Pipelining

Question

A 5 stage pipeline system is in operation with clock cycle of n ns. If the clock per instruction CPI for non-pipelined system is 5,and Instruction per clock for pipeline is 5,and pipeline efficiency is 70% what is the speed up factor?

 

Please explain the Soultion and concept briefly i am little bit confused.

Comments

but bro efficiency is 70% only....given

so speedup=stages*efficiency....that is 5*0.7=3.5

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@ Na462 since they have given for pipline clock cycle time is n ns.if it is uniform that is on each stage we require n ns of clock cycle.so for non pipeline clock cycle time must be 5n ns.They must have given wrong.

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Yea thats what i am saying thnku Sir

So sir answer should be

Speedup = (I * 5 * 5*n) / (I * 0.20 * n) = 125

so Speedup factor = 125*0.70 = 87.5

Am i right sir?

Answer 1

 time taken by pipelined processor = n sec for 1 clock cycle (ideally we are considering that instruction are executed in 1 clock cycle)  and 5 instruction are executed in one cpi (mentioned at last line of question)

means time taken to execute 5 instruction in pipelined processor is n sec

  and for sequential processor = 5 x 5 x n ( 5 x n  represent time per instruction in non pipelined processor multiplied by 5 instruction)

speed up = 25 n / n => 25

since efficiency is 70% speed up factor become 25 x (70/100) => 17.5

Comments

Time taken to Non pipeline processor = 5* n sec

Time taken to Pipeline processor = n sec.

Speed up = non pipeline / pipeline

   Speed up =  5

Speed up factor = 0.70*5 = 3.5

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@sachin486 plz explain why u take 

Time taken to Non​Non pipeline processor  = 5*5*n

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i dont know what i was thinking that i didnt give proper explanation but i redited the answer u can recheck it again and ask me if still have doubt