Art of computer systems performance analysis techniques for experimental design measuremet
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Transcript of Art of computer systems performance analysis techniques for experimental design measuremet
�
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
�
��� � 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�
�
��� incomplete
��� Can be done
�
��� 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�
�
��� 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
�
��
�
�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 � ��
�
� �� 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
�
��� 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�
�
��� 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 ����
��
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�
��
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�
��
��� 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�
��
�� 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�
��
��� incomplete
��� incomplete
��
��� 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 � ��
��
���� 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�
��
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�
��
���� 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��
��
����
pdf � fx� �dF x�
dx
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a
�m��Xi�
x�a�i
i�
�� e�x�a
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a � i�
�
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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
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m� ���am
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� mZ �
e�x�adx
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i�
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Variance � �� �Z �
x� ���fx�dx
��
m� ���am
Z �
x� am��xm��e�x�adx
��
m� ���am
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��
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m� ���am
Z �
xm��e�x�adx
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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 �
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� a
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�a� ��
���
�a
a� �
Variance � �� �Z �
�x� ���fx�dx
�Z �
�x� a
a� ���ax�a���dx
Integrating by parts
��
� a
�x� a
a� ���x�a
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� �Z �
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a� ��x�adx
��
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��
sa
a� ���a� ��
a� �
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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�
��
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�
��
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�
��
���� 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� � ��
��
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 �
��
���� 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������
�����
����� �� ����� ��������� ������
��
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 ������� ������� �
��
���� 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�
��
���� 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 �
���������������
� ����� �� ����� �� ����� �� ����� �� ����� �� ����� �� ����� �� ����� �
��������������
��
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 ������ �������
��
���� 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�
��
���� 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
��
��� 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�
��
���� 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�
��
��� 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
��
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������� � ������
��
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�
��
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�
��
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���� � ����
��
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 ���� ����� ���� ����� ����� �����
��
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
��
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�
��
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� ������ ���� ������� �������
��
Table ����� Interactions
B A� A� A�B� ������� ������� ������� ������� �������� ��������B� ���������������� ������������ ������������B� �������������� �������������� ����������������B� �������������� ���������������� ��������������B� ���������������� ���������������� ��������������
Table ����� Interactions
A� A� A�A� �������� �������� ��������� ���������A� ��������� ���������
��
���� 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 � ���
��
���� 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�
��
���� 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 ��
��
���� 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� � �
��
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�
��
Table ����� Random Numbers Generated by the LCG xn � ��xn�� ��� mod ���
n xnDecimal Binary
� �� ��
� ��� � ���
� ���� � ����
� ������ ����� �����
� ������ ������ �����
� ���� ��� �����
� ��� �� �����
� ����� ������ �����
��� ��� ���
� ����� �� �����
�� ���� ��� �����
�� ������ ������ �����
�� ����� ����� ��
�� ����� ������ �������
�� ����� ������ �����
�� ����� ��� ����
�� ����� ���� ���
�� ������ ���� ����
� ������ ���� ��
� ������ ���� ����
��
���� 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�
��
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�� �������� �������� ��������� ��������
��
���� 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�
��
��� 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��
��
���� 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���
��
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���
��
��� 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 � ��
��
���� 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� �
��
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�
��
��� 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�
��
���� 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
��
���� 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
��
���� 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�
��
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