Tutorial 6

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TUTORIAL 6 Tutorial 6

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Tutorial 6. Tutorial 6. Overview. Predicting Performance Simulation vs Analytical Models Mathematical Modelling of Distributions Operational Variables Operational Laws Tutorial 5 Solutions are posted on eLearning Question 2 Review – Average Request time and saturation Tutorial 6. - PowerPoint PPT Presentation

Transcript of Tutorial 6

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TUTORIAL 6Tutorial 6

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Overview

Predicting Performance Simulation vs Analytical Models Mathematical Modelling of Distributions Operational Variables Operational Laws

Tutorial 5 Solutions are posted on eLearning Question 2 Review – Average Request

time and saturation Tutorial 6

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PREDICTING PERFORMANCETutorial 6

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Question 1

Given Parameters: τ = 10*60 = 600s (observation period) C0 = 1200 tasks (total requests completed) Cdisk = 2400 accesses Sdisk = 0.1 s (mean disk service time)

Required Equations: a) Ui = ? (Ui = Sdisk x Xdisk) = 0.1 x (2400/600) = 0.4 (40% utilization) b) Ddisk = Udisk / X0 = 0.4 / (1200/600) = 0.2 c) Accesses per second = (2400/600 = 4 per scond) Time between each access is therefore ¼ =0.25 s.

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Question 2

Average number of visits to each device per arriving packet.

Given: information required to work out average number of requests per second at each resource: B= 0.5*A (requests per second at B) A = 20 + 0.75*B (requests per second at A) Solving simultaneously: A = 32, B=16 For 20 requests there are 32 visits at A: 1.6 visits

at a per request. For 20 requests there are 16 visits at B: 0.8 visits

at b per request.

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Question 3 (a)

Given Parameters: N=20 Z=5s Ra=0.1s (requests of type A made 50% of the time) Rb=3s (requests of type B made 50% of the time) Xa =?

Required Equations: R= (N/X) – Z (“Interactive Response Time Law”) Ravg = (0.5)(0.1) + (0.5)(3) = 1.55 1.55 = (20/Xavg) – 5 Xavg = 20/6.55 = around 3 completions per second Xa = around 1.5 completions per second since half of

the requests are of type A.

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Question 3 (b)

What’s the effect if half of the clients submit type A request, and half submit type B requests?

Use Response time law again: For type A: Ra=(Na/Xa) – Z 0.1=(10/Xa) – 5 Xa=around 2 For type B: Rb=(Nb/Xb) – Z 3=(10/Xb) -5 Xb=1.25

More short requests can be completed per second because they are not held up by long type B requests.

Less type B requests completed because there are less clients to service them.

As an architect you can trial many system configurations until performance metrics suit your needs.

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PART B

The resource with the largest service demand / highest utilization corresponds to the bottleneck resource.

Often we need to identify the resource that represents this bottleneck and take appropriate action to remove this bottleneck: add an additional disk, or multi-core CPUs, faster disks, faster CPUs, etc.

Chapter 5 of “Performance By Design” goes through a practical database example where bottlenecks are identified and removed in an iterative fashion