The probability for this is

Here,

j

generates at

Gammaj_M (MR sx)

is the state encountered by the arriving i t e m ,

as

and o c -

pq, j .

Taking total probabilities over all possible states, Fw(X) By c o l l e c t i n g powers of

x } 0, i f the

one arrives at

oo 1 - P + ~ pq, j G a m m a j _ M ( M Rs x ) j=M

=

x , or by taking a detour over m o m e n t g e n e r a t i n g functions,

one

can si~nplify this sum to Fw(X)

=

1

-

P e

-(M

-U)Rsx

for

This is a distribution of the form already suggested in chapter 4,

x

>

0

°

50 . Its e x p e c t a t i o n

page

is r

=

PT

W

/(M-U)

=

a

a result which is consistent with the g e n e r a l relations of chapter The p e r c e n t i l e s of the distribution are -1 F w (p) = - log((1-p)/P)T and

0

for the other

R N

S

p . If

s/(M

, W

8 .

-U)

if

p~

1 -P,

T s, U, and a bound for s o m e p e r c e n t i l e are g i v e n ,

to be chosen as the s m a l l e s t i n t e g e r for which

Fwl(p)

is less than the bound.

M

has

The n e x t

e x e r c i s e shows an a p p l i c a b l e t e c h n i q u e .

Exercise. 48

With the data of e x e r c i s e

4%

choose

p e r c e n t i l e condition stated in the case study,

M

page

for e a c h l o c a t i o n so as to satisfy the 5 . Find the l i n e l o a d per t e r m i n a l .

Solution. 48

The stated condition is

Fwl(.9)

~

20 sec . With

Ts

=

90 sec,

M must be

chosen such that (2.3o + log p) / (

M - g ) <

2O / 9O

=

.222 .

This i n e q u a l i t y for a transcendental function is satisfied by trying several

M ~. U .

101

E.g.,

for

U

p a g e 172, M-U

=

6.24,

t h e v a l u e of

-logP

c a n b e r e a d f r o m t h e t a b l e in a p p e n d i x

by i n t e r p o l a t i o n b e t w e e n t h e t a b l e e n t r i e s for

= M-7 , with fixed

U = 6, M - U

= M - 6 , and

M . T h e n e c e s s a r y c o m p t a t i o n s are in t a b u l a r form

:

M-6

M-U

M-7

9

3

2.76

2

1.630

-.24(1.630

-

.955)

=

1.47

.30

10

4

3.76

3

2.290

-.24(2.290

- 1.506)

=

2.10

• 053

M

=

10

is c h o s e n for t h e l o c a t i o n

( 2 . 3 0 + l o g P) / ( M - U )

KOP.

In the s a m e way t h e f o l l o w i n g

further results are f o u n d :

Location

GE

FRA

KOP

LON

MIL

PAR

M

14

17

!0

17

12

15

=

The total,

including three further locations,

is

117

terminals.

Again,

terminals•

a c h a n g e of

Ts

has an a p p r o x i m a t e l y l i n e a r e f f e c t .

T h e a v e r a g e c o n t r i b u t i o n of a t e r m i n a l to t h e l i n e l o a d c a n now b e d e t e r m i n e d . p e a k hour m e s s a g e r a t e line service time

Ts

Ra =

9033 h -1

1 . 1 9 sec

U Each of t h e

=

=

2.98

16

of p a g e

2 . 5 sec -1 33

of p a g e

16

With the

, and the expected

, t h e t o t a l l i n e use is

.

117 t e r m i n a l s c o n t r i b u t e s

c o n t r i b u t i o n a s s u m e d for e x e r c i s e

=

31.

2.8%

line utilization.

According to

T h i s is s l i g h t l y less t h a n t h e

t h e t a b l e of p a g e

71

it p e r m i t s up to

terminals per line.

However,

t h e b o u n d of

row if t h e

T

<

3 sec

a s s u m e d for t h a t t a b l e m a y a p p e a r s o m e w h a t too n a r -

9 0 - t h p e r c e n t i l e of response t i m e s is at

lysis w o u l d r e q u i r e k n o w l e d g e of t h e v a r i a n c e c h a p t e r 14 . As a s i m p l e a s s u m p t i o n , single l i n e e a c h ,

lines.

as discussed in

FRA, LON,

a n d PAR

by a

T h e f i n a l d e c i s i o n depends on t h e

a n d w i l l a t t e m p t to m i n i m i z e t h e cost of t h e l i n e

as i n d i c a t e d in t h e survey of this s e c t i o n . 7

98 ). k close a n a -

in t h e p o l l i n g system,

w h i l e t h e o t h e r l o c a t i o n s share lines.

Later chapters will assume

(see p a g e

it s e e m s p o s s i b l e to serve

g e o g r a p h i c a l s i t u a t i o n a n d t h e l i n e tariffs, network,

Vw

12 sec

,

U = 7

M

As a result,

- log P ( U , M - U )

C

Section COMPUTER

2

CENTER

ANALYSIS

The number of decisions required for the design of a c o m p u t e r center is considerably higher than for the c o m m u n i c a t i o n s network. -

This is because

the c o m p u t e r center has to h a n d l e several functions,

storage and r e t r i e v a l of data,

the essential ones being the

the control of the c o m m u n i c a t i o n s network and its data flow,

and the processing of data. -

the d e v i c e s that are a v a i l a b l e for c h o i c e differ not only in operation speed but

also in their l o g i c a l structure.

Channels for data transmission are m o r e or less dependent on

the central processing unit (CPU) ; control units m a y be l i n k e d to one or m o r e channels ; storage devices m a y or m a y not require a seek m o t i o n before data can be t r a n s m i t t e d ; the control of the c o m m u n i c a t i o n s network m a y be a function of the CPU or be p a r t i a l l y shifted to control units ; etc. -

a v a r i e t y of p r o g r a m m e d techniques are in use to e x e c u t e the functions of the

c o m p u t e r center. methods,

e.g.

Different methods for data r e t r i e v a l and transmission constitute the access

for sequential files, indexed files, t e l e c o m m u n i c a t i o n s .

Supervisor programs

and priority disciplines l e a d to further distinctions. As a consequence, this section

2

w h i l e the network design was c o v e r e d almost c o m p l e t e l y in section

rather discusses some s e l e c t e d problems of c o m p u t e r center analysis,

1 ,

- main-

ly those in which probabilities p l a y a m a j o r role. Tfiey can be classified by functions of the c o m p u t e r center to which they are r e l a t e d :

Function

I , Communications

control.

Design data are the line network specification,

control procedures,

length distribution and

arrival rate of messages. The decisions concern the

core buffer provision; storage and r e t r i e v a l of messages not in

process; t e l e c o m m u n i c a t i o n s access m e t h o d and control programs. Problems for analysis are

103

the core buffer requirements,

(chapter

9)

storage and retrieval of messages as discussed under the next function. Function

2 , Storage and retrieval of data.

Design data are the volume of permanent and temporary data sets, the number of accesses per application or control function, and the total rate of application and control functions. The decisions concern the types of storage devices, layout of data sets on the devices, required number of devices and their assignment to control units and channels. Problems for analysis are Access rates per d e v i c e and per data set, distribution of seek times and transmission times,

(chapter

10)

(chapter

11)

utilization of channels, devices, data sets, wait times for channels, devices, data sets, response times for data sets. Function

3 , Processing in the CPU.

Design data are the contributions to CPU t i m e use of the applications and control functions, the rates with which these forms of use occur, and the interference from channels. The decisions concern the CPU type, channel types, required numbers of these, and the core requirements. Also, a supervisor program, and processing priorities have to be chosen. Problems for ~

are

the total CPU utilization, the utilizations and wait times per priority cIass,

(chapter

12)

(chapter

12)

and, by putting together results from all three functions, the overall message response t i m e s .

It should again be noted that programmed computations for Computer Center Analysis have been conceived, following the above order, and have been used successfully.

104 Chapter

9 :

Core buffer requirements.

Message transmission over a t e l e c o m m u n i c a t i o n line, once it is started, continues to the end of the message irrespective of how the characters are received in the computer center. It is necessary to provide enough space in core memory of the center so that character by character can be stored there as they arrive. Otherwise, data might be lost. The length of an input message is usually not known to the computer center when a transmission starts. A simpIe provision would then be to atlocate for each line enough core storage to hold the longest possible message. In the case study, Input

B, the longest messages have

150 characters. Some

30

additio-

nal characters of core memory are required to store an identification of the message and a time stamp, arranged in a message control header. The total buffer space is then for the 7

lines which are at least required (see page 101 )

However, the average message length is only about

7 • 180

=

1260 characters.

60 characters for the combined input

and output. Also, the fraction of time during which a iine actually needs its buffer is given by the productive line utilization, which may be about

40 %

(see again page 101 ).

Finally, the buffer use begins with a single characters and builds up to the full message length during an input transmissions; similarly, it decays during an output transmission. This again reduces the actual use of the individual character position of the core buffer. Since the times for line control and transmission are long in comparison to CPU execution times, a more flexible buffer allocation can be applied. It provides a pool of buffer space for all lines. A line which starts a transmission occupies the required space out of this pool. As it needs more, for input of longer messages, it occupies it at an appropriate later time. If it needs less, during output, or after a complete transmission when the whole message has been stored, it gives space back into the pool for use by other Iines. The effect of such dynamic allocation schemes is the subject of this chapter.

i05

The probability assumptions. The message length is assumed to be a random variable ties and moments as considered in chapter At any time

1,

nch

pages 13

with independence, probabili-

ft.

t, for each t e l e c o m m u n i c a t i o n line there is a number

the computer center, requiring buffer space. This number is on. It grows linear with

t

from

0

During output, it decays from some

to some length nch

to

nch

0

nch]t

of characters in

if no transmission is going

during an input transmission.

0. The values

nch[t

for different lines are

assumed to be independent random variables. The above assumption neglects the buffer requirements other than caused by the transmissions. This is acceptable as an approximation if the times to allocate or free space in the buffer pool are small in relation to the transmission times. The additional requirements caused by the storage and retrieval of messages are discussed in exercise 52 . The figure shows a possible pattern of

nchlt

for a half-

duplex line on which input and output are mutually exclusive. Note the analogy to the pattern of remaining service times on page 47 A last assumption concerns the organization and allocation of buffer poot space. The pool consists of a number

M

of cells, each of

L

characters length. In each ceil, some space

will be occupied by identifiers and other control information. The rest is available to store the message proper. Typical sizes of control information are ceil. Meaningful The number number

L

b(t)

%@ for

then are at least

40

to

50

20

two functiom bi(nchlt) , bo(nch[t )

b0

30

characters per

characters.

of p o d cells allocated to a line at any time that time. A value

to

t

depends only on the

which applies if no transmission is going on, and

referring to input and output transmissions, describe this

number of cells. As an example, consider an allocation scheme which always alIocates at

106 least one cell to each line ; then hold the current

nchlt

b0

=

1 . During input, one more cell than required to

is allocated. During output, cells the content of which has been

transmitted are freed. If each cell can hold bi(x ) t bJx)

40

characters of a message,

is 1 more than the smallest integer not Iess than is the smallest integer not less than

x / 4 0 , and

x/40. The figure shows the

b(t)

re-

suiting from the previous figure and the above example. A cell size of

L

=

assumed, and

60 characters was nch[t , L b(t) I

are shown with the same scale.

Expectation and moments of the buffer allocation. The first step in the analysis is to find the probability distribution of

n c h l t . As indicated

above, this random variable behaves very similar to the remaining service times discussed in chapter

3. Its values vary linear between 0 and a random variable

nch

with known pro-

babilities. The only difference is that the base of each interval of linearity is not self, but the transmission time

it-

nch Tch. Thus, the reasoning of pages 52 ft. now leads to

E( nch]t ) k E( nch[t )

=

Fehlt (x)

=

=

Nchlt

=

2 Tch Ra E( nch ) / 2

Tch Ra E(nckh)

and

Here,

nch

/

(k + 1)

for k = 1,2 . . . .

X

U

=

Tch Ra Nch

As stated on page

T c h R a 0~(1 - Fch(Y) ) dy

is the expected line utilization (in the long run) by the trans-

mission of messages proper. For arises with the value

1 - U +

x

=

0, the probability that no character requires buffering

1 - U .

52 , both input and output contribute an individual summand, based on

their individual arrival rate and message length distribution, to the utilization and to the moments and distribution of ed separately.

nch]t. For numerical evaluation, these summands will be treat-

107 As a next step, the moments of

b(t)

probability distribution of

by the general rule

nch[t

which is a function of (see e.g.

nchlt

are found from the

Brunk 1)

oa

E(f(v) )

=

~/~ f(x) d Fv(X) -GO

The differential

dFchlt(x)

is

1 - U

for

x = 0, and

Tch Ra (1 - Fch(X) ) dx

x > 0. The contributions of input and output are separated since different Thus,

for

bi(x), bo(X) apply.

CO

E(b(t) k )

=

b 0 (I - U) + Tch ( Ra, iO~ bi(x)k (I

Fch, i(x) ) dx

CO

+ Ra, o ~0b o ( x ) k (1 Since

bi(x ), bo(X )

and the

- Fch ,o(x) ) dx ).

are piecewise constant in most applications, the

Fch(X)

above integrals can be evaluated as finite sums of products. Exercise

80

shows an a p p l i c -

able technique. The values

k = 1

and B

2 =

are of p r a c t i c a l interest, leading to the expected number E(b(t)

)

of cells a l l o c a t e d to a line, and to the variance V(b(t))

=

E(b(t) 2 )

-

B2

of this number.

Probabilities to exhaust the buffer pool, Since independence of the allocations for different lines is a plausible assumption, the e x pected total allocation and its variance are found by adding the values

B

and

V(b(t) ) for

the individual lines. The criterion for the choice of a number size

ML

M

of cells in the pool, and hence of the pool

in characters, usually is a small probability to exhaust it. Two approximations

for this probabiiity can be suggested. The first is based on the expected total allocation only, and considers the pool as a system of

M

servers, as discussed in chapter

by the expected allocation. Thus,

8. The expected server use of chapter

the probability

P

8

is given

to exhaust all servers can be found

108 from the tables of appendix

C

, page 172 . The other method uses the expected total

allocation and its variance, and approximates the distribution of the allocated number of cells by a normal distribution, as is plausible for the sum of many independent contributions from the different lines. A table of the standard normal distribution is then sufficient to find the probability to exhaust any given

M. Exercise

51

It should be noted that the choice of the cell size tion and the

M

also influences the expected a l l o c a -

to be chosen. TriM-and-error computations may be necessary to m i n i m i z e

the buffer pool size following exercise

h

applies both approximations.

L M. Examples of such computations for the case study are reported 51.

Exercises. 49 of

Define L

b0, bi(x ) and

bo(X) for the following allocation scheme:

One poot cell

characters is always allocated to each line. A further cell is allocated to a line that

starts input. As the arriving characters fill the last but one allocated ceil, a further one is allocated. Each first cell can hold

L - 30

message characters, the consecutive ceils hold

L - 20 characters since they require less control information. For output messages, consecutive cells are allocated as required to hold the whole message. Each cell is freed, except for the last, 50

as its content has been transmitted.

Assume the data of the case study for Input

a line utilization by messages proper of 51

B and Output, the line type

60.

Find the expected total allocation for the case study data as used in exercise

48,

M

B

of servers which has less than

and

1%

ed. Then find the variance of the total allocation, and a similar

V(b(t) ) for

L

and

=

page 100 , and a number

40 %. Compute

II

probability to be exhaustM

with the approxima-

tion by a normal distribution. 52

Starting from the results of exercise

if each message occupies its buffer for an extra

60, which is the effect on 50

B

V(b(t) )

msec after input transmission, or before

output transmission. Note that these time intervals add their contribution into E( b(t~ ).

and

B

and

109

Solutions. 49

T h e p e r m a n e n t a l l o c a t i o n is

the v a l u e of

2. It steps up to

L -20

=

3

40 characters,

at

b0

=

t.

x = L -30

i.e.

at

The input a l l o c a t i o n starts at =

30.

x = 0 with

Further increases occur in intervals

x = 70, 110, 1 5 0 . . .

For output a l l o c a t i o n ,

it is assumed that the longer control i n f o r m a t i o n of

held in the last cell.

T h e n the a l l o c a t i o n is for all

x

30

characters is

one c e l l less than the input a l l o c a -

tion. A f o r m a l expression,

useful in p r o g r a m m e d computations,

is GO

bi(x )

=

1 + bo(X )

with a relation expression of v a l u e •50

0

S i n c e the i n c r e m e n t s of

c o m p u t e d in e x e r c i s e

or

Fch(X)

= 1

l

~[ (x k=0

+

as used on p a g e

8

at the possible message lengths w e r e already

2 , it is c o n v e n i e n t to express the

by these i n c r e m e n t s .

~_ L - 3 0 + k ( L - 2 0 ) )

A partial integration proves that,

B

and

E(b 2)

found on p a g e 107

for both input and output,

O21

.p 0

b(x) k ( 1 - Fch(X ) ) dx

=

,~(~b(y) 0 0

k dy)

dFch(X ) ,

where the i n d e f i n i t e integrals are still fairly s i m p l e expressions, X

X

~/~ bi(y) dy 0

=

x

+

(3O

~/~ bo(Y ) dy 0

S i n c e the longest possible message has k < (180 - L ) / ( L - 2 0 ) . Furthermore, x bo(y)2 ,/" dy 0

=

x J" 0

= 4x

bi (y)2

dy

oo

x+

=

x

+

~-" m a x (0, x - L + 30 - k(L - 2 0 ) ) . k=0

x = 150 characters,

the integrals over

the sum is considered only for

b2

can be expressed as

( 2 k + 3 ) m a x (0, x - L + 30 - k(L - 20) )

k=0 +

m ~_ (2 k+ 5) max(0, k=O

x - L + 30 - k(L - 20) ) .

The following t a b l e shows the first few values of these functions. the i n d e f i n i t e integrals are l i n e a r functions of x

=

(Input)

,2~ b(y) dy 0,jX~b(y) 2 dy 0

viz.

Between the table values,

x.

0

30

70

110

180

0

60

180

340

0

120

480

1120

(Output)

0

30

70

540

0

g0

110

2120

0

g0

190

!I0

T h e i n d e f i n i t e integrals are now e v a l u a t e d for the a c t u a l l y occurring message lengths, m u l t i p l i e d with the r e l a t i v e frequencies of o c c u r r e n c e for these lengths. summed.

and

T h e s e products are

T h e following t a b l e shows the necessary values.

Input

Output s

x

=

124

36

60

72

96

108

120

dFch

1.005

.437

.003

.065

.169

.310

.011

~(y)dy

148

78

150

188

284

332

540

174

390

512

896

1088 2120

J~(y)2dy196 Exercise

4, p a g e

=

48

72

.328

.336

.336

203.9

24

66

114

68.4

623.0

24

lO2

198

108.7

16 , resulted in the arrival rates

Increased by the error overhead Ra, o

24

~..dF

c'

.133 see -1 for e a c h of U = .366 • 6 3 . 3 "

Tch

of page 7

lines. =

33,

Ra, i

=

~..dF

5740 / h o u r , Ra, o = 3293/hr.

the values are

Ra, i

=

.233 sec -1 and

T h e resulting l i n e u t i l i z a t i o n is

.348

Substitution into the f o r m u l a e of p a g e 107 results in B

=

1.50

E(b 2)

=

3.06

V(b)

--

. 797

for e a c h line. 51

For

7

lines,

the e x p e c t e d use of buffer ceils is

If the m u l t i s e r v e r m o d e l with Markov property is applied, searched at which is

U

=

10.5

for the

M

is just not yet successfuI.

1200 characters,

are required.

tionned on p a g e 1 0 4 . a l l o c a t i o n times, information,

In fact,

10.5

60

II

Thus,

M

=

20

- log(1%) = log 100, U = 12,

ceils, with a total size of

This is scarcely less than with the constant a l l o c a t i o n m e n the Markov m o d e l assumes an e x p o n e n t i a l distribution of c e l l

and is thus somewhat pessimistic.

Also, the extra space required for control

together with the assumed extra c e l l for idIe or input times,

7 • Tch..

5.58 .

of p a g e 172 must be

U = 10, M = U + 9, and

e x p e c t e d number of data characters that require buffering, has the v a l u e

table

with a t a b l e v a l u e of at least

4. 60 . Interpolation b e t w e e n the entries for

M = U + 7

1 0 . 5 , its v a r i a n c e

366 • 5010 / 2

=

is significant.

E(nchlt ) , expressed on p a g e

9 1 . 5 characters.

The 106 ,

This is m u c h less than the

= 630 characters which on the a v e r a g e are a l l o c a t e d out of the pool.

111 !f a normal distribution of the number of allocated cells is taken as an approximation, its expectation is

10.5 , its variance 10.5

Thus, a pool of

16

+

5.58 . Its

99-th percentile is at

2. 326 " 5 . ~ ' ~

=

ceils with a total size of

16. O1 .

960 characters would be chosen.

A programmed procedure was used to do similar computations for different cell sizes

L.

Some of the results were for

L

=

a pool size (in characters) of

80

34

44

52

60

68

120

1170

!020

924

936

960

1020

1560

=

is too small a cell size, hence

The two end values are high for obvious reasons.

L

makes the required number of cells excessive. L

=

30

120

keeps the required number of cells

low but leaves too much unused space within each cell. On the other hand, all results for

34 < L ~

68

were between

cating that the pool size is not very sensitive to the choise of

L

924

and

1020, i n d i -

within this range. For a

case with only one or two possible message lengths it would be more sensitive, because then only a few cell size~ fit well with the message lengths. 52

The m a x i m a l number of cells allocated to a message is

bo(nch )

bi(nch )

upon input, and

upon output. The expectations of these number are GO

~/" b(x) 0

d Fch (x)

with the corresponding allocation functions study data and

L

=

b

and length distributions

60 , the expectations equal

If this allocation is held for an extra time

Th

3. 561

Fch . With the case

for input, and

2.008

for output.

per message, the expected allocation

B is

changed by the amount T h ( Ra, i . 3.561 + R

a,o

For

Th

=

.05

sec, this amounts to an increase of

will increase by the same order of magnitude.

