Question

Consider the pipelined RISC-V version with 5 stages discussed in class that does not handle data hazards (i.e., the programmer is responsible for addressing data hazards by inserting NOP instructions where necessary). Suppose that (after optimization) a typical n-instruction program requires an additional 0.3∗n NOP instructions to correctly handle data hazards.

i. Suppose that the cycle time of this pipeline without forwarding is 200ps. Suppose also that adding forwarding hardware will reduce the number of NOPs from 0.3∗n to 0.06∗n, but increase the cycle time to 250ps. What is the speedup of this new pipeline compared to the one without forwarding?
ii. Different programs will require different amounts of NOPs. How many NOPs (as a percentage of code instructions) can remain in the typical program before that program runs slower on the pipeline with forwarding?
iii. Repeat (ii); however, this time let x represent the number of NOP instructions relative to n. (In (ii), x was equal to 0.3 ) Your answer will be with respect to x.
iv. Can a program with only .075∗n NOPs (in the no-forwarding case) possibly run faster on the pipeline with forwarding? Explain why or why not.
v. At minimum, how many NOPs (as a percentage of code instructions) must a program have before it can possibly run faster on the pipeline with forwarding?

Answer 1

To calculate the speedup of a RISC-V pipeline with forwarding versus without it, we compare the total execution times incorporating NOPs. Forwarding reduces the number of NOPs but increases the cycle time. We also determine the threshold percentage of NOPs for which forwarding is beneficial.

Step-by-step explanation:

The question addresses the concept of instruction pipelining in a RISC-V processor and the effect of adding forwarding hardware on the pipeline's performance. To optimize for data hazards, NOP (No Operation) instructions may be inserted, and forwarding hardware can be used to mitigate these hazards. To compute speedup, we evaluate the pipeline's performance with and without forwarding.

1. Without forwarding, we expect (1 + 0.3)*n instructions due to NOPs inserted to handle hazards. At 200ps per cycle, the total time is (1 + 0.3)*n*200ps.
2. With forwarding, we expect (1 + 0.06)*n instructions, at a new cycle time of 250ps. Thus, the total time is (1 + 0.06)*n*250ps.
3. To find the speedup, we compare the execution times of both scenarios: Speedup = Time without forwarding / Time with forwarding.
4. For part ii and iii, we set up an inequality to determine the threshold of NOPs added at which the total execution time with forwarding becomes higher than without it.
5. To answer part iv, we plug in the given NOP rate (0.075*n) into the execution time equation with forwarding to compare it with the original system.
6. Lastly, in part v, we derive an inequality to find the minimum percentage of NOPs required for the forwarding pipeline to be beneficial.