hw1-solutions

You cannot use the direct execution times in the

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Unformatted text preview: dup_enhanced) Here the application is 40% parallelizable so the Fraction_enchanced is 0.4 Speedup = 1 / (0.01 + 0.99 / 2) = 1.98 1.17c Given that 40% of the first application is parallelizable, how much overall system speedup would you observe if you parallelized it? Here both programs must be considered. You are going to enhance 40% of the first application that uses 80% of the resources. So the net fraction being enhanced is 0.4 * 0.8. The fraction not being enhanced is all of the second application (1.0 * 0.2) plus 60% of the first application (.6 * .8 ). Speedup = 1 / ( (1.0 * 0.2) + (0.6 * 0.8) + (0.4 * 0.8) / 2) = 1.19 1.17d Given that 99% of the second application is parallelizable, how much overall system speedup would you observe if you parallelized it? Again, both programs must be considered. The fraction being enhanced is 99% of the second application that is 20% of the total, or .99 * .2. Speedup = 1 / ( (1 (.99 * .2)) + (.99 * .2) / 2) = 1.11 Bonus: What if you parallelize both applications? The enhanced fraction will be .8 * .4 for the first application plus .2 * .99 for the second application. CSCE 4610/5610 Assignment 1 Answers 5 Speedup = 1 / ( (1 (.8 * .4 + .2 * .99)) + (.8 * .4 + .2 * .99) / 2) = 1.35 CSCE 4610/5610 Assignment 1 Answers 6...
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This note was uploaded on 05/21/2012 for the course CSCE 4610 taught by Professor Kavi during the Spring '12 term at North Texas.

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