advanced systems lab exercises from the book – viski

Advanced Systems Lab Exercises From the BookAus VISki

Inhaltsverzeichnis

1 Recommended Exercises from the Book1.1 1.11.2 12.11.3 12.71.4 13.21.5 14.21.6 14.31.7 16.11.8 17.11.9 18.11.10 30.31.11 30.41.12 31.11.13 31.21.14 31.31.15 31.41.16 33.11.17 33.21.18 33.31.19 33.5

2 Errata

Recommended Exercises from the Book

R. Jain: The Art of Computer Systems Performance Analysis, Wiley.

1.1

The measured performance of two database systems on two different work-loads is shown in Table 1.6.Compare the performance of the two systems and show thata. System A is better.b. System B is better.TABLE 1.6 Throughput in Queries per Second

Workload 1 Workload 2

A 10 30

Advanced Systems Lab Exercises From the Book – VISki http://wiki.vis.ethz.ch/Advanced_Systems_Lab_Exercises_From_the...

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B 30 10

Solution:a.

Workload 1 Workload 2 Average

A 33% 300% 166.5%

B 100% 100% 100%

b.

Workload 1 Workload 2 Average

A 100% 100% 100%

B 33% 300% 166.5%

--Kermit 14:29, 22. Jan. 2011 (CET)

12.1

A distributed system has three file servers, which are chosen independently and with equal probabilitieswhenever a new file is created. The servers are named A, B, and C. Determine the probabilities of thefollowing events:a. Server A is selectedb. Server A or B is selectedc. Servers A and B are selectedd. Server A is not selectede. Server A is selected twice in a rowf. Server selection sequence ABCABCABC is observed (in nine successive file creations)

Solution:a. P(A) = 0.33b. P(A∨B) = 0.66c. P(A∧B) = 0 (because they are never selected at the same time)d. P(¬A) = P(B∨C) = 0.66e. P(AA) = 0.33 * 0.33 = 0.1089f. P(ABCABCABC) = 0.33⁹ = 4.6411e-05--Kermit 17:45, 22. Jan. 2011 (CET)

12.7

The execution times of queries on a database is normally distributed with a mean of 5 seconds and astandard deviation of 1 second. Determine the following:a. What is the probability of the execution time being more than 8 seconds?b. What is the probability of the execution time being less than 6 seconds?c. What percentage of responses will take between 4 and 7 seconds?


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d. What is the 95-percentile execution time?

Solution:Wikipedia: standard deviation (http://en.wikipedia.org/wiki/Standard_distribution#Cumulative_distribution_function)

a.

b.

c.

d.

95% of the execution times are between 3.04s and 6.96s. --Kermit 11:18, 23. Jan. 2011 (CET)

13.2

Answer the following for the data of Exercise 12.11:a. What is the 10-percentile and 90-percentile from the sample?b. What is the mean number of disk I/O’s per program?c. What is the 90% confidence interval for the mean?d. What fraction of programs make less than or equal to 25 I/O’s and what is the 90% confidence interval forthe fraction?e. What is the one-sided 90% confidence interval for the mean?

Data: {23, 33, 14, 15, 42, 28, 33, 45, 23, 34, 39, 21, 36, 23, 34, 36, 25, 9, 11, 19, 35, 24, 31, 29, 16, 23, 34,24, 38, 15, 13, 35, 28}

Sorted Data: {9, 11, 13, 14, 15, 15, 16, 19, 21, 23, 23, 23, 23, 24, 24, 25, 28, 28, 29, 31, 33, 33, 34, 34, 34,35, 35, 36, 36, 38, 39, 42, 45}

Solution:a. round( (n-1)*(p/100) + 1 )th element from sorted sequence.10-percentile > 4th element > 1490-percentile > 30th element > 38b. have fun doing this manually... wolfram alpha says 26.91

c. the formula to calculate CI:

looking up t in students t-distribution, value is for N=30 (for 2-sided CI it is one to the right in the book(wikipedia is more clear on that))

the values are slightly off from the master solution

d. p:=16/33=0.485α = 0.1,s = 9.49,n = 33

CI=

e. see c. for e formula:


