Disk average the mean service times are for the cpu
Performance Analysis
Peter Harrison & Giuliano Casale
▶ Lectures: Wednesdays 9:00–11:00, in 144; Tutorials: Wednesdays 11:00–12:00, in 144.
▶ 18 lectures, 9 tutorials, 2 pieces of assessed coursework.
Example 1: A simple transaction processing (TP) server
A transaction processing (TP) system accepts and processes a
service rate
Q: If both the arrival rate and service rate are doubled, what
▶ The arrival rate is 15tps
Example 3: A simple multiprocessor TP system
Consider our TP system but this time with multiple transaction processors
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▶ The arrival rate is 16.5 tps
▶ The mean service time per transaction is 58.37ms
Q: By how much is the system response time reduced by adding one processor?
q
∝1
γ
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Example 5: A simple computer model
Consider an open uniprocessor CPU system with just disks
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▶ Each submitted job makes 121 visits to the CPU, 70 to disk 1 and 50 to disk 2 on average
▶ The mean service times are 5ms for the CPU, 30ms for disk 1 and 37ms for disk 2Q: What is the effect of replacing the CPU with one twice the speed?
∝1
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Example 7: A multiprogramming system with virtual
▶ Each job page faults at a rate determined by the following lifetime function:
Life-time function example
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Q: What number of batch jobs keeps the system throughput at its maximum and at what point does thrashing occur?
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Disk 1
Introduction
Computer systems are
We also see these characteristics in queues of customers in a bank
or supermarket, or prices on the stock exchange.
| {Xt ∈ Ω|t ∈ T}, each defined on some sample space Ω (the same for each) for a parameter space T. |
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▶ T and Ω may be either discrete or continuous
▶ T is normally regarded as time
The Poisson process is a renewal process with renewal period
(interarrival time) having cumulative distribution function F and
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Memoryless property of the (negative) exponential distribution
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Proof.
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▶ e.g. when you get to a bus stop, how long will you have to wait for the next bus?
▶ If the renewal process is Poisson, R has the same distribution as S by the memoryless property
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Therefore
▶ From it we can derive the distribution function of the
interarrival times (i.e. negative exponential) and the Poisson distribution for Nt (the number of arrivals in time t)
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Derivation of the interarrival time distribution
| P(S > t + h) = P = P(S > t)P(no arrival in (t, t + h])�(S > t) ∧ (no arrival in (t, t + h])� |
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by the memoryless property. Let G(t) = P(S > t). Then: G(t + h) = G(t)P(no arrival in (t, t + h]) = (1 − hλ)G(t) + o(h)
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