2 votes 2 votes 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? $3.88$ $1.88$ $3.68$ $2.723$ GATE tbb-mockgate-1 pipelining co-and-architecture + – Bikram asked Jan 16, 2017 • retagged Jan 9, 2020 by Arjun Bikram 618 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 2 votes 2 votes 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 Bikram answered Jan 16, 2017 • selected Jan 22, 2017 by Bikram Bikram comment Share Follow See all 2 Comments See all 2 2 Comments reply VASUGANESHAN commented Dec 28, 2017 reply Follow Share Sir, In the question 'execution begins with each new clock period' So I considered, there must be some stalls. Where am I wrong... 0 votes 0 votes Psy Duck commented Sep 8, 2022 reply Follow Share everywhere 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes 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 ] Srividyaketharaju answered Feb 4, 2020 Srividyaketharaju comment Share Follow See all 0 reply Please log in or register to add a comment.