" 2. 008 )

=

1. 0 9 8 T h

5.5 %. The required buffer pool size

112

Chapter

10 :

Service times of direct-access storage devices.

A major part of the time necessary to process a message is often spent on accesses to a data base. The desire to process messages as soon as they arrive calls for a possibility to access the data base 'randomly', i . e .

at a random point within the space in which it is stored.

Direct-access storage devices with magnetic recording on different media have been devised as economic solutions. Typical forms are magnetically covered drums or cylinders, disks, and strips that can be placed on a drum. Such devices need a physical motion to enable storing and reading of data, and possibly a positioning of read heads and strips. The physical motion is in all cases provided by a rotation with practically constant angular speed. Any particular data pass under the read heads only in intervals of a full rotation time. Thus, if requests for data transmission arise asynchronous with the rotation, the~c execution is delayed by half a rotation time on the average. In fact, a probability distribution of the delay times can be assumed which is uniform between

0

and a full rotation time.

The positioning of read heads is also known as seeking. For disk storage, it consists in a radial motion of a set of read heads. The time of motion depends on the radial distance travelled. For strip storage, it consists of a motion of read heads along the drum surface, and possibly a concurrent rotation of the cell which contains all strips. This rotation takes the sense of the shorter angular distance to the desired strip. A general concept of a cylinder is introduced for the sets of data which are accessible with the same position of read heads. The different cylinders and positions are then numbered such that adjacent positions have adjacent numbers. A seek motion is then described by the cylinder number

m

at which it ends, and the cylinder number

n

at which it starts.

The following tables give some characteristics of typical devices. Capacity (characters) Drum

Subcells

Strips

4.10 6

Disk

2.9.10 7

Strip

4-10 8

Cylinders

2000

char/see

17.5

1.2.10 6

200

25

3.12.10 5

10000

50

5.5-10 4

1

200

Rotation (msec)

113

S e e k t i m e s for a t y p i c a l disk storage depend on the distance Distance

d

Seek t i m e (msec)

d

= I m - n l of cylinders:

0

1

7

8

10

37

60

91

199

0

25

32

33

29

50

66

70

135

A v e r a g e seek t i m e s for a t y p i c a l m a g n e t i c strip storage are for a m o t i o n b e t w e e n cylinders of the s a m e strip

95 m s e c

of different strips in the s a m e subcell

175 m s e c

in subcells with a c y c l i c distance of

E.g.,

if the m o t i o n starts at cylinder

cylinders

m =

2-5

6-50

95

175

is (in msec) Additional

1 subcell

250 m s e c

50 subcells

800 m s e c

100 subcells

400 m s e c

n = 1, the t i m e to r e a c h 51 - 100 . . .

4951 - 5000 . . .

250

9951 - 10000

400

250

200 m s e c are required to restore a current strip before a new one is accessed.

This t i m e does not contribute to the message processing t i m e if it i s ' c o m p l e t e d before the next data r e f e r e n c e arises. All t i m e s not quoted a b o v e should be found by l i n e a r i n t e r p o l a t i o n .

Probability assumptions. The t i m e s for data transmission are d e t e r m i n e d by the transmission speed of the d e v i c e , quoted in the t a b l e of the previous p a g e in c h a r a c t e r s / s e e , mitted. amounts,

and by the a m o u n t of data trans-

T h e i r probability distribution is simply r e l a t e d to a known distribution of data as discussed in chapter

2, p a g e

26

.

The delays by rotation are also random variables, transmission times.

i n d e p e n d e n t of e a c h other and of the

T h e i r uniform distribution was m e n t i o n n e d above.

T h e positioning t i m e s

tp

in a s e q u e n c e of accesses are again random variables.

d e n c e of e a c h other and of the other a b o v e - m e n t i o n n e d t i m e s is assumed. distribution of the

tp

ing occurs from c y l i n d e r

is d e t e r m i n e d by the probabilities n

to cylinder

m.

Pro, n

Indepen-

T h e probability

that a p a r t i c u l a r p o s i t i o n -

114 Expectation and moments of the seek time distribution. The positioning time for given

m,n

is a function

f(m,n)

of these two cylinder numbers.

Its expectation and moments are therefore found as tk

E( p ) where

M

M Z" m=l

=

M Z n=l

;k=l,2

fk(m, n) Pm, n

....

is the total cylinder number of the device. The expected positioning t i m e 9. 2 k = 1. The variance Vp can be found as E(C.t,) - Tp

results for

T

P

For actual computation of these sizeable double sums it is advisable to first collect all terms with the same value tance

d

=

f(m,n). With disk storage, all values of

m -n

for the same dis-

are the same, and can be described by a function

E( t k ) P Typical values for

f(m,n)

f

assumptions about the

M-1 ~ gk(d) d=l

=

and

g

Pm, n

g(d). Then

M-d m~=l ( Pm, m+d + Pm+d,m) =

were stated on the previous page. The effect of several is discussed in the following exercises.

Exercises. 53

Assume that only the first

N

cylinders of a disk storage are used, that each

access adresses any one of these cylinders with equal probability, and that the positions for consecutive accesses are independent of each other. Find the probabilities sums which are factors of N = 1. . . . . M

and each

gk(d). Show that the sequence of all values k, can be computed from the values

Give the expected seek time and its standard deviation for based on the 54

g(d)

N

gk(d) =

Pm, n

and the

N2 E('t k ) / 2 with

2 M

, for additions.

2, S, 10, 50, 100, 200

of page 113 .

Assume that the ITEM data set is stored on one magnetic disk device. Use the

relative access frequencies to find probabilities that a particular cylinder is accessed, and the multiplication rule for independent random variables to find the

P m , n ' Compare the

expected seek times for two possible layouts: Cylinders

91 - 110

for the most frequently accessed items, 1 - 60 and 141 - 200

for the least frequently accessed items; or

1 -20 ,

81 - 2 0 0

respectively.

115 55

Assume equal probabilities to access any of

a d e p e n d e n c e of consecutive accesses which makes

200 cylinders in a disk storage,

Pm, n

=

0

if

m,n

but

belong to the

same half of cylinders. This situation arises if two data sets on the s a m e device, stored in cylinders

I -100

and

101 - 2 0 0 ,

are accessed a l t e r n a t i v e l y . Find the other

the simplest possible i n d e p e n d e n c e assumption, and express and

N = 200

of exercise

T

53. C o m p a r e with the result for

Pm, n

by the results for P N = 200.

under

N = 100

Solutions. 53

T h e probabilities to access one of the

Pm, n

1/N2

In fact,

for

cylinders are all equal to

m ~ N, n ~ N , while they are zero for all other M-d ( Pm, m+d + P m + d , m ) = 2 (N - d) / N2 m=l

this results if any

Furthermore,

N

N

m,n.

l/N, and the Therefore,

consecutive cylinders are the only ones used.

k E( tp )

=

S.(k)

=

N-1 Z d=l

gk(d) • 2 (N - d) / N 2

=.

2 S(Nk)/N 2 .

N-1 For the sums

N2E(tk)/2

=

gk(d) ( N - d ) d=l

the recursion

s(k)N+I

=

s(k)

+

~N

g k(d)

d=l is evident,

so that ali

been found by another

S(k) M

can be found by

M

additions, once all sums

N ~ gk(d) d=l

had

additions.

For the seek times of page 113 , some results are stated fn the following table. Number

N

of cylinders used

Expected seek t i m e

Tp

Standard d e v i a t i o n V ~ p

2

5

10

50

100

200

12.5

20.9

25.2

35. I

45.8

62.3

(msec) i 12.5

I0.5

8.7

9.2

14.9

24.2

(msec)

The expected seek t i m e iS sensibly reduced by small

N .

54

5

With the access frequencies stated on page

linder are 40% / 60

50% / 20 =

=

.025

, the access probabilities per c y -

for the most frequently used

.00667 for the next group, and

20

cylinders,

.000833 for the least frequently used ones.

116 For the first, containing

c e n t e r e d layout these probabilities are arranged as as a s e q u e n c e of values 60 values . 00083,

values . 00083.

For the second,

60 values .0067, The

Pm,n

30 values . 0067,

20 values .028,

80 values .0067,

a s y m m e t r i c layout the a r r a n g e m e n t shows

Pm

and 60

20 values . 025,

and 120 values . 00083 .

are found as products

PmPn • In the expression for

is the probability that the distance

d

occurs.

the factor

E(t~,

of

g~d)

S o m e of these probabilities are i n c l u d e d in

the foUowing results of a p r o g r a m m e d c o m p u t a t i o n : Layout

Tp

Vp

msec

msec 2

probability of the distance 5

20

100

1 (centered)

42.9

259.1 = 16.12

.0268

.0156

.0015

2 (asymmetric)

49.3

548.8 = 23.42

.0285

.0106

.0015

T h e a s y m m e t r i c layout results in an increased e x p e c t e d seek t i m e b e c a u s e the short distances occur with slightly lower probabilities. T h e array of probabilities

58

size.

Pm, n

is partitioned into four square blocks of equal

T h e two blocks which contain the diagonal d e m e n t s

tions within one data set, h e n c e h a v e all zero e l e m e n t s . pair

m,n

is assumed to be e q u a l l y probable.

Thus,

Pro, m

correspond to seek m o -

In the two other blocks,

the r e m a i n i n g

Pm,n

each

are all e q u a l to

2/N 2 The resulting array can be composed from the arrays, only the first (last) h a v e the probabilities

N = M/2

cylinders are used,

4 / M 2 ; these arrays are denoted by

all cylinders are used i n d e p e n d e n t l y , In fact,

=

Tp

which is

26 %

so that

83, arising if

so that the occurring motions p(1,100)

, 2,0n0 ) F. ( 1m

=

and p(101,200) 1 /M 2

2 .(1,200) i, m , n

( _(1, :00) p(101,200) i' m , n + m,n

) / 2

:

2 T

( T

) / 2

=

2 • 62.8

Pro, n Therefore,

discussed in e x e r c i s e

200) -

45.8

100) + - P msec

:

78.8 m s e c

m o r e than for independent accesses to the s a m e data.

7 9 . 6 r o s e , arises if accesses a l t e r n a t e b e t w e e n cylinder

1

"

A similar result,

and the other cylinders.

viz.

117

Alternating accesses to an index and data. Alternating accesses to one cylinder and to the r e m a i n i n g cylinders, of the previous exercise,

as discussed at the end

occur frequently in data r e t r i e v a l applications.

cylinder contains an index,

i.e.

the desired data are stored.

T h e index is p a r t i a l l y read,

There,

the single

i n f o r m a t i o n about the l o c a t i o n within the d e v i c e at which

is positioned for the data access.

and interpreted,

b e f o r e the d e v i c e

Examples for the case study are considered in the next

chapter. Assume that the index is in cylinder cylinders

1

noted by

Pm

Then,

through for

m 0 , and that the data occupy the r e m a i n i n g of the

N. T h e probability that the index w i l l point to a cylinder m = 1, 2 . . . . .

the joint probabilities

N. This probability is

Pm, n

0

are z e r o if neither

for

:

Pm, m 0

=

m = m0

nor

n = m 0. T h e y are

Pm,n'

viz.

Pm / 2 :

T h e probabilities to c o m e to the index from position use of the index,

is d e -

m : m0 .

different from z e r o only in one row and in one c o l u m n of the array Pm 0 , m

m

m,

and to r e a c h position

m

are both equal to h a l f the probability that a d a t u m is in cylinder

after a m.

Mean and m o m e n t s of the seek t i m e are then reduced to the sum over only one index N

E(tk)

=

~

)Pm

gk( m - m 0

"

m=]

If all data cylinders h a v e the s a m e probability a v e r a g e of the values Two choices of

g

m0

k

1/(N-l)

to be accessed,

this sum is just an

w h i c h enter it.

h a v e b e e n considered for p r a t i c a l a p p l i c a t i o n ,

viz.

m0 :

1

which

places the index at one end of the data set; and an i n d e x in the center of the data set, which for odd E(t k )

N

amounts to

is t h e a v e r a g e of all

m0

=

gk(d)

gives some results for this case.

(N-1)/2.

with

d

With the index at the end of the data,

from

1

through

N-l.

T h e following t a b l e

For c e n t e r e d index, read it at h a l f the data set size. 2

5

i0

50

100

200

(msec)

12.5

21.4

26.4

40.5

54.0

79.6

Standard d e v i a t i o n IVVP (msec)

12.5

10.8

9.1

8.5

15.9

30.5

Number

N

of cylinders used

E x p e c t e d seek t i m e

T

p.

118

Chapter

11 :

Queuing for data accesses.

A c o m p u t e r center which serves as a data base for m a n y independent users is bound to c r e a t e a v a r i e t y of queuing situations. quests asynehronously,

For e f f i c i e n t operation,

as far as this is possible.

p e t e for the central processing unit

(CPU)

and for the data transmission channels. action at a t i m e ,

Thus,

it is essential to treat individual rethe user messages do not only c o m -

as a server, but also for the data storage devices

Both devices and channels can only be devoted to one

and other requests for service arriving during this t i m e h a v e to wait.

T h e typical structure is a queuing situation on two levels. Tasks which are in exectition by the CPU c a l l for data accesses as they are necessary for the application considered.

T h e case study,

page

4

, describes such requirements for different

message processing tasks. T h e data accesses are e x e c u t e d by one or several d e v i c e s which hold the data set. S i n c e each access adresses a unique p i e c e of data, device.

Thus,

it can only be serviced by one particular

the devices constitute a set of independent single servers.

accesses is split,

according to usually known r e l a t i v e frequencies,

T h e total stream of

into separate arrival

streams for these servers. The service t i m e of each d e v i c e is a sum of transmission t i m e , sibly a seek t i m e .

Depending on the data organization,

e a c h such t i m e once, data layout,

or several times.

known as an index,

rotational delay,

and pos-

a single data access m a y require

T h e l a t t e r situation occurs if a description of the

must first be read from the s a m e d e v i c e .

access is always required for the data themselves.

an extra

Another type of data l a y o u t i m p l i e s that

the data found in a first access m a y not yet be the desired data, s a m e d e v i c e where the desired data can be found.

Then,

Then,

but point to an area on the

an extra delay arises only for a

certain fraction of the data. A further contribution to the d e v i c e service t i m e is caused by the second l e v e l of queuing, that of d e v i c e s c o m p e t i n g for a channel.

In fact,

several devices are usually a t t a c h e d to one

channel as a single server, or m a y be switched to any one of a few channels which form a

119 multi-server system as discussed in chapter

8.

The service time of a channel is, for the devices quoted on page 112 , the sum ofrotatior~al delay and data transmission time. These time constitute the essential part of channel u t i l i zation. The amount of control information which also passes the channel, viz. device adresses, desired seek positions, and responding interrupts, is usually small in comparison with the data transmitted. Devices which sense their rotational position permit to reduce the part of channel service time which is rotational delay. They try to use the channel only when the data are almost under the read heads. This implies a small risk to delay the transmission by one or even several full rotations because the channel may be busy when the device tries to use it. Thus, the device service time, considered in the first level of queuing, might be increased. This chapter discusses both levels of queuing under a common point of view.

Queues with finite population. A common property of the queuing situations for both channels and devices is the finite number of individuals that can ever place a request for service. These individuals form a finite population of items. The number of devices attached to a channel is finite. If all of them have placed a request for data transmission, by sending an appropriate interrupt signal, say, no further requests are possible until a device completes its transmission. With a wait time

T

w

FIFO

discipline, the expected

for any device cannot exceed the sum of expected service times for the

competing devices. The queuing system is statistically stable by its very nature. This was already concluded for the Markov queuing process with finite population in chapter cise 44 , page 95 . Also, the resulting

Tw

POISSON arrivals for the same utilization

U.

8, exer-

turn out to be considerably smaller than with

The number of tasks in a CPU is also finite for reasons of memory space. It may, however, be considerable. The effect of a limited number is noticeable whenever the utilization of the serving device is high~ it is also noticeable if the number of tasks goes up as high as 20.

120

Only for still higher numbers, the POLLACZEK-KHINTCHINE formula could be considered as approximation. Another effect of the finite population is that the number of arrivals depends on the number of services completed. Hence, even though the individuals might desire to enter the system at a high rate, they cannot do so unless the previous service is complete. Thus, the 'throughput' of the queuing system is not only determined by the individuals' desire to reenter the system. It should be noted that the 'conversational' use of terminals, discussed in chapter

6, is also

generated by a finite population, in this case consisting of one single user.

The probability assumptions. As for all preceding queuing models, the service times

ts, m

independent random variables with a common distribution The population size

N

are assumed to be mutually

Fs(X).

is considered constant.

The arrival process is no longer a POISSON process but is described by the intervals

tr

bet-

ween the end of service for one item, and the t i m e when it returns into the queue with a request for service. The return times

tr

of each item, and of the different items, are

assumed to be mutually independent with a common exponential distribution Fr(X)

=

1 - e-Rrx.

This assumption, introduced already on page 78 the service is a repair, and

tr

, corresponds with practical experience if

the time to the next failure of a repaired device. The

finite-population queues were in fact first discussed for such applications. (Takacz 1'2) In the context of this course, the assumption is used because it is the only one maintaining a certain Markov property of the arrival process (see Section

8.) In fact, it is the only

assumption leading to formally convenient results. As discussed at the end of chapter these results may be pessimistic in a practical design situation.

5,

121

Relations b e t w e e n u t i l i z a t i o n and queuing variables. For e a c h i t e m

n ,( n = 1 , 2 . . . . .

als is the sum of a t i m e to the queue.

Thus,

tq

N)

of the population,

the t i m e b e t w e e n c o n s e c u t i v e arriv-

spent in the queuing system,

a time interval

(0,T)

passes of the i t e m through the server,

and a t i m e

tr

before it returns

b e t w e e n any two arrivals which covers

ns(0,T)

and h e n c e as m a n y i n t e r - a r r i v a l t i m e s of the i t e m ,

can be written as T

=

ns(0, T) ( t q , m + tr, m ) m~ =- l

=

ns(0, T) ( tq

B

If statistical stability can be assumed, i t e m has the l i m i t e x p e c t a t i o n Next,

T

q

the a v e r a g e i n t e r - a r r i v a l t i m e +

consider the arrivals of any of the

in the long run,

is

N

+ tr )

T

T/ns(0, T)

of the one

r N

items.

For reasons of s y m m e t r y ,

t i m e s as high as for the i n d i v i d u a l i t e m ,

their number,

resulting in a l i m i t e x p e c -

tation of the i n t e r v a l t i m e b e t w e e n arrivals of any i t e m of 1 = --~ ( T q + T r )

Ta

Using the g e n e r a l first-order r e l a t i o n of p a g e Tq

=

Nq T a

37

,

,

the above equation can be rewritten as (N

- Nq)

/ Tr

T h e a v e r a g e arrival rate of the items is i t e m s no___~tin the queuing system. system desire to return,

= Rr

I/T a =

=

t/T r

Ra

m u l t i p l i e d by the e x p e c t e d number of

This result is plausible since e x a c t l y the i t e m s not in the

with an individual rate of

Rr,

T h e server u t i l i z a t i o n has the l i m i t e x p e c t a t i o n N Ts U

=

Ts/T a

Tq + T r

which in fact depends on the e x p e c t e d t i m e spent by e a c h i t e m in the queuing system. Inversely,

all l i m i t e x p e c t a t i o n s of the queuing variables can be expressed by

assumed data,

viz. Tq

=

N Ts / U

-

Tr

,

U

and the

122 with the other l i m i t expectations following from the general first-order relations as Tw

=

Tq - T s

=

T s ( N / U - 1) - T r

Nq

=

Tq / T a

=

N

U Tr/T s

Nw

=

Nq - U

=

N

U( T r / T s + 1)

The remaining problem is to find the l i m i t utilization

U(T r, T s, N)

in its dependence on

the assumed data. This can be called the throughput problem because it brings the actually possible utilization, or 'throughput', of the server in relation to the utilization 'desired' by the items as individuals. In fact, U As an example, consider

N

=

1.

expected time spent in the system is

N Rr T s

is always less than this desired value.

If there is only one item, it will never wait, and the Tq

=

Ts.

Thus,

T U(T r , T s,1) The same expression for

U

-

s Ts + Tr

=

RrT s / ( 1 + RrT s) <~ RrT s .

was already derived in exercise 42 of chapter

8

under the

additional assumption of an exponential distribution for the service times. For general

N,

U(T r , T s,N)

can be found (Takacz 1) by the method discussed in Section

of this course. In fact, this method could be presented in the context of this chapter. However. it is described in a seperate section because it applies to a l i queuing models used so far. This chapter only states the final formula for

U, and uses some of its numerical re-

suits for several application exercises. The numerical results are contained in the tables of appendix

C , pages 170 and 171.

Exercises. 56

Assume, for the processing described in the Case Study, a system with a single

data channel through which pass all input messages when logged, all requests when stored and retrieved before they enter processing all accesses to the data base as required by the application processing all output messages when stored and retrieved before being sent over a line. Since these operations are not in all detail determined by the Case Study data, make the

123 following additional assumptions

:

The data sets of queued messages, logged messages and ITEMS are placed on three different disk devices. The CUSTOMER data are stored on two further devices. Queued and logged messages are written sequentially. In retrieving a queued message, only the two most recently written cylinders need be considered. The required adressing information is held in core memory. This information is usually called an index. Retrieval and storage of ITEM data and CUSTOMER data requires large indices, partially stored on the disk devices themselves. Some of this information must first be read, before the actually desired location becomes accessible. Thus, at least two data transmissions, and possibly two positioning motions are required to read or write data. Only when an updated record is written, it can be assumed that the corresponding adressing information was found during a preceding read and has been saved in the CPU. Assume that the ITEM data are stored in One track in each cylinder, i . e .

