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Page 1: Solution Manual

Solution Manual

for

The Art of ComputerSystems Performance Analysis

Techniques for Experimental Design�Measurement� Simulation� and Modeling

By

Raj Jain

Professor of CIS

���� Neil Avenue Mall� ��� Dreese Lab

Columbus� OH ���������

Internet Jain�Cis�OhioState�Edu

Tentative Publication Date August ����

Copyright ���� Raj Jain

Please do not copy without Author�s written

permission�

Copy No� For

Page 2: Solution Manual

��� � Compare the ratio with system A as the base

System Workload � Workload � AverageA � � �B ���� � ����

Considering the ratio of performance with system A as base� we con�clude that system B is better�

� Compare the ratio with system B as the base

System Workload � Workload � AverageA � ���� ����B � � �

Considering the ratio of performance with system B as base� we con�clude that system A is better�

Page 3: Solution Manual

��� incomplete

��� Can be done

Page 4: Solution Manual

��� a� Measurements� Run your favourite programs and pick the one thatruns them faster�

b� Use measurements and simulations of various network conguirations�

c� Measurement�

d� a� Analytical modelling

b� Analytical modelling and simulations�

c� Extensive simulations and modelling�

��� a� � Response time for commonly used programs�

� Failure rate rate of crashing��

� Storage capacity�

� User�friendliness�

b� � Query response time�

� Failure rate�

� Storage capacity�

� Usability�

c� � Capacity�

� Response time�

� Failure rate�

d� � Response time�

Page 5: Solution Manual

��� The following information is from SPEC Standard Performance EvaluationCorporation� home page�

CPU benchmarks

CINT��� current release Rel� ���� Integer benchmarks contains

Name Applicationespresso Logic Design

li Interpretereqntott Logic Designcompress Data Compression

sc Spreadsheetgcc Compiler

CFP��� current release Rel� ����

Floating point benchmark suite contains

Name Applicationspice�g� Circuit Designdoduc Simulation

mdljdp� Quantum Chemistrywave� Electromagnetism

tomcatv Geometric Translationora Optics

alvinn Roboticsear Medical Simulation

mdljsp� Quantum Chemistryswm��� Simulationsu�cor Quantum Physicshydro�d Astrophysicsnasa� NASA Kernelsfpppp Quantum Chemistry

More information about these benchmarks can be found in http���performance�netlib�org�performaweb page�

��� A C program to implement sieve workload�

��

� seive�c � Program to implement sieve workload

��

Page 6: Solution Manual

�include �stdio�h�

�define MaxNum ��� �� List all primes upto MaxNum ��

�define NumIterations � �� Repeats procedure NumIterations times ��

�define TRUE �

�define FALSE

void main�void�

int IsPrime�MaxNum����

int i�k�Iteration� �� Loop indexes ��

int NumPrimes� �� Number of primes found ��

printf��Using Eratosthenes Sieve to find primes up to �d�n�� MaxNum��

printf��Repeating it �d times��n��NumIterations��

for �Iteration � �� Iteration �� NumIterations� Iteration���

�� Initialize all numbers to be prime ��

for �i � �� i �� MaxNum� i���

IsPrime�i� � TRUE�

i � ��

while �i�i �� MaxNum�

if �IsPrime�i��

�� Mark all multiples of i to be nonprime ��

k � i � i�

while �k �� MaxNum�

IsPrime�k� � FALSE�

k � k � i�

� �� of while k ��

� �� of if IsPrime ��

i � i � ��

Page 7: Solution Manual

� �� of WHILE i�i ��

NumPrimes � �

for �i � �� i �� MaxNum� i���

�� Count the number of primes ��

if �IsPrime�i��

NumPrimes � NumPrimes � ��

printf���d primes�n��NumPrimes��

� �� of for Iterations ��

�� The following can be added during debugging to list primes� ��

�� for �i � � i � MaxNum� i���

if �IsPrime�i�� printf���d�n��i�� ��

The result of running the program

Using Eratosthenes Sieve to find primes up to ���

Repeating it � times�

�� primes

�� primes

�� primes

�� primes

�� primes

�� primes

�� primes

�� primes

�� primes

�� primes

Page 8: Solution Manual

��� a� Cannot compare systems o�ering di�erent services�

b� � Metric response time�

� Workload Favourite programs Word processor� spreadsheet�

c� Metric response time� functionality Workload A synthetic programwhich tests the versions using various operating system commands�operating system services�

d� � Metric Response time� reliability� time between failures

� Workload A synthetic program generating representative �oppydrive I�O requests

e� � Metric size of code� structure of code� execution time

� Workload A representative set of programs in C and Pascal�

Page 9: Solution Manual

��� a�

�tCPU ��

n

nXi��

tCPU ���

�� �����

�nI�O ��

n

nXi��

nI�O ��� ���

�� �����

s�xs ��

n� �

nXi��

xsi � �xs��

��

n� �

��nXi��

x�si

�� n�x�s

����� �� ������

�� �����

Similarly�

s�xr ��� ���� ���� �� ����

�� �������

b� Normalize the variables to zero mean and unit standard deviation� Thenormalized values x�s and x�r are given by

x�s �xs � �xssxs

�xs � �����

����

x�r �xr � �xrsxr

�xr � ���

������

The normalized values are shown in the fourth and fth columns ofTable b�

The other steps are similar to example ����

Page 10: Solution Manual

��

Observation Variables Normalized Variables Principal FactorsNo� xs xr x�s x�r y� y�

� �� ���� ����� ����� ����� ������� �� ��� ����� ������ ����� ������ � �� ����� ������ ������ ������ � �� ������ ������ ������ ������ � �� ������ ������ ������ ������ � �� ������ ������ ������ ������� � �� ������ ������ ������ ������Px �� ����� ����� ����� ����� �����Px� ��� ��������� ����� ����� ����� �����

Mean ���� ����� ����� ����� ����� �����StandardDeviation

���� ������ ����� ����� ����� �����

The correlation between CPU time tCPU� and number of I�O�s nI�O�

is ������ The principal factors y� and y� are

�y�y�

��

� �p�

�p�

�p�� �p

� ���tCPU�����

����nI�O

�����

����

��

The rst factor explains ������������������ or ��� of total variation�

��� There is no unique solution to this exercise� Depending upon the choice ofoutliers� scaling technique� or distance metric� di�erent results are possible�all of which could be considered correct� One solution using no outliers� rangenormalization to ����� and Euclidean distance starts with the the normalizedvalues shown in the following

Program CPU time I�O�sTKB ���� ����MAC ���� ����COBOL ���� ����BASIC ���� ����Pascal ���� ����EDT ���� ����SOS ���� ����

BASIC� Pascal� EDT� COBOL� SOS� MAC� and TKB join the dendro�gram at distances of ����� ����� ����� ����� ����� and ����� respectively�

Page 11: Solution Manual

��

Other possibilities are to discard TKB and MAC as outliers� normalizeusing the mean and standard deviation� and transform I�O�s to a logarithmicscale� All these and similar alternatives should be considered correct provideda justication is given for the choice�

Page 12: Solution Manual

��

��� a� Hardware monitor as software monitor cannot measure time�

b� Software monitor becuase with hardware it is di�cult to monitor soft�ware events�

c� Software monitor� Program reference is a software event

d� Hardware monitor� Virtual memory reference is a hardware event�

e� Hardware monitor� Software interferes with time measurements�

f� Software monitor� Database query is high�level software� event�

��� Let us choose a network card our computer subsytem� Then the quantitiesthat can be monitored using the di�erent monitors are as follows

a� Software monitor� Total number of packets received� total number ofpackets sent� Number of error packets� Total bytes sent�

b� Hardware monitor� Record of all tra�c to the card using promiscuousmode��

c� Firmware monitor� The card be programmed to monitor tra�c onlyfrom a particular node�

