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
Explanation : For non pipelined system time required
= 2.5 + 1.5 + 2.0 + 1.5 + 2.5 = 10
for pipelined system
= Max (stage delay) + Max(Latch delay)
= 2.5 + 0.5 = 3.0
speedup = time in non-pipelined system/time in
pipelined system
= 10/3 = 3.33
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