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CI=26.91-1.31*(9.495/sqrt(33))=(24.7;26.91)orCI=26.91+1.31*(9.495/sqrt(33))=(26.91;29.08)again slightly off.--Kermit 12:15, 23. Jan. 2011 (CET)

14.2

For the disk I/O and CPU data in Example 14.1, find a linear formula to predict the number of disk I/O’sgiven the CPU time. Answer the following questions about this regression:a. Which parameters are significant?b. What percentage of variation is explained by the regression?c. What is the expected number of disk I/O’s for a program with a CPU time of 40 milliseconds?d. What bounds would you put on your answer in c if you wanted to take less than 10% chance of error on asingle program to be measured tomorrow?e. Repeat d for the case that you want to take a less than 10% chance of error on the mean of a large numberof programs to be measured tomorrow?

Data: {(14, 2), (16, 5), (27, 7), (42, 9), (39, 10), (50, 13), (83, 20)}

Solution:

a. only b1 is significant. (as we have an obvious (0,0) origin, only the slope is relevant for a linear model ofthe data - this is confirmed by the very small b0.)b. SSR indicates the variation that was explained by the regression.

, i.e., 97% of the variation is explained by the

regression.

c.

d.

wiki_qm1.png

Dem Autor des Lösungsvorschlags stellt sich eine Unklarheit betreffend der Prüfungsfrage oder derenLösung:How to compute s?Hilf mit, die Qualität dieser Seite zu verbessern, indem du eine eindeutige Lösung einfügst oder aufder Diskussionsseite dieses Themas deine Meinung äusserst.

e. *beep*--Kermit 15:24, 23. Jan. 2011 (CET)


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14.3

The memory size of the seven programs mentioned in the disk I/O and CPU time data were also measured.The memory size (in kilobytes) and CPU time (in milliseconds) pairs observed are {((70, 2), (75, 5), (144,7), (190, 9), (210, 10), (235, 13), (400, 20)}. Analyze the data using a simple regression model to predictCPU time as a function of the memory size.

Solution:

--Kermit 15:24, 23. Jan. 2011 (CET)

16.1

The performance of a system being designed depends upon the following three factors:1. CPU type: 68000, 8086, 802862. Operating system type: CPM, MS-DOS, UNIX3. Disk drive type: A, B, CHow many experiments are required to analyze the performance ifa. There is significant interaction among factors.b. There is no interaction among factors.c. The interactions are small compared to main effects.

Solution:a. 3³ = 27

b. simple design:

c. 9 = 3²(Fractional factorial design due to little interaction)--Kermit 15:35, 23. Jan. 2011 (CET)--Flwidmer 14:44, 25. Jan. 2011 (CET)

17.1

Analyze the 2³ design shown in Table 17.10a.Quantify main effects and all interactions.b. Quantify percentages of variation explained.c. Sort the variables in the order of decreasing importance.

Solution:a.b.

I A B C AB AC BC ABC y

1 1 1 1 1 1 1 1 100

1 1 1 -1 1 -1 -1 -1 15

1 1 -1 1 -1 1 -1 -1 40


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1 1 -1 -1 -1 -1 1 1 30

1 -1 1 1 -1 -1 1 -1 120

1 -1 1 -1 -1 1 -1 1 10

1 -1 -1 1 1 -1 -1 1 20

1 -1 -1 -1 1 1 1 -1 50

385 -15 105 175 -15 15 215 -65 Total

48.125 -1.875 13.125 21.875 -1.875 1.875 26.875 8.125 Total/8

11596.9 28.125 1378.125 3828.125 28.125 28.125 5778.125 528.125 Variation

100 0.24 11.88 33.0 0.24 0.24 49.82 4.55 Variation %

c. BC, C, B, ABC, A, AB, AC--Kermit 11:54, 24. Jan. 2011 (CET)

18.1

Table 18.11 lists measured CPU times for two processors on two workloads. Each experiment was repeatedthree times. Analyze the design.