For the software monitor� using the quantities one could measure theaverage packet size� error rate in packets� For the hardware monitor� therecord of tra�c can be used to measure time between packet arrivals� Forthe rmware monitor� the number of packets received from a particular nodecan be measured�

Page 13: Solution Manual

��

�� a� Those with the largest number of terminal reads�writes per CPU sec�ond�

a� Find the average number of disk reads�writes per second of programX� and the maximum rate that the disk can support� The ratio givesyou the number of copies of program X that can run simultaneously onthe disk drive�

a� Find the mode of typical data recorded by the log and compare thatwith data of the benchmark� If they are close� then the benchmark isrepresentative�

a� I�O bound programs � those with high �disk I�O�s per CPU second�should be chosen for I�O optimization�

Page 14: Solution Manual

��

��� incomplete

��� incomplete

Page 15: Solution Manual

��

��� a� Bar chart� as intermediate values have no meaning�

b� Line chart� as intermediate values have meaning�

c� Bar chart� no meaning for intermediate value�

d� Line chart� intermediate values have meaning�

��� a� a� Axes lables are not self�explanatory�

b� Scales and divisions are not shown

c� Curves are not labelled�

b� a� Y�axis is not labelled�

b� No Y�axis scales and division are not shown�

c� Y�axis minimum and maximum are not appropriate�

c� a� Scales and divisions are not shown�

b� Too many curves�

c� Curves are not individually labelled�

d� a� Order of bars is wrong�

b� Y�axis scales divisions not shown

��� incomplete

��� FOM � ��

��� FOM � ��

Page 16: Solution Manual

��

���� Raw Execution Time is a LB Lower is better� parameter�

� Compare the ratio with system A as the base

Benchmark System A System B SystemI ���� ���� ����J ���� ���� ����K ���� ���� ����

Average ���� ���� ����

Considering the ratio of performance with system A as base� we con�clude that system A is better�

� Compare the ratio with system A as the base

Benchmark System A System B SystemI ���� ���� ����J ���� ���� ����K ���� ���� ����

Average ���� ���� ����

Considering the ratio of performance with system B as base� we con�clude that system B is better�

� Compare the ratio with system A as the base

Benchmark System A System B SystemI ���� ���� ����J ���� ���� ����K ���� ���� ����

Average ���� ���� ����

Considering the ratio of performance with system C as base� we con�clude that system C is better�

Page 17: Solution Manual

��

System A System BTest Total Pass � Pass� a ax ���� x� b by ���� y

Total a � b ax � by �ax�by�a�b

Test Total Pass � Pass� c cu ���� u� d dv ���� v

Total c� d cu� dv ����cu�dv�c�d

���� Consider two systems A and B with two experiments�

System A is better than B based on the individual experiments� if the fol�lowing conditions are satised

���� x � ���� u

and���� y � ���� v

These conditions simplify to

x � u and y � v

System A is better than B based on total both the experiments�� if

���ax� by�

a � b�

���cu� dv�

c� d

which simplies toax � by�

a� b�

cu� dv�

c� d

If any of the above conditions are satised then the percentages can be usedfor system A�s advantage�

Page 18: Solution Manual

��

���� a� The servers are chosen independently with equal probablility� thereforethe probability that server A is chosen P A� � �

��

b� P AorB� � P A� � P B� � P AandB�� Only one server at a time isselected� so P AandB� � �� Thus P AorB� � �

��

c� �� Only one server at a time is selected�

d� P �A� � �� P A� � ���

e� Successive selections are independent� so we can multiple their proba�bilities� Thus P AA� � �

�� �

�� �

��

f� All nine events are independent� each with probablility ��� therefore the

probability that they occur in sequence is ����

���� The distribution is geometric� The mean of a geometric distribution withpmf � � p�x��p is � � �

p� The variance �� � ��p

p�� The standard deviation

� �p��pp

� The Coe�cient of variation COV � ���p�� p�

���� The mean of a Poisson distribution with pmf �x e��x

x is � � �� The variance

�� � �� The COV � ���p��

���� From ����� we know that the mean and variance of a Poisson distributionwith pmf � � p�x��p are equal to �� x and y are independent randomvariables�

a� Meanx � y� � Meanx� �Meany� � ���

b� V arx � y� � V arx� � V ary� � ���

c� Meanx � y� � Meanx� �Meany� � ��

d� V arx� y� � V arx� � V ary� � ���

e� Mean�x� �y� � �Meanx� � �Meany� � ���

f� V ar�x � �y� � �V arx� � ��V ary� � ���� COV �pV ar�Mean �

�p��

Page 19: Solution Manual

��

����

pdf � fx� �dF x�

dx

�e�x�a

a

�m��Xi�

x�a�i

i�

�� e�x�a

�m��Xi�

x�a�i

a � i�

�xm��e�x�a

m� ���am

Mean � � �Z �

xfx�dx

�Z �

xme�x�a

m� ���amdx

��

m� ���am

Z �

xme�x�adx

Integrating by parts

��

m� ���am

h�axme�x�a

i�

�am

m� ���am

Z �

xm��e�x�adx

�am

m� ���am

Z �

xm��e�x�adx

�amm�

m� ���am

Z �

e�x�adx

� mZ �

e�x�adx

� amh�e�x�a

i�

� am��� ����

� am

Variance � �� �Z �

x� ���fx�dx

��

m� ���am

Z �

x� am��xm��e�x�adx

��

m� ���am

Z �

xm�� � �amxm � a�m�xm���e�x�adx

Page 20: Solution Manual

��

�a�m� ��m� �a�m� � a�m�

m� ���am

Z �

xm��e�x�adx

� a�m

Mode is the maximum possible probability�

fx� is maxium whendfx�

dx� �

dfx�

dx�

m� ���amm� ��xm��e�x�a� xm��e�x�a� � �

xm��e�x�a

m� ���amm� ��� x�a� � �

Therefore mode occurs at x � am� ��

C�O�V ��

��

pa�m

am�

�pm

����

pdf � fx� �dF x�

dx� ax�a���

Mean � � �Z �

�xfx�dx

�Z �

�ax�adx

� a

�x�a��

�a� ��

���

�a

a� �

Variance � �� �Z �

�x� ���fx�dx

�Z �

�x� a

a� ���ax�a���dx

Integrating by parts

Page 21: Solution Manual

��

� a

�x� a

a� ���x�a

�a

���

� �Z �

�x� a

a� ��x�adx

��

a� ���� �

�x� a

a� ��x�a��

�a � �

���

��

�a � �

Z �

�x�a��dx

��

a� ���� �

a� ����

�a � �

�x�a��

�a � �

���

���

a� ����

a� ��a� ��

��a� �� a � �

a� ���a� ��

�a

a� ���a� ��

C�O�V ��

��

sa

a� ���a� ��

a� �

a

��q

aa� ��

� �aa� ����

���� The pdf for normal distribution is given by�

fx� ��

�p��

e��x����

���

Here � � � and � � �� Hence�

fx� ��p��

e��x����

For pdf values for x � �� � � � � � are tabluated below�

Page 22: Solution Manual

��

x fx�� ��������� ��������� ��������� ��������� ��������� ��������� ��������� ��������

