• edited by
936 views
0 0 votes
Consider a 5 stage pipeline that allows overlapping of all instructions except branch instructions. The target of branch instructions is not available until the branch instruction is completed. Let each stage delay is 20 ns and there are 30% branch instructions.
What is the performance gain of pipeline over non-pipeline

1 Answer

0 0 votes
CPI=1+0.3*4(stall due to branch instruction)=1.12

instruction execution time in non pipeline processor=20*3=100ns

instruction execution time in non pipeline processor=1.12*20=22.4ns;

speedup=100/22.4=4.46
Position:
Show:

Related questions

0 0 votes
2 2 answers
2.1k
2.1k views
Harshit Bajpai asked Jan 14, 2019
2,094 views
A hypothetical 5 stage pipeline processor is designed in which branch is predicted at 3rd stage and each stage takes 1 cycle to compute its tasks. If f is the probability...
0 0 votes
1 1 answer
2.4k
2.4k views
Swarnava Bose asked Jul 2, 2022
2,350 views
In a pipeline the maximum ideal speed-up is 5. Let the percentage of unconditional branches in a set of typical program be 5% and that of conditional branches be 10%. If ...
1 1 vote
0 0 answers
2.2k
2.2k views
Cpt.Nemo143 asked Nov 17, 2018
2,184 views
Given a non-pipelined architecture running at 1GHz, that takes 5 cycles to finish an instruction. You want to make it pipelined with 5 stages. The increase in hardware fo...
0 0 votes
1 answers 1 answer
967
967 views
anjali007 asked Jan 23, 2019
967 views
A hypothetical 5 stage processor is designed in which branch is predicted at 3 stage and each stage takes 1 cycle to compute its task. If f is the probability of an instr...