599 views
0 votes
0 votes
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?

1 Answer

Best answer
1 votes
1 votes

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

selected by

Related questions

0 votes
0 votes
1 answer
1