Total ��������

a� P X � �� � � � P X � �� � � �P�i�� fi� � ��� xxx answer

in the book is wrong�

b� P X � �� �P�

i�� fi� � ����� xxx answer in the book is wrong�

c� f�� � f�� � f�� � f�� � �������� � ����� xxx answer in thebook is wrong�

d� P x � � � z��� � Here � ���� From appendix table A��� z��� ������� Hence x � � � ������ � � ����� seconds

��� a� The distribution is not skewed� nor is the data categorical� so we usethe Mean�

b� The total number of packets makes sense� also the distribution is notskewed� so we use the Mean�

c� The distribution is skewed� so we use the Median�

d� The keywords constitute categorical information� hence we use theMode�

���� a� CPU type is a category� so we would use Mode to summarize it�

b� Memory size is typically skewed � most personal compters have ap�proximately the same amount of memory� but a few users have lots ofmemory � so the Median is the best choice�

c� Disk type is a category� so we would use the Mode�

Page 23: Solution Manual

��

d� Number of peripherals is skewed� so the Median is a good choice�

e� Using the same logic as for memory size and number of peripherals� wechoose the Median�

���� Since the ratio of maximum to minimum is very high� use the median� Thegeometric mean can also be used if a logarithmic transformation can bejustied based on physical considerations�

����� Arithmetic mean since the data is very clustered together not skewed� andymax�ymin ratio is small�

����� Use SIQR since the data is skewed�

����� Range � � to �� Variance � �� ���percentile � �� � ��������� � �thelement � �� ���percentile � �� � �������� � ��th element � �� Semi�Interquartile RangeSIQR� � Q��Q�

�Q � � �th element � �� Q � � ��th

element � �� SIQR � ��� a� ��� Coe�cient of Variation � ����

Use the coe�cient of variation or standard deviation� since the datais not skewed�

����� The normal quantile�quantile plot for this data is shown in Figure ���� ofthe book� From the plot� the errors do appear to be normally distributed�

Page 24: Solution Manual

��

���� The normal distribution has the linearity property� Hence� the means getadded� when sum of two normal distribution are taken� The variance is givenby

� �

s���

n�����

n�

a� From central limit theorem N�� ��pn� is the distribution of the sam�

ple means� Here � � �� Hence the distribution is N�� ��pn�

b� mean � �� � �q��n� ��n �

q��n� Hence the distribution is

N�q��n� xxx answer in the book is wrong�

c� mean � ��� � ��� � �q��n� ��n �

q��n� Hence the distribution

is N���q��n��

d� mean � ���� � �� � �� � �q���n� ���n � ��

p�n� Hence the

distribution is N�� ��p�n��

e� The sum of square of normal variates has the chi�square �n� distri�bution�

f� The sum of the variance has ��n� distribution�

g� The ratio of two chi�square distribution has an F distribution� Hereboth numerator and denominator have the same chi�squre distribution�Hence� F n� n� is the distribution for the ratio of variances�

h� If x is normal variate and y � ��� then x�qy�� where � is the degrees

of freedom� has a t distribution� Here x��� is normal variate� sx has�n� distribution� The degrees of freedom is n� hence x����sx�

pn�

has tn� distribution�

���� The numbers in the sorted order is f �� ��� ��� ��� ��� ��� ��� ��� ��� ��� ������ ��� ��� ��� ��� ��� ��� ��� ��� ��� ��� ��� ��� ��� ��� ��� ��� ��� ��� ��� �����g� There are n � �� numbers�

a� The ���p�percentile is xp � x���n���p��� Therefore x����������� � x� ��� and x����������� � x� � ��

Page 25: Solution Manual

��

b� Mean � ����P��

i�� xi � ������ � �����

c� s� � �n��

Pni��xi � �x�� � �������� There s �

p������� � �������

A ��� condence interval for the mean � ��������������������p��

� ������ ������

d� Number of programs with less than or equal �� I�O�s � ��� Fraction� ����� � ����

A ��� condence interval for the fraction � p� z�����

sp�� p�

n

� ������ z

s������� ������

��

� ������ ������������

� ����� �����

xxx answer in book is wrong� gives ��� C�I�s�

e� One�sided condence interval for mean is given by�x� �x � z���s�

pn� or �x� z���s�

pn� �x��

��������������������p��� ������ or ������ ��������������������

p���

������ ������ or ������ �����

���� The standard deviation for the codes are sRISC�I � �������� sZ�� ���������sV AX������ � �������� sPDP����� � �������� sC�� � �������� The ���

condence interval is given by �x � t�������s�p��� since n � �� is less than

��� The condence intervals are CIRISC�I � �������� ��������� CIZ�� ��������� ��������� CIV AX������ � �������� ��������� CIPDP����� � ������� ���������CIC�� � ������� ��������� xxx answer in the book is wrong�

Let us choose RISC�I and Z���� as the two systems�

a� The condence intervals for both the processors include �� so if theprocessors are not di�erent�

b� incomplete� question not clear �

Page 26: Solution Manual

��

���� incomplete refer math book to use Langrange multiplier technique�

���� A linear model to predict disk I�O�s as a function of CPU time can bedeveloped as follows

For this data n � �� !xy � ����� !x � ��� !x� � ���� !y � ���� !y� � ��� ���� �x � ������y � ������ Therefore�

b� �!xy � n�x�y

!x� � n�x���

����� �� ����� �����

���� �� ������� ������

b � �y � b��x � ������ ������� ���� � ������

The desired model is

Number of disk I�O�s � ������ � ������CPU time�

SSE � !y��b!y�b�!xy � ��� �������������������������� � ������

SST � SSY� SS� � !y� � n�y�� � ��� ���� �� ������� � �������

SSR � SST� SSE � �������� ������ � �������

R� �SSR

SST�

�������

�������� ������

Thus� the regression explains ��� of CPU time�s variation�

The mean squared error is

MSE �SSE

Degrees of Freedom for Errors�

������

�� �����

The standard deviation of errors is

se �pMSE �

p����� � �����

sb� � se

��

n�

�x�

!x� � n�x�

����� �����

��

��

������

���� �� ����� ����

����� ������

sb� �se

�!x� � n�x������

�����

����� �� ����� ��������� ������

Page 27: Solution Manual

��

The ��� condence interval for b is

������� ������������� � ������� ������ � �������� �������

Since this includes zero b is not signicant�

The ��� condence interval for b is

������� ������������� � ������� ������ � ������� �������

a� Only b� is signicant�

b� ��� xxx answer in book is wrong it said ��

c� yexpected � ������ � ������ � �� � �������

d�

s�y�p � �����

�� �

��

��� ������

���� �������

����� �������

The ��� condence interval for a single prediction � ��������� ��������������

� ��������� �������

� ������� �������

xxx answer in book is wrong ������� ������� �

e�

s�y�p � �����

��

��

��� ������

���� �������

����� �������

The ��� condence interval for predicted mean � ��������� ��������������

� ��������� �������

� ������� �������

xxx answer in book is wrong ������� ������� �

Page 28: Solution Manual

��

���� For the data n � �� !xy � ��� ���� !x � ����� !x� � ���� ���� !y � ��� !y� � �����x � ������� �y � ����� Therefore�

b� �!xy � n�x�y

!x� � n�x���

��� ���� �� ������� ����

���� ���� �� ��������� ������

b � �y � b��x � ����� ������� ������ � ������

The desired model isCPU time in milliseconds � ������ � ������� memory size in kilobytes�

