Speedup = 1 / ((1-f) + f/s)
= 1 / ((1-0.6) + 0.6/2)
= 1 / (0.4 + 0.3)
= 1.43
Given that 95% of the second application is parallelizable, how much speedup would this application observe if run in isolation?
Speedup = 1 / ((1-f) + f/s)
= 1 / ((1-095) + 0.95/2)
= 1 / (0.05 + 0.475)
= 1.90
Given that 60% of the first application is parallelizable, how much overall system speedup would you observe if you parallelized it, but not the second application?
System Speedup = 1 / ((1-f) + f/s)
= 1 / (0.25 + 0.75/1.43)
= 1 / (0.25 + 0.52)
= 1.29
How much overall system speedup would you achieve if you parallelized both applications, given the information in parts (a) and (b)?
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