GATE CSE 2014 Set 1 | Question: 43

Question

Consider a $6$-stage instruction pipeline, where all stages are perfectly balanced. Assume that there is no cycle-time overhead of pipelining. When an application is executing on this $6$-stage pipeline, the speedup achieved with respect to non-pipelined execution if $25$% of the instructions incur $2$ pipeline stall cycles is ____________

Comments

@vaishnavi30

 “If the stages are perfectly balanced, then the time per instruction on the pipelined machine is equal to Time per instruction on non pipelined machine”

This is true only for first instruction . 

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@abir_banerjee can u share a referance where it’s written that it’s true only for first instruction ?

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Assume that there is no cycle-time overhead of pipelining.

This means that we are assuming that introducing a pipeline into our currently non-pipelined system keeps the clock period unaffected, right?

Answer 1

Assume we are doing it for 100 instructions

Non pipelined instructions

every instruction will take 6 cycle’s each so a total of  6*100 =600

 

Pipelined

25% has 2 stall cycles when 2 stall cycles they will have a cpI=3

Stall cyle is the part of the cycle where no operation is performed

 

75* of instructions when done on pipeline will have a cpi =1

 

so total cycles for pipeline = 25*3 + 75*1 =150 cycles for 100 instructions

 

speedup achieved = cycles for non pipeline/cycles for pipeline =600/150   =4