Pipeline and Vector Processing Q.6

0. We have two designs D1 and D2 for a synchronous pipeline processor. D1 has 5 pipeline stages with execution times of 3 nsec, 2 nsec, 4 nsec and 3 nsec while the design D2 has 8 pipeline stages each with 2 nsec execution time. How much time can be saved using design D2 over design D1 for executing 100 instructions?

Answer
Comment
Array

Option : B
Explanation : Total execution time = (k + n – 1) * maximum clock cycle
Where k = total number of stages and n = total number of instructions
For D1 :
k = 5 and n = 100
Maximum clock cycle = 4ns
Total execution time = (5 + 100 – 1) * 4 = 416
For D2 :
k = 8 and n = 100
Each clock cycle = 2ns Total execution time = (8 + 100 – 1) * 2 = 214
Thus, time saved using D2 over D1
= 416 – 214
= 202

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