40

cylinders, an index to which is held in core.

the part of a cylinder read in one rotation, contains an

index to the individual blocks of 1/6

track length which can be found within the same cy-

linder. Thus, one seek, but two transmissions are required for retrieval and storage. The CUSTOMER data are stored in two times

200

cylinders of two devices. In both devices

one cyiinder contains an index to the cylinder and track of the desired data. Thus, two seek motions and two transmissions are required. An index in core selects a track of the index. Data transmissions for queuing and logging occur in full buffer pool ceils of as were discussed in chapter

60

characters

9. Index data are scanned during the read, starting at a parti-

cular point of the track, until the desired entry is found. This takes a time with approximately the same uniform probability distribution as assumed for the rotational delay. ITEM and CUSTOMER data are transmitted in blocks of 1/6 stant transmission time of 1/6

track length. This takes a con-

rotation t i m e , i . e . 4.2 msec.

The data layout is not quite realistic, as the following exercise problem will show : Find the percentage of storage capacity used in the devices for ITEM and CUSTOMER data,

124

57

With the assumptions of exercise

56

,

find the expectation and variance of trans-

mission times for the different operations of each device, the occurrence rates of these operations, and the overall service t i m e parameters for the channel. page 171, with a population size of

N

=

Use the table of appendix C

5 , to find approximately the expected t i m e

waited for the channel service. g8

Based on exercise

57 , find approximately the effect of channel wait times on

the service times of the different devices. Note that not only a data transmission as discussed above may have to wait until the channel is available. A seek motion is also started by a short adress transmission through the same channel and may therefore be delayed if the channel is busy. These adress transmissions, however, add very l i t t l e to the channel u t i l i z a tion,

and are handled with high priority. Thus, results of chapter

approximate this additional delay.

can be used to

Variances of wait times can be approximated from the

exponential distribution suggested on page 59

ff

50 already.

Assume that a finite population of

3

tasks may issue requests for access to the

ITEM data. With the device service t i m e considered in exercise

58,

find t h e expected t i m e

which a request is waiting before it can use the ITEM device.

Solutions. 86

Current sizes of the ITEM and CUSTOMER data sets are stated on the first page of

the Case Study report. For the ITEM data the current size is million characters.

Incremented by

5%

8.42 • 106 .( 1.05 )8 which is

14 %

18,000 • (10 + 180 ) = 8.42

during three years, the size will reach =

8.96

m i l l i o n characters,

of the total capacity of the device, hence

20 %

of the capacity of

40

cylinders. A growth factor of the CUSTOMER data set is not stated in the Case Study. If the

9%

yearly increase of the worMoad is also applied to the CUSTOMER data volume, and the additional locations are assumed to bring new customers proportional to the workload, a future size of

47,000 • (20 + 350) • (1.09) 3 • ( 1 + 5 3 , 2 0 0 / 1 5 5 , 1 0 0 )

=

3.08.

107

125

characters results. This is slightly more than the capacity of on__~edevice. The capacity of two devices is used to about

80 %. A natural design alternative would therefore be to store

both ITEM and CUSTOMER data on only two devices. The choice of another data layout, however, changes the access times and their probabilities. 87

The solution of the next three exercises is prepared by giving a tabular arrange-

ment of the rates with which data input and output occur, and of the corresponding service times for devices and channel. Table

1

shows the rates for the message processing tasks,

accesses to data. vidual

and the corresponding rates for

The tasks are stated in the order in which they are performed for an indi-

message. Rates are given per hour as on page TASK

16

.

INPUT / OUTPUT

function

rate

operation

Log a message

5740

WRITE sequential

Store a message

5740

Retrieve request

3293

DATA SET

rate/task

identifier

total rate

1

LOG

5740

WRITE small set

1

QUEUE

5740

READ smali set

1.74

QUEUE

5740

Process request CR1 2216

READ indexed

1

ITEM

2216

Process request CR2

READ indexed

1

CUSTOMER

READ indexed

8.45

ITEM

8340

WRITE updated

8.45

ITEM

8340

WRITE updated

i

CUSTOMER

988

988

988

Process request CR3

63

WRITE indexed

1

CUSTOMER

63

Process request CR4

26

READ indexed

1

CUSTOMER

26

WRITE updated

1

CUSTOMER

26

Store a message

3293

WRITE small set

1

QUEUE

3293

Retrieve a message

3293

READ small set

1

QUEUE

3293

On the next page, rences per hour,

the table shows for each device and access operation the rate of occur-

and the expectation and variance of all contributions to the service time.

The sums in the last two columns include an expected rotational delay of the variance of this delay,

equal to

252/12

=

52 msee 2

12.5 msec, and

for a uniform distribution.

126 DEVICE

ACCESS

identifier

operation

rate

Seek t i m e

Transmission

Sum + rotation

Ts

Vs

T

Ts

Vs

msec 2

msec

msee 2

msec

msec 2

.3

0

12.8

52

.3

0

25.3

208

/hour m s e c LOG

WRITE

5740

QUEUE

WRITE

9033

READ

9033

total

18066

12.5

READ index

10556

27.42)

ITEM

data

READ

(each)

156

73

27.4

73

29452

S

.3

0

25.3

208

12.53)

52

52.4

177

4.3

0

16.8

52

4.3

0

44. 2

125

7.24

34.2

37.3

363

index

5071 54.04)

253

12.53)

52

79.0

357

data

5071 54.0

253

4,3

0

70.8

305

62.55)

590

4.3

0

79.3

642

54.0

253

12.5

52

79.0

357

32] 54.0

253

4.3

0

70.8

305

76.3

446

WRITE updated WRITE new data total transmissions

CHANNEL

V

s a m e as previous line

8340

total

156

10556

WRITE

CUSTOMER1 , 2

12.51)

S

507

32

i

15841

7.09

32.8

74492 ]

3.33

26.7

Notes concerning the c o m p u t a t i o n of t a b l e entries: t)

It is assumed that

2

cylinders are in actual use at any t i m e .

T h e table entries

are read from the results of e x e r c i s e 58 , p a g e 115 . 2)

T h e particular access probabilities for the ITEM data w e r e a p p l i e d to

40 cylinder~

3)

T h e t i m e to scan an index track has a p p r o x i m a t e l y the s a m e probabilities as the rotational delay.

4)

Read from the t a b l e of p a g e 117 .

5)

This is a seek from data to data.

Hence,

the t a b l e of p a g e 115 applies.

All totals of e x p e c t a t i o n s w e r e found by linear c o m b i n a t i o n in proportion to the a p p l i c a b l e rates,

as described oh p a g e

second m o m e n t s .

15

. T o t a l v a r i a n c e s are found by linear c o m b i n a t i o n of the

127

T h e a r r i v a l r a t e t o t h e c h a n n e l as server is

74492 / 3600

:

2 0 . 7 sec -1.

The parameters

of its s e r v i c e t i m e distribution are found f r o m those of t h e transmission t i m e s by a d d i n g t h e v a l u e s for t h e r o t a t i o n a l d e l a y : For t h e d e v i c e under c o n s i d e r a t i o n , t h e c h a n n e l is d e v o t e d to t h e d e v i c e also during t h e t i m e of r o t a t i o n a l delay.

It is only w i t h d e v i c e s w h i c h sense

t h e i r own r o t a t i o n a l position t h a t this c h a n n e l use c a n b e m o s t l y e l i m i n a t e d .

T h e c h a n n e l u t i l i z a t i o n results as

U

=

20.7

• 1 5 . 8 . 10 -3

=

.3275.

A further p a r a -

m e t e r for r e a d i n g t h e t a b l e s o f t h e a p p e n d i x is t h e r a t i o 2 T s / Vs With

N

=

;

3.16

5 , t h e a c h i e v a b l e throughput r a t i o R

=

U /(NRrTs)

=

.92

is a p p r o x i m a t e l y r e a d by i n t e r p o l a t i o n in t h e t a b l e o f p a g e Tr and,

=

N R Ts / U ,

by substitution in t h e f o r m u l a of p a g e 122 , Tw

= =

171 . It p e r m i t s to find

T s ( N (1 - R ) / U .221

Ts

=

(or by r e a d i n g p a g e 170 directly)

- 1) 3.5

=

T s ( .4 /.3275

msec .

If t h e c h a n n e l is not h e l d during t h e t i m e of r o t a t i o n a l d e l a y , U

=

2 T s /V s

- 1)

t h e c o r r e s p o n d i n g results a r e

.069 =

so t h a t t h e t a b l e s of p a g e s 170,

.4 171 are not s u f f i c i e n t . A s e p a r a t e c o m p u t a t i o n r e s u l t e d in

R

=

.9848

Tw

=

T s ( .076 / .069

=

.llT s

=

1 86 m s e c ,

showing a r e m a r k a b l e r e d u c t i o n .

58 mission.

T h e transmission o f seek adresses has a n o n - p r e e m p t i v e priority o v e r the data t r a n s T h e adresses t h e m s e l v e s are s m a l l a m o u n t s of d a t a t r a n s m i t t e d over t h e c h a n n e l .

C o m p a r e d w i t h t h e p r o b l e m data, channel.

t h e y result in a very s m a l l a d d i t i o n a l Utilization of the

With the n o t a t i o n of c h a p t e r 7, p a g e

79 , two u t i l i z a t i o n s

c o n s i d e r e d w h i c h h a v e t h e a p p r o x i m a t e values U1

=

0

;

U2

=

U .

U 1 , U2

h a v e to be

128

T h e result of c h a p t e r data,

?

about the expected wait times

was b a s e d on t h e a s s u m p t i o n of POISSON arrivals.

be a reasonable approximation that the ratio arrivals,

Tw, t

for seek adresses,

For a f i n i t e p o p u l a t i o n ,

Tw, 1 / Tw, 2

Tw, 2 for

it s e e m s to

is t h e s a m e as for POISSON

viz. Tw, I

=

(1

-S2)Tw,

T h e e x p e c t e d w a i t t i m e of data, a f f e c t e d by t h e a d d i t i o n a l ,

Tw,2

2

=

(1

-U)Tw,

2 .

' as found in t h e previous e x e r c i s e ,

very s m a l l u t i l i z a t i o n caused by t h e adresses.

is n o t m u c h

Thus,

Tw, 1

can

b e a p p r o x i m a t e d as Tw,1

=

( I - .3275)

=

149 T s

• .221 T s =

2.3

reset.

T h e s e w a i t t i m e s are further a d d i t i v e c o n t r i b u t i o n s to t h e d e v i c e s e r v i c e times. below shows how often t h e y o c c u r w i t h i n t h e d i f f e r e n t o p e r a t i o n s . occurrences,

The table

W i t h t h e s a m e n u m b e r s of

t h e v a r i a n c e s of w a i t t i m e s c o n t r i b u t e to t h e v a r i a n c e of t h e d e v i c e s e r v i c e

time. A p p r o x i m a t e v a l u e s for t h e v a r i a n c e s of w a i t t i m e s c a n b e found from t h e a p p r o x i m a t e distribution Fw(X )

=

1 - U e -Ux/T

w h i c h was a l r e a d y used in c h a p t e r Vw

page

50

2 Tw ( 2/U

=

For t h e c h a n n e l u t i l i z a t i o n of

4,

w

U

=

.3275,

Vw, 1

=

27 m s e c 2 ,

Vw,2

=

62 m s e c 2 .

. T h i s d i s t r i b u t i o n has t h e v a r i a n c e 1)

.

this r e l a t i o n is

Vw

=

2 5.1 Tw , i.e.

T h e e x a c t results w o u l d b e s m a l l e r s i n c e t h e e x p o n e n t i a l a p p r o x i m a t i o n p e r m i t s u n l i m i t e d wait times,

w h i l e t h e f i n i t e p o p u l a t i o n keeps a l l w a i t t i m e s b e l o w t h e sum of

N - 1

set-

vice times. T h e t a b l e b e l o w also shows t h e a c c u m u l a t e d times.

e x p e c t a t i o n ond v a r i a n c e of d e v i c e s e r v i c e

The indexed operations have relatively long service times because they hold the de-

v i c e u n t i l two accesses h a v e b e e n c o m p l e t e d . form of l i n e a r c o m b i n a t i o n as used on p a g e

T o t a l s per d e v i c e w e r e found by t h e s a m e 15

.

129

T a b l e of w a i t o c c u r r e n c e s and d e v i c e s e r v i c e t i m e s ,

DEVICE

operation

rate

LOG

WRITE

5740

QUEUE

any

18066

ITEM

READ

Ts(msec )

V s ( m s e c 2)

1

16.3

114

1

1

31.3

297

10556

1

2

78,5

380

8340

1

1

50.0

214

65.9

507

WRITE u p d a t e d total CUSTOMER1 , 2 (each)

18896

READ

507

2

2

161,4

840

WRITE u p d a t e d

507

1

1

85,1

733

32

2

2

161,4

840

124.4

2245

WRITE new total

59

1046

T h e arrival r a t e to t h e

T h e d e v i c e u t i l i z a t i o n is value R

=

8.56,

adresses data

U

=

ITEM d e v i c e

as server is

5.25 • .0659

=

so that the t a b l e o f a p p e n d i x C ,

18896 / 3600

. 3 4 6 . T h e ratio

=

T 2 / Vs

5 . 2 5 sec -1 has t h e

p a g e 171 , is read for a throughput ratio of

. 8 6 6 . T h e e x p e c t e d w a i t t i m e is c o m p u t e d as

T

W

=

Ts ( 3..134/.346

-

.158 T

s

1) =

10.4

msec .

:

76.3 msec .

T h e e x p e c t e d d e v i c e response t i m e is then Tq

:

T s + Tw

The r e l a t i o n b e t w e e n u t i l i z a t i o n and return rate.

For a g i v e n return r a t e

Rr

=

1/T r

and p o p u l a t i o n s i z e

N , the l i m i t e x p e c t a t i o n

U

of

the server u t i l i z a t i o n can b e c o m p u t e d as a r a t i o n a l expression i n v o l v i n g the LAPLACE t r a n s form

LFs(S )

of t h e distribution o f s e r v i c e t i m e s ( T a k a e z l ) .

T h e expression is

NR r T s U (Rr , N , F s ( x ) )

= N Rr T s +

where

P0

ver i d l e .

is the l i m i t p r o b a b i l i t y that an i t e m , Section

3

discusses why this

P0

P0

w h e n finishing its s e r v i c e ,

a p p e a r s in this c o n t e x t .

l e a v e s the s e r -

T h e v a l u e of

P0

can

130

be computed

as P0

with constants

c',

recursively

l

cj

As s e c t i o n

3

N-1 1 / ~ cj j=0

=

d e f i n e d by

that,

1 - LFs(JR r)

J

LFs( J R r ) the expressions

given

j

j

for

LFs( j R r )

, for

1 - LFs( j R r )

j = 1,2 .....

j = 0, 1, 2 . . . . .

n o n e of t b e m

Fs(X )

viz.

U

is thus d e t e r m i n e d .

Two special

=

1 - e-X/Ts

a re-

that at

c a s e s a r e of

1

-

;

hence

LFs( j Rr )

1 + j Rr T s

s i m p l i f i e s to

cj

= =

In o r d e r to s e e t h e r e l a t i o n t h e e x p r e s s i o n for

,

, t h e LAPLACE t r a n s f o r m is v e r y

1 + Ts s ej

N-1

places

is t h e n t h e p r o b a b i l i t y

1

T h e r e c u r s i o n for t h e

,

to e a r l i e r results.

distribution

LFs(S )

N-1

a request within a service time.

distribution,

interest since they are related

simple,

cj-1

i t e m s a r e no__i in t h e s y s t e m ,

items places

For a n y g i v e n s e r v i c e t i m e

With an exponential

1 , and

N - j

quest for s e r v i c e w i t h i n o n e s e r v i c e t i m e . l e a s t o n e o u t of t h e

=

=

discusses i n d e t a i l ,

are the probabilities

c~

(N - j) R r T s (N-I)'

cj_l

(R r T s)j

for

=

1 -U

-

N-1

.

/ (N - j - 1) '

o f t h i s r e s u l t to t h e r e s u l t of c h a p t e r

3 - U

j = 1,2 .....

Pq, 0 ' v i z .

with the

U

8,

page

96

, consider

from the previous page

P0 N R r T s + P0

If t h e a b o v e

cj

are substituted here,

the resulting

expression

1 -U N-I

NRrTs~

c'j

+

1

j=O can also be written as

Pq, 0 with

c

= 0 if one takes

1

and Ra[ j

=

1/

N 2~2 c. j=0 ]

cj+ 1 = (N-j) R r T scj . This is the s a m e =

(N-

j)R r

,

and

Rslj+ 1

=

I/T s

result as found on p a g e

for all j .

96

,

131

T h e second special case is that of a constant service t i m e . LFs(JR r)

=

e - J Rr Ts . T h e recursion for the

cj' with

c

:

e

Rr T s

N - j J

=

appearing on p a g e 74

eTs/Tr

(

cj

ci

LFs(S)

Now,

=

e - S T s , and

is then

c' j-1

-1)

This recursion shows a certain r e s e m b l a n c e with the one

for the c o n v e r s a t i o n a l use of terminals.

Conversational t e r m i n a l use

apparently i m p l i e s a f i n i t e population of users.

The throughput ratio for g i v e n u t i l i z a t i o n . T h e previous exercises showed that the u t i l i z a t i o n , for the analysis.

rather than the return rate,

is a datum

It is then necessary to solve the e q u a t i o n U

N Rr T s

=

N Rr T s + P0(Rr, N, Fs(X ) ) with g i v e n

U

for the unknown v a l u e

cedure must be applied. to infinity.

Thus,

T h e range of

Rr . For g e n e r a l

N

and

Fs(x ) , an i t e r a t i v e p r o -

Rr , as known at this stage,

an i t e r a t i v e procedure a i m i n g at

Rr

m a y be from

U / ( N Ts)

itself m a y cause stability problems.

It is safer to first solve for the throughput ratio

R

=

=

1 / ( N R r T s + p0 ( R r, N, Fs(X) )

.

NR r T s The i t e r a t i o n

U

R (i+l) :

1/(

, N, Fs(X) )

U/R (i) + P0 ( R(i)NT

)

S

has in fact proven its n u m e r i c a l stability if started with

R(0)

=

1 - U , or with the

~mewhat closer a p p r o x i m a t i o n R (°)

:

-

I

u I

+(N'1)/2

It was a p p l i e d to G a m m a - d i s t r i b u t i o n s of the s e r v i c e times, LFs(J Rr) if

=

1 / (1+

for which

j Rr T s / n ) n

n

is the i n d e x of the G a m m a - d i s t r i b u t i o n . A good a p p r o x i m a t i o n to a g i v e n Fs(X) is 2 found if n is chosen equal to T s / V s , as suggested in the previous exercises. Integer

results in a rational c o m p u t a t i o n . Once

R

is d e t e r m i n e d ,

the return rate is found as

Rr

=

U / (R N Ts),

and its inverse as

132

Tr

=

R N Ts / U .

S u b s t i t u t i o n of this expression i n t o t h e f o r m u l a e of p a g e 122 p e r m i t s to c o m p u t e t h e l i m i t e x p e c t a t i o n s of t h e four q u e u i n g v a r i a b l e s as Tq

=

T ~,N ( I - R ) / U

Tw

=

Ts(N(I

N

:

N (1 - R )

:

N (1 - R ) - U

N T h e expression for

q w

Tw

-R) / U

1

)

was a l r e a d y used in t h e p r e v i o u s exercises.

p r o v i d e s a f u r t h e r i n t e r p r e t a t i o n of t h e r a t i o

R , viz.

ion of i t e m s t h a t a r e not in t h e queuing system.

T h e expression for

N

q

as t h e l i m i t e x p e c t a t i o n of t h e f r a c t -

133

Chapter

12 :

Final remarks on the Computer Center Analysis.

All t i m e delays discussed so far were times for data transmission, or times waited for such transmissions. This may appear surprising in the analysis of data processing_systems, but it is natural in view of two facts : Data transmissions which involve m e c h a n i c a l devices, or long-distance communications, are often slow in relation to electronic processing. Thus, the delays caused by transmissions are often the major part of response times. On the other hand, data processing in its physical realization is again a form of data transmission, through special circuitry. It differs from the transmissions mentionned above by its speed and by its complexity. In fact, the possible structures of a CPU are manyfold. It may be strictly sequential in its actions, or split into parts which concurrently give different forms of service, e . g . tion handling,

instruc-

data retrieval and storage in core memeory, and data processing proper. AIso,

the CPU may or may not be involved in data transmissions through the channels discussed in the previous chapter. One cannot,

therefore, speak of a typical CPU structure which could be

assumed for the case study. The following remarks show rather some typical reasonings that may be a p p l i c a b l e in the analysis. Also, it is not possible to quote typical values for the times which the CPU spends on the different services required by a particular application. Some values for the processing proper were quoted in the Case Study, but are chosen arbitrarily. In practice, an evaluation of actual programs is required to find such design data. Apart from the processing which is determined by the application, considerable CPU t i m e is spent on control functions. Every single message transmission must be prepared by some CPU action. Similarly, every storage or retrieval of data requires CPU t i m e to set up adresses and instructions for the channel, to interpret index information, and to select desired data from a block of retrieved data. The control programs which handle these functions may differ widely in structure and efficiency. Hence, in p r a c t i c e the actually used programs must be evaluated.

134

T h e f o l l o w i n g v a l u e s o f CPU s e r v i c e t i m e s are a r b i t r a r i l y c h o s e n as d a t a for a few further exercises

:

CPU t i m e r e q u i r e d to

r e c e i v e or send a m e s s a g e

50

msec

store or r e t r i e v e d a t a on disk

10

msec

start or restart a processing task

5

msec.

