Question

Consider a 6-stage instruction pipeline, where all stages are perfectly balanced. Assume that there is no cycle-time overhead of pipelining. When an application is executing on this 6-stage pipeline, the speedup achieved with respect to non-pipelined execution if 25% of the instructions incur 2 pipeline stall cycles is ________.

Answer 1

Data:

Instruction pipeline = 6 stage

25% of the instructions incur 2 pipeline stall cycles

Formula:

Speed up = \(\frac{{Execution\;time\left( {non\;pipeline} \right)}}{{Execution\;time\;\left( {pipeline} \right)}}\)

Time with pipeline  =1+ stall freqency×stall cycle
Calculation:

Time without pipeline = 6 stages=6 cycles

Time with pipeline = =1+.25 × 2 =1.5

Speed up = \(\frac{{6}}{{1.5}} = 4\)