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This is self study, but not homework. I am reviewing some slides I found online and have come across the following question.


Question:

If the latency of integer multiply is $3$ and the cycles/issue is $1$ then

  1. How fast can 10 independent int mults be executed? $$t_1 = a_1*b_1 \quad t_2 = a_2*b_2 \quad t_3 = a_3*b_3 \quad \cdots \quad t_{10} = a_{10}*b_{10}$$
  2. How fast can $10$ sequentially dependent int mults be executed? $$t_1 = a_1*b_1 \quad t_2 = t_1*b_2 \quad t_3 = t_2*b_3 \quad \cdots\quad t_{10} = t_9*b_{10}$$

Attempt:

Obviously the sequentially dependent case will take longer, because each multiply must wait for the previous multiply to finish. I'm not sure exactly how to interpret latency and cycles/issue in this context. My attempt.

  1. Each multiply takes $3$ cycles, but we can start a new multiply each cycle (can we?), so we need $3 + 10 - 1 = 12$ cycles?

  2. Naively, this would take $30$ cycles if we wait for the previous one to finish. It seems that we could do better though. For instance

    • $t_1$ first
    • $b_2*b_3$
    • wait a cycle
    • $t_2 = t_1* b_2$
    • $t_3 = t_1*(b_2*b_3)$

So the first $3$ mutliplys can be done in $7$ cycles instead of $9$. I think I just need to see a problem worked out and I'll be able to pick up whats going on.

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  • $\begingroup$ cs.stackexchange.com/q/80859/755 $\endgroup$ – D.W. Mar 30 '20 at 5:52
  • $\begingroup$ @D.W. Thank you, that is helpful. What is the relationship between throughput and cycles/issue? $\endgroup$ – knrumsey Mar 30 '20 at 7:10

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