Exercises. 60

Use t h e m e s s a g e rates a n d d e v i c e a c c e s s rates as in e x e r c i s e s

57,

58 , p a g e 125

a n d 129 , and t h e a b o v e CPU t i m e s to c o m p u t e the CPU u t i l i z a t i o n by l i n e c o n t r o l and storage accesses. 61

Use t h e task rates as in e x e r c i s e

s t a t e d in t h e C a s e Study,

page

4

57, p a g e

125,

a n d t h e CPU processing t i m e s

, to c o m p u t e the CPU u t i l i z a t i o n by a p p l i c a t i o n p r o -

cessing, 62

S e p a r a t e t h e CPU u t i l i z a t i o n into t h e c o n t r i b u t i o n s from l i n e control,

messages,

a p p l i c a t i o n h a n d l i n g of c u s t o m e r request types 1 , 2 , 3 a n d 4.

It c a n b e f o r e -

seen t h a t t h e s e f u n c t i o n s are h a n d l e d on d i f f e r e n t l e v e l s of priority.

Solutions. 60

T h e results are a r r a n g e d in t h e f o l l o w i n g t a b l e :

CPU t i m e r e q u i r e d to

time

r a t e of o c c u r r e n c e hour-1

sec-1

msec

utilization %

i

i

Receive messages

5740

1.59

50

7.97

Send m e s s a g e s

3293

0.91

50

4.57

Log m e s s a g e s

5740

1.59

I0

1.59

18066

5.02

i0

5.02

Access ITEM a l o n e

2216

0.62

10

0.62

Access CUSTOMER

1976

0.55

10

0.55

t h e n ITEMs

16680

4.68

I0

4. 63

0.02

I0

0.02

Store and r e t r i e v e m e s s a g e s

Access CUSTOMER a l o n e

89

l o g g i n g of

135

The total from this type of CPU utilization is then 61

25

% .

A similar table can be given for the application processing:

CPU time to process

rate of occurrence hour -1

i

time

see -1 i

utilization

msec i

% ,

a request while logged

3293

0.91

30

2.73

a request of type 1

2216

0,62

40

2.48

a request of type 2

988

0.27

40

1.08

8340

2,32

30

6.96

988

0,27

35

0.95

89

0.02

15

0.03

(per ITEM : ) (finally : ) a request of types 3 or 4

The total from this type of CPU utilization is 62

14.2 %.

Arranged by the possible priority classes, the c o l l e c t e d utilizations are : Task

CPU utilization for control

for processing

total i

12.54

-

12.54

Message queuing

5.02

-

5.02

Logging

1.59

2.73

4.32

Request type 1

0.62

2.48

3,10

Request type 2

5.18

8.99

14.17

Request types 3, 4

O. 02

0.03

0.05

Line control

The overall total of the CPU utilization is

about

40 % .

Waiting for the CPU as server. The need for processing of data by the CPU arises at times which are virtualiy independent of the CPU itself. Transmission of a message requires CPU service mainly for line control and buffer allocation. The times when this service is required are mostly determined by the t e l e c o m m u n i c a t i o n

136

lines. Some of the service must be given fast enough to avoid a loss of data travelling over a line. The service time is determined by the particular line control program chosen. For the use in the following exercises, it is assumed that such high-priority service arises for each input message in addition to the sists in 80 msec

3

50 msec

independent service intervals of

processing time assumed earlier, and con-

4 msec each. Also, it is assumed that the

line control time for each message arise in two independent service intervals of

25 msec each. Another process which generates random times at which CPU service is required, is the storage and retrieval of data. It is usual that a processing task starts some data access, but then suspends further action until the data have been transmitted. The random seek times and transmission times of storage devices determine when further CPU action besomes possible. The total CPU service for a request, say, is therefore split into several shorter contiguous intervals which are determined by the application considered. These intervals have to be considered as the CPU service times in the sense of Queuing Theory. A model of the queuing situation with the CPU as server must also consider priorities. The results of chapter

7

are then applicable to find l i m i t expectations for the wait times in the

different priority classes. With the Case study, the first few priority levels Fast receive processing,

Line control,

Message queuing,

Logging

suggest thernselves quite naturally. Priority assignment for the processing proper can be based on several arguments. It may be satisfactory to assure the smallest average expected wait time for all requests together. As discussed in exercises

35, 39 , pages

82.

88

,

this may require the comparison of seve-

ral possible priority assignments. On the other hand, the response time for a particular request type may be critical. Then, this request type must be preferenced. Another choice is necessary between non-preemptive and preemptive priorities. For the Fast receive processing, preemptive priority-is mandatory. However, the average expected wait time may be increased if preemptive priorities are used on all levels. Exercise

40

page 87 gave some indications when this effect will occur. Also, preemption of a CPU service requires further CPU time to save, and later restart, an interrupted task.

on

137

Exercises. 63

Assume the four priorities indicated above, and three further priority classes for the

requests of type

1 , of type

2 , and of types

3 and 4

together.

The contiguous service intervals for Logging and request processing are formed of the processing which precedes a data access, the CPU t i m e required for the data access itself, and the processing which may be overlapped with the data access, At the end of such service intervals, the CPU will be free for other work. Collect for all priority classes the occurrence rate of service intervals, their expected duratton

Ts

and their second moment

E(t ) , approximated by

ximation corresponds to a standard deviation of

.816 T s , and is in line with the

range of processing times stated in the case study page Apply the formulae of chapter

7

2

1.1 T s. Note that this appro+_ 40 fro

4

for preempt-resume priorities to compute expected wait

times for all priority classes. 64 classes

With the data of exercise 63 , compute expected wait times for the priority 4 through 7 , if only the first three priority classes preempt the others.

Solutions. 63

The data for the assumed queuing system with priorities are coilected in the fol-

lowing table. CPU service times are taken as assumed on pages 134 and 136 , or as estimated in the Case study page tion sums

Sn

of chapter

4.

The accumulation of

7 . The accumulation of

Ra,n Ts, n

Ra, n E ( t : n ) / 2

results in the u t i l i z a provides the expect-

ed remaining service times under a preemptive discipline. In the case of logging and request processing, several t i m e contributions arise with different rates. They enter into the accumulation by simple addition. For utilizations, this is quite plausible. The general rule was discussed in chapter

4, page

52

138 Table of arrival rates and service t i m e moments. n

Priority class

items /hour

services /item

9~

services /sec

Ts, n En(ts)

Sn

Ts[t, n

msec msec 2

%

msec

1.9

.04

688

14.5

1.77

10

ii0

19.5

2.04

1.59

10

110

1

.91

30

990

23.8

2.58

2216

1

.62

30

990

2216

1

.62

20

440

26.9

3.02

988

1

.27

50

2750

8340

1

2.32

30

990

8340

1

2.32

20

440

988

1

.27

45

2228

41.1

5.36

89

1

.025

20

440

89

1

.025

5

41.2

5.37

1

Fast receive processing

5740

3

4.78

4

2

Line control

9033

2

5.02

25

3

Message queuing

9033

2

5.02

4

Logging

5740

1

3293 5

6

7

Request type 1

Request type 2

Request types 3 , 4

With the

Sn

and

17.6

27.5

Tslt, n ' the formulae for expected wait times under the preempt-resume

discipline of page 86 result in the following values : Priority class

n

I

I

Expected wait t i m e (msec) I

64

2

3

4

5

6

7

.04

2.59

4.66

8.39

13.2

22.5

24.2 I

1 - 3

preempt those of weaker prio-

If only the services for the priority classes

rities, the same formulae as above can be used with two modifications : the expected remaining service t i m e the other hand, the summand

I

1

Tslt, n

Sn-1 Ts, n

to be replaced by the smaller term

Tslt, 7

is now equal to

=

] average 8.07

For classes

4 - 7

5.37 msec. On

generated by the preemptions during service, is for the classes

$3 Ts, n

n = 4

- 7. The resulting

values are now : Priority class

i

n

Expected wait t i m e (msec)

I

1

I

i

.04

2

3

4

5

6

7

2.59

4.66

12.9

16.0

19.8

19.6

average 8.08

139 Waiting for a program as server. Data processing is controlled by programs. Each unit of work, like the processing of one request, requires that a program is devoted to it for the time of processing. In fact, what is actually devoted the particular unit of work, is core storage within or associated with the program, such as branch adresses within the code, and temporary data storage. If this associated storage exists once for a program, the program can work on only one work unit at a time. On the other hand, programs can be designed so that they are not m o dified during use, and have the associated storage available in any required number of copies. Such programs are called re-entrant. They act like an unlimited number of copies of the same program. While the control programs for data storage and telecommunications use to be re-entrant, application programs are often available in a finite number of copies only, or even in only one copy. As requests for processing arise at random times, they compete for the program copies as servers, and may have to wait before processing can start. The t i m e for the service given by a program is a sum of times for data accesses, times of CPU processing not overlapped with data accesses, and times waited for the CPU, in an order which is determined by the application considered. E x e r c i s e 65 d i s c u s s e s t h e - C a s e Study request type

2

as an example.

The times waited for program service can then be estimated using the POLLACZEK-KHINTCHINE formula, if a single program copy is assumed, or the results on multiple servers of chapter

8

in the case of multiple program copies. If re-entrant programs are assumed, no

such wait times occur. Note that a change in the number of program copies affects the size of the f i n i t e population waiting for service by the data storage devices, and hence the expected wait time during data accesses. More program copies reduce the time waited for program service, but increase the population size for data accesses and hence the times waited there.

140

Exercises, 65

Find t h e e x p e c t e d p r o g r a m s e r v i c e t i m e for t h e processing of request t y p e

a d d i n g e x p e c t a t i o n s c o m p u t e d in p r e c e d i n g exercises. t h e v a r i a n c e of p r o g r a m s e r v i c e t i m e ,

Also,

2

by

find an a p p r o x i m a t e v a l u e for

t a k i n g i n t o a c c o u n t a t l e a s t t h e v a r i a n c e of t h e n u m -

ber of ITEMs p e r request. 66

Find t h e e x p e c t e d t i m e w h i c h a request of t y p e

g r a m if this p r o g r a m exists in o n e copy,

2

w a i t s for t h e p r o c e s s i n g p r o -

or in two copies.

Solutions. 65

E x p e c t a t i o n s for t h e c o n t r i b u t i o n s to t h e processing t i m e s a r e

for t i m e s to a c c e s s CUSTOMER a n d ITEM d a t a : 162 + 86 + 8 . 4 5 ( 2 .

t0.4

+ 78.5 + 50.0 ) msec

=

1. 510 sec

=

• 408 sec

=

. 3 7 4 sec .

for not o v e r l a p p e d CPU t i m e s : 25

+

45 + 8 . 4 5 (

20 + 2 0 )

msec

for t i m e s w a i t e d b d f o r e CPU s e r v i c e : ( 1 + 8.45 ) • 2"

19.8

msec

2 . 2 9 2 sec .

T h e t o t a l e x p e c t e d s e r v i c e t i m e is t h e n T h e e x p e c t e d n u m b e r of ITEM a c c e s s e s per request of t y p e

2

was found in c h a p t e r

1

as

8 . 4 5 . Not o v e r l a p p e d CPU t i m e s a r e s p e c i f i e d in t h e C a s e Study a n d on p a g e 134 . D a t a a c c e s s t i m e s w e r e t h e result of e x e r c i s e s

58 a n d 59 , p a g e 124 . Note t h a t t h e t i m e s to

w a i t for a CUSTOMER data a c c e s s are v e r y s m a l l d u e to t h e low u t i l i z a t i o n o f d e v i c e s . t i m e w a i t e d b e f o r e CPU s e r v i c e was t a k e n from e x e r c i s e

64,

The

p a g e 138 , since t h e p a r t i a l

p r e e m p t i o n discussed t h e r e g i v e s t h e m o r e f a v o r a b l e result for p r i o r i t y class

6 .

T h e v a r i a n c e of t h e processing t i m e is t h a t of a sum of r a n d o m v a r i a b l e s w i t h i n d e p e n d e n t l y v a r y i n g n u m b e r of s u m m a n d s .

W i t h t h e f o r m u l a of a p p e n d i x D , p a g e 180, it is found as

t h e sum of v a r i a n c e s of t h e t i m e s r e l a t e d to t h e CUSTOMER

plus

8 . 4 5 t i m e s t h e sum of v a r i a n c e s ai" t h e t i m e s r e l a t e d to t h e ITEMs

plus

1 3 . 7 4 t i m e s t h e s q u a r e d sum of e x p e c t e d t i m e s r e l a t e d to an ITEM , where

13.74

is t h e v a r i a n c e of t h e n u m b e r of ITEMs per r e q u e s t of t y p e

2.

141

T h e c o n t r i b u t i n g e x p e c t a t i o n s and v a r i a n c e s w e r e found in e a r l i e r e x e r c i s e s , w i t h t h e e x c e p tion of t h e v a r i a n c e s for w a i t t i m e s spent b e f o r e d a t a a c c e s s e s and CPU s e r v i c e . For t h e s e , approximations of the form Then,

=

T

2

( 2/U - 1 )

W

w i l l b e used as a l r e a d y on p a g e 128 .

to t h e e x p e c t e d t i m e w a i t e d for an ITEM a c c e s s o f

a variance of class

Vw

at

.346

utilization

about 520 m s e c 2 results. For t h e t i m e w a i t e d for CPU s e r v i c e in priority

6 , w i t h t h e e x p e c t a t i o n of

is about

10.4 msec,

1 9 . 8 m s e c , and

S6

=

. 4 1 1 , t h e sesulting v a r i a n c e

1530 m s e c 2.

T h e v a r i a n c e of t h e p r o c e s s i n g t i m e t h e n results as 840

+

8.45

( 380

13.74

i.e.

733

+

63

+

( 78.5

214 +

+

203

+

2.

50.0

+ 520

+

2.

2.

1530

+

+

40

+

40

+

2.

I0.4

+

20

+

20

+

1530 ) 2.

+

1 9 . 8 )2

msec 2

as Vs

=

. 7 6 5 sec 2 .

Most o f this v a r i a n c e is c a u s e d by t h e r a n d o m n u m b e r of ITEMs p e r request. rows in t h e a b o v e s u m m a t i o n c o n t r i b u t e only about

6 %.

66

=

With t h e arrival r a t e

service time

Ts

=

Ra

=

988 hour -1

.2'744 sec -1

T h e first two

and t h e e x p e c t e d

2. 292 sec, t h e e x p e c t e d n u m b e r of p r o g r a m s in use is U

=

Ra T s

=

.629 .

T h e POLLACZEK-KHINTCHINE f o r m u l a results in an e x p e c t e d w a i t t i m e of Tw

=

2. 292 ( 1 + . 7 6 5 / 2 . 2 9 2 2 )

=

2 . 2 3 sec .

Two p r o g r a m s w i t h t h e s a m e e x p e c t e d t o t a l use service times, C,

.629 / 2(1-.629)

U , and e x p o n e n t i a l distributions of t h e

h a v e a p r o b a b i l i t y to b e both in use w h i c h is r e a d f r o m t h e t a b i e in a p p e n d i x

p a g e 172 , as P

=

4s.e

%.

T h e c o r r e s p o n d i n g e x p e c t e d w a i t t i m e is Tw

=

.486 • 2.292 / ( 2 - . 6 2 9 )

=

.81

sec .

This result is p e s s i m i s t i c , s i n c e t h e a c t u a l s e r v i c e t i m e distribution for t h e p r o g r a m has a v a r i a n c e c o n s i d e r a b l y less t h a n t h a t of an e x p o n e n t i a l distribution w i t h t h e s a m e e x p e c t a t i o n . T h e c o m p a r a b l e w a i t t i m e w i t h one p r o g r a m is

2 . 2 9 2 • . 629 / . 8 7 1

=

3 . 8 8 see.

142 Message response times. Finally, the quantity which is usually taken as a criterion of performance, is the message response time as defined on page

5

of the Case Study. From the results of the earlier

exercises, the expected response time for a request of type

2

can be collected. For other

types of requests, they can be found by analogous computations. Expected response times are sums of expected times

Tq

spent in different queuing systems

as indicated in the following table. For message queuing and logging, re-entrant programs were taken as a simplifying assumption. One processing program is assumed. Server

Tw

Ts

Tq

in seconds

Line, input

3.0

1.35

4.35

Message queuing

.18

.18

Logging

.03

.03

2.23

2.29

4.52

.42

.25

.67

5.65

4.10

9.75

m|

Processing of type 2 Line, output i

Total

This expected response time applies to

30 %

of the requests. The other types require less

processing, bringing their expected response time close to the cessing. The average will be about

sec.

5.23 sec for other than pro-

7 sec. As discussed on page 98 , the design require-

ment is a 90-th percentile of the response times at or below

12 sec.

Each response t i m e is the sum of a considerable number of independent random variables. The distribution of response times will therefore be approximately normal. The above expectation and percentile then permit a variance of response times up to the square of ( 12 - 7 ) / 1.28

=

3.85 see , i . e .

up to

waited for line input, and for processing of type

15

sec 2 . To this, the variance of times 2 , will give the major contributions.

In fact, the above case is on the borderline. In order to safely satisfy the response time requirement, the above analysis should be reviewed with at least a second copy of the processing program for type

2 , but also with some more or faster communication lines.

143

Conclusions. The reader who, by now, has fought his way through a long series of exercises, requiring a considerable amount of numerical computations, will be asking himself two questions of p r a c t i c a l importance : Whether a similar series of computations is feasible in any design situation, and whether the results are precise enough to base a design upon them. Both questions can be answered positively, however with some comments. The amount of computation for the analysis of one possible configuration, if done from scratch makes a programmed computation mandatory. In actually executing the computations for the Case Study, a conversational computing facility was used. It is operated from a k e y board terminal which also prints the resuIts. This facility has proven to be a most adequate tool, as it would be for any engineer engaged in a design process. The computations can be reduced by use of the tables contained in Appendix

C

and in

some parts of the text. With these aids, the analysis of one configuration can be handled with the slide rule in an a c c e p t a b l e time. There is, however, the need to compare different design possibilities. Within the design process, analysis serves as a tool which one intends to use repeatediy. The considerations of this last chapter on computer center analysis showed this need most d e a r l y . In such circumstances, the analysis should be programmed so as to be computed with varying parameters. The precision of results is influenced by three aspects : the precision of data, the exactness of the m a t h e m a t i c a l or probability model, and the nature of the results. Design data are statistics from a l i m i t e d number of observations. Their inherent uncertainty was, to some extent, discussed in chapter chapter

1

and, for service times and utilizations, in

3. Neither the systems designer nor the user of a system should expect results more

reliabIe than the data provided. The m a t h e m a t i c a l models presented in this course contain certain assumptions, especially the probability models of Queuing Theory. Some of these assumptions can be well ascertained,

144 e.g.

some of the service t i m e distributions. Others would be difficult to prove. E.g. in

assuming independence of random variables, one has often to reason with the general nature of the process producing them. arrival process,

A particularly frequent assumption is that of a POISSON

- chosen since it simplifies many results of Queuing Theory. This assumption

is made even if some dependence of arrivals is evident. If this dependence is in the sense of mutual inhibition, the POISSON process may be considered as a pessimistic assumption, i n creasing e.g. predicted response times over the actual values. Finally, the m a t h e m a t i c a l nature of results is often that of expectations or similar parameters from probability theory. This implies a particular understanding what 'precision of results' can mean. In fact, there is no possibility to check the results by means of a simple experiment. Even in a long run of observations, random results arise that may deviate from probability predictions. One can only state that such results would have a small probability to occur.

Thus, ~most results of Computer Systems Analysis cannot, and do not, claim to be of high precision. ( Note that a certain precision, in numbers of decimal digits, has been m a i n tained throughout the text. This has the only purpose to suppress a further source of uncertainty, and simplifies the identification of numerical results.) The value of analysis lies above all in the ease with which different designs can be compared. Quantitative results should be interpreted more critically. It is mainly as a base for this critique that the course gives so many details about the probability assumptions and the derivation of results.

The whole section

3

has thee same purpose.

Section

FURTHER

RESULTS

3

OF

QUEUING

THEORY

The results of Queuing Theory which were presented and applied in Sections

1

and

2

are

mainly concerned with the l i m i t expectations of the queuing variables. These results permit a simple derivation, but are fairly general. The effort to find distributions of queuing variables as in the simulation of chapter by the iteration method of chapter

3 , and

5 , was considerable. Thus, for the application exercises

a very crude approximation of distributions was suggested. However, the Computer Center Analysis with its nested queuing systems made evident that at least the variance of variables in one queue has some effect on the variables in another, related queue. Formulae for this variance would be useful in this context. At the same time, the design criteria are frequently stated as bounds on percentiles of the queuing variables. Therefore, a knowledge of the distribution of queuing variables, or again at least of its first two moments, is of practical value. Chapter

8

showed that the Markov property of a queuing process gives an easy access to

its theory. A queuing process has this property for any time if both the inter-arrival times and the service times have an exponential distribution which does not vary with time,

- a

somewhat restrictive assumption. Kendall 1 noticed that the Markov property still exists at the end times of services if only the inter-arrival times have the exponential distribution. The analysis of a queuing process for these points in time carries almost as far as for the Markov processes of chapter ter

13

is devoted to this analysis. From its results, chapter

14

8 . Chap-

leads to some distributions

of queuing variables, and to expressions for their moments. Within this course, the method of imbedded Markov chains is preferred to other powerful m e thods because it lends itself easily to programmed computations, producing numerical evidence on convergence rates and correlations which is of high practical interest.

146 Chapter

18

:

Imbedded Markov chains of some. queuing processes.

Assume a queuing system with one server, and the inter-arrival times or return times with an exponential distribution as on page 120 . Reconsider the queuing processes with Markov property and the notation of page by

t 0, t 1, t 2 . . . .

