GATE | GATE-CS-2015 (Set 1) | Question 65
Consider a non-pipelined processor with a clock rate of 2.5 gigahertz and average cycles per instruction of four. The same processor is upgraded to a pipelined processor with five stages; but due to the internal pipeline delay, the clock speed is reduced to 2 gigahertz. Assume that there are no stalls in the pipeline. The speed up achieved in this pipelined processor is __________.
Speedup = ExecutionTimeOld / ExecutionTimeNew ExecutionTimeOld = CPIOld * CycleTimeOld [Here CPI is Cycles Per Instruction] = CPIOld * CycleTimeOld = 4 * 1/2.5 Nanoseconds = 1.6 ns Since there are no stalls, CPInew can be assumed 1 on average. ExecutionTimeNew = CPInew * CycleTimenew = 1 * 1/2 = 0.5 Speedup = 1.6 / 0.5 = 3.2
Refer http://www.cs.berkeley.edu/~pattrsn/252F96/Lecture2a.pdf for more information on this topic.
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