Workload Processor A Processor B

I (41.16, 39.02, 42.56) (63.17, 59.25, 64.23)

J (51.50, 52.50, 50.50) (48.08, 48.98, 47.10)

Solution:

W P WP y Mean y Error

1 1 1 1 (41.16, 39.02, 42.56) 40.9133 6.35707

1 1 -1 -1 (63.17, 59.25, 64.23) 62.2167 14.98746

1 -1 1 -1 (51.50, 52.50, 50.50) 51.5 2

1 -1 -1 1 (48.08, 48.98, 47.10) 48.0533 1.76827

202.8633 3.7566 -18.0366 -24.93 Total

50.71 0.94 -4.51 -6.23 Total/4

745.82 10.58 243.99 466.13 Variation 25.1128

100 1.42 32.71 62.499 % Variation 3.367

SSE = (40.9133-41.16)² + (40.9133-39.02)² + (40.9133-42.56)² + (62.2167-63.17)² + (62.2167-59.25)² +(62.2167-64.23)² + (51.5-51.5)² + (51.5-52.5)² + (51.5-50.5)² + (48.0533-48.08)² + (48.0533-48.98)² +(48.0533-47.1)²


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= 6.35707 + 14.987 + 2 + 1.76827 = 25.1128

30.3

Which queueing system would provide better performance: an M/M/3/300/100 system or an M/M/3/100/100system?

Solution:Both will provide the same performance. Increasing buffers beyond the population size has no effect.--Kermit 14:53, 24. Jan. 2011 (CET)

30.4

During a 1-hour observation interval, the name server of a distributed system received 10,800 requests. Themean response time of these requests was observed to be one-third of a second.a. What is the mean number of queries in the server?b. What assumptions have you made about the system?c. Would the mean number of queries be different if the service time was not exponentially distributed?

Solution:a. E[n] = λ * E[w] = 10800/3600 * 1/3 = 1b. Job flow balance. (That all jobs that are sent into the system get a response.)c. Doesn't have an influence.--Kermit 15:47, 24. Jan. 2011 (CET)

31.1

Consider a single-server system with discouraged arrivals in which the arrival rate is only λ/(n + 1) whenthere are n jobs in the system. The interarrival times, as well as the service time, are independent andidentically distributed with an exponential distribution. Using a birth-death process model for this system,develop expressions for the following:a. State probability pn of n jobs in the systemb. State probability ρ0 of the system being idlec. Mean number of jobs in the system, E[n]d. Effective arrival rate λ’e. Mean response time E[r]

Solution:

a.

With

b. As we know: and with the previous formula for pi we see that this corresponds to a poisson

distribution, and therefore p0 = e − ρ

c. As already figured out, the probabilities correspond to a poisson distribution with parameter ρ, hence


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d. We know that thus we can solve:

Using the formula: we get:

e. Little's Law: we directly get:

-- Davids 15:52, 25. Jan. 2012 (CET)

31.2

Consider an M/M/∞ system with an infinite number of servers. In such a system, all arriving jobs beginreceiving service immediately. The service rate is nµ when there are n jobs in the system. Using abirth-death process model for this system, draw a state transition diagram for this system and developexpressions for the following:a. State probability pn of n jobs in the systemb. State probability p0 of the system being idlec. Mean number of jobs in the system, E[n]d. Variance of number of jobs in the system, Var[n]e. Mean response time E[r]

Solution:a. same as 31.1.ab. same as 31.1.b, e − ρ

c. same as 31.1.c, E[n] = ρd. Knowing it is a poisson distribution (compare with 31.1), we know that Var[n] = ρe. E[r] = E[n]/λ = λ/μ/λ = 1/μNote the difference to 31.1: Here the effective arrival rate is the same as the normal arrival rate!-- Davids 16:06, 25. Jan. 2012 (CET)

31.3

The average response time on a database system is 3 seconds. During a 1-minute observation interval, theidle time on the system was measured to be 10 seconds. Using an M/M/1 model for the system, determinethe following:a. System utilizationb. Average service time per queryc. Number of queries completed during the observation intervald. Average number of jobs in the systeme. Probability of number of jobs in the system being greater than 10f. 90-percentile response timeg. 90-percentile waiting time

Solution:


