PIPELENE PROBLEM

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PIPELENE PROBLEM

Postby sukantadasgupta » Wed Nov 07, 2012 10:27 pm

In a 32 bit machine to execute an instruction the following steps are carried out: Fetch,decode,execute and store.each of which takes one clock period. In a pipelined execution of a four step task a new instruction is read and its nano second and there are 100 instruction in sequence then what is the speedup ratio of pipeline processing system over an equivalent non pipeline processing system?
sukantadasgupta
 
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Re: PIPELENE PROBLEM

Postby sukantadasgupta » Wed Nov 07, 2012 10:35 pm

Please reply with correct answer. if i wrong
it's my answer.
1 clock period = 1 cycle
1 clock period each. so tota 4 cycle.
in non pipeline each process would complete all tasks at a time. so in 4 stage (4*4)=16 cycle for one process.
for 100 process:- 100*16=1600 for non pipelene.

in pipeline (k+n-1)*d [k=stage, n = processes, d = time]
so, (4+100-1)*4 = 412
so speed up = 1600/412 = 3.88
sukantadasgupta
 
Posts: 3
Joined: Thu Apr 12, 2012 10:17 pm
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