Overview of MVA and Techniques _Anil Kumar a(Cse)

8/8/2019 Overview of MVA and Techniques _Anil Kumar a(Cse)

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OVERVIEW OF MEAN VALUE

ANALYSIS AND RELATEDTECHNIQUES

BY

ANIL KUMAR A(1RV09SCS03)

M.Tech., CSE., I SEM

RVCE, BANGALORE

Presentationon


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OUTLINE

Analysis of Op en Queueing Networks

Mean value analysis

App roximate MVA

Balanced job bounds

Problems

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Used to re p resent transaction p rocessingsystems, such as airline reservation systems,or banking systems.Transaction arrival rate is not de pendent onthe load on the com puter system.Arrivals are modeled as a Poisson p rocess witha mean arrival rate .Exact analysis of such systems



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Assum p tion: All devices in the system can bemodeled as either fixed-ca pacity servicecenters (single server with ex ponentiallydistributed service time) or delay centers(infinite servers with ex ponentially distributedservice time).



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Inputs:X = external arrival rate, system through put

Si = service time per visit to the i th device

Vi = number of visits to the i th deviceM = number of devices

Out puts:

Q i = mean number of jobs at i th deviceR i = res ponse time of the i th deviceUi = Utilization of the i th deviceN = mean number of jobs in the system



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C ont(....)Total service demands: D i = SiViDevice utilization: Ui = XDiDevice through puts: X i = XViDevice res ponse times:

Si/ (1 - Ui ) Fixed ca pacity centersR i =

Si

Delay C enters

Ui/ (1 - Ui ) Fixed ca pacity centersQ i =

Ui DelayC

enters12/19/2010 11:17 PM MVA and related Techniques 6


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Mean-Value Analysis (MVA)

Mean-value analysis (MVA) allows solvingclosed queueing networks in a manner similarto that used for o pen queueing networksIt gives the mean performance. The variancecom putation is not possible using thistechnique. Initially limit to fixed-ca pacityservice centers.



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C ont(....)

Given a closed queueing network with N jobs,the res ponse time of the i th device

R i (N) =Si[1+Q i(N- 1)]

Here, Qi( N-1) is the mean queue length at ith devicewith N-1 jobs in the network.

It assumes that the service is memory less



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C ont( .)

Given the res ponse times at individualdevices, the system res ponse time using thegeneral res ponse time law is:

The system throughput using the interactiveresponse time law is:



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C ont( .)The device through puts measured in terms of jobs per secondare:

Xi(N)= X(N) ViThe device queue lengths with N jobs in the network using

Little's law are:Q i(N) = Xi(N) R i(N) = X(N) Vi R i(N)

Res ponse time equation for delay centers is sim p ly:R i(N) = Si

Earlier equations for device through puts and queue lengthsapp ly to delay centers as well.

Q i(0)=0



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App roximate MVASchweitzer's a pp roximation:Estimate the queue lengths with N jobs and

com puting the res ponse times and

through puts. The values so com puted can beused to re-com pute the queue lengths.

Assumes that as the number of jobs in anetwork increases, the queue length at eachdevice increases p ro portionately.



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C ont ( .)



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C ont ( .)

If the new values of Q i are not close to the oldvalues continue iterating.

If they are sufficiently close, we sto p .



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Balanced Job BoundsA system without a bottleneck device is calleda balanced system.Balanced system has a better performancethan a similar unbalanced system Allows getting two sided bounds on performance

An unbalanced system's performance can

always be im p roved by re p lacing thebottleneck device with a faster device.Balanced System: Total service time demands

on all devices are equal.12/19/2010 11:17 PM MVA and related Techniques 14


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Balanced Job Bounds



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Problems34.1) A transaction p rocessing system can bemodeled by an o pen queueing network. Thetransactions arrive at a rate of 0.8transactions per second, use 1 sec of C PU time, make 20 I/O s to disk A and 4 I/O to disk B. Thus, the total no. of visits to the CPU is 25 .The disk service times are 30 and 25 msec .

Determine the average no of transaction inthe system and average res ponse time.



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Sol:Given data:arrival rate = through put of system (X) = 0.8 transaction /secdemand at C PU (D CPU )= 1 secdemand at disk A (D A )= no of I/ O S X service time for each I/ O

= 20 X 0.03 =0.6 sec

demand at disk B (D B )= 4 X 0.025 = 0.1 sec

Q i =Ui / (1- Ui) = XDi / (1- XD i)Q C PU =0.8 X 1 / (1-0.8) = 4Q B =0.8 X 0.6 / (1-0.8 X 0.6) = 0.923Q A =0.8 X 0.1/ (1-0.8X 0.1) = 0.0869average transactions in the system = i=1n Q i= 4 + 0.923 + 0.0869 = 5.009



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ii)

res ponse time R i = Si /(1 Ui )res ponse time R cpu = 0.04 /(1 0.8)= 0.2 sec

res ponse time R A = 0.03 /(1 0.8 X 0.6)

= 0.0576res ponse time R B = 0.025 /(1 0.8 X 0.1)

= 0.0271average res ponse time = R iVi

= 0.2X25 + 0.0576 X20 + 0.0271X4= 6.2604 sec



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34.4) A 2 ho p com puter network with a flowcontrol window size of n is re p resented by the

closed queueing network shown in fig .Assuming that the network is balanced in thateach com puter takes the same service time Sto p rocess a packet and that each packet makesone visit to each queue, determine thenetwork through put and res ponse time as afunction of n for n= 1 5 using MVA. Write an

exp ression for X(n) and R (n)

1 2 3

s s s

N packets



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Each packet is serviced by the 3 com puters .hence the through put is when there is one packet is

X1= 1/3S.

The packet s pent a time of S in each ho p The res ponse time

R 1 =3Ssimilarly

R 2= 4S X2= 2/4SR

3= 5S X3= 3/5SR 4= 6S X4= 4/6SR 5= 7S X5= 5/7S



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34.7) C onsider a balanced, closed queueingnetwork consisting of M fixed ca pacity servicecenters(no terminals). The service demand oneach center is D i =D/M. For this systema) Using MVA develo p an ex p ression for the

through put and res ponse time of the system as afunction of the number of jobs N in the n/w

b) Write ex p ression for balanced job bounds of thesystem through put and res ponse times for thissystem.



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Overview of MVA and Techniques _Anil Kumar a(Cse)

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