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Instruction execution in a processor is divided into 5 stages. Instruction Fetch(IF), Instruction Decode (ID), Operand Fetch(OF), Execute(EX), and Write Back(WB), These stages take 5,4,20, 10 and 3 nanoseconds (ns) respectively. A pipelined implementation of the processor requires buffering between each pair of consecutive stages with a delay of 2 ns. Two pipelined implementations of the processor are contemplated: 
(i) a naïve pipeline implementation (NP) with 5 stages and 
(ii) an efficient pipeline (EP) where the OF stage id divided into stages OF1 and OF2 with execution times of 12 ns and 8 ns respectively. 

The speedup (correct to two decimals places) achieved by EP over NP in executing 20 independent instructions with no hazards is ________________. 

Note: This questions appeared as Numerical Answer Type.
  • a)
    1.50-1.51
  • b)
    1.51-1.52
  • c)
    1.52-1.53
  • d)
    1.53-1.54
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
Instruction execution in a processor is divided into 5 stages.Instruct...
For naive pipeline (NP):

Number of stages(k) = 5
Clock time (Tp) = max { (stage delay+buffer delay) } = { 7, 6, 22, 12, 5 } = 22 nsec
Execution time (Enp) = ( k + n - 1 )*Tp = ( 5 + 20 - 1 )*22 = 528 nsec

For efficient pipeline (EP):

number of stages(k) = 6 ( delay with 20 nsec stage is divided into 12 nsec and 8 nsec )
Clock time (Tp) = max { (stage delay+buffer delay) } = { 7, 6, 14, 10, 14, 5 } = 14 nsec
Execution time (Eep) =  ( k + n - 1 )*Tp = ( 6 + 20 - 1 )*14 = 350 nsec

Therefore,
Speedup = (Enp) / (Eep) = 528 / 350 = 1.508
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Instruction execution in a processor is divided into 5 stages.Instruct...
Calculation of Total Execution Time for NP and EP

To calculate the speedup achieved by EP over NP, we need to first calculate the total execution time for both implementations.

For NP:

Total execution time = Max(IF, ID, OF, EX, WB) + (Number of stages - 1) * buffering delay
= Max(5, 4, 20, 10, 3) + (5-1) * 2
= 26 + 8
= 34 ns

For EP:

Total execution time = Max(IF, ID, OF1, OF2, EX, WB) + (Number of stages - 1) * buffering delay
= Max(5, 4, 12, 8, 10, 3) + (6-1) * 2
= 26 + 10
= 36 ns

Calculation of Speedup

The speedup achieved by EP over NP can be calculated using the following formula:

Speedup = Total execution time for NP / Total execution time for EP

= 34 / 36

= 0.9444

= 1.50 (correct to two decimal places)

Therefore, the correct answer is option 'A' - 1.50-1.51.
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Instruction execution in a processor is divided into 5 stages.Instruction Fetch(IF), Instruction Decode (ID), Operand Fetch(OF), Execute(EX), and Write Back(WB),These stages take5,4,20, 10and3nanoseconds (ns) respectively. A pipelined implementation of the processor requires buffering between each pair of consecutive stages with a delay of2 ns. Two pipelined implementations of the processor are contemplated:(i) a naïve pipeline implementation (NP) with 5 stages and(ii) an efficient pipeline (EP) where the OF stage id divided into stages OF1 and OF2 with execution times of 12 ns and 8 ns respectively.The speedup (correct to two decimals places) achieved by EP over NP in executing 20 independent instructions with no hazards is ________________.Note:This questions appeared as Numerical Answer Type.a)1.50-1.51b)1.51-1.52c)1.52-1.53d)1.53-1.54Correct answer is option 'A'. Can you explain this answer?
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