Question

Consider a 5-stage instruction pipeline, where each stage takes 2 ns time. Intermediate buffers are required after each stage and the buffer delay is 0.5 ns. 1000 instructions executed in this pipeline, among those 20% instructions incur 2 pipeline stall cycles. The speedup achieved using this pipeline with respect to corresponding non-pipelined execution is ___________

Answer 1

1 instruction execution time is non-pipeline = 2 + 2 + 2 + 2 + 2 = 10 nsec
1000 instruction execution time is non-pipeline = 1000 * 10 = 10,000 nsec
pipeline cycle time = max( 2, 2, 2, 2, 2) + 0.5 = 2.5 nsec
Total cycles to execute 1000 instructions in pipeline
= (5 + 1000 - 1 ) + 0.2 * 1000 * 2
= 1004 + 400
= 1404
Total pipeline time = 1404 * 2.5 = 3510
$\text{Speedup =}\frac{10,000}{3510}=2.849$