Unformatted text preview: 0.65 = 0.887 1.13c What is the speedup of the Opteron over the Itanium 2? The answer to part b is the performance ratio of the Itanium to the Opteron. In this case, the Opteron is faster since the number is < 1.0. If we invert that number we get the ratio of execution times of the Opteron over the Itanium. 1 / 0.887 = 1.128 CSCE 4610/5610 Assignment 1 Answers 4 1.17a Given that 40% of the first application is parallelizable, how much speedup would you achieve with that application if run in isolation? Notice that this part is for the first application in isolation, the second application is not involved in this calculation. From the equation on page 47 (fifth edition): Speedup = Exec_time_old / Exec_time_new Speedup = 1 / ( (1 Fraction_enhanced) + Fraction_enhanced / Speedup_enhanced) Here the application is 40% parallelizable so the Fraction_enchanced is 0.4 Speedup = 1 / (0.6 + 0.4 / 2) = 1.25 1.17b Given that 99% of the second application is parallelizable, how much speedup would this application observe if run in isolation? Speedup = 1 / ( (1 Fraction_enhanced) + Fraction_enhanced / Spee