SSE � ������� SST � ��������� SSR � ��������� R� � ������MSE � ������� se � ������� sb� � ������� sb� � �������

The ��� condence intervals of b and b� are �������� ������� and������� �������� the intercept is zero but the slope is signicant�

���� Elasped time � ����� number of days� � ������ the ��� condence intervalsfor the regression coe�cients are ����� ����� for the intercept and ���������� for the slope� both are signicant� Note Calculations are similar tothe solution for exercise �����

���� Elapsed time � ������������number of keys�� R� � ������ the condenceintervals of the coe�cients are ������������ and ������������� respectively�Note Calculations are similar to the solution for exercise �����

���� Number of disk I�O�s � ������ � ����� � number of keys�� R� � ������ the��� condence intervals for the coe�cients are �������� ������� and ������������ b is not signicant� Note Calculations are similar to the solutionfor exercise �����

���� Time � ���������� � ������ record size�� R� � ������ Both parametersare signicant� However� the scatter plot of the data shows a nonlinearrelationship� The residuals versus predicted estimates show that the errorshave a parabolic trend� This indicates that the errors are not independentof the predictor variables and so either other predictor variables or somenonlinear terms of current predictors need to be included in the model�

Page 29: Solution Manual

��

���� yyy question not clear� no x�� but solution talks about x�� �

a� R � ���� R� � ������� So ������ of variance is explained by theregression�

b� Yes

c� x�

d� x�

e� x�� x�� and x�

f� Multicollinearity possible

g� Compute correlation among predictors and reduce the number of pre�dictors�

Table ���� Time to Encrypt a k�bit Record after log tranformation�

����

log�k� Uniprocessor Multiprocessor����� ����� ���������� ����� ���������� ����� ���������� ����� �����

Let us use the following model

y � b � b�x� � b�x�

where y is logtime�� x� is the key size and x� is a binary variable�x� � � � multiprocessor and x� � ��uniprocessor�

In this case

X �

���������������

� ����� �� ����� �� ����� �� ����� �� ����� �� ����� �� ����� �� ����� �

��������������

Page 30: Solution Manual

��

XTX �

��� � ������ �

������ ������ ������� ������ �

��

C � XTX��� �

��� ����� ������ ����������� ����� ������ � ���

��

XTy �

��� ������

������������

��

The regression parameters are

b � XTX���XTy � ������� �������������T

The regression equation is

logtime� � ������ � ����� logkey size�� �����x�

R� � ������� The condence intervals of the coe�cients are ������������� ����� ������ and ������ �������

Page 31: Solution Manual

��

���� Each of the factors have � levels�

a� A full factorial experiment is necessary if there is signicant interactionamong factors� Hence� number of experiments is �� �� � � ���

b� If there is no interaction among factors� then simple design can be used�The number of experiments is � � �� �� � �� �� � �� �� � � �

c� A fractional factorial experiment can be used if the interaction is smallamong factor� Hence� number of experiments is ���� � �

xxx answer in book is wrong� The answers for b� and c� are interchanged�

Page 32: Solution Manual

��

���� The sign table for this data is given below

I A B C AB AC BC ABC y� �� �� �� � � � �� ���� � �� �� �� �� � � ���� �� � �� �� � �� � ��� � � �� � �� �� �� ���

� �� �� � � �� �� � ��� � �� � �� � �� �� ��� �� � � �� �� � �� ��� � � � � � � � ����� �� ���� ���� ��� �� ��� �� Total������ ����� ������� ������� ������ ����� ������ ����� Total��

SST � ��q�A � q�B � q�C � q�AB � q�AC � q�BC � q�ABC�

� ������� � ������� � ������� � ������ � ������ � ������� � �������

� ������ � �������� � �������� � ������ � ������ � �������� � �������

� ���������

a� q � ������ qA � ����� qB � ������� qC � ������� qAB � ������qAC � ����� qBC � ������ and qABC � ����

b� The portion of variation explained by the seven e�ects are ����������������������������������������� ������������ ���������������� ����������������������� ������� ������������������ �������� and ������������������������ respectively�

c� Sorting according to their coe�cient values� the factors with decreasingorder of importance are BC� C� B� ABC� A� AB� AC

Page 33: Solution Manual

��

��� Let A indicate workload and B indicate Processor�s used�

I A B AB y Mean �y� �� �� � ������ ������ ������ ������� � �� �� ������������������ ������� �� � �� ������������������ ������� � � � ������������������ ������������� ������ ������ ������� Total����� ������ ����� ������ Total��

The e�ects are ������ ����� ������ and ������ The e�ect of workloads������ is not signicant� Interactions explain ������ of the variation�

Page 34: Solution Manual

��

���� The following sign table with I � ACD as the generator polynomial is usedto analyze the ���� design�

I A B C AB D BC BD y� �� �� �� � � � �� ��� � �� �� �� �� � � ���� �� � �� �� � �� � ��� � � �� � �� �� �� ���� �� �� � � �� �� � ��� � �� � �� � �� �� ��� �� � � �� �� � �� ��� � � � � � � � ����� ��� �� ���� �� ���� �� ��� Total������ ������ ����� ������� ����� ������� ����� ������ Total��

a� q � qACD � ������ qA� qCD � ����� qB � qABCD � ������� qC � qAD �������� qAB � qBCD � ������ qAC � qD � ����� qBC � qABD � ������and qABC � qBD � ����

b� ������ ������� ������� ������ ������ ������� �����

c� BC� C� B� BD� A� AB� D� Higher order interactions are assumedsmaller�

d� The generator is I � ACD� The confoundings are

I � ACD� A � CD� B � ABCD� C � AD� D � AC� AB � BCD�BC � ABD and ABC � BD

e� I � ABCD may be better since its resolution will be IV�

f� RIII� since the generator is I � ACD�

���� Yes� I � ABC is a ����

III design� yes� I � AB is a ����

II design� yes� I � ABCD

is a ����

IV design�

Page 35: Solution Manual

��

��� Rewrite the given equation as

j �rXi��

aXk��

aikjyik

We know thatj � �y�j � � � �y�j � �y��

Expanding the terms for �y�j and �y�� we get the following equation�

j ��

r

rXi��

yij � �

ar

rXi��

aXk��

yik

Collecting the terms we get

j � �

r� �

ar�

rXi��

yij � �

ar

rXi���i��j

aXk��

yik

Comparing the coe�cients of yij we get

aikj �

�r� �

ra� k � j

��ra� Otherwise

xxx answer in book is wrong� it gives ��ar instead of ���ar�

The variance of eij can be written as

��e�j

� ��e

rXi��

aXk��

aikj

��e�j

�r��

r� �

ra

��

� ar � r��

ra

�����e

ra�a� ��� �

ra�a� ��

���e

�a� ��

ra�a� �� � ��

���e

�a� ����

e

ar

Page 36: Solution Manual

��

Table ���� Computation of E�ects for the Scheme versus Spectrum Study

Row Row RoWorkload Scheme�� Spect��� Spect���� Sum Mean E�eGarbage Collection ����� ����� ����� ������ ����� ����Pattern Match ����� ����� ����� ����� ����� �����Bignum Addition ������� ������� ������� ������ ������ �����Bignum Multiplication ����� ����� ����� ����� ����� �����Fast Fourier Transform ����� ����� ����� ����� ����� ������ ����Column Sum ������ ������� �������� �������Column Mean ������ ������ ������ ������Column e�ect ������ ����� ������

