Question

Consider a version of the MIPS 5-stage pipeline processor 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 0.4*n nop instructions to correctly handle data hazards. Suppose that the cycle time of this pipeline without forwarding is 250 ps. Suppose that adding forwarding hardware decrease the number of NOPs but will increase the cycle time to 300 ps.

(a)Suppose that the pipeline with forwarding requires an additional 0.05*n NOP instructions rather than the additional 0.4*n NOP instructions without forwarding. What is the speed up with forwarding over without forwarding?

Answer 1

The speed up with forwarding over without forwarding is 83.33%.

Step-by-step explanation:

The speed up with forwarding over without forwarding can be calculated using the formula:

Speed up = (Cycle time without forwarding) / (Cycle time with forwarding)

Given that the cycle time without forwarding is 250 ps and the cycle time with forwarding is 300 ps, we can substitute these values into the formula:

Speed up = (250 ps) / (300 ps) = 0.8333

Therefore, the speed up with forwarding over without forwarding is approximately 0.8333, or 83.33%.