89 . Denote

the times at which the server ends a service and is ready to start a

further service. At such times

t i , the future behaviour of the system depends on the further

service times, which are assumed to be independent of the history before

t i . Also, the

times remaining to the future arrivals are independent of the history. Therefore, the queuing system has the Markov property for each

t i.

Kendall 1 suggested to consider the state of the system only at the service end times

ti .

These state values form the imbedded Markov chain of the queuing process. The matrix iteration technique of exercise

41, page

90 , permits to find l i m i t probabilities of the

imbedded chain. In a further step, the l i m i t state probabilities for any time

t

are found.

The transition probabilities of the imbedded chain are the basis for this analysis. They depend on the nature of the queuing process, and of numerical values of its parameters. Three types of queuing processes will be discussed which correspond to three queuing models used earlier in this course. In fact, still other types of queuing processes permit analysis with the same method (see Chang 1) . For all of the discussed cases,

nq(ti) , the state of the queuing system at the time

t i , is

understood as the number of items left behind in the system by the leaving item. Thus, the possible states are

0

and the positive integers. The transition probabilities are probabilities

to have the same or another state at the next service end time. Since at that time, exactly one item will leave the system, at1 transition probabilities a r e p r o b a b i l i t i e s to have a certain number of arrivals within one service time. The following exercises give some examples and properties of such probabilities. Exercises. 67

Denote by

p a , j l k (x)

the probability that, given the system state

nq(ti)

= k ,

147

exactly Pa

j

items arrive in the t i m e interval from

are conditional probabilities.

Consider

bilities

x

t.

1

to

t. + x . If

x

1

as any service time,

is random,

these

and find the total proba-

(2O

Pa, j~k

=

~

Pa, jlk (x) d Vs(X)

-133

to have exactly

j

arrivals in one service time,

for the two cases of POISSON arrivals and

of arrivals from a finite population. 68

For the two cases of exercise

67, find the expected number of arrivals in one

service time.

Also show that the probabilities

ing function

Ga[k(Z )

form

of the service time distribution.

LFs(S)

of the probabilities

Pa,0[k

Pa, j[k

to have no arrivals,

and the generat-

can be expressed by the LAPLACE trans-

Both

G

and

LF

are defined in appendix

D

pages 178, 179. 69

show' that

For the finite population,

N -k Pa, j - l [ k + l

Pa, j[k holds for all

N -k-j+ j

1 Pa, j - l [ k

j ~ 0.

Solutions. 67

The POISSON arrival probabilities were defined in chapter

population,

I - e -Rrx

1 . For the finite

is the probability that a particular item arrives,

e -Rrx

the proba-

bility that it does not arrive. With these values, binomial probabilities are formed. Thus for the case of

POISSON arrivals

finite population

Pa, jl k(x)

(Ra x) ] -Ra x ~j: e

(

=

To form the

Pa, j!k

N-k J ) (1 - e "Rrx) j e -Rrx(N-k-j)

requires only a substitution into the integral stated above. Explicit

expressions can only be given for an individual service time distribution. constant service time =

Pa, j[k For Pa,0lk

~

=

Ts

I

results in

(Ra Ts)] j :

e -Ra Ts

0 , the conditional probabilities =

I

As one example,

LFs(Ra)

( N-k ) (1 - e -RrTs) j e -RrTs(N-k-j) j

Pa,0lk(X)

are very simple,

I "LFs(Rr(N-k) )

so that explicitly

148 The expected number of arrivals in one service time can be found using that

68

for the case of

POISSON arrivals

finite population

J Pa, jlk (x)

Ra x p a , j _ l [ k (x)

(N - k ) ( 1 -e-Rrx) Pa, j - l l k - 1 (x)

=

sum over all j , v i z .

(N - k) (1 - e-Rr x)

Rax

, and the

is absolutely conver-

gent for all x . Therefore, with a change in the order of summation and integration, the expected values are

j GO - Ra x d Fs(X)

GO

(N - k ) ~

-00

(1 - e -Rrx) dFs(X )

-(30

:

Ra T s

(N - k) ( 1 - L F s ( R r )

The generating functions of the conditional arrival probabilities Appendix

can be read from

D , page 178 , as

Ga,k(X,Z) :

I

e ' ( 1 - z ) Rax

I

Their integration with respect to

Ga~k(Z)

Pa, jlk (x)

)

!

=

Fs(X)

(

z + e'RrX(1-z)

) N-k

results in the generating functions for the total pro-

i

Lrs((1-z) Ra)

j=0 ( J )

(l-z) j LEs(JR r)

These generating functions play a role in the determination of limit probabilities. 69

If one intends to compute all transition probabilities of the queuing process with

finite population, the formulae stated in the exercise permit a recursive computation, starting from the probabilities for

j

=

0

which were stated at the end of exercise solution 67.

The relation is caused by a simple property of the binomial probabilities (1 -e'Rrx) j

=

(1 - e - R r x ) j - l (

the product of probability powers in ducts with

j-l[k+l

and

be written as

,

Pa, jIk (x) can be split into the difference of the pro-

j-l{k. Furthermore, the binomial coefficient of this product

N -k (

1 - e -Rrx)

Pa, j[k (x)" Since

N -k ) -

j

j

The integration with respect to

N -(k+l) (

Fs(X)

) , and also as

j -1

(

can

N - k N - k - j+1 )

j -1

j

can then be applied to each summand separately.

As a final consequence, all transition probabilities can be expressed by the values which the LAPLACE transform

LFs(S)

takes on for the integer multiples of

Rr .

149

The transition probabilities. For a queuing system with one server, the transition probabilities of the i m b e d d e d Markov chain are closely r e l a t e d to the p r o b a b i l i t i e s of arrivals in one service t i m e . that

In fact, given

nq( t i ) = k, the number of arrivals in the next service t i m e d e t e r m i n e s the state

nq( ti+ 1 ). If

k

=

0 , state

j

will be reached if one i t e m arrives, and if further

rivals occur in the service t i m e of the first i t e m . Pq, j]0

=

the i t e m l e a v i n g service is If

k

at

ti+ 1 , l e a v i n g

j

ar-

Thus,

Pa, j}l

In a system with finite population,

j

for

j =

'9,1, 2 . . . . .

the targest number of items that can be left behind by =

N - 1.

is different from zero, one i t e m is i m m e d i a t e l y in service. It w i l l l e a v e the system j i t e m s behind if Pq, j[k

=

j < k - 1

In consequence,

are

the m a t r i x

arrived in its service t i m e .

Pa, j - k + l l k

Obviously after only one service t i m e , bilities with

j - (k-l)

j

for

k ~ 0

cannot be less than

and

j

Thus,

= k-l,k .....

k - 1 . All transition p r o b a -

zero. Pq

of transition p r o b a b i l i t i e s for any pair of times

ti, ti+ 1

appears as

P

q

Pa, 0[1

Pa, 0 [t

Pa, l ] l

Pa, 111

Pa, 0[2

Pa, 211

Pa, 211

Pa, 112

Pa, 013

Pa,3[1

Pa, 311

~a, 212

Pa, 113

,

.

0

0

0

...

0

0

...

0

...

Pa, 014 " ' "

.

It has two i d e n t i c a l first columns; only one e l e m e n t above the diagonal in e a c h c o l u m n is different from zero, m a k i n g N

by

N

P

an ' a l m o s t triangular' matrix; and it has the finite size of

e l e m e n t s for the queuing system with finite population.

Some numerical examp-

les are considered in the next exercise. For the queue in front of a p o l l e d t e r m i n a l , matrix terminal.

Pq

an i m b e d d e d Markov chain with a similar

is defined by considering the t i m e s

ti

at which a polling message reaches the

The number of messages which is then found at the t e r m i n a l ,

is chosen as the

150

state

nq(ti) . The transition probabilities are then probabilities to have a certain number of

arrivals in the t i m e of a polling cycle

t c (instead of the service times considered above.)

If at one polling t i m e the queue is empty, the given state is ed in the next c y c l e t i m e to produce the next state polt, chapter distribution

6

assumed a distribution

Fc[l(x)

Fcl0(x )

after a positive poll. Thus,

k = 0. j

arrivats are requir-

j . For this situation of a negative

of the next c y c l e times, different from the for POISSON arrivals

130

Pq, jl0

=

Pa, jl0

=

¢~ pa, j(x) d Fc[0(x ) -03

and this first column of In fact, for any state

P k

q :/

cycle t i m e with distribution

is different from the second column. 0, the next state is F c l l ( x ). Thus,

for

j k

if there are ~

j - k+ 1

arrivals in a

0 ,

OO

Pq, ilk

=

This is the same pattern as in the

Pa, j - k + l i k Pq

=

d~

-GO

Pa, j-k+1 (x) d Fc#l(X )

stated above, with only

Fs(X)

replaced by

Fcll(X).

Thus, the imbedded chain for three queuing models has been defined, and will be analyzed in the sequel. As a note of general importance, it should be seen that the usual queuing disciplines do not affect the transition probabilities of the imbedded chain. This statement holds as long as the discipline does not create another service t i m e distribution (resp. cycle t i m e distribution) for given state

k , than assumed above. Thus, the disciplines

FIFO, LIFO, random selection

result in the same probabilities for the imbedded chain. However, a discipline which always selects from waiting items the one with the shortest service t i m e would m a k e the distribution of service times dependent on the state. The m i n i mum of

k ,~ 0

1 - (1 -Fs(X))k with varying

service times with the common distribution as stated in Appendix

Fs(X)

has the distribution

D, page 179 . This distribution may vary strongly

k .

Limit probabilities of the imbedded chain. As discussed in chapter

8 , the state probabilities for the times

tl, t2 ....

are found from

151

those assumed at

to

by repeated matrix multiplication. If the elements of Pq are known

by their numerical value, this iteration can be handled easily by programmed computations. The next exercise shows some typical results. Another problem is whether the probabilities so generated do converge towards l i m i t probabilities which are characteristic for the queuing system, and should therefore be independent of the initial probabilities at t i m e

t O.

With POISSON arrivals, the condition

Ra T s ~

i

of page 88 is sufficient for this form

of convergence. The proof is classic in the theory of Markov chains ( e . g . but involved since the matrix that

Ra Ts ~

1

Pq

Saaty 1, Takacz 1)

is of infinite size. The following derivations will show

is necessary.

The condition can be interpreted as stating that, given a state state of the Markov chain is

k - 1+

Ra T s

~

k. Exercise

k 68

#

0 , the expected next

gave the necessary c o m -

putations for this interpretation. It indicates a stabilizing effect in the queuing system. For a finite population, convergence of the iteration is assured by the fact that the value is the dominant eigenvalue of the matrix Pq.

Appendix

D , page 181 references some

related theorems of matrix analysis. In order to find the l i m i t probabilities

p'~, j

from the matrix

Pq , its eigenvector defined

by the linear equations p*q

=

Pq p*q ,

abbreviated as a matrix equation as on page 89 ments of

Pq are known by their numerical value, the almost triangular shape of Pq per-

mits a simple procedure : Assume any value for express the all

already, has to be determined. If the e l e -

p~,j

(j = 1,2 . . . . )

p'~,j (j = 0 , 1 , 2 . . . . )

by

Pq, 0 ; use one equation after the other to

Pq, 0 ; finally n o r m a l i z e to probabilities by dividing

by their sum. A similar technique was already applied in chapter

8 . The reader may use the results of the next exercise to work out an example. The same technique could be applied for a general analysis. There, however, it is more efficient to first aim at the generating function

G*q (z)

of the l i m i t probabilities. This

generating function arises if the above matrix equation is l e f t - m u l t i p l i e d by a row matrix

152 with the elements

1, z, z 2 . . . . . Z pq*

Denote this row matrix by

Z . Then

G~l(z) •

=

On the right side of the equation for the eigenvector, the product the particular form of the matrix ZP q The generating functions ed in exercise

Z Pq

appears. Due to

Pq, this product is the row matrix with the elements

= Ga|k(Z )

of arrival probabilities for given state

k

were discuss-

68, page 1 4 8 .

In the case of POISSON arrivals, the

Galk(Z)

for

k

/

0

are all equal. Then,

ZPq

can be written as the sum of I and

o

Gall(Z)

°al°(Z)

-

z

Gall(Z)

Z .

o

0

The matrix equation

I

"'"

p~

=

Pq pq*

therefore ira-

Z

plies

:

- call(Z))' P~l,o

(Galo(z)

+ Call(Z)

Z

c-~(z)

Z

Thus,

Gq(z)

:

z GalO(Z) z

Gall(Z )

P*

q, 0

Gall (z)

-

relates the generating function to its value

Pq, 0

=

G~(0)

for a particular

z, viz. O.

(30

In such a situation, the normalizing relation to determine

G~(1)

=

~

p*

=

j=o q'j

I

must suffice

Pq, 0 " In fact, with the functions

Calo(Z )

--

found in exercise

68,

6all(z)

=

LFs((l-z) R a)

=

I - (l-z) R a T s +O((l-z) 2)

z - LF((I-z) R a T s) * Pq, 0

=

lim z-~l

(z - i) LFs((1-z) R a T s) z - I + (l-z) R a T s + O((1-z) 2)

lim z-.~ 1

1

results uniquely. Since

p* q, 0

(z-1)(i-

O(l-z)

)

Ra T s

cannot be negative, the condition

Ra T s ~

1

is necessary.

153

By an analogous reasoning, p*

q, 0

1

=

Ra Tcl 1

1

Ra ( Tel I - Tcl 0 )

results as the probability that a polling message finds no message at the terminal. here,

Ra

denotes the arrival rate to the particular terminal,

Ra, m

or

Ra / M

Note that

a quantity denoted by

in chapter

6 . With this in mind, the above equation can be rearrang-

I - p*q, 0

=

* 0 Tc~0 Ra ( Pq,

=

R

ed as

reconfirming the result

for

p

T

a

of chapter

* 0 ) Tc|l ) (I -Pq,

+

,

c

6, page

67

The other state probabilities of the imbedded chain are not of practical interest, and will not be determined from the generating function. However, they are uniquely determined by it. For the finite population, the result of Takacz 1 is reported. He finds G*(z)a

=

N-1 (~". c j ) / c ' n

Pq,* 0 N'n~ 2 (Nn-1) (1 - z ) n z N - l - n =

where the

cj

are defined, as in chapter

cj At

1 - LFs(J Rr)

J

LFs(J Rr)

z = 1 , only the summand with (1)

=

as a relation which determines

11 , by

N - j -

1 p* q, 0

n =

,

l=n

=

0

c~

=

c] _I

1

, and for

j,

1,2 . . . . . N - 1 .

is not zero, leaving

N-1 Pq, 0 j~=0 c'.j equal to the

P0

already introduced in chapter

11,

page 129.

Examples for the rate of convergence: Repeated matrix multiplications as a numerical technique to find an eigenvector were already suggested by yon Mises 1, due to their relatively favorable convergence. With Markov chains, the iteration arises from the problem, but it has the same convergence properties.

154

A fast c o n v e r g e n c e i m p l i e s that the later state probabilities are not much i n f l u e n c e d by the assumed i n i t i a l probabilities, so that the l i m i t probabilities in fact describe the essential properties of the queuing system. This aspect was already discussed in chapter p r a c t i c a l interest is the e x p e c t a t i o n of an average of states l i m i t value soon, i f there is a fast convergence.

5 . Of most

nq(ti) . It again reaches its

The following exercises provide some n u -

m e r i c a l e v i d e n c e of typical situations.

Exercises. 70

C o m p u t e the transition probabilities

up to adequate

j

N =

, with such p a r a m e -

Pq, j~k

and

k , for the

four cases of constant service times, and POISSON arrivals, or a f i n i t e population of ters that the l i m i t u t i l i z a t i o n U 71

=

.5 , or

U

.8

pq(ti)

for

i

items

equals

respectively.

Starting with the probabilities

c o m p u t e the

5

=

pq(t0)

for an i n i t i a l l y empty queuing system,

1; 5, 10, 20, 50

, and their expectations.

Solutions. 70

Expressions for the

Pq, j[k

were derived in exercise

arrivals, the values appearing in any c o l u m n of =

Pa, j

u j e-U/j,

are the POISSON probabilities

' '

shifted into the right places. With the given to s a v e

Pq

67, page 174 . For POISSON

U, the values are (written in a row, in order

)

space

0

1

2

3

4

5

6

7

.5

.6065

.3033

.0758

.0126

.0016

.0002

.0000

.0000...

.8

.4493

.3595

.1438

.0383

.0077

.0012

.0002

.0000...

U

=

j

The m a t r i x

=

Pq

In order to have

for a f i n i t e population of U = .8 , or .8 , Rr Rr T s

where

R

=

N = 5

items can be c o m p u t e d if

has to be chosen properly, i . e . U/NR

...

Rr

is given.

so that

,

is the throughput ratio t a b u l a t e d in Appendix

C, page 171 . Using that table,

155

R

the values of

for g i v e n

N

and

U

can be read• T h e values for constant service t i m e 2 T s / Vs--~ co.

are not shown e x p l i c i t l y but can be found by e x t r a p o l a t i o n as l i m i t values with They are

for

U

=

R

=

hence

•5

.8687

• 8

7161

Rr T s

.868?

=

.1151

, 8 / 5 " .7161

=

.2235

.5/

These values are now substituted into the expressions for

Pa, j~k

=

5 •

derived in e x e r c i s e

67.

The recursive c o m p u t a t i o n i n d i c a t e d in exercise 69 m a k e s this substitution r e l a t i v e l y simple. As a result,

for

U

.5

=

p

q

whereas for

U

=

0

0

0

0

•

308

• 308

.708

056

• 056

259

•794

005

• 005

032

.194

.891

000

.000

00l

.012

•109

409

.409

0

0

0

410

.410

511

0

0

154

.154

384

.640

026

•026

096

.320

.800

002

•002

008

.040

.200

631

0

,8

p

--

q

71

0

631

0

A p r o g r a m m e d c o m p u t a t i o n produced the following e x a m p l e s from the sequence of

state probabilities after POISSON arrivals,

U

i =

services.

Initially,

pq, 0(t0)

=

1

is assumed•

•5 Nq(t i)

i

Pq,0(ti )

pq, l(ti)

1

.607

.303

.076

013

.002

.000

5

.512

,326

.i18

033

.008

.002

.000

10

.502

,325

.122

037

.010

.003

.001

.000

•

20

.500

.324

.123

038

.011

.008

.001

.000

• 749

50

no c h a n g e . . .

• 500 .707

741

.750

T h e individual p r o b a b i l i t i e s c o n v e r g e to the third d e c i m a l p l a c e within less than

20

itera-

156

tions. The rate of convergence for the expectations average of the first

20

expectations, i . e .

Nq(ti)

is slightly less favorable. The

the expectation of an average of the first

20

states

n (t.) , is .73 . This value is 3°]o below the l i m i t expectation• As the number of q 1 observations is increased, this bias decreases with the reciprocal value of the number of observations. This behaviour was already discussed in chapter POISSON arrivals,

U

=

5 .

.8

Nq(t i)

i

Pq, 0(ti)

Pq,l(ti )

1

.449

.359

.144

.038

.008

.001

.000

5

.282

.319

.210

.i09

.049

.020

.007

.003

.001

.000

1.434

10

.243

.288

.208

.125

.089

.036

.018

.008

.004

.002

1.748

20

.218

.265

.200

.129

.079

.047

.027

.016

.009

.005

2.040

50

.204

.249

.192

.128

.083

.053

.034

.021

.013

.008

2.305

50

iterations,

0.800

Apparently, the increased utilization slows down the convergence. Now, after Pq, 0

is still off its l i m i t value

l i m i t value

2.4

by

1 - U

=

.2

by

2°]0. The expectation differs from its

4%, and the average of the first

50

expectations is about

20% less

than this l i m i t value. Thus, the l i m i t values would have to be used with caution. However, a queuing system with POISSON arrivals will rarely be utilized to

80%0. A reason-

able ratio between service times and wait times can practically not be assured with this utilization. For this, several exercises of the earlier chapters may serve as examples. Finite population,

U

=

.5 Nq(t i)

i

pq, 0(ti)

pq, l(ti)

1

.631

.308

.056

.005

.000

.435

5

.576

.336

.078

.009

.000

.520

I0

.576

.337

.079

.009

.000

•

521

, and no further changes .

Here, convergence is much faster than with POISSON arrivals. The l i m i t values describe the queuing system very closely. Note also, that the l i m i t value

Pq, 0

=

(1 - U ) / R

is always greater than

1 - U . The

157

probability to l e a v e an e m p t y system behind at the end of a s e r v i c e is g r e a t e r than the p r o bability to find the system e m p t y at a random t i m e . F i n i t e population,

U

=

.8

Nq(t i)

i

pq, 0(ti)

Pq, l ( t i )

1

.409

.410

.154

• 026

002

5

.286

.405

.236

• 065

007

i, I01

10

.229

.403

.242

• 068

OOq

I. 121

20

.279

.403

.242

• 069

007

1. 122

, and no further changes.

the c o n v e r g e n c e is fast,

and the l i m i t v a l u e s describe

Even at this considerable u t i l i z a t i o n ,

•

801

•

the queuing system closely• T h e s a m e c o m p u t a t i o n s w e r e carried out with the assumption of an e x p o n e n t i a l distribution of the s e r v i c e times.

This slowed down the c o n v e r g e n c e to the e x t e n t that in all four cases,

very closely t w i c e as m a n y iterations w e r e required for the s a m e d e g r e e of a p p r o x i m a t i o n to the l i m i t values.

Lag correlations in the i m b e d d e d chain. In chapters

1

and

3,

c o v a r i a n c e s and correlation w e r e discussed for random variables

w h i c h o c c u r with a certain d i s t a n c e or ' l a g ' within the s a m e process.