���� The computation of e�ects for Scheme versus Spectrum study is given intable ����

SSY �Xij

y�ij � ��� �������

SS� � ab�� � �� �� �������� � �� �������

SSA � bXj

�j � �� ��������� � ������� � ��������� � �������

SSB � aXi

��i � �� ��������� � ��������� � ��������� � ��������� � ���������

� �� �������

SST � SSY� SS� � ��� �������� �� ������� � ��� �������

SSE � SST� SSA� SSB � ��� �������� �������� �� ������� � ��������

The di�erences of the e�ects of di�erent processors are������������ �������� ������ � ����� � ������ and ����� � ����� � �������

MSE �SSE

a� ��b� ��

MSE ���������

�� ���� ��� �������

se �pMSE �

p������� � ������

Page 37: Solution Manual

��

Table ���� Computation of E�ects the Intel iAPX ��� Study

System Workload Row Row RowNo� ��� Sieve Puzzle Acker Sum Mean E�ect� ����� ����� ����� ����� ������ ����� �������� ����� ����� ����� ����� ������ ����� �������� ����� ����� ����� ����� ������ ����� �������� ����� ����� ����� ����� ������ ����� �������� ����� ����� ����� ����� ������ ����� �������� ����� ����� ����� ����� ������ ����� ������� ����� ����� ����� ����� ����� ����� �������� ����� ����� ����� ����� ������ ����� �������� ����� ����� ����� ����� ������ ����� �������� ����� ����� ����� ����� ������ ����� �������� ����� ����� ����� ����� ������ ����� �������� ����� ����� ����� ����� ������ ������ ������

Column Sum ����� ������ ������ ������ �������Column Mean ����� ����� ����� ����� �����Column E�ect ������ ������ ����� �����

The standard deviation for the di�erences � se�p� � ������������ �

������

The condence intervals for the di�erences are

������ ���� � ������ � ������ �������� � �������� ������

������� ���� � ������ � ������� �������� � �������� �������

������� ���� � ������ � ������� �������� � ������� �������

The processor are not signcantly di�erent� The plots of residul errorsdoes not show any trend and the plot of normal quantile�quantile plot doesappear linear� But since the ymax�ymin is large a multiplicative model shouldbe used�

���� The table after the log transformation is shown in table �����

The ANOVA for Scheme versus Spectrum study is given in table �����which agrees with the table ����� given in the book�

Page 38: Solution Manual

��

Table ���� ANOVA Table for the Intel iAPX ��� Study

Compo� Sum of �Variation DF Mean F � F �nent Squares Square Comp� Tabley �������y�� �������y � y�� ������� ������ ��Workload ������� ����� � ���� ������ ���System ������ ����� �� ��� ���� ���Errors ���� ���� �� ����

se �pMSE �

p���� � ����

���� After logarithmic transformation� the table for computing e�ects is shownin table �����

The ANOVA table for RISC Code size study is given in table �����

The condence intervals for e�ect di�erences are shown in table �����

a� ����� variation is explained by the processors

b� ������ variation is due to workloads�

c� Yes� several processor pairs are signicantly di�erent� at ��� con�dence level�

���� After log transformation� the table for computing e�ects including �����column� is given in table �����

The ANOVA table including ����� column is given in table �����

The condence intervals of e�ect di�erences for RISC code study in�cluding column ����� is given in table ������

a� ����� variation is explained by the processors�

b� ������ variation is due to workloads�

c� Yes� several processor pairs are signicantly di�erent at ��� condencelevel�

Page 39: Solution Manual

��

Table ���� Computation of E�ects for the RISC Code Size Study

Processors Row Row RowWorkload RISC�I Z���� VAX������� PDP������ C��� Sum Mean E�ectE�String Search ���� ���� ���� ���� ���� ����� ���� ����� �

F�Bit Test ���� ���� ���� ���� ���� ����� ���� �����H�Linked List ���� ���� ���� ���� ���� ����� ���� �����K�Bit Matrix ���� ���� ���� ���� ���� ����� ���� �����I�Quick Sort ���� ���� ���� ���� ���� ����� ���� ����Ackermann���� ���� ���� ���� ���� ���� ����� ���� �����Recursive Qsort ���� ���� ���� ���� ���� ����� ���� ����Puzzle Subscript� ���� ���� ���� ���� ���� ����� ���� ����Puzzle Pointer� ���� ���� ���� ���� ���� ����� ���� ����SED Batch Editor� ���� ���� ���� ���� ���� ����� ���� ����Towers Hanoi ��� ���� ���� ���� ���� ���� ����� ���� �����Column Sum ����� ����� ����� ����� ����� ������Column Mean ���� ���� ���� ���� ���� ����Column E�ect ���� ���� ����� ����� �����

Table ���� ANOVA Table for RISC Code Size Study

Compo� Sum of �Variation DF Mean F � F �nent Squares Square Comp� Tabley ������y�� ������y � y�� ����� ������ ��Workload ���� ����� � ���� ���� ����System ����� ������ �� ���� ������ ����Errors ���� ����� �� ����

se �pMSE �

p���� � ����

Page 40: Solution Manual

��

Table ���� Condence Intervals of E�ect Di�erences in the RISC Code SizeStudy

RISC�I Z���� VAX������� PDP������ C���RISC�I �����������y ����� ����� �����������y ����� �����Z���� ����� ����� ����� ����� ����� �����

VAX������� �����������y ������ �����yPDP������ ������ �����y

y � Not signicant

Table ���� Computation of E�ects for the RISC Code Size Study

Processors Row RowWorkload RISC�I ����� Z���� ������ ����� C��� Sum Mean EE�String Search ���� ���� ���� ���� ���� ���� ����� ���� ��F�Bit Test ���� ���� ���� ���� ���� ���� ����� ���� �H�Linked List ���� ���� ���� ���� ���� ���� ����� ���� �K�Bit Matrix ���� ���� ���� ���� ���� ���� ����� ���� �I�Quick Sort ���� ���� ���� ���� ���� ���� ����� ����Ackermann���� ���� ���� ���� ���� ���� ���� ����� ���� �Recursive Qsort ���� ���� ���� ���� ���� ���� ����� ����Puzzle Subscript� ���� ���� ���� ���� ���� ���� ����� ����Puzzle Pointer� ���� ���� ���� ���� ���� ���� ����� ����SED Batch Editor� ���� ���� ���� ���� ���� ���� ����� ����Towers Hanoi ��� ���� ���� ���� ���� ���� ���� ����� ���� �Column Sum ����� ����� ����� ����� ����� ����� ������Column Mean ���� ���� ���� ���� ���� ���� ����Column E�ect ���� ����� ���� ����� ����� �����

Page 41: Solution Manual

��

Table ���� ANOVA Table for RISC Code Size Study

Compo� Sum of �Variation DF Mean F � F �nent Squares Square Comp� Tabley ������y�� ������y � y�� ����� ������ ��Workload ���� ����� � ���� ���� ����System ����� ������ �� ���� ������ ����Errors ���� ����� �� ����

se �pMSE �

p���� � ����

Table ����� Condence Intervals of E�ect Di�erences in the RISC Code SizeStudy

���� Z���� ������ ����� C���RISC�I ���������� �����������y ����� ����� �����������y ����� ���������� ������ ������ ������ �����y ������ ������ ������ �����yZ���� ����� ����� ����� ����� ����� ����������� �����������y ������ �����y����� ������ �����y

y � Not signicant

Page 42: Solution Manual

��

Table ����� Computation of E�ects for exercise ����

A Row Row RowB A� A� A� Sum Mean E�ectB� ������ ������ ������ ������� ������ �������B� ������ ������ ������� �������� ������� ������B� ������ ������ ������ ������� ������ �������B� ������ ������ ������� ������� ������� ������B� ������ ������� ������� �������� ������� ������Column Sum ������� ������� ������� ��������Column Mean ������ ������ ������� ������Column e�ect ������� ������� ������

