XML CO527 Anonymous Questions and Answers Keyword Index

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Keyword reference for speedup

Question 17 (2010):

Submission reference: IN2019

Ok, so I've been checking out the past papers, and I was looking at the 2008 paper. For Question 5, the last part. The speed for when a process is via pipeline is kt+(n-1)t, but for this question it is asking to compare 2 different processors each with a different number of stages (k) is 8 or 3 but instructions (n) is 10 for both.

For the speed of the processor:
2GHz = 8t + 9t = 17t
1GHz = 3t + 9t = 12t

but the ts are different... so we would perhaps scale up? for 1 GHz now 12*2t, so:
2GHz= 17t
1GHz= 24t

Would this work and then 24/17 = 1.4 something... Or is something a bit more complicated taking into consideration more the fact that the first processor is twice as fast, but has 8 stages, and the 2nd processor is half the speed of the first, but only has 3 stages?

So perhaps something like...
(24/3)/(17/8)=
8/2.125 = 3.61

I am really unsure how to tackle this, its a tricky question and one I really wouldn't have expected given the quizzes and the practice lecture examples... Am I over-complicating or misinterpreting the question?

Any advice would be much appreciated, Anonymous ;)

Answer 17:

I think you're over-complicating this. The question asks you to compute the speed-up factor for two things, the 2 GHz 8-stage pipeline and the 1 GHz 3-stage pipeline, and then compare them.

I would start by computing the speed-up factor for both, using the given formula:

S = n.k.t / (k.t + (n-1).t)

You can obviously cancel out the ts here, giving a simpler:

S = n.k / (k+n-1)

For the first, plug in k=8 and n=10, which gives a speed-up factor of S=4.7 (roughly). For the second, plug in k=3 and n=10, which gives a speed-up factor of S=2.5. You could then state that the first architecture has a speed-up of almost twice the second. Given that the question is asking you to compare speed-up factors, and not absolute computation times, I'd probably stop at that point.

If the question wanted you to compare the execution times for the two architectures, it would state that explicitly. In that case, you would use just the bottom half of the formula, i.e. "k.t + (n-1).t", as the time required for n tasks through a k long pipeline at t time units per step. Dealing with nano-seconds (doesn't matter what order of magnitude you pick here, but that would be the most obvious), for the first architecture you plug in k=8, n=10 and t=0.5 to give 8.5 nano-seconds. For the second, plug in k=3, n=10 and t=1 to give 12 nano-seconds. To get the comparison, just divide these, which gives you the 1.4 you had. However, the question isn't asking this..

Keywords: speedup , speed-up-factor