These c o v a r i a n c e s h a v e

a significant e f f e c t on the c o n f i d e n c e in a v e r a g e s observed in the process. The a b o v e n u m e r i c a l results g i v e some e v i d e n c e of the correlation.

In fact,

the l i m i t v a l u e

of the correlation b e t w e e n

the states at t~¢o c o n s e c u t i v e t i m e s

t i, ti+ 1

t h e transition p r o b a b i l i t i e s

Pq

In the case of POISSON

and the l i m i t p r o b a b i l i t i e s

arrivals as discussed in e x e r c i s e

cI

=

67,

as derived in the next e x e r c i s e , V*(nq) results as

.85

c o r r e l a t i o n is

e1

for =

this correlation has the v a l u e

Nh(Z-u)/V*(nq)

I

pq.

can be found from

,

From the l i m i t p r o b a b i l i t i e s found above, U

0.56

= for

.5 , U

and =

.5,

5.4 and

for cI

U =

=

the v a r i a n c e

. 8 . T h e resulting

0.91

for

U

=

.8 .

158

The discussion of chapter

3

showed already that a correlation of

0.9

for the lag

1 , and

consequentially high correlations for the other tags, have a severe effect on the confidence in averages.

Exercise. 72

Derive the expression for the lag correlation

c1

in a queuing system with

POISSON arrivals and a service time distribution independent of the state.

Solution. 72

The joint probability to have state

k

at

ti

and

state

j

at

ti+ 1

is found as

the product Pq, ilk Pq, k(ti ) of a conditional probability, a n t e of the states

n ~ ti) mand ~" k=O

The sum over Exercise for

k/0.

68

J

and the probability that the condition holds. Then, nq(ti+l)

the covari-

is defined as

~ k j pq, jl k pq, k(ti) j=O

Nq(ti) Nq(ti+l) .

of the two inner factors is the expectation of the next state,

showed that this expectation equals

k - 1 + Ra T s

given

k.

k - Pq, O

This makes the covariance equal to O3

k--O

Nq(t i) Nq(ti+l)

k ( k - P'~t, 0 ) Pq, k(ti ) (l)

Since

Nq(ti+ 1)

=

Pq, 0(ti ) Nq(ti)

+

It" ( k - P ~ , o k=0

Pq, k(ti )

+ Pq, 0(ti ) - P~t, 0

the covariance is finally E

(nq2(ti) )

Nq(ti) ( Pq, * 0 =

Taking limit values for correlation

cI

+

Nq(ti)

+

V (nq(ti) )

i.-Ip~o, and dividing by the variance

as used on the previous page.

Pq, 0(ti ) - P*q, 0 ) Nq(t i) pq, 0(ti) . V*(nq), one arrives at the

159 Chapter

14

:

In this chapter,

Limit utilization and wait time distributions,

two final problems are discussed. The first problem is the relation between

the probabilities for the imbedded chain and for any time

t. It is sufficient to study this

problem with respect to the l i m i t probabilities only, since these are intended for application. A very simple result arises for POISSON arrivals, viz. identity of both sets of l i m i t probabilities. The somewhat less trivial relations for the system with finite population will be reported from Takacz 1. Their main result, the throughput ratio chapter

R , was already discussed in

11.

Secondly, the distribution of wait times, and its derivation from state probabilities, is considered. The important result of this consideration is the fact that the distribution of wait times depends also on the queuing disci~tine. The results of a simulation are used to give numerical evidence of this fact. Then, for a FIFO discipline in the classical single-server system, the wait time distribution is analyzed. This shows the POLLACZEK-KHINTCHINE formula as one result in a series of similar formulae for higher moments of the l i m i t distribution of wait times. These considerations intend to give the reader a basis from which he can start the analysis of further similar problems whenever a particular application problem might require this.

Limit probabilities at a random time. A famous argument of KHINTCHINE which Takacz 1 develops in all m a t h e m a t i c a l detail, shows a simple relation between the limit probabilities the l i m i t probabilities

pq, j

of the imbedded chain, and

for the state of the system at any time

Consider arrivals which find the system in state state

pq, j

t .

j . They bring the system from state

j + 1 . At some later time, a leaving item will cause a transition from state

to state into state

j

to

j + 1

j . Other states may have been passed in between. Anyhow, the number of arrivals j , and of departures leaving state

j behind cannot differ by more than

they are observed in one contiguous time interval.

1

if

160

Hence,

the l o n g - r u n rate of arrivals into state

rate of departures l e a v i n g state

j

behind,

Pq, j

=

j , which is

Ral j pq, j , and the l o n g - r u n

Ra pq, j , are equal.

w h i c h is

T h e relations

P~t, J Ra / Ra[j

p e r m i t to find the l i m i t probabilities for any t i m e from those of the i m b e d d e d chain. For a system with POtSSON arrivals,

all

Ra[ j

are equal to

the corresponding probability in the i m b e d d e d chain. find the system e m p t y ,

Ra , thus all

In particular,

pq, j

equal to

the probability

Pq, 0

to

and h e n c e the server not in use, has the l i m i t v a l u e Pq, 0

=

1 - Ra T s

=

1 -U.

A system with f i n i t e population leads to pq, j This d e t e r m i n e s all The relation for Ra

=

pq, j

j = 0

=

including

pq, j Ra / ( N

- j) Rr

for

j = 0,1 . . . . .

Pq, N ' since the sum of the

can be written as an equation for

pq, j

U , since

N-1 .

has to be

Pq, 0

=

1 .

1 - U , and

U / T s . T h e equation 1 - U

=

Pq, 0

U/NRrTs

=

NRrTs/(NRrTs

is e q u i v a l e n t to U as already used in chapter Thus,

11,

+

Pq, 0 )

page 129.

all l i m i t e x p e c t a t i o n s of the queuing variables are d e t e r m i n e d by the

Pq, 0

of the

i m b e d d e d Markov chain.

S o m e e v i d e n c e about wait t i m e distributions. T h e problem of finding a l i m i t distribution of w a i t t i m e s from the results about the i m b e d d e d chain is introduced by s o m e n u m e r i c a l e v i d e n c e . the s a m e queuing system as in chapter

3

A c o m p u t e r program was used to s i m u l a t e

under different disciplines.

the FIFO discipline was assumed, as e x p l a i n e d on p a g e 'last-in-first-out'

or

LIFO,

38

While in chapter

8

, two further disciplines known as

and ' r a n d o m s e l e c t i o n ' w e r e simulated.

T h e l a t t e r choses with

equal probabilities a m o n g the w a i t i n g i t e m s if there are m o r e than one at the end of a service.

181

~

e-.9 I

4,t

* : LIFO d i s c i p l i n e

¢~- -~.

*F

,R

I

*F

¢$ b-.8

Random selection

t I

¢5

I t l t

4~ ¢-.6 i

RT ,F

I

FIFO d i s c i p l i n e

average wait time

for a l l t h r e e d i s c i p l i n e s

*R

-F E m p i r i c distributions ~.,4

of s i m u l a t e d

w a i t

I l

for t h r e e d i s c i p i i n e s .

Fc/x) t i m e s

S~F, R ¢,_ 4~ 4.-

t2 4" ¢$ -t~. 1

0.5

1

1.5

2

162 The empiric distributions are shown on the previous page• Averages and sample variances of 1000 sample values were for the discipline FIFO random selection LIFO

t---~

V(tw)

.656

.693

.667

1.352

.646

2.507

The averages show some differences. In view of the confidence intervals discussed in chapter 3 , these differences can be attributed to random effects, and are n0t-significant. In fact, they are very small since all three simulations were based on the same sequence of arrival times. The difference in shape of the empiric distributions is noticeable. I.t is also reflected in the difference of sample variances which range from a smallest value, under FIFO discipline, to a value about

4

the estimate of

times as large under LIFO discipline. This variance enters as a factor in V(tw)

, hence its square root in the estimated confidence interval for

~-w"

On the other hand, the lag correlations computed with the procedure STANAL2 of appendix E page 182 , are largest under FIFO discipline, and less for the two other discipline. This is a plausible result, since the two latter disciplines remove the strong dependence between two consecutive items, if the wait line contains more than one item. Some correlations were for lag

1

2

3

6

FIFO

.79

.62

.49

• 30

random selection

.25

.27

.24

LIFO

.12

.15

.07

Discipline

20

30

16

• 05

. O0

• 18

14

• 09

.04

, O0

07

• 03

.06

10

The difference in correlations somewhat outbalances the mcrease of confidence intervals caused by the variances. With the approximation described on page 42 variance and

90% confidence intervals of

FIFO

V(F-w)

=

.007

, the estimated

{-- were for W

and

1.645 V ( ~ w )

=

•

14

random selection

.011

.18

LIFO

,008

.15

163

Wait t i m e distributions. L i m i t distributions for the wait times have been derived for different queuing disciplines. S o m e of these results are quoted by Saaty 1. A starting point for this derivation is given by the l i m i t p r o b a b i l i t i e s of the i m b e d d e d Markov chain. For the FIFO discipline, the reasoning is p a r t i c u l a r I y simple. i m b e d d e d chain to l e a v e p r o b a b i l i t i e s that

j

j

Here,

the p r o b a b i l i t i e s of the

i t e m s behind at the end of a service,

are the same as the

i t e m s arrived during the t i m e which the l e a v i n g i t e m spent in the

queuing syszem. Assume thai: all such times of exercise

68

tq

have the same (limit) distribution

can be a p p l i e d with

probabilities of arrivals in a t i m e Fq(X) still unknown.

F

tq

q

instead of

=

(1-z) Ra

and

z

= =

LFq(S) As an e x a m p l e ,

E.g.,

the result

with POISSON arrivals the L F q ( ( l - z ) Ra ) , with

But the above e q u i v a l e n c e of arrival p r o b a b i l i t i e s with p r o b a b i l i t i e s in

L F q ( ( l - z ) Ra ) s

s

have the g e n e r a t i n g function

the i m b e d d e d chain i m p l i e s equal generating function,

or, with

F

Fq(X). Then,

consider the

Gq(Z)

1 - s/R a :

Gq(Z)

viz.

, ,

G* ( 1 - s/R a ) q

found on page 152 , with a service t i m e distribution

independent of the state, viz. (z -1) G* (z) q This

Gq(Z) leads

z -

L F s ( ( 1 - z ) Ra) LFs(!l-z)Ra)

0

to LFq(S)

=

( 1 - U ) LFs(S) / ( 1

Ra ( 1 -LFs(s ) ) / s )

Finally, the t i m e s spent in the queuing system are sums of a w a i t t i m e , service time. Therefore,

and an independent

the m u l t i p l i c a t i o n rule for the LAPLACE transforms

LFq(S)

=

LFw(S) LFs(S)

=

( 1 - U ) / ( 1 - R a ( 1 - LFs(S)) / s )

applies, with the result that LFw(S)

determines the w a i t t i m e distribution c o m p l e t e l y by arrival rate and service t i m e distribution.

164

Exercises. 79

Find the series for

Ra ( 1 - LFs(S) ) / s

LFw(S) =

1 / ( 1 + c l s - c2s2/2 + O(s 3) )

=

by powers of

s

up to

O(s 3) . Use

1 - c l s + (c 2 + 2 c 2) s2/2

+

O(s 3)

to find the first two m o m e n t s and the v a r i a n c e of the w a i t t i m e distribution derived at the end of the previous page. 74

Consider the data of e x e r c i s e

times.

19 , page 49 , and c o m p u t e the v a r i a n c e of wait 2 T w ( 2/U 1 ) of the a p p r o x i m a t i n g e x p o n e n t i a l

C o m p a r e it to the v a r i a n c e

distribution suggested on p a g e 75

50 .

Find the first two m o m e n t s and the v a r i a n c e for the t i m e s

tq

spent in the

queuing system.

Solutions. 73

T h e suggested relation b e t w e e n the power series of a function and its r e c i p r o c a l

is easily proven by a m u l t i p l i c a t i o n with the d e n o m i n a t o r , If the power series of

LFs(S)

is g i v e n as

and c o l l e c t i o n of powers of s .

1 - T s s + E(ts2) s2/2 - E(t$) s3/6 + O(s 4) ,

the d e v e l o p m e n t of the expression quoted in the e x e r c i s e is Ra ( 1 -LFs(S)) / s

=

a aT s

This expression is now subtracted from

Ra E ( t s2) s /2 1, and

1 - U

% E@ LFw(S ) cI

and

c2

=

1 / ( 1 + - 2 (1-u)

can now be identified.

way,

+

R a E(t~) s 2 / 6

+ O(s 3) .

is divided by the result to find

Ra E(t ) s

- 3 (1-u)

s2/2

+

Interpreting the power series of

O(s 3) LF .(s) W

) . in the usual

Ra E(ts2) Tw

=

cI

=

(POLLACZEK-KHINTCHINE) , 2 (1-u)

and

E(tw2)

=

c2

+

Vw

=

2 E(t ) - T w

T h e v a r i a n c e of w a i t t i m e s is then

2c~

w2

=

2 c2 + Cl

T 2 w

=

Ra E(tsa) 3 (l-u)

Note that this v a r i a n c e is at least equal to

the standard d e v i a t i o n is at least as

great as the e x p e c t e d w a i t t i m e ,

=

-

T 2 . Thus, w m a i n l y because t w

+

0

has considerable probability.

165

74

As p a r a m e t e r s of the s e r v i c e t i m e distribution, Ts

=

1 . 1 9 sec ;

Vs

=

0.307 sec 2

;

In order tO avoid a c o m p u t a t i o n of a fourth m o m e n t , from the f o r m u l a at the end o f p a g e of

(23 - 1 ) / ( 2 2

-1)

suited f r o m e x e r c i s e E(t 3)

=

= 8

7/3

31

,this

E(ts2) =

19

used

1 . 7 5 sec 2.

as would be required to find

E(t 3)

E(ts3) is a p p r o x i m a t e d by adding an overhead

of the o v e r h e a d for

, p a g e 33.

exercise

E(t 2)

to the third m o m e n t which r e -

Thus,

2.46 + (7/3)'.0.84

2.46

=

3 . 0 sec 3

w i l l be used. With a l i m i t u t i l i z a t i o n Then, Vw

U

=

.75 , T w

=

2.172

+

=

2.17 sec

3.0 / 1.19

was already found in e x e r c i s e

=

7 . 3 sec 2

=

q. 8 sec 2

19.

T h e a p p r o x i m a t i o n was 2 . 1 7 2 ( 2 / . 75

1 )

a slightly m o r e p e s s i m i s t i c v a l u e if a p e r c e n t i l e c o m p u t e d from it is h e l d below s o m e bound. 75

T h e first order r e l a t i o n T

q

=

T

w

holds i n d e p e n d e n t of the discipline. and p r e c e d i n g w a i t t i m e s , Vq

+

T

s

As a c o n s e q u e n c e of the i n d e p e n d e n c e of s e r v i c e t i m e s

their v a r i a n c e s are a d d i t i v e , =

Vw

+

so that

Vs •

T h e second m o m e n t is found by 2 Tq

E(t ) =

+

V

q

,

The n u m e r i c a l data used in the previous e x e r c i s e result in T and Here,

V

q q

= --

3 . 3 6 sec 7 . 6 sec 2

the standard d e v i a t i o n is s o m e w h a t less than the e x p e c t a t i o n

Tq,

but still c o n s i d e r a -

ble. If both e x p e c t a t i o n and v a r i a n c e of a distribution are known,

its p e r c e n t i l e s can be a p p r o x i -

m a t e d by those of a G a m m a distribution with the s a m e e x p e c t a t i o n and variance.

The index

of the G a m m a distribution is found by dividing t h e squared e x p e c t a t i o n by the variance. t a b l e of Appendix D, p a g e 177 , m a y be helpful in this context.

The

APPENDICES

A : Remarks concerning the notation. 13 : C o l l e c t e d f o r m u l a e of Queuing Theory. C : Tables for some f o r m u l a e of Queuing Theory, D : Mathematical reference material. E : C o m p u t i n g procedures.

167

Appendix

A

:

Remarks concerning the notation.

The use of arithmetic operators and parentheses, exponents, functions magnitude

O(dx), and subscripts

The letters

i, j, k, re, n; 0, 1 , . .

e for the base of natural logarithms,

f(x), g(x), orders of

follows the usual rules. t for t i m e , and

T for the duration of a

t i m e interval have this meaning throughout the text. Constant natural numbers

M, N , real numbers

and ratios

R , are defined in the context.

Subscripts

m,n

are preferred for finite range,

Random variables are in general denoted by moments by

c, c ' , c " , c 0, c t . . . .

E(V~, and the variance by

i, j, k

probabilities

p

or

P,

for infinite range.

v, their probability distribution by

Fv(X), their

V(v). LFv(s) denotes the LAPLACE transform of Fv(X)

GO

,{'~e "sx dFv(x ) . F v l ( p ) denotes the inverse of a distribution, e . g .

the 100 p - t h percentile.

-00

For random variables with (non-negative) integer values, the probabilities

Prob( v = j )

are

Since random times and (integer) numbers arise in many contexts, these are denoted by

t or

O3

denoted by

Pv, j" Their generating function is

Gv(Z)

= ~0

n

with an identifying subscript. The subscripts used are :

a

:

Pv, j z j "

' b e t w e e n arrivals', 'of arrivals'

c

:

'of polling cycles', 'in a polling c y c l e '

ch :

'per character',

e

:

'between errors', 'of errors'

p

:

'to poll a t e r m i n a l ' , 'to position'

and, specific for Queuing Theory,

q

:

'in the queuing system'

r

:

'to return into the system'

s

:

'of services'

w

:

' w a i t e d ' , 'of waiting items'

'of characters'

In order to avoid t w o - l e v e l subscripts, the corresponding probabilities, distributions, LAPLACE transforms, and generating functions show only the identifying subscript sub. The expectation by

E(tsub)

is abbreviated by

Tsu b,

E(nsub)

by

Nsu b, the variance

V(tsub)

Vsu b. Rsu b denotes a corresponding rate or intensity.

The identifying subscript is in some cases extended by a condition, e . g . to random v a r i a b l e considered at GIVEN t i m e The number of servers used at t i m e expectation

t

sublt

for a

t.

is a discrete random variable

u(t), with the l i m i t

U .

A horizontal bar over a random variable denotes an average or t i m e average.

168 Appendix

B

:

C o l l e c t e d f o r m u l a e of Queuing Theory.

Sufficient condition for statistical stability : R T a

M

is less than the number

s

of servers .

General relations b e t w e e n l i m i t expectations : U T

q

Nw

=

R T a s

=

T

:

Ra T w

Nq

+

w

Nw

Ts

U

+

=

Ra T q

L i m i t expectations of w a i t t i m e s

: ii|

One server and any service t i m e distribution

POISSON arrivals.

Ra E(t 2) T

=

=

w

2 (liU) M

W

N

Ts ( 1 +

w

with

i t e m s returning with rate N(1 Tw

=

The

Ts( c'.

U (POLLACZEK-KHINTCHINE) 2 (1-U)

servers and e x p o n e n t i a l distribution

P Ts Ivl - C

--

V ....~S ) Ts

Rr.

P

=

Fs(X) .

Fs(X ) .

M-1 + (M - V ) (M - 1) ' ~ uJ-M/j:

1 / [1

One server and any

Fs(X ) .

-R) U

- 1)

with

R

=

:i

1/

N-1 j~=0 ci ) "

(I+NRrTs

are defined by a recursion on p a g e 130 , and

U

=

R.

NRr T s .

J

M (

Compute

servers and e x p o n e n t i a l distribution Nq

as the e x p e c t a t i o n of the l i m i t probabilities stated on p a g e

T h e n use the above g e n e r a l relations with

Approximation for

N

Ra

=

(N - Nq) R r . )

polled terminals with POISSON arrivals. One line, any

NTp/T s + T

Fs(X ) .

U (I+Tp/T

s) ( i + Vs/T 2)

= w

211

- U(1

+Tp/Ts) ]

For conversational t e r m i n a l use, apply the f o r m u l a e of p a g e

74 .

Fs(X) .

96 .

169

Limit e x p e c t a t i o n s of w a i t t i m e s under priority disciplines. I

II

POtSSON arrivals for e a c h priority class. One server. different priority classes

n

=

1.....

N.

Any s e r v i c e t i m e distributions for the

T h e g e n e r a l relations hold within e a c h class.

N o n - p r e e m p t i v e priority d i s c i p l i n e : Ra E(ts2) Tw, n

for

=

n = 3., . . . . N.

2 ( 1 - Sn _ l ) ( 1 - S n ) P r e e m p t - r e s u m e priority discipline : n

Ra, m Em(ts2) m=l Tw, n

= (

+ 2 (1

Sn-1 Ts, n )

/

(1 -S n_l)

-Sn) for

P r e e m p t i o n by classes up to

nO

only :

n =1 .....

As p r e c e d i n g case for

N. n {[ n O , but

R a E(ts2) Tw, n

= (

Sn 0 Ts, n ) / ( 1 - Sn _ l ) 2(1

-Sn) for

n ~

no .

V a r i a n c e of l i m i t distributions. FIFO discipline.

POISSON arrivals. 2

V V

W

q

=

Ra E(t$)

T

+ w

=

3 (l-U)

Vw

+

Vs

Approximation. 2 Vw

T,,~ ( -

P

1

Note that

P

=

U

for a single server.

170

Appendix

Table of the l i m i t expectation

C

:

Tables for some formulae of Queuing Theory.

Tw / T s ,

finite population of

N

of wait times in a single-server system with

,,

,

,

items. Service t i m e distribution is

.2

.3

.4

N

T2/V~

U : ,1

1

any

No waiting occurs since the i t e m is the only user of the server.