Table ����� Interactions

B A� A� A�B� ������ ������ �������B� ������� ����� �����B� ������ ������ �������B� ������ ������� ������B� ������� ������� ������

���� The table for computing e�ects is given in table ������ The interactions aregiven in table ������ The Analysis of Variance table is given table ������The ��� condence intervals for e�ects are given in table ������ The ���condence intervals for the interactions are given in table ������ The ���condence intervals for the e�ect di�erences are given in table ������

a� Yes� All processors are signicantly di�erent from each other�

b� �����

c� All e�ects and interactions are signicant�

Page 43: Solution Manual

��

Table ����� ANOVA Table exercise ����

Compo� Sum of �Variation DF Mean F � F �nent Squares Square Comp� Tabley ����������y�� ����������y � y�� ���������� ������ ��A ���������� ����� � ��������� �������� ���B ��������� ����� � ��������� ������� ���Interactions ��������� ����� � �������� ������ ���Errors ������ ���� �� ����

se �pMSE �

p���� � �����

Table ����� Condence Intervals for E�ects

Para� Mean Std� Condencemeter E�ect Dev� Interval� ������ ���� ������� �������AA� ������� ���� �������� ��������A� ������� ���� �������� ��������A� ������ ���� ������� �������BB� ������� ���� �������� ��������B� ������ ���� ������� �������B� ������� ���� �������� ��������B� ������ ���� ������� �������B� ������ ���� ������� �������

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Table ����� Interactions

B A� A� A�B� ������� ������� ������� ������� �������� ��������B� ���������������� ������������ ������������B� �������������� �������������� ����������������B� �������������� ���������������� ��������������B� ���������������� ���������������� ��������������

Table ����� Interactions

A� A� A�A� �������� �������� ��������� ���������A� ��������� ���������

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���� The experiment number � maximizes TI ������ so TI is high for A � ���B � ��� and E � ��� The experiment number � maximizes TB ������ so�TB is high for A � ��� B � �� and D � ���

���� The throughputs are ranked according to decreasing value of TI and it isseen that� TI is high for A � ��� B � ��� and E � ��� Similarily when thethroughputs are ranked according to decreasing value of TB it is seen that�TB is high for A � ��� B � �� and D � ���

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���� a� yt� � t � ��� Countinuous state� deterministic� dynamic� linear� andunstable

b� yt� � t� Continuous state� deterministic� dynamic� nonlinear� andunstable

c� yt � �� � yt� � "� " is not an integer� Discrete time� continuousstate� deterministic� dynamic� linear� and unstable

d� nt� �� � �nt� � � Discrete time� deterministic� dynamic� linear� andunstable

e� yt� � sinwt� Continuous time� continuous state� dynamic� nonlinear�and stable

f� �yt � �� � �yt� � " Discrete time� continuous state� probabilistic �dynamic� linear� and unstable

���� a� Since the number of factors this is best modeled by a Trace�drivensimulation�

b� The known distribution can be used to generate the events� Hence thisis best modeled by Discrete�event simulation�

c� The value of � is independent of time� Hence Monte Carlo simulationis best suited to nd the value of ��

���� The unit time approach is a time�advancing mechanism to adjust the sim�ulation clock� In this approach the time is incremeneted in small intervalsand checks are done at each increment to see if there are any events whichhave to be scheduled� This approach is generally not used� since unnecessaryincrements and checks are done during idle time�

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���� a� This is expected when the system is underloaded� Make sure that thesystem is in underloaded region

b� This is quite common when the system is overloaded� Make sure thatthe system is in overloaded region

c� This is expected�

d� This is uncommon and would require validation�

e� This is rare and would require serious validation e�ort�

���� The transient interval using the truncation method is �� since � is neitherthe maximum nor the minimum of the remaining observations� However�this is incorrect� since the actual transient interval seems to be ��

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���� This is multiplicative LCG� Hence the maximum Maximum period is �l�� ����� � �� a must be �i� �� that is � or ��� The seed must be odd�

���� The values of ��n mod �� for n � �� � � � � �� are ��� ��� ��� ��� ��� �� �� ������ ��� ��� ��� ��� �� ��� �� ��� �� ��� �� ��� ��� ��� �� �� ��� ��� ��� ��� �� Thesmallest n that results in � is ��� Yes� �� is a primitive root of ���

���� �� �� �� �

���� ����

���� x� � �������������

����

q � m div a � �� div �� � �

r � m mod a � �� mod �� � �

No� the seqence generated with and without Schrage method are di�erent�Since� q � � and r � � do not satisfy the condition r less than q�

����

q � m div a � �� div �� � �

r � m mod a � �� mod �� � �

Since� q � � and r � � do not satisfy the condition r � q� this cannot beimplemented using Schrage�s method�

��� a� Primitive

b� Primitive

c� Not primitive� Since � � x� � x���� x�� � �� x��

d� Primitive

���� a� ���

b� �� Since x� � ��x� � x� �� � x� � �

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Table ����� Tausworthe method

Step� Copy seed Y� ������ ������ ������ ������ ������Step� ��bit Right shift Y� ������ ������ ������ ������ ������Step� Xor Y� � Y� � Y� ������ ������ ������ ������ ������Step� ��bit Left shift Y� ������ ������ ������ ������ ������Step� Xor� Y� � Y� � Y� ������ ������ ������ ������ ������

c� ���

d� ��� Since x�� � x�� � ��x��� �� � x��� �

���� The characteristic polynomial is x� � x � �� In this case r � �� q � �� andq � r � �� We need a ��bit right shift and ��bit left shift� The initial seed isX � �������

The sequence of calculations are show in table ������ Hence the rstve ��bit numbers are ���������� ���������� ���������� ���������� ����������

����� In both cases� the additive parameter c should be replaced by c mod m�

����� The rst �� numbers and their binary representations are listed in ta�ble ������ From the table it can be seen that the period of the lth bit is�l�

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Table ����� Random Numbers Generated by the LCG xn � ��xn�� ��� mod ���

n xnDecimal Binary

� �� ��

� ��� � ���

� ���� � ����

� ������ ����� �����

� ������ ������ �����

� ���� ��� �����

� ��� �� �����

� ����� ������ �����

��� ��� ���

� ����� �� �����

�� ���� ��� �����

�� ������ ������ �����

�� ����� ����� ��

�� ����� ������ �������

�� ����� ������ �����

�� ����� ��� ����

�� ����� ���� ���

�� ������ ���� ����

� ������ ���� ��

� ������ ���� ����

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���� The nal seed value is ���������� The nal random number is ���������

Table ����� Chi�Square Test on ������ Numbers

Cell Observed Expected Observed�Expected��

Expected� ��� ������ ������ ���� ������ ������ ��� ������ ������ ��� ������ ������ ���� ������ ������ ���� ������ ������ ��� ������ ������ ��� ������ ������ ���� ������ �����

