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# GATE | Gate IT 2005 | Question 44

• Last Updated : 19 Nov, 2018

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, 2 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?

(A) 214 nsec
(B) 202 nsec
(C) 86 nsec
(D) – 200 nsec

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

Thus, option (B) is correct.

Please comment below if you find anything wrong in the above post.

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