1 4 16 64

0.050 0.032 0.027 0.026

0. I01 0.066 0.057 0.055

0.154 0.103 0.089 0.086

0.209 0.143 0.125 0.120

0.268 0.188 0.165 0.159

0.333 0.240 0.212 0.204

0.408 0.301 0.268 0.259

0.500 0.380 0.341 0.331

0.627 0.496 0.450 0.437

1 4 16 64

0.069 0.044 0.037 0.036

0.143 0.093 0.080 0.076

0.223 0.148 0.127 0.122

0.311 0.210 0.182 0.175

0.409 0.282 0.246 0.237

0.520 0.368 0.324 0.312

0.653 0.474 0.420 0.406

0.820 0.614 0.550 0.532

1.063 0.827 0.750 0.729

1 4 16 64

0.085 0.054 0.046 0.044

0.181 0.116 0.099 0.095

0.291 0.189 0.162 0.156

0.417 0.276 0.238 0.229

0.563 0.381 0.331 0.318

0.738 0.511 0.447 0.430

0.955 0.678 0.598 0.577

1.240 0.908 0.809 0.783

1.676 1.277 1.153 1.120

1 4 16 64

0.092 0.058 0.049 0.047

0.199 0.127 0.108 0.104

0.324 0.209 0.179 0.172

0.472 0.309 0.266 0.255

0.649 0.433 0.375 0.360

0.867 0.591 0.515 0.496

1.144 0.801 0.703 0.678

1.520 1.098 0.975 0.942

2.113 1.589 1.430 1.388

I0

1 4 16 64

0.098 0.061 0.052 0.050

0.214 0.135 0.115 0. II0

0.352 0.225 0.193 0.185

0.520 0.337 0.289 0.277

0.727 0.479 0.413 0.397

0.989 0.665 0.577 0.555

1.333 0.919 0.804 0.774

1.815 1.292 1.141 1.102

2.600 1.929 1.730 1.677

15

1 4 16 64

0.102 0.064 0.055 0.052

0.225 0.142 0.121 0.116

0.375 0.239 0.204 0.195

0.562 0.361 0.309 0.296

0.799 0.520 0.447 0.429

1.109 0.735 0.635 0.609

1.529 1.038 0.903 0.869

2.142 1.500 1.319 1.272

3.183 2.326 2.076 2.010

20

1 4 16 64

0.104 0.065 0.056 0.053

0.231 0.146 0.124 0.119

0.388 0.246 0.210 0.200

0.586 0.374 0.320 0.306

0.841 0.543 0.466 0.446

1.181 0.775 0.668 0.641

1.655 1.111 0.963 0.926

2.365 1.637 1.434 1.382

3.609 2.610 2.322 2.247

POISSON arrivals. Any service t i m e distribution

1 4 16 64 constant t s :

0. iii 0.069 0.059 0.056 0.056

0.250 0.156 0.133 0.127 0.125

0.429 0.268 0.228 0.218 0.214

0.667 0.417 0.354 0.339 0.333

.5

Gamma T / / V s .

:

.6

.7

.8

.9 Tw =

0.

POLLACZEK-KHINTCHINE formula .

1.000 0.625 0.531 0.508 0.500

1.500 0.937 0.797 0.762 0.750

2.333 1.458 1.240 1.185 1.167

4.000 2.500 2.125 2.031 2.000

9.000 5.625 4.781 4.570 4.500

171

T a b l e of t h e t h r o u g h p u t r a t i o

R

=

of a s i n g l e - s e r v e r s y s t e m w i t h

U / (N Rr Ts)

I

f i n i t e p o p u l a t i o n of

N

items.

S e r v i c e t i m e d i s t r i b u t i o n is

N

Ts2/Vs U = . 1

1

any

.9000

.8000

.7000

.6000

.5000

.4000

1 4 16 64

.9475 .9484 .9486 .9487

.8899 .8934 .8943 .8945

.8270 .8346 .8366 .8372

.7582 .7714 .7750 .7760

.6830 .7030 .7088 .7103

1 4 16 64

.9644 .9652 .9654 .9655

.9238 .9272 .9280 .9283

.8777 .8852 .8873 .8878

.8252 .8387 .8424 .8433

.9783

.9226 .9286 .9302 .9306

.8867

10

20

.3

.4

.5

.6

.7

.8

.9

.3000

.2000

.I000

.6000 .6281 .6365 .6387

.5071 .5445 .5561 .5592

,4000 .4479 .4635 .4677

.2675 .3267 .3475 .3532

.7652 .7863 .7923 .7938

.6960 .7264 .7353 .7376

.6144 .6561

.5146 .5697

.6686

.5868

.6719

.5913

.3812 .4519 .4751 .4813

.9017

.8437 .8619 .8669 .8682

.7915 .8187 .8264 .8284

.7264 .7651 .7763 .7792

.6417 .6947 .7106 .7147

.5186 .5903 .6126 .6185

1 4 16 64

!.9791 9791

.9528 .9554 .9560 .9562

1 4 16 64

9844 9849 9850 9850

.9657 .9678 .9683 .9685

.9432 .9482 .9495 .9498

,9159 .9252 .9276 .9283

.8822 .8976 .9018 .9028

.8400 .8636 .8701 .8718

.7856 .8199 .8297 .8322

.7120 .7603 .7744 .7781

.5999 .6672 .6877 .6931

1 64

9890 9894 9895 9895

.9757 .9773 .9777 .9778

.9594 .9632 .9642 .9645

.9392 .9465 .9484 .9489

.9136 .9260 .9293 .9302

.8806 .9001 .9054 .9067

.8367 .8657 .8737 .8758

.7748 .8167 .8287 .8~]8

.6759 .7363 .7543 .7591

1 4 16 64

9927 9929 9930 9930

.9887 .9848 .9851 .9851

.9725 .9752 .9759 .9761

.9583 .9637 o9651 .9654

.9400 .9493 .9518 .9524

.9157 .9306

,8820 .9049

.8824 .8667

.7490 .8004

.9346

.9112

.9356

.9128

.8763 .8788

.8194

1 4 16

9945 9947 9947 9947

.9877 .9885 .9888 .9888

.9792 .9813 .9819 .9820

.9683 .9725 .9736 .9736

.9540 .9614 .9638 .9638

.9346 .9467 .9500 .9508

.9071 .9261 .9313 .9326

.8654 .8945 .9026 .9047

.7926 .8375 .8505 .8539

.9789

4 16

15

.2

G a m m a Ts2/V s •

64

.8979

.9010

.8t54

£

As

N~oo

,

the limit of

For low u t i l i z a t i o n A utilization

U

U,

R

is

1

for any

T s / V s , a n d for a l l

R is a p p r o x i m a t e l y e q u a l to

close to 1 , for given

U <

1 .

1 - U/N .

N , can only be achieved with a high return rate

Rr , and t h e r e f o r e i m p l i e s a low t h r o u g h p u t r a t i o

R.

172

T a b l e of the probability

P

to find all servers busy,

in a

M-server system with i

POISSON arrivals and the same exponential distribution of service times for all servers.

I

:

Probability values in

% .

U/M

is the expected utilization per server. i

g

=

U/M

.1

.2

.3

.4

.5

.6

i

i

.7

.8

I

i

6 7

20.00 6.67 2.47 .96 .38

30.00 13.85 7.00 3.70 2.01

40.00 22.86 14.12 9.07 5.97

50.00 33.33 23.68 17.39 13.04

60.00 45.00 35.47 28.70 23.62

70.00 57.65 49.23 42.87 37.78

64.72 59.64 55.41

90.00 85.26 81.71 78.78 76.25

.00

.16 .06 .03 .01 .00

1.11 .62 .35 .20 .12

4.00 2.71 1.85 1.27 .88

9.91 7.62 5.90 4.60 3.61

19.66 16.51 13.95 11.86 10,13

33.60 30.07 27.06 24.45 22.17

51.78 48.59 45.76 43.22 40.92

74.01 72.00 70.15 68.45 66.87

.04 .01 .00

.43 .21 .10 .05 .03

2,25 1.42 .90 .58 .37

7.47 5.57 4.19 3.17 2.41

18.39 15.39 12.97 10.99 9.36

36.88 33.45 30.49 27.90 25.61

64.00 61.45 59.13 57.02 55.08

.04

.65

4.39

17.29

47.14

12 14 16 18 20

30

i

Negative natural logarithms of use. M-U

U

The entries for =

i

I0.00 1.82 .37 .08 .02

8 9 10

II :

.9

1

P <

2

P .

e -10

3

=

M -U

I

80.00 71.11

I

is the excess of servers over the expected

4.5.10"5 4

are omitted. 5

6

7

8

9

i

i

5.565

0.674 0.590 0.532

2.398 1.749 1.443 1.256 1.127

3.892 2.818 2.311 2.002 1.788

3.016 3.280 2.829 2.519

7.397 5.337 4.3 7 3.740 3.321

9.371 6.772 5.509 4.731 4.194

8.312 6.762 5.801 5.135

9.950 8.099 6.946 6.144

9.516 8.163 7.217

6.0 7.0 8.0 9.0 10.0

0.488 0.454 0.426 0.402 ~.383

1.030 0.955 0.894 0.843 0.800

1.630 1.506 1.407 1.324 1.254

2.290 2.111 1.967 1.848 1.748

3.011 2.771 2.577 2.417 2.282

3.795 3.486 3.237 3.032 2.859

4.641 4.257 3.948 3.693 3.478

5.547 5.083 4.708 4.400 4.140

6.512 5.962 5.519 5.153 4.844

12.0 14. 0 16.0 18.0 20.0

0.350 0.325 0.305 0.288 0.274

0.730 0.676 0.633 0.596 0.566

1.142 1.055 0.985 0.927 0.878

1.587 1.462 1.362 1.280 1.210

2.066 1.900 1.767 1.657 1.565

2.582 2.369 2.199 2.059 1.942

3.134 2.871 2.660 2.487 2.343

3.724 3.405 3.150 2.942 2.767

4.351 3.972 3.670 3.423 3.217

30.0

0.224

0.462

0.712

0.977

1.257

1.552

1.863

2.190

2.534

1.0 2.0 3.0 4.0 5.0

1.099 0.811

173

T a b l e of t h e l i m i t e x p e c t a t i o n

Tw

of t h e t i m e a m e s s a g e w a i t s a t o n e of

N

cyclically

• ml

polled terminals with equal load. Cases

A : Conversational arrivals, C : T 2 /V s

Poll t i m e r a t i o N

4

Tp/T s

=

D :

.300

,

b o t h w i t h POISSON arrivals.

b o u n d for t h e p r o d u c t i v e u t i l i z a t i o n

1.022 1.242 1.293 1.446

1.217 1.661 1.745 1.996

1.455 2.383 2.521 2.939

1.753 3.945 4.202 4.972

2.160 9.996 10.703 12.793

2. ~00

1.386 1.448 1.461 1.501

1.611 1.770 1.799 1.886

1.887 2.219 2.269 2.419

2.231 2.901 2.982 3.227

2.675 4.083 4.219 4.629

3.274 6.660 6.917 7,687

4.165 16.694 17.424 19.585

5.200

2.053 2.137 2.150 2.190

2.362 2.580 2.609 2.695

2.747 3.201 3.250 3.399

3.237 4.148 4.228 4.471

3.883 5.793 5.928 6.336

4.781 9.384 9.642 10.413

6.165 23.378 24.118 26.318

7.800

2.720 2.827 2,840 2.879

3.113 3.391 3.419 3.505

3.605 4.183 4,232 4.380

4.240 5.396 5.476 5.718

5.087 7.505 7.640 8.047

6.284 12.111 12.368 13.141

8.165 30.054 30,801 33.022

10.400

3.387 3.516 3.530 3,569

3.863 4.201 4.230 4.316

4.463 5.166 5.215 5.363

5.241 6.645 6.725 6.966

6.290 9.219 9.353 9.759

7.786 14.838 15.095 15.869

10.165 36.727 37,478 39.714

13.000

C D A B

C

D A B

C D A B

C D

In this a n d t h e f o l l o w i n g t a b l e s ,

.3

.4

t h e l i m i t of

.5

Tw/T s

.6

.7

0.769 .

0.858 0.962 0.992 1,081

B

N/2

;

Vs

0.719 0.759 0.772 0.812

A

.2

T2 =

both with constant service time;

A C

29

4 ,

POISSON a r r i v a l s ,

U = .1

D

16

=

B :

Ts .

Case

B

12

T h e v a l u e s are g i v e n i n units of

for low u t i l i z a t i o n s ,

.769

U~

0 ,

is

times the poll time ratio .

For c o n v e r s a t i o n a l a r r i v a l s , lization remains finite,

t h e e x p e c t e d w a i t t i m e at t h e m a x i m a l

a n d is

T a b l e s for t h e poll t i m e ratios

possible productive uti-

N(Tp + Ts) / 2 .

. 1 5 0 , . 075 , . 025 , . 0 1 0

on t h e f o l l o w i n g p a g e s .

174

A Conversational, N

Case l

Poll t i m e ratio 4

A t3

C D 8

A B

C D 12

A B

C D 16

A B

C D 20

A B

C D

B,C,D

POISSON arrivals with .2

U=.I

0.150

.3

.4

2 Vs/T s

=

.5

0 , 1/4 , 1 .6

bound for the p r o d u c t i v e u t i l i z a t i o n

;

.7

.8

.870

0. 870 .

0.386 0.399 0.412 0.452

0.487 0.519 0,548 0.634

0.606 0.674 0.722 0.869

0.748 0.889 0.965 1.197

0.919 1.217 1.335 1.698

1.130 1.785 1.977 2.560

1.403 3.019 3.365 4.404

1.793 7.781 8.698 11.389

2.300

0.720 0.738 0.751 0.790

0.866 0.906 0.934 1.018

1.044 1.126 1.173 1.315

1.266 1.435 1.509 1.733

1.551 1.910 2.025 2.376

1.927 2.740 2.929 3.499

2.457 4.562 4.910 5.954

3.315 11.664 12.629 15.465

4.600

1.054 1.076 1.089 1.128

1,242 1.295 1.323 1.406

1.476 1.583 1.629 1.769

1.774 1.988 2.061 2.281

2.164 2.612 2,725 3.071

2.696 3.704 3,891 4.457

3.477 6.104 6.453 7.501

4.818 15.470 16.458 19.372

6.900

1.388 1.415 1.428 1.467

1.618 1.684 1.712 1.795

1.906 2.040 2.086 2.225

2.277 2. 542 2.614 2.832

2.770 3.315 3.428 3.771

3.457 4. 670 4.856 5.419

4.488 7. 644 7.995 9.046

6.318 19.250 20.252 23.213

9.200

1.721 1.754 1.767 1.806

1.993 2.074 2.101 2.184

2.336 2.497 2.543 2.682

2.780 3.096 3.168 3.385

3.375 4.020 4.133 4.473

4.213 5.637 5.822 6.383

5.494 9.184 9.536 10.588

7.812 23.019 24.029 27.022

11.500

,,,,

Poll t i m e ratio 4

8

U=.9

0. 930 .

0.220

0.302

0.399

0.514

0.652

0.821

1.036

1.327

1.808

0.224

0.311

0.420

0.569

0.786

1.135

1.784

3.416

15.675

C D

0.237 0.276

0.339 0.425

0.468 0.615

0.644 0.875

0.902 1.256

1.314 1.862

2.078 2.969

3.983 5.680

18.160 25.340

A

0.388 0.392 0.404 0.443

0.493 0.499 0.527 0.610

0.623 0.636 0.682 0.823

0.784 0.822 0.894 1.115

0.990 !.095 1.266 1.546

1.260 1.539 1.712 2.241

1.632 2.377 2.666 3.541

2.192 4.524 5.101 6.821

3.324 20.998 23.719 31.581

0.555 0.559 0.572 0.611

0.682 0.690 0.717 0.799

0.841 0.855 0.901 1.039

1.042 1.081 1.152 1.368

1.306 1.413 1.522 1.853

1.663 1.954 2.123 2.642

2.175 2.978 3.264 4.131

2.990 5.613 6.195 7.932

4.815 25.896 28.733 36.976

0.721 0.727 0.740 0.779

0.870 0.880 0.908 0.989

1.057 1.076 1.121 1.258

1.297 1.342 1.412 1.625

1.615 1.734 1.841 2.168

2.054 2.372 2.540 3.052

2.699 3.582 3.866 4,728

3.766 6.694 7.280 9.027

6.308 30.663 33.566 42.046

0.888 0.895 0.908 0.947

1.058 1.071 1.098 1.180

1.272 1.296 1.341 1.477

1.550 1.604 1.673 1.884

1.920 2.057 2.163 2.485

2.439 2.792 2.959 3.466

3.215 4.186 4.470 5.328

4.532 7.772 8.360 10.115

7.803 35.373 38.320 46.958

C A B

C D

A B

C D 20

bound for the p r o d u c t i v e u t i l i z a t i o n

A

D

16

;

B

B

12

O. 075

A B

C D

175

Cases

A Conversational,

N

Case [

Poll t i m e ratio 4

A

B,C,D

U = .1 0.025

.2 ;

POISSON arrivals with

.3

.4

Vs/T ?

.5

=

.6

bound for the p r o d u c t i v e u t i l i z a t i o n

0 , i/4 , 1

.7

.8

.9

O. 976 .

C D

0.109 0.109 0.122 0.161

0.179 0.178 0.207 0.295

0.261 0.266 0.316 0.469

0.358 0.387 0.466 0.707

0.474 0.559 0.679 1.045

0.616 0.824 1.006 1.559

0.795 1.278 1.565 2.432

1.030 2.234 2.733 4.240

1.378 5.605 6.814 10.422

8

A B C D

0.166 0.164 0.177 0.216

0.245 0.238 0.266 0.351

0.342 0.331 0.379 0.526

0.463 0.458 0.534 0.766

0.617 0.642 0.757 1.111

0.818 0.927 1.103 1.642

1.090 1.423 1.702 2.552

1.486 2.487 2.980 4.470

2.159 6.367 7.602 11.272

12

A B C D

0.222 0.222 0.232 0.271

0.309 0.309 0.327 0.411

0.417 0.417 0.447 0.590

0.555 0.555 0.610 0.836

0.735 0.735 0.846 1.191

0.977 1.042 1.213 1.740

1.320 1.582 1.855 2.691

1.843 2.752 3.240 4.718

2.804 7.070 8.322 12.040

16

A

0.277 0.277 0.288 0.326

0.372 0.372 0.389 0.472

0.491 0.491 0.517 0.657

0.643 0.643 0.689 0.911

0.845 0.845 0.939 1.277

1.122 1.162 1.330 1.848

1.525 1.747 2.017 2.841

2.161 3.022 3.507 4.975

3.393 7.749 9.013 12.767

0.383 0.333 0.344 0.382

0.435 0.435 0.452 0.534

0.563 0.563 0.587 0.726

0.730 0.730 0.770 0.988

0.952 0.952 1.036 1.3681

1.261 1.286 1.451 1.961

1.717 1.916 2.183 2.997

2.458 3.295 3.777 5.237

3.953 8.416 9.689 13.469

B

B

C D

20

A B

C D Poll t i m e ratio 4

A

O. O1

; bound for the p r o d u c t i v e u t i l i z a t i o n

O. 990

C D

0.076 0.076 0.088 0.128

0.142 0.142 0.171 0.262

0.219 0.228 0.280 0.438

0.311 0.346 0.428 0.675

0.421 0.513 0.637 1.009

0.555 0.766 0.952 1.511

0.723 1.190 1.480 2.352

0.942 2.048 2.546 4.045

1.260 4.727 5.860 9.262

8

A B C D

0.099 0.099 0. II0 0.149

0.171 0.171 0.192 0.280

0.258 0.258 0.300 0.454

0.367 0.367 0.447 0.690

0.505 0.536 0.657 1.024

0.685 0.793 0.976 1.530

0.929 1.228 1.514 2.379

1.279 2.116 2.609 4.098

1.855 4.945 6.077 9.475

12

A

0.122 0.122 0.131 0.170

0.197 0.197 0.215 0.301

0.290 0.290 0.232 0.474

0.409 0.409 0.471 0.710

0.564 0.565 0.683 1.045

0.772 0.827 1.007 1.555

1.065 1.275 1.557 2.414

1.508 2.195 2.684 4.162

2.287 5.165 6.295 9.690

0.144 0.144 0.153 0.192

0.223 0.223 0.238 0.324

0.321 0.321 0.348 0.496

0.447 0.447 0.498 0.732

0.614 0.614 0.713 1.070

0.844 0.866 1.043 1.584

1.175 1.327 1.606 2.454

1.692 2.282 2.766 4.233

2.647 5.385 6.514 9.907

0.166 0.166 0.176 0.214

0.248 0.248 0.263 0.347

0.351 0.351 0.374 0.520

0.484 0.484 0.526 0.757

0.661 0.661 0.745 1.097

0.908 0.908 1.082 1.617

1.271 1.383 1.658 2.499

1.853 2.373 2.852 4.311

2.966 5.604 6.733 10.124

B

B

C D 16

A B

C D 20

A B

C D

176 Appendix

D

:

Mathematical reference material.

S o m e probability distributions. N o r m a l distributions arise as the l i m i t distributions v a r i a b l e s with c o m m o n distribution. T h e standard n o r m a l distribution F'(x)

and has e x p e c t a t i o n

(Central L i m i t T h e o r e m ,

F(x)

for the sum of

N

random

see Brunk I )

is defined by its density

e -x2/2 / ~

0 , variance

F( has e x p e c t a t i o n

=

( N--~ m )

, for all real

x ,

1 . With a shifted origin and a c h a n g e in scale,

(x-E(v)) / V~/'~ )

E(v) , v a r i a n c e

V(v) , and is still c a l l e d a n o r m a l distribution.