�� ���� ������ �����Total ����� ������� �����

The computed statistic is ������ The ����quantile of a chi�square variatewith nine degrees of freedom is ������ The sequence doest not passes the testat ����

���� Fifteen random numbers generated using the given LCG are �� ��� ��� ����� ��� �� �� ��� �� �� �� ��� �� ��

The normalized numbers obtained by dividing by �� are �������� ���������������� �������� �������� �������� �������� �������� �������� �������� ���������������� �������� �������� ��������

Table ����� shows a sorted list of these numbers and di�erences� Usinthe maximum values obtained from the table� K�S statistics can be computedas follows

K� �pnmaxj�j

n� xj

��p��� ������� � ������

K� �pnmaxj�xj � j � �

n

��p��� ������� � ������

K� � �������K� � ������� Both values are less thanK����������������The sequence passes the test�

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Table ����� Computation for the K�S Test

j xjjn� xj xj � j��

n

� ������� ������� �������� ������� ������� �������� ������� ������� �������� ������� ������� �������� ������� ������� �������� ������� ������� �������� ������� ������� �������� ������� ������� �������� ������� ������� �������

�� ������� ������� ��������� ������� ������� ��������� ������� ������� ��������� ������� ������� ��������� ������� ������� ��������� ������� ������� �������

Max ������� �������

Table ����� Autocovariances for the Random Sequence for exercise ����

Lag Autocovariance St� Dev� ��� Condence Intervalk Rk of Rk Lower Limit Upper Limit� ��������� �������� ��������� ���������

p� �������� �������� ��������� ��������� ��������� �������� ��������� ��������� ��������� �������� ��������� ��������� �������� �������� �������� ��������

p� ��������� �������� ��������� ��������� �������� �������� ��������� ��������� �������� �������� ��������� ��������� ��������� �������� ��������� ���������

p�� �������� �������� ��������� ��������

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���� The autocovariance and condence intervals for the serial autocovariancesat lags � to �� are given in table ������ Three autocovariance at lag �� �� ��are signicant�

���� The pairs generated by the rst generator lie on the following two lines witha positive slope

xn ��

�xn�� �

��

�k� k � �� �

The distance between the lines is ���p�� The pairs generated by the second

generator lie on the following two lines with a negative slope

xn � ��xn�� � ��k� k � �� �

The distance between the lines is ���p�� Both generators have the same

��distributivity�

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��� a� Inverse transformation Generate u � U�� ��� If u � ���� then x �p�u� otherwise x � ��

q��� u��

b� Rejection Generate x � U�� �� and y � U�� ��� If y � minx� �� x��then output x� otherwise repeat with another pair�

c� Composition The pdf fx� can be expressed as a weighted sum of aleft triangular density and a right triangular density�

d� Convolution Generate u� � U�� �� and u� � U�� ��� Return u� � u��

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���� a� Geometric� Geometric distribution is used to model number of at�tempts between successive failures�

b� Negative binomial� It can be used model the numberof failures beforethe mth success�

c� Logistic� � xxx�

d� Normal� The mean of large set of uniform distribution is a normaldistribution�

e� Lognormal� The product of large set of uniform is a lognormal distri�bution�

f� Pareto� This is used t power curves�

g� Poisson� Sum of two Poisson�s is a Poisson distribution�

h� Chi square� Variances of normal population has chi�square distribution�

i� F � Ratio of variances of normal population has F distribution�

j� Exponential� Exponential is used to model memoryless events�

k� Erlang�m� Sum m memoryless servers can be represented by Erlang�mdistribution��

l� Binomial� This models the successes in n independent and identicalBernoulli trails�

m� Beta� This is used to model ratio of random�variates�

���� a� Sum of normal distribution is normal� � � ��� � ��� �q��� � ��

Hence it is a N�� �� distribution� ��� quantile is ����� from appendixtable A��

b� Sum of variances is chi�square distribution� There are � normal variates�hence it is a ��� distribution� ��� quantile is ����� from appendixA���

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c� Ratio of two chi�square variates is a F distribution� Since both numer�ator and denominator have two variates the degrees of freedom is � forboth� Hence it is a F �� �� distribution� ��� quantile is ���� fromappendix A���

d� Ratio of normal to square root of chi�square distribution is a t distinc�tion� Number of degrees of freedom is �� Hence it is a t�� distribution���� is ����� from appendix A���

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��� Erlang�k arrivals� general bulk service� ve servers� ��� waiting positions����� population size� and last come rst served preemptive resume service�

��� Because it provides �� waiting positions for a population of only �� Also�since there are �� servers and only �� waiting positions� two servers have nowaiting positions�

��� Both will provide the same performance� Increasing bu�ers beyond thepopulation size has no e�ect�

��� E�n� � �E�r� � �����

� ��� �� Job �ow balance was assumed� Service time

distribution has no e�ect�

��� If k � � then Ek becomes a exponential distribution� Hence Ek�M�� can becalled a Poisson process if k � ��

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���� a� The probability pn for birth�death processes is given by

pn ���� �n��

���� �n p� n � �� �� � � � �

Here �n � �n��

and �n � �� Therefore

pn � n

n�p�where �

b� The sum of the probabilities should be �� Therefore�

p ��

� �P�

n��n

n

p � e��

c�

E�n� ��Xn��

npn � p�Xn��

n n

n�

E�n� � e�� e� �

d� �� � ���� e���

e� E�r� � E�n���

� ��

���e���

���� a� The probability pn for birth�death processes is given by

pn ���� �n��

���� �n p� n � �� �� � � � �

Here �n � � and �n � n�� Therefore

pn � n

n�p�where �

b� p � e�� derivation is similar to previous exercise solution�

c� E�n� �

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d� Var�n� � E�n��� E�n��� �P�

n�� n� �n

n p � � � e�� � ��e� � �

Var�n� �

e� E�r� � E�n��

� ��

���� a� � ����� � ���b� E�s� � ��� � E�r��� � � ��� ���� � ��� � ��� secondc� � � � � ����� ������� � ��� queries per minute

d� E�n� � ���� � ���

������ �

e� P n � ��� � �� � ������ � �����f� r� � E�r� ln���� � ��� secondsg� w� � E�r� ln��� � � ���� seconds

���� a� � �m�

� ��������

� ���b� p � ����

c� � � ������

� ��������� � ����

d� E�n� � � � ��� � ���� ������� ���� � ����e� E�nq� � ���� ������� ���� � ����

f� E�r� � ��

�� � ���

������

�� ������ second

g� Var�r� � ������� second� use forumla �� of Box �����h� w� � ������ second use formula �� of Box �����

���� a� � � ���� � ��� � ���������� � ���b� p � � ����c� � � ����d� E�n� � ��

����� � request per drive

e� E�nq� ����

����� ��� request per drive

f� E�r� � �������

� ��� secondg� Var�r� � E�r��� � ���� ��� � ���� second�

h� w� � ���� second use formula �� of Box �����

���� Yes� With the new system� the ���percentile of the waiting time will be zero�

���� Yes� since with � � ������� � ������� average waiting time is ���� minutesand the ���percentile of waiting time is zero�

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��� a� p � ���� use formula � of Box �����p� � ����� p� � ����� p� � ����� p� � ������ use formula � of Box �����b� E�n� � ��� requests use formula � of Box �����c� E�nq� � ������ requests use formula � of Box �����d� Var�n� � �� � p� � �� � p� � �� � p� � �� � p� � ����� � ���e� �� � ��� pB� � ���� ������� � �� requests per secondf� Loss rate� �pB � ��� ������ � ��� requests per secondg� U � �� pB� � ����� ������� � ����h� E�r� � E�n���� � ������ � ������ second