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a. ρ=5/6b. using E[r] = (1/μ) / (1-ρ)E[s] = 1 /μ = E[r]*(1-ρ) = 0.5sc. x = 50/0.5 = 100d. E[n] = E[nq]+ρ = E[w]*λ+ρ=(E[r]-E[s])*(ρ*μ)+ρ=(3-0.5)*(5/6*2)+5/6 = 5e. P[n>=10] = p10 = rho^11 = (5/6)^11 = 0.1346f. q-percentile of E[r] = E[r] * ln(100/(100-q)) => 90-percentile = E[r]*ln(10) = 6.9sg. E[wq] = E[r]-E[s] = 2.5s. q-percentile of E[wq] = (E[wq]/rho) * ln(100rho/(100-q)) = 6.36--Mbruggmann 18:07, 24. Jan. 2011 (CET)

31.4

A storage system consists of three disk drives sharing a common queue. The average time to service an I/Orequest is 50 milliseconds. The I/O requests arrive to the storage system at the rate of 30 requests persecond. Using an M/M/3 model for this system, determine the following:a. Average disk drive utilizationb. Probability of the system being idle, p0c. Probability of queueing,d. Average number of jobs in the system, E[n]e. Average number of jobs waiting in the queue, E[nq]f. Mean response time E[r]g. Variance of the response timeh. 90-percentile of the waiting timeSolution:a. μ (per disk) = 1/0.05, λ (per disk) = 30/3 = 10 => ρ = λ/μ = 0.5b. pain-in-the-ass formel (aus textbook box 31.2) p0=0.21c. ( ((m*ρ)^m) / (m!*(1-ρ)) ) * p0 = 0.236d./e./f./g./h. > richtige formel aus box 31.2 auswerten--Mbruggmann 11:24, 25. Jan. 2011 (CET)

33.1

During a 10-second observation period, 400 packets were serviced by a gateway whose CPU can service200 pps. What was the utilization of the gateway CPU?

Solution:Throughput = X = 400/10 = 40Servicetime = S = 1/200U = X*S = 40/200 = 20%--Kermit 15:22, 25. Jan. 2011 (CET)

33.2

The throughput of a timesharing system was observed to be five jobs per second over a 10-minuteobservation period. If the average number of jobs in the system was four during this period, what was theaverage response time?Solution:

R = Number of Jobs / Throughput = Q/X = 4/5--Kermit 15:42, 25. Jan. 2011 (CET)

33.3


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During a 10-second observation period, 40 requests were serviced by a file server. Each request requires twodisk accesses. The average service time at the disk was 30 milliseconds. What was the average diskutilization during this period?

Solution:X = 40/10 = 4Vdisk = 2Xdisk = X*Vdisk = 8Sdisk = 0.03Udisk = Xdisk * Sdisk = 8 * 0.03 = 0.24 = 24%--Kermit 15:42, 25. Jan. 2011 (CET)

33.5

For a timesharing system with two disks (user and system), the probabilities for jobs completing the serviceat the CPU were found to be 0.80 to disk A, 0.16 to disk B, and 0.04 to the terminals. The user think timewas measured to be 5 seconds, the disk service times were 30 and 25 milliseconds, and the average servicetime per visit to the CPU was 40 milliseconds. Using the queueing network model shown in Figure 32.8,answer the following for this system:a. For each job, what are the visit ratios for CPU, disk A, and disk B?b. For each device, what is the total service demand?c. If disk A utilization is 60%, what is the utilization of the CPU and disk B?d. If the utilization of disk B is 10%, what is the average response time when there are 20 users on thesystem?

Solution:a. V_CPU=1/0.04=25, V_A=0.8/0.04=20, V_B=0.16/0.04=4 (Divide visit probability by exit probability)b. D_CPU=25*40ms=1, D_A=20*30ms=0.6, D_B=4*25ms=0.1c. U_CPU=1, U_B=0.1d. N=20, X=U_B/D_B = 1 Job/s, R = N/X - Z = 20/1 - 5 = 15s

Errata

Several solutions in the book contain errors. Here's a complete list of errors as PDF(http://www1.cse.wustl.edu/~jain/books/ftp/errors_all.pdf)

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advanced systems lab exercises from the book – viski

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