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Consider an instruction pipe containing 4 stages as follows:
FI=100NS
DI=150NS
EI=300 NS
WO=200 NS
1O instructions are exexuted in a pipe lining system. What is the speed up factor?
asked in CO & Architecture by Active (2.3k points) | 39 views
1.923 ??
please explain , they are given different answer?

Twithout pipelinig = 10*(100 + 150 + 300 + 200) = 7500ns

Tpipelining = 1*300*4 + 9*300 = 3900ns // since 1st instr will use all 4 stages and rest 9 uses only one stage delay

speedup = Twithout pipelinig/Tpipelining = 1.923

is it 2.17?

1 Answer

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Best answer

Stages (k) = 4

Cycle time for Pipeline = max(100,150,300,200) , i.e 300 ns

total #f cycles in pipeline processor = (k+n-1) = (4+10-1) = 13 cycles {n = #f instructions}

total time in pipeline = 13 * 300 =3900 ns

Cycle time for Non-Pipeline = 100+150+300+200 =750 ns

total time in non- pipeline = 10*750 =7500 ns

Speedup(s) = Execution time of non pipeline/Execution time of pipeline

s=7500/3900 = 1.923

 

answered ago by Loyal (4.1k points)
selected ago by

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