���� The probability pn for birth�death processes is given by

pn ���� �n��

���� �n p� n � �� �� � � � �

pn �

�����������

pm �n

�Kn

�� � n � m

p n

�Kn

�n m mm m � n � K

where � �m�

Average throughput �� �PK��

n� K � n��pn � �K � E�n��U � ��

m�� K � E�n��

E�r� � E�n���

� E�n��K�E�n��

���� pn �

�����������

�Kn

�m �np � � n � m�

Kn

�n ��nmm

m p m � n � B

where� � �m�

Average throughput �� �PB

n�K � n��pn � � K � E�n�� K � B�pB�where pB is the probability of B jobs in the system�U � ��

m�� K � E�n�� K �B�pB�

E�r� � E�n���

� E�n��K�E�n��K�B�pB�

Page 61: Solution Manual

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���� a� Job �ow balance� Number of job arrivals is not equal to number ofdepartures since some jobs are lost�

b� Fair service

c� Single resource possession

d� Routing homogeneity

e� No blocking

f� Single resource possession

g� One�step behavior

Page 62: Solution Manual

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���� X����������� S������� U � XS�����������

���� n��� X��� n � XR � R���� second�

���� X��������� Vdisk���Xdisk � XVdisk��� Sdisk������� Udisk � XdiskSdisk��� ���� � ���� � ���

���� Xprinter����������� Vprinter��� X � Xprinter�Vprinter� ���� � ��

jobs�minute

���� a� VCPU � ������ � ��� VA � �������� � ��� VB � ��������� � ��b� DCPU����� � �� � �� DA����� � �� � ���� DB� ������� � ���c� Uk � XDk � X����������� UCPU��� UB����d� Uk � XDk � X����������� R � N�X � Z � ����� � � �� seconds

���� xxx The data used here looks like that from problem ����� a� CPU sinceit has the most demand�b� Rmin � D� �D� �D� � ��������� � ���c� X � �� U� � �� ��� � ���d� Dmax��� Uk � XDk � X � � job�seconde� R � maxfD�NDmax � Zg � Dmax � R�Z

N� ���� We need at least a

��� faster CPU� disk A would be just OK�f� D����� Dmax��� Z��� � X � minf N

���� �g� R � maxf���� N � �g

���� a� D������D� � ����D� � ��� � disk Ab� D������ D������D� � ��� � disk Ac� D����D� � ����D����� � CPUd� D����D� � ����D��� � disk B

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���� X � ��� D� � �� D� � �� � ���� � ���� D� � � � ���� � ���� Qi �Ui�� � Ui� � XDi�� � XDi�� Q� � ��� � ��� � ��� � �� � �� Q� ���� � ����� � ����� � ������ Q� � ��� � ����� � ����� � ����� Qavg �P

iQi � � � ����� � ����� � �����Ri � Si���Ui� � Si���XDi�� R� � ������� ���� � ���� R� � ������������ � ������� R� � ������������� � ������� Ravg �

PRiVi � �������

������� �� � ������� � � ������ seconds�

���� xxx depends on data of ���� which is wrong��

Response Time System Queue LengthsN CPU Disk A Disk B System Throughput CPU Disk A Disk B� ����� ����� ����� ����� ����� ����� ����� ������ ����� ����� ����� ����� ����� ����� ����� ������ ����� ����� ����� ����� ����� ����� ����� ������ ����� ����� ����� ����� ����� ����� ����� ������ ����� ����� ����� ����� ����� ����� ����� �����

���� xxx depends on data of ���� which is wrong��

Itera� Response Time System Queue Lengthstion No� CPU Disk A Disk B System Throughput CPU Disk A Disk B

� ����� ����� ����� ������ ����� ����� ����� ������ ����� ����� ����� ������ ����� ������ ����� ������ ����� ����� ����� ������ ����� ������ ����� ������ ����� ����� ����� ������ ����� ������ ����� ������ ����� ����� ����� ������ ����� ������ ����� �����

���� Each packet is serviced by the � computers� Hence the throughput is whenthere is one packet is X� � ���S� The packet spents a time of S in each hopso the response time is R� � �S� SimiliarilyX� � ���S� R� � �SX� � ���S� R� � �SX� � ���S� R� � �SX� � ���S� R� � �SIn general R � n � ��S� X � n

n���S� n � �

���� If there are h hops in the network� then the each packet has to get servicedby h hops and has to wait a time of nS hops before getting serviced� Hence�

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the response time is R � n � h�S� Arguing similarily the throughput isX � n

n�h�S� Power X�R � n

n�h��S� is maximum when n�h����n�h�n ��� n � h

���� xxx depends on data of ���� which is wrong��The balanced job bounds are

N

� � ��� � N � ������ ���N������N�����

� XN� � min

��

N

� � ��� � N � �� ��������

max�N � �� ��� � N � ��

���

��� � �

�� RN� � ����N�������

N � �����

N � ����� � �

Response Time ThroughputLower Upper Lower Upper

N BJB MVA BJB BJB MVA BJB� ����� ����� ����� ����� ����� ������ ����� ����� ����� ����� ����� ������ ����� ����� ����� ����� ����� ������ ����� ����� ����� ����� ����� ������ ����� ����� ����� ����� ����� ������ ����� ����� ����� ����� ����� ������ ����� ����� ����� ����� ����� ������ ����� ����� ����� ����� ����� ������ ����� ����� ����� ����� ����� �����

�� ����� ����� ����� ����� ����� �����

���� a� X � ND��N��

M�� NM

DM�N���� R � Q�X � N�X � DM�N���

M

b� Substituting Davg � Dmax � D�M and Z � � in Equations �����and ������ we get the same expressions as in Exercise ���� for balancedjob bounds�

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Table ����� Computing the Normalizing Constant for Exercise ����

n yCPU � �� yA � � yB � �� � � �� �� �� ��� ��� ��� ���� ���� ���� ����

���� DCPU � ����� �� � �� DA � ����� �� � ���� DB � ������ � � ���� Forscaling factor choose � ����� � ��� This results in yCPU � ��� yA � ��yB���

The probability of exactly j jobs at the ith device is

P ni � j� � P ni � j�� P ni � j � ��

�yji

GN�GN � j�� yiGN � j � ���

P QCPU � �jN � �� ��

��������� ��� ���� � �����

P QCPU � �jN � �� ���

�������� ��� ��� � �����

P QCPU � �jN � �� ����

������� ��� �� � �����

P QCPU � �jN � �� �����

������ ��� �� � �����

xxx the book answer is di�erent� P QCPU � njN � �� for n � �� ���� � are ������ ������ ������ and ������ respectively���

The throughputs for N � �� �� � are

X�� � GN � ��

GN�� ��� �������� � �����

X�� � ��G��

G��� ��� ������ � �����

X�� � ��G��

G��� ��� ���� � �����

xxx in book are di�erent�

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���� xxx data in ���� is wrong�� P QCPU � njN � �� for n � �� �� �� � are������ ������ ������ and ������ respectively�

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���� xxx answer depends on data of Exercise ����� X � ������ ������ and �����for N � �� �� �� respectively� R � ������ ������ ����� for N � �� �� ��respectively�

���� xxx answer depends on data of Exercise ����� The service rates of the FECare ������ ������ and ������ respectively� for one through three jobs at theservice center