Test by Bikram | Computer Organization and Architecture | Test 1 | Question: 24

Question

Consider two pipelines A and B. Pipeline A has $8$ stages with uniform stage delay of $2$ns. Pipeline B has $5$ stages with uniform stage delays of $3$ns. Time saved (in ns) by pipeline A compared to pipeline B to execute $100$ instructions is _____.

Answer 1

in pipeline A, first instruction executes in 16ns (8stage*2ns). after that each 1 instruction get completed in 2ns.

so total delay is 1*16 + 99*2 = 214ns

in pipeline B, first instruction executes in 15ns =(3+3+3+3+3). after that each one instruction get completed in 3ns. (as delay of pipeline is maximum delay of any stage)

so total delay is 1*15 + 99*3 =  297 + 15 = 312ns

so time saved is 312 - 214 = 98 ns

Comments

For pipeline B, it is (3+3+3+3+3) and not (3+2+1+3+2) as once pipelining is done, stage delays are made uniform.