A computer spends 25% of its time accessing a hard disk. It spends 20% of the time doing floating point. The hard disk is replaced by two disks operating in parallel and the floating point unit is replaced by one four times faster. The speedup is given by
So, the speedup for the disk is Sdisk = 2/(2 × 0.75 + 1 ? 0.75) = 2/1.75 = 1.429.
The speedup for the floating?point unit is Sfloating?point = 4/(4 × 0.80 + 1 ? 0.80) = 4/1 = 4/3.4 = 1.176. The total speedup ratio is the product of the individual speedups, which is 1.429 × 1.176 = 1.681. Is this answer correct?
No. The answer is incorrect. You cannot apply the speedup formula to a problem twice and then multiply the two speedup factors.
The enhanced time is given by Disk/diskSpeedup + FP/FPspeedup + remainder = 0.25/2 + 0.20/4 + 0.55 = 0.125 + 0.050 + 0.550 = 0.725. The speedup is the reciprocal of this 1/0.725 = 1.380 (which is less than the incorrect calculation).
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