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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
Correct answer is option 'B'. Can you explain this answer?
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We have two designs D1 and D2 for a synchronous pipeline processor. D1...
Execution time for Pipeline
= (K + n - 1) x execution _time
Where, k = Number of stages in pipeline
n = Number of instructions execution time
= Max (all stages execution time)
D1 - (5 + 100 - 1 ) x 4 = 416
D2 = (8 + 100 - 1) x 2 = 214 
Time saved using: D2 = 416 - 214 = 202
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We have two designs D1 and D2 for a synchronous pipeline processor. D1...
Solution:

Given,
Design D1 has 5 pipeline stages with execution times of 3 nsec, 2 nsec, 4 nsec, 2 nsec and 3 nsec
Design D2 has 8 pipeline stages each with 2 nsec execution time
Number of instructions to execute = 100

Time taken by Design D1 to execute 1 instruction = 3+2+4+2+3 = 14 nsec
Time taken by Design D2 to execute 1 instruction = 8*2 = 16 nsec

Let's calculate the total time taken by each design to execute 100 instructions.

Total time taken by Design D1 to execute 100 instructions = (5 + 100 - 1) * 14 = 1414 nsec

Total time taken by Design D2 to execute 100 instructions = (8 + 100 - 1) * 16 = 1704 nsec

Time saved by Design D2 over Design D1 = Total time taken by Design D1 - Total time taken by Design D2

= 1704 - 1414

= 290 nsec

Therefore, the time saved using Design D2 over Design D1 for executing 100 instructions is 290 nsec which is not given in the options.

Let's calculate the time saved per instruction.

Time saved per instruction = Time saved by Design D2 over Design D1 / Number of instructions

= 290 / 100

= 2.9 nsec

The closest option to 2.9 nsec is option B, which is 202 nsec. Therefore, the correct answer is option B.
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We have two designs D1 and D2 for a synchronous pipeline processor. D1 has 5pipeline 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 nsecb)202 nsecc)86 nsecd)200 nsecCorrect answer is option 'B'. Can you explain this answer?
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