P e r c e n t i l e values of the standard n o r m a l distribution are g i v e n in the t a b l e below. G a m m a - d i s t r i b u t i o n s w i t h an i n t e g e r index

n

v a r i a b l e s with c o m m o n e x p o n e n t i a l distribution. POISSON process (see chapter 1 ).

apply to the sum of Thus,

n

i n d e p e n d e n t random

they h a v e a close r e l a t i o n to the

Also, G a m m a - d i s t r i b u t i o n are r e c o m m e n d e d (Martin 1) as

a p p r o x i m a t i n g distributions for n o n - n e g a t i v e random variables.

Then,

n

is not restricted

to integers. T h e distribution

Gamman(X )

is defined,

Gamma,n(X) and r e l a t e d to the has the v a l u e Gamman(X )

=

G a m m a function

(n - 1) '.

for

n ~ ' 0 , by its density

x n-1 e ' X / 1-1 (n) T "t (n)

=

,

for

o3 ~/~ x n-1 e -x d x 0

x>O , which for integer

of a f a c t o r i a l .

has e x p e c t a t i o n and v a r i a n c e Gamman(X)

=

n . 1

-

For i n t e g e r e -x

n-1 m~__ 0= x m / m :

can be found by f o r m a l i n t e g r a t i o n of the above density. Gammal(x )

=

1

e-X

n ,

T h e e x p o n e n t i a l distribution is

n

177

Gamma-distributions, cont'd With a change in scale,

Gamma n ( n x / E(v) )

has expectation

g(v)

and variance

E(v) 2 / n , and is still called a Gamma-distribution. The choice of E(v)

and

n

=

variance

E ( v ) 2 / V(v)

now leads to the

Gamma-distribution

with expectation

V(v). The third moment of this distribution is E(v) 3 ( 1

+

3

n

+

2 / n2 ) .

Percentile values of Gamma-distributions are given in the table below.

Table of percentiles for With given

E(v)

tion. Read

y

=

=

F(x)

G a m m a - and normal distributions.

and

V(v) , find

n

=

E(v)2/V(v) , or use last row for normal distribu-

from the table. Compute the percentile as 0.01

,u,

0.05

0. i

E(v) + y ~ .

O. 25

O. 5

O. 75

O. 9

O. 95

O. 99

i

-0.707

-0.704

-0.696

-0.635

-0.386

0.228

1.206

2.009

3.9~5

-0.990 -1.309 -1.588 -1.801

-0.949 -1.163 -1.317 -1.421

-0.895 -1.038 -1.128 -1.182

-0.712 -0.735 -0.732 -0.723

-0.307 -0.227 -0.164 -0. I17

0.386 0.490 0.555 0.596

1.303 1,336 1.340 1.333

1.996 1.940 1.877 1.820

3.605 3.280 3.023 2.828

16 32 64 128

-1.955 -2.064 -2.141 -2.196

-1.491 -1.538 -1.571 -1.593

-1.216 -1.238 -1.252 -1.261

-0.712 -0.703 -0.695 -0.690

-0.083 -0.059 -0.042 -0.029

0.622 0,639 0.650 0.658

1.323 1.313 1.305 1.299

1.774 1.739 1.713 1.694

2.686 2.582 2.508 2,455

normal

-2.326

-1.645

-1.282

-0.674

0.674

1.282

1,645

2.326

0.5

0

Uniform distributions are sometimes assumed for random variables with known range. The distribution val

F(x)

=

(0, 1), and has expectation

x

for

0 ~t x

1/2 , variance

~_ 1

is uniform over the inter-

1/12 , With a shifted origin and a change

in scale, F( is uniform over

(x O,xl)

(x-

X o ) / (x 1 -Xo)

and has expectation

)

for

(x 0 + Xl) / 2

x0 ~

x ~ variance

x1 (x 1 - Xo)2 / 12 .

178 Some discrete probabilities. Binomial probabilities apply to the n u m b e r of successes in the same probability

p

N

i n d e p e n d e n t trials, each with

of success (see Brunkl). Their values are

Pn

=

( N ) pn ( 1 - p )N-n

for

n

~

0, 1 . . . . . N

,

n

i.e.

the summands in the d e v e l o p m e n t by powers of

(p + q)

with

q

=

of the

N-th power of the b i n o m i a l

1 - p . Binomial probabilities have e x p e c t a t i o n

POISSON probabilities arise as the l i m i t probabilities the p a r a m e t e r

p

E(v)

=

Np

Pn

=

( N ~ co)

N p, v a r i a n c e N p ( ] - p )

of b i n o m i a l probabilities if

is held constant (see chapter 1). Their values are g(v)n e-E(v) / n :

They have e x p e c t a t i o n and v a r i a n c e

for

n

=

0, 1

. . . . . .

E(v) .

G e o m e t r i c probabilities apply to the n u m b e r of i n d e p e n d e n t trials, each with the same probability

p

of success, which is required to have the first success. Pn

They h a v e expectation

=

)n-1

(1 -p

1/p , variance

p

for

Their values are n

=

1,2 .....

(1 - p) / p2

Generating and m o m e n t generating functions. In general,

an a n a l y t i c function is called g e n e r a t i n g function of a sequence of numbers if

the coefficients of one of its power series developments are equal to or in a simple relation with the sequence of numbers. T h e sequence and the function d e t e r m i n e each other uniquely. For discrete probabilities

Pn ' n

=

0,1 . . . . .

(3O

Gv(Z)

=

~" n=0

Pn z

the generating function

Gv(Z)

is defined as

n

I

It is a n a l y t i c at least for ]z[ < 1 . The series m a y be finite. Examples are for Binomial probabilities

POISSON probabilities

Gv(Z ) =

Gv(Z) =

(1 - p +

pz) N

e - ( 1 - z ) E(v)

..

G e o m e t r i c probabilities Gv(Z ) :

p / ( 1 -(l-p) z)

179 As m o m e n t g e n e r a t i n g function for general distributions LAPLACE transform

130

LFv(S)

=

./~

e-SX dF (x) V

-00

(see W i d d e r l ) .

Fv(X) , this course uses the

Its power series

LFv(S)

=

1

+

~ j=l

(-s) j E(v J) / j :

makes the m o m e n t - g e n e r a t i n g property evident• For constant transform is

v

E(v) , the LAPLACE

e -sE(v) • Other e x a m p l e s are for

Normal distribution

G a m m a n ( n x/E(v) )

eV(V) s2 - E(v) s

1 / ( 1 + E(v) s )n

Uniform distribution e-S x 0 _ e-S x I

n

From the LAPLACE transform E(vJ) of its

=

LFv(S) , the =

s (x 1 - x0) j - t h m o m e n t of

Fv(X)

is found as the v a l u e

(-1) j LF(vJ) (0)

j - t h derivative at s = O, (if this derivative exists) .

For the discrete probabilities m e n t i o n e d above, LFv(S)

=

so that the g e n e r a t i n g function E(v j) E(v)

Gv(Z ) =

=

Gv(e'S ) , is another m o m e n t g e n e r a t i n g function. In fact,

dJ Gv(e-S) I (-1) j -. ds j s = 0 G'(1) ,

E(v 2)

=

so that

G"(1) + G'(1)

,

etc.

Distributions of some functions of i n d e p e n d e n t r a n d o m variables. Let

v1 , v2

denote two i n d e p e n d e n t r a n d o m variables which h a v e the joint probability

distribution (see Brunk I ) vI

and

of the

v2

Fl(x) F2(Y) . Then the following distributions for functions of

can be found by integration of the joint distribution over appropriate regions

x-y-plane: CO

Sum of variables

Prob ( v 1 + v 2 ~ x )

=

¢'Fl(X

-y)

dF2(y )

-O3 130

Difference

Prob ( v I - v 2 ~ x )

=

¢" E l ( X + y ) dF2(Y ) . -(30

'convolution' .

180 Maximum

Prob ( max(v 1 , v 2 ) ~

Minimum

Prob ( m i n ( v 1, v2) ¢ x )

The sum

=

Fl(X ) V2(x )

=

1 - (1-Fi(x))
))

has a distribution (as stated above) the LAPLACE transform of which is

v 1 + v2

equal to

x )

LFI(S ) LF2(s ) . This i m p l i e s the relations

and

g(v l+v 2)

=

g(v 1)

+

E(v 2)

V(v l+v 2)

=

V(v 1)

+

V(v 2)

All above relations can be g e n e r a l i z e d to any n u m b e r

N

of i n d e p e n d e n t random variables.

Distributions derived from conditional distributions. If a random v a r i a b l e has the c o n d i t i o n a l distributions v a r i a b l e has the v a l u e value

k

variable

is v

Pk

Fvlk(X ) , given that the c o n d i t i o n i n g

k , and if the probability that the c o n d i t i o n i n g variabIe has the

, k = 0,1 . . . . .

then the total probability d i s t r i b u t i o n of the random

is Fv(X)

co '~k=O

=

Pk Fv[k (x) "

This is a consequence of the definition of

Fv[k(X)

as a conditional probability.

All m o m e n t s of Pk"

Also,

F (x) are l i n e a r c o m b i n a t i o n s of the conditional moments, with the factors v LFv(S) is a s i m i l a r c o m b i n a t i o n of the LFvlk(S).

As a particular case,

v

consider a v a r i a b l e

=

LFv(S)

=

k

F(x) , which is i n d e p e n d e n t of

variables with c o m m o n distribution LFvlk(S)

which is the sum of

LF(s)

mutually independent k, Then

k

as stated above, and

where Fv(X) i.e.

hence

G(z) are

co k ~ Pk LF(s) k=0

is the g e n e r a t i n g function of the E(v)

=

- LF'(0)

=

V

=

G(LF(s) )

Pk " As a consequence, - G'(1) LF'(0)

the m o m e n t s of

,

the expected n u m b e r of summands m u l t i p l i e d by the expectation of each s u m m a n d ; E(v 2)

=

V(v)

=

LF;(0)

=

G"(1) L F ' ( 0 ) 2 +

G'(1) LF"(0)

(G"(1) + G'(1) - G'(1) 2 ) LF'(0) 2

+

,

etc.

G'(1) (LF"(0) - LF'(0) 2 ),

181

This variance is a sum of two constituents, viz. the variance of the number of summands, multiplied by the square of the expected summand value, and

the e x p e c t e d number of summands, multiplied by the variance of each summand.

Iterated m u l t i p l i c a t i o n s with stochastic matrices. The matrices and matrix multiplications considered in this course are defined in the text. A general property of iterated matrix multiplications is the convergence of the (appropriately normalized) result to an eigenvector of the matrix, if the matrix has a dominant simple eigenvalue, (see v. Mises 1, Bodewigl). The matrices of transition probabilities which are considered in this course have the p a r t i cular property that all their columns have the same sum of elements, viz.

1 . Such m a -

trices are generally c a l l e d 'stochastic' matrices, since they arise typically in stochastic, i . e . probabilistic problems. A left m u l t i p l i c a t i o n of a stochastic matrix with a row of all ones produces a row of all column sums, i . e .

reproduces the row of all ones. Hence, this row is a left eigenvector of

the matrix, to the eigenvalue

1.

This is also the dominant eigenvalue. Due to a theorem

of L e v y - H a d a m a r d (see Bodewigl), all eigenvalues of a matrix

P

l i e in the union of all

circles (including the peripheries) of the complex plane which are centered at a diagonal e l e m e n t of

P

and have the absolute sum of non-diagonal elements of the correspond-

ing column as radius. For the stochastic matrices of our context, all diagonal elements are probabilities different from

0

and

1, and the remaining sums complement the diagonal elements to

1, Thus,

all L ¢ v y - H a d a m a r d circles l i e within the unit circle and touch it only at the point Thus, the real number

1

+ 1 ,

is the only eigenvalue of that magnitude, and dominates all

other eigenvalues. The result of the iteration is determined by the matrix only, and independent of the initial values assumed.

182

Appendix : A

ARRPROC

procedure

E

which

:

Computing

generates

and

Procedures sorts

i000

random

arrival

times. ARRPROC: PROC OPTIONSIMAIN};

DCL (J,K) FIXED,RANOOM ENTRY;D(LO00)=IO00; DCL D|IO001 FLOAT(151; DO J:=i TO 9 9 9 ; D(JI=LOOO*RANOC)M; O0 K=I TO J; IF DIJ}
CONT;

END;

DO J = lOGO TO 2 BY - 1 ; D { J I = D { J I - D { J - I ) ; END; CALL PLOTDI{D~'RANDOM ARRIVALS',95E-2);CALL STANAL2ID); DCL N(IO) I N I T ( ( I O ) ( O } ) FIXED,NN FIXED; DO J=l TO IO00;NN=HAX(I,CEILIIO*IL-EXPI-D(J)})));NINN)=NINNI+I;END; PUT SKIP EDIT('COUNTS IN PERCENTILE INTERVALS',N)(A,IO F | 6 ) ) ; PUT LIST ('CHISQUARE ~,SUM(LE-2*IN-IOO)*~2) ) SKIP; STANAL2 : A

procedure

stochastic

~r

statistical

analysis

to the

second

order

of any

process.

STANAL2: PROC(O); DCL C(~OI,RHO(4OI,(AD(Ot#O~O:Ir2)~DX(2),D(*),R,S)FLOAT | 1 5 ) , (ItJ,K,LDIFIXED; LD=DIM|D~I); J=AO; AD IS [0 HOLD K-TH SAMPLE MOMENT OF LD-[ (IJ)-SH[FTED VALUES, I* C(I) SAMPLE COVARIANCE OF LAG I , RHO AN ESTIMATE OF THE CORRELATION *I l*

FIND

DO I = [

AO : * I AD=O; AD|O~O,I)=SUMIDI;AD(O~O,2)=SUMID**2};

TO J ;

DX|II=D|HBOUND(D,I|÷I-I);DX(2)=DXIII**2;AO(I,O,*}=AO([-L,O,*)-DX; DX(I}=DILBOUND(D,I)'[+I);DX(2}=DXII)**2; ADII,I,*)=ADII-I,MIN(I-I,I),*)-DX;END; DO I=O TO J; A D ( I , ~ , * ) = A D I I , * , * ) / ( L O - I ) ; END; I* FIND C : *I DO I = I TO J; S=O; DO K=LBOUND(D,1I TO HBOUNDID,I)-I; S=S÷D(K}*D(K+[}; END; C I I ) = S / ( L D - I ) ; END; /*

ESTIMATE THE CORRELATIONS : *I DO [ = I TO J; R H O ( I ) = I C ( I ) - A D I I , O , 1 } * A D ( I , L , I ) } I SQRT((ADII,O,2)-ADIIsO,II**ZI~(AD(I,I,2|-AD(I,L,II**2I~;ENO;

/*

PUT OUT ESTIMATES : */ PUT EDITI'AVERAGE IS ' , A D ( O ~ O t L ) , ' S A M P L E VARIANCE IS | t A D ( O v O ~ Z ] - A D I O , O , L | * ~ 2 | (2 ( C O L I I | , A , C O L I 2 [ ) , F ( 1 2 , 3 ) ) ) ; PUT S K I P I 2 ) EDIT [ ' E S T I M A T E S A R E t , ' F O R THE EXPECTED VALUE : 't A D ( O , O ~ I | , ' F O R THE STANDARD DEVIATION OF THIS ESTIM&TE SQRTIIADIO,O,2I-AD(OtO~II**2|*II+Z*SUM(RHO)I/LD}, 'FOR THE CORRELATION OFW,'LAG I : ','RHOII) = 'I (A,2 |COLI[I,A,CDLIZII,FIIP,3II,SKIP,A,SKIP,A,AI; DO I=1 TO J; PUT SKIP E D I T ( I , R H O I I I I ( F I 5 1 , F ( I Z , 3 ) I ; E N D I END SIANAL2; END ARRPR~C;

183 MACK

: A procedure t e rminal s,.

computing

the

PROCEDURE COMPUTING WAITTI~E

/*

MACK:

/*

wait

time

expected

at conversational

AT CONVERSATIONAL TERMINALS

*/

PROC(M,TR,TS,TP,U,TW,NS,TCI;

M TR

= : NUMBER OF TERMINALS ON ONE LINE : : EXPECTED INTERVAL BETWEEN RESPONSE TO ONE AND ARRIVAL OF NEXT MESSAGE,

TS

.=;

TRANSMISSION

TP

=:

TRANSMISSION TIME FOR A POLL MESSAGE

TIME

FOR A MESSAGE

RETURNS U T~ NS TC DCL

=: =: =: =:

LINE UTILIZATION BY MESSAGE TRANSMISSION EXPECTED WAIT TI('4E AT A TERMINAL EXPECTED NR OF POSITIVE P(3LLS IN A CYCLE EXPECTED TIME OF A POLLING CYCLE

(I,J,M) FIXED,(TR,TS,TP,U~TW,NS,TC,A,B) FLOAF, (GIG:2} INITIAL(I,O,0)yTOtTItT2 INITII),VR) FLOAT BIN(53);

TO=TS/TR;A,TO=EXP|TO); TI=MmTP/TR;BtTI=EXP|TI); DO I = I . T O M; T2 = T 2 ~ ( T I - I ) * ( M - I + I ) / I ; TI = TI~TO; DO J=O TO 2 ; G(J) = G ( J | + T 2 * I * * J ; END; END; NS = G I I I / G ( O ) ; VR = G ( ? ) / G I O ) - NS*NS ; TC = M*TP+NS*TS;

U = NS*TSITC

;

TW = . 5 * ( T C + V R * T S * * 2 1 T C I ;

END MACK;

QSIM

: A

procedure

simulating

a single-server

queue

/*

S I M U L A T I O N OF A SINGLE SERVER QUEUE WITH RANDOM ARRIVALS OF RATE RA = I , SERVICE TIMES UNIFORMLY D I S T R I B U T E D (.4,1I, AND F I F O D I S C I P L I N E , FOR T = (O,IO00) */ QSIM: PROCEDURE O P T I O N S ( M A I N ] ; DECLARE RANDOM ENTRY, J F I X E D , (ATOIO:I500) INIT(OI / ~ SPACE FOR AT .1, (ATII) I * ARRIVAL TIMES ./, DTJTI / * DEPARTURE TIMES */ INITIO), TS(1) / * SERVICE TIMES */)CTL)FLOATII5); DO M : [ TO i O ; DO

DO

I = I TO 1500 WHILE I A T O | I - I | < I O 0 0 I ; A T O ( I } = A T O ( I - 1 1 - LOG(|RANDOM)}; END; I = I - 2 ; ALLOCATE ATIDTtTS; J = I TO I ; A T | J ! = A T O ( J ) ; T S ( J ) = .4+.6*RANDOM; D T ( J ) = M A X ( D T ( M A X ( 1 , J - I ) ) , A T ( J } ) ÷ T S ( J } ; END;

D T = D T - A T ; CALL P L O T D I I D T , I T I M E S IN QUEUING S Y S T E M ' , 9 5 E - 2 ) ; CALL S T A N A L 2 ( D T I ; D T = D T - T S ; CALL P L O T D I I D T , ' ~ A I T TIMESV,g5E-21; CALL S T A N A L 2 ( D T ) ; PUT L I S T ( ' N U M B E R IS ' , I , , UTILIZATION IS ',SUMiTS)IIOOQ); FREE A T , D T T T S ; END;

*/

184

SPITZER

: A procedure for approximate tion discussed by Spitzer I

SPITZER: PROC OPTIONSIMAIN); ON UNDERFLOW; DCL |FSAI-I6*DIlIO:D1} FSA2{-I6*DI/IO:D1-L) FWO(O:Y*D1) DFWO{YmDI} FW(O:Y*DI)

GET /* DO

I* /* l* /~ /*

iterative

solutions

of an

integral

equa-

VALUES OF F_SUB_S-A * I , ARITHMETIC MEANS TO THE RIGHT * / , INITIAL VALUES OF F_SUB_W * / , DIFFERENCES TO THE LEFT * / , NEW VALUES OF F_SUB_W * I ) CTL FLOAT{15},

( l y J , M ) FIXED; D A T A ( Y ~ D I ) ; D = I / D I ; ALLOCATE FSAvFSA2tFWOtDFWOtFW; STW :

O;

DO

PREPARATION OF THE INTEGRAND *I I = -16mDL/lO TO - D I ; FSA(I} : ( I * D + 1 . 6 } * , 2 / 2 . 4 ; END; I= - 0 1 + I TO 4 m D l / l O ; F S A I I ) = . 5 * D * I + . b S ; END; I= 4 . D 1 / I 0 + I TO D I ; FSA(1) = I - (ImD-1)mm2/2.4; END; I= -16mDl/10 TO D I - I ; F S A 2 ( 1 ) = . 5 ~ ( F S A ( I } + F S A ( I + I ) ) ; END;

DO

RECURSION OF F_SUB_W ml FWO = I ; DO M=I TO 50; J = l TO YmDI; DFWO{J)=FWO{J}-FWO{J-I); FND;

DO DO

/# DO DO

I* /* PUT

NUMERICAL INTEGRATION *I J=O TO Y * D I ; IF J < DI THEN FW(J)=FSA(J}~FWO(O); ELSE FW(J)=FWO(J-D1); I = M A X { I , J - O I + I ) TO M I N { Y * O I , J + I 6 * D I I I O ) ; FW|J} = FWIJ) + FSA2(J-I}mDFWO(1); END; END; EXPECTATIONS TW : SUM(I-FW)*D;

*I STW = STW + TW;

O U T P U T OF ONE D I S T R I B U T I O N */ EOIT{'ITERATION ',M,'FW IS 'I{FW{I) DO .I=O TO Y~DI BY Y m O l l 2 0 ),'EXPECTATION'~ T~,'AVERAGE EXPECT~TION',STWlM) (SKIP,AtF(5ITSKIPtA,21 F{5,2},SKIP~2 ( A ~ F ( 7 , 3 } ) ) ; TEST FOR CONVERGENCE IF ALL(ABS(FW-FWOI
SPITZEX:

*I THEN GOT~ S P I T Z E X ;

END S P I T Z E R ;

All procedures

use

the

PL/I

programming

language.

FWO = FW;

END;