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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 it's execution begins with each new clock period. If each clock period is of $20$ $nanosecond$ and there are $100$ instructions in sequence then what is the speedup ratio of pipeline processing system over an equivalent non-pipeline processing system?

  1. $3.88$
  2. $1.88$
  3. $3.68$
  4. $2.723$
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Pipeline  system processing is (K + n - l) tp
K = 4, n = 100, tp = 20 given
 = (4+ 100 -1)20 = 2060ns ,
 speed ratio = 8000 / 2060 = 3.88
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Tp=20ns, n=100, k=4

ETnonpipelined = k*n*tp = 4*100*20 ==> 8000

ETpipelined = (k+n-1)tp = (4+100-1)20 = 2060

S = Performance of pipelined processor /
        Performance of Non-pipelined processor

speedup =  ETnonpipelined/ETpipelined = 8000/6920 = 3.88     [performance = 1/Execution time ]

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