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Solution Manual for The Art of Computer Systems Performance AnalysisTechniques for Experimental Design, Measurement, Simulation, and Modeling

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By Raj Jain Professor of CIS 2015 Neil Avenue Mall, 297 Dreese Lab Columbus, OH 43210-1277 Internet: Jain@Cis.Ohio-State.Edu

Please do not copy without Author's written permission.Copy No. For

Tentative Publication Date: August 1997 Copyright 1997 Raj Jain

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1.1

Compare the ratio with system A as the base System Workload 1 Workload 2 Average A 1 1 1 B 0.33 3 1.66 Considering the ratio of performance with system A as base, we conclude that system B is better. Compare the ratio with system B as the base System Workload 1 Workload 2 Average A 3 0.33 1.66 B 1 1 1 Considering the ratio of performance with system B as base, we conclude that system A is better.

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2.1 incomplete 2.2 Can be done

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3.1

a. Measurements. Run your favourite programs and pick the one that runs them faster. b. Use measurements and simulations of various network con guirations. 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. Query response time. Failure rate. Storage capacity. Usability. Capacity. Response time. Failure rate. Response time.

3.2

b.

c.

d.

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4.1 The following information is from SPEC (Standard Performance Evaluation

Corporation) home page. CPU benchmarks CINT92, current release: Rel. 1.1. Integer benchmarks contains Name Application espresso Logic Design li Interpreter eqntott Logic Design compress Data Compression sc Spreadsheet gcc Compiler CFP92, current release: Rel. 1.1. Floating point benchmark suite contains Name Application spice2g6 Circuit Design doduc Simulation mdljdp2 Quantum Chemistry wave5 Electromagnetism tomcatv Geometric Translation ora Optics alvinn Robotics ear Medical Simulation mdljsp2 Quantum Chemistry swm256 Simulation su2cor Quantum Physics hydro2d Astrophysics nasa7 NASA Kernels fpppp Quantum Chemistry More information about these benchmarks can be found in http://performance.netlib.org/performa web page./* * seive.c : Program to implement sieve workload * */

4.2 A C program to implement sieve workload.

6#include #define MaxNum 8191 #define NumIterations 10 #define TRUE 1 #define FALSE 0 void main(void) { int IsPrime MaxNum+1] int i,k,Iteration int NumPrimes /* List all primes upto MaxNum */ /* Repeats procedure NumIterations times */

/* Loop indexes */ /* 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 = 1 Iteration 10) = 11 = (5=6)11 = 0:135 f. r90 = E r] ln 10] = 6:9 seconds g. w90 = E r] ln 10 ] = 6:36 seconds;

31.4 a. = m = 3;

= 0:5 b. p0 = 0:21 3 (3 c. % = 3!(1 0:05)5) 0:21 = 0:24 : d. E n] = 3 0:5 + 0:5 0:24=(1 ; 0:5) = 1:74 e. E nq ] = 0:5 0:24=(1 ; 0:5) = 0:24 1 f. E r] = 20 1 + 3(10:24:5) = 0:0579 second 0 g. Var r] = 0:00296 second2 (use forumla 14 of Box 31.2) h. w90 = 0:0287 second (use formula 19 of Box 31.2);

30 (1=0:05)

31.5 a. = 30=3 = 10 ) = 10=(1=0:05) = 0:5;

b. p0 = = 0:50 c. % = 0:50 d. E n] = 1 0:05:5 = 1 request per drive e. E nq ] = 10:502:5 = 0:5 request per drive f. E r] = 11=205 = 0:1 second 0: g. Var r] = (E r])2 = 0:1 0:1 = 0:01 second2 h. w90 = 0:16 second (use formula 20 of Box 31.1); ;

31.6 Yes. With the new system, the 90-percentile of the waiting time will be zero. 31.7 Yes, since with = 0:167=2 = 0:0833, average waiting time is 0.18 minutesand the 90-percentile of waiting time is zero.

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31.8 a. p0 = 0:22 (use formula 4 of Box 31.3)

p1 = 0:34 p2 = 0:25 p3 = 0:13 p4 = 0:0629 (use formula 5 of Box 31.3) b. E n] = 1:5 requests (use formula 6 of Box 31.3) c. E nq ] = 0:0629 requests (use formula 7 of Box 31.3) d. Var n] = 12 p1 + 22 p2 + 32 p3 + 42 p4 ; (1:5)2 = 1:3 e. = (1 ; pB ) = 30(1 ; 0:0629) = 28 requests per second f. Loss rate= pB = 30 0:0629 = 1:9 requests per second g. U = (1 ; pB ) = 0:5(1 ; 0:0629) = 0:47 h. E r] = E n]= = 1:5=28 = 0:0535 second0 0

31.9 The probability pn for birth-death processes is given bypn = pn = >8 > > > p0 (m )n > < > > p0 n > :

0 1 1 2!

n 1 p0 n;

n = 1 2 :::

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where = m Average throughput = PK=01(K ; n) pn = (K ; E n]) n U = m0 = (K ; E n]) E r] = E n] = (KE n] n]) 0 E0 ; ;

K n

K n !

0 n K ! n!( )n mm > > > : m! p0 m n B n where, = m . Average throughput = PB=0(K ; n) pn = (K ; E n] ; (K ; B )pB ) n where 0pB is the probability of B jobs in the system. U = m = (K ; E n] ; (K ; B )pB ) ] E r] = E n] = (K E nE nK B)pB) 0 ] (0 ; ; ;

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