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Operational Laws and Mean Value Analysis Nirdosh Bhatnagar 1. Introduction Operational technique can be used to analyse both computer and networking systems. We …rst introduce basics of operational analysis. This technique is then extended to open and closed queueing networks. The closed queueing network considered has single servers at its queueing centers. 2. Operational Analysis Basics

The word operational means measurable. The analysis presented in this chapter assumes that the performance metrics of a computer can be directly inferred by measurable quantities. The measurable quantities are called operational quantities. The laws which relate operational quantities are called operational laws. Operational laws for computer systems will be studied in this note. We …rst introduce some notation. Notation: M =Number of queueing centers in the network. T = Finite observation period. Ai = Number of arrivals at queue i (device i) during observation period. Bi = Total busy period of queue i during observation period. Ci = Number of service completions at queue i during observation period. Di = Total service demand by a customer at device i. Qi = Queue length at device i (including the job in service) Ri = Response time per visit to the ith device. Si = Average service time per customer visit to queue i during observation period. Ui = Utilization of queue i during observation period. Vi = Average number of visits to queue i by a customer before it leaves the system during observation period. Xi = Throughput of queue i during observation period. i = Arrival rate of queue i during observation period. pij = The transition probability of a job moving to the jth queue after service completion at the ith queue. The following variables correspond to the entire system under observation. D = Sum of service demands on all devices. Dmax = Service demand on the bottleneck device. N = Number of jobs in the system. Q = Number of queued jobs in the system. R = System response time. X = System throughput. Z = Think time.

1

Operational Laws and Mean Value Analysis

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2.1.

Observations. The following relationships can be readily noted. Arrival rate Mean service time Utilization Throughput = = = = Ai T Bi Si = Ci Bi Ui = T Ci Xi = T i =

Utilization Law: The utilization of device i is also given by Ui = Xi Si Forced ‡ law: If during the observation period T is such that the number of job ow arrivals at each device is the same as the number of job completions, that is, Ai = Ci we say that the device i satis…es the assumption of job ‡ balance. If the observation ow period T is long, the di¤erence (Ai Ci ) is usually small compared to Ci : It will be exact, if the initial queue length at each device is the same as the …nal queue length. Let each job make Vi requests for the ith device in the system. Let C0 be the number of completed jobs which go outside the system, then Ci = C0 Vi Let the system throughput be X: Then X= Then Xi Xi Ci T = XVi = C0 T

This equation is the forced ‡ow law. It is applicable whenever the job ‡ balance ow assumption is true. Also Ui = X i Si = XVi Si Di = Vi Si we have Ui = XDi

De…ning

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Di is the total service demand on the device for all visits of a job. Consequently, the device with the highest Di has the highest utilization and is the bottleneck device. Let Dmax be the service demand of the bottleneck device, then Dmax = max fDi g i Visit ratios and transition probabilities: Visit ratios are one way of specifying the routing of jobs in a queueing network. Another way is to specify the transition probabilities pij of a job moving to the jth queue after service completion at the ith queue. The visit ratios and transition probabilities are equivalent in the sense, that given one we can always …nd the other. In a system with job ‡ balance, ow Cj =
M X i=0

Ci pij

where M is the number of queueing centers in the network. The subscript 0 is used to denote visits to the outside link. Thus, pi0 is the probability of a job exiting from the system after completion of service at the ith device. Dividing both sides of the equation by C0 ; we get M X Vj = Vi pij i=0 Since each visit to the outside link is de…ned as the completion of the job, we have V0 = 1

The above two equations are called the visit ratio equations, and can be used to get visit ratios from the transition probabilities. A unique solution is always possible provided the network is operationally connected, that is each device in the system is visited at least once by each job. Little’ law: This law holds under the assumption of job ‡ balance. Let Qi be the s ow mean number of jobs in the device i: Then Qi = i Ri

= X i Ri

Response time law: Let Q be the total number of jobs in the system. Then Q = XR Also Q =
M X i=1

Qi

XR Since Xi = XVi ; we have

=

M X i=1

X i Ri

R=

M X i=1

Ri Vi

Operational Laws and Mean Value Analysis

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This equation is called the general response time law. The law states that the total time spent by a job at a queueing center is the product of response time per visit and the number of visits. Further the total time in the system is equal to the sum of total times at various queueing centers. Example. Problem Statement: We consider a time sharing system. See Figure 32.8 of the text-book by Raj Jain. In a time sharing system, accounting log data produced the following user pro…le for user programs. Each program requires 5 seconds of CPU time and makes 80 I/O requests to disk A and 100 I/O requests to disk B. The average think time of the users was 18 seconds. From the device speci…cations, it was determined that disk A takes 50 milliseconds to satisfy an I/O request and disk B takes 30 milliseconds per request. With 17 active terminals, disk A throughput was observed to be 15.70 I/O requests per second. The average queue length was observed to be 8.88, 3.19, and 1.40 jobs at the CPU, disk A, and disk B respectively. 1. Compute the CPU visit ratio. 2. Determine the service demand of all the devices. 3. Find the system, the CPU, and disk B throughput. 4. Find device utilizations. 5. Determine the response time of jobs at CPU and disks A and B. 6. Compute the response time for the timesharing system. 7. The transition probabilities form CPU to other queueing centers. Solution: We will …rst list the information which is given. DCP U VA Z SA N XA Also QCP U = 8:88; QA = 3:19; QB = 1:40 1. Since the jobs visit the CPU before going to disks and terminals, the CPU visit ratio is VCP U = VA + VB + 1 = 181 2. The service demands of the disks A and B are: DA DB = VA SA = 4 seconds = VB SB = 3 seconds = = = = = = 5 seconds 80; VB = 100 18 seconds 0:05 seconds, SB = 0:03 seconds 17 terminals 15:7 I/O requests per second

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3. Since XA = XVA ; we have X= 15:7 = 0:1963 jobs per second 80

Also the relationships XB = XVB and XCP U = XVCP U yields

XB XCP U

= 0:1963 x 100 = 19:6 requests per second = 0:1963 x 181 = 35:5 requests per second

4. The utilization law yields UCP U UA UB = XDCP U = 0:1963 x 5 = 0:9815 = XDA = 0:1963 x 4 = 0:7852 = XDB = 0:1963 x 3 = 0:5889

5. Using Little’ law, the device response times are s RCP U RA RB = = = QCP U XCP U QA = XA QB = XB 8:88 = 0:250 seconds 35:5 3:19 = 0:203 seconds 15:7 1:40 = 0:071 seconds 19:6 =

6. The system response time is R = RCP U VCP U + RA VA + RB VB = 0:25 x 181 + 0:203 x 80 + 0:071 x 100 = 68:6 seconds

The system response time is 68.6 seconds. It could also have been computed from the following equation N = X (R + Z) where N = 17; X = 0:1963; and Z = 18: This equation is actually the Little’ law s applied to the complete system. 7. The transition probabilities are now computed. Denote the outside node by 0: In this example pA;CP U pB;CP U Trivially p0:0 = 0: Then 1 = V0 = = = 1; pA;0 = 0 1; pB;0 = 0 X i Vi pi;0

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In this equation i takes on values 0; A; B; and CP U: We have 1 pCP U;0 = VCP U pCP U;0 1 1 = = = 0:005525 VCP U 181

pCP U;0 is the exit probability. Also p0;A p0;B Then VA = X i = pA;A = pB;A = 0 = pA;B = pB;B = 0

Vi pi;A

= VCP U pCP U;A yields pCP U;A = = Therefore pCP U;A = 0:442: Similarly VB = X i VA VCP U 80 = 0:442 181

Vi pi;B

= VCP U pCP U;B yields pCP U;B = = VB VCP U 100 = 0:552 181

In summary pCP U;0 = 0:005525; pCP U;A = 0:442; and pCP U;B = 0:552: 2.2. Bottleneck Analysis of an Interactive System. An interactive system consists of a set of terminals, which issue requests for service at the di¤erent queueing centers, and the responses come back to the terminal. After a think time of Z, the users submit the next request. Let the mean system response time for these requests be R: Also let the number of users be N; and the system throughput be X: Then using Little’ law, we have s N = X (R + Z) We will obtain bounds on the system throughput and the response time. Let the number of queueing centers in the system be M: Let the total service demand, at each of the queueing centers be Di ; 1 i M: The utilization of device i; is Ui ; and Ui = XDi :

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Response time

Slope = Dmax Bounds

D 0 Number of users Bounds Slope = 1/(D+Z) -Z 1/Dmax Throughput

Number of users

Figure 1: Typical asymptotic bounds

Therefore, the device with the highest utilization will be the bottleneck device. This will happen to the device with the highest total service demand Di : Denote this highest value by Dmax . Let the corresponding device be denoted by b: Therefore Dmax Ub = max fDi g i = XDmax

The highest utilization of a device is equal to 1: Therefore XDmax This implies X Observe that Ri Vi we have Di ; 1 i 1 Dmax D; where D = N (D + Z) PM i=1 1

M: Then R N (R + Z)

Di : Consequently

X=

Therefore X min

1 N ; Dmax (D + Z)

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A lower bound on throughput X; is next obtained. Smallest throughput occurs when a total of N D time units is spent for servicing N users. This implies N (N D + Z) These two inequalities can be written as N (N D + Z) X min 1 N ; Dmax (D + Z) X

Inversion of the above relationship yields max Dmax ; Simpli…cation gives max fD; N Dmax Zg R ND The above discussion can be summarized in the following observation. Observation: Suppose, we have an interactive system with N users, each with a think time of Z time units. There are M queueing centers, each with a total service PM demand of Di time units, where 1 i M: Let Dmax = maxi fDi g ; and D = i=1 Di : Let the system throughput and the system response time be X and R respectively. Then N (N D + Z) max fD; N Dmax Zg Example. X R min ND 1 N ; Dmax (D + Z) (D + Z) N (R + Z) N (N D + Z) N

Consider the con…guration described in the earlier example:

1. Compute bounds on the system throughput and the response times. 2. How many terminals can be supported on this timesharing system, if the response time has to be kept below 100 seconds. Solution: 1. Recall that DCP U = 5; DA = 4; DB = 3; and Z = 18: Then D Dmax Then the asymptotic bounds N (12N + 18) max f12; 5N 18g For example, if N = 30; then 0:0794 X X R min 12N R 360: 1 N ; 5 30 = DCP U + DA + DB = 12 = 5

0:2 and 132

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2. The speci…ed bound on the response time is 100 seconds, then 100 That is 100 23:6 (5N N 18) R max f12; 5N 18g

Therefore the system can support no more than 23 terminals in order to meet the given response time constraints. Algorithms are developed for open and closed queueing networks. 3. Open Queueing Networks

Open queueing network models are used to represent transaction processing systems such as airline reservation systems or banking systems. 3.1. Assumptions. The transaction arrival process is a Poisson process, with a mean arrival rate of : The queueing centers are single server systems with exponentially distributed service time. The delay centers have in…nite number of servers and have exponentially distributed service times. Assume job ‡ balance, that is the throughput of the system ow is equal to the arrival rate. 3.2. Notation. The input and output variables are:

Input variables. The following are the input variables. X = External arrival rate, or system throughput. M =Number of queueing centers in the network, excluding terminals. Si = Average service time per customer visit to queue i during observation period. Vi = Average number of visits to queue i by a customer before it leaves the system during observation period. Output variables. The output variables are: Di = Total service demand by a customer at device i. Qi = Queue length at device i (including the job in service) Ri = Response time per visit to the ith device. Ui = Utilization of queue i during observation period. Q = Number of queued jobs in the system. R = System response time. 3.3. Algorithm. Under the assumption of job ‡ balance, X = : Total service ow demand Di ; the device utilization Ui ; and device throughput Xi are given by Di Ui Xi = V i Si = XDi = XVi

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The device response times are: Ri =
Si (1 Ui )

Si

Fixed-capacity centers Delay centers

The device queue lengths are: Qi =
Ui (1 Ui )

Ui

Fixed-capacity centers Delay centers

System response time R; and the number of queued jobs in the system Q is: R =
M X i=1

Ri Vi

Q =

M X i=1

Qi

3.4. Example. See Figure 32.1 of the text-book by Jain. It represents a queueing model of a …le server consisting of a CPU and two disks, A and B. Measurements on a distributed system with 6 clients system making requests to the …le server produced the following data: Observation interval = 3,600 seconds Number of client requests = 10,800 CPU busy time = 1,728 seconds Disk A busy time = 1,512 seconds. Disk B busy time = 2,592 seconds Number of visits (I/O requests) to disk A = 75,600 Number of visits (I/O requests) to disk B = 86,400 Solution: X = 10800=3600 = 3 client requests per second VA = 75600=10800 = 7 visits per client request to disk A VB = 86400=10800 = 8 visits per client request to disk B VCP U = 1 + VA + VB = 1 + 7 + 8 = 16 visits per client request to CPU DCP U = 1728=10800 = 0:16 second of CPU time per client request DA = 1512=10800 = 0:14 second of disk A time per client request DB = 2592=10800 = 0:24 second of disk B time per client request SCP U = 0:16=16 = 0:01 second per visit to CPU SA = 0:14=7 = 0:02 second per visit to disk A SB = 0:24=8 = 0:03 second per visit to disk B The utilizations are UCP U = 1728=3600 = 0:48 UA = 1512=3600 = 0:42 UB = 2592=3600 = 0:72 The throughputs are XCP U = XVCP U = 48 CPU requests per second XA = XVA = 21 disk A requests per second

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XB = XVB = 24 disk B requests per second The response times are RCP U = SCP U = (1 UCP U ) = 0:01= (1 0:48) = 0:01923 second RA = SA = (1 UA ) = 0:02= (1 0:42) = 0:03448 second RB = SB = (1 UB ) = 0:03= (1 0:72) = 0:1071 second The server response time R = VCP U RCP U + VA RA + VB RB = 16 x 0:01923 + 7 x 0:03448 + 8 x 0:1071 = 1:406 seconds 4. Closed Queueing Networks

We discuss closed queueing networks. Closed queueing networks with single server queueing centers can be solved via a technique called mean-value analysis (MVA). This technique computes mean values of the performance metrics. It is assumed that the …xed capacity service centers and delay centers have exponentially distributed service times. In a closed queueing network with N jobs, the response time of the ith device is given by Ri (N ) = Si (1 + Qi (N 1)) where Qi (N 1) is the mean queue length at the ith device with (N 1) jobs in the network. This equation suggests that, since Qi (0) = 0, the Ri (:)’ can be computed s recursively. After computing Ri (N ) the following quantities can be computed R (N ) = X (N ) =
M X i=1

Vi Ri (N )

N (R (N ) + Z) Xi (N ) = X (N ) Vi for i = 1; 2; 3; : : : ; M Qi (N ) = Xi (N ) Ri (N ) = X (N ) Vi Ri (N ) If a device is a delay center (in…nite servers), there is no waiting before service, and therefore the response time is equal to the service time. The queue length in this case denotes the number of jobs receiving service. For such devices Ri (N ) = Si : The meanvalue algorithm is now described. 4.1. Notation. The input and output variables are:

Input variables. The following are the input variables. M =Number of queueing centers in the network, excluding terminals. N = Number of jobs in the system. Z = Think time. Si = Average service time per customer visit to queue i during observation period. Vi = Average number of visits to queue i by a customer before it leaves the system during observation period. Output variables. The output variables are: Qi = Queue length at device i (including the job in service) Ri = Response time per visit to the ith device. Ui = Utilization of queue i during observation period. Xi = Throughput of device i:

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R = System response time. X = External arrival rate, or system throughput. 4.2. Mean Value Algorithm. The MVA in pseudo-Pascal. The comments are speci…ed inside brackets in italics. Initialization: for i = 1 to M do Qi = 0 Iterations: for n = 1 to N do begin for i = 1 to M do begin Si (1 + Qi ) (Fixed capacity) Ri = Si (Delay centers) end (end i loop) (Compute system response time and system throughput) PM R = i=1 Vi Ri X = n= (Z + R) (Compute the average number of jobs at the ith device) for i = 1 to M do Qi = XVi Ri end (end n loop) (Compute device throughputs and utilizations) for i = 1 to M do begin Xi = XVi Ui = XSi Vi end (end i loop) 4.3. Example. Refer to Figure 32.8. Each user makes 10 I/O requests to disk A and 5 I/O requests to disk B. The service times per visit to disk A and disk B are 300 and 200 milliseconds, respectively. Each request takes 2 seconds of CPU time, and the user think time is 4 seconds. The number of clients is 20. Analyse this system using MVA. Solution: We …rst list the given parameters. N SA VA DCP U = 20; Z = 4 = 0:3; SB = 0:2 = 10; VB = 5 = 2

The following quantities are readily inferred. DA VCP U SCP U Initialization: Number of users: N = 0 = 3; DB = 1 = 1 + VA + VB = 16 DCP U = 0:125 = VCP U

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Device queue lengths: QCP U = QA = QB = 0 Iteration 1 : Number of users: N = 1 Device response times: RCP U RA RB = SCP U (1 + QCP U ) = 0:125 (1 + 0) = 0:125 = SA (1 + QA ) = 0:3 (1 + 0) = 0:3 = SB (1 + QB ) = 0:2 (1 + 0) = 0:2

System response time is: R = VCP U RCP U + VA RA + VB RB = 16 x 0:125 + 10 x 0:3 + 5 x 0:2 = 6 seconds System throughput: X= Device queue lengths: QCP U QA QB = XRCP U VCP U = 0:1 x 0:125 x 16 = 0:2 = XRA VA = 0:1 x 0:3 x 10 = 0:3 = XRB VB = 0:1 x 0:2 x 5 = 0:1 1 N = = 0:1 (R + Z) (6 + 4)

Iteration 2 : Number of users: N = 2 Device response times: RCP U RA RB = SCP U (1 + QCP U ) = 0:125 (1 + 0:2) = 0:15 = SA (1 + QA ) = 0:3 (1 + 0:3) = 0:39 = SB (1 + QB ) = 0:2 (1 + 0:1) = 0:22

System response time is: R = VCP U RCP U + VA RA + VB RB = 16 x 0:15 + 10 x 0:39 + 5 x 0:22 = 7:4 seconds System throughput: X= Device queue lengths: QCP U QA QB = XRCP U VCP U = 0:1754 x 0:15 x 16 = 0:421 = XRA VA = 0:1754 x 0:39 x 10 = 0:684 = XRB VB = 0:1754 x 0:22 x 5 = 0:1929 2 N = = 0:1754 (R + Z) (7:4 + 4)

The iterations can be continued for higher values N: This can be easily implemented on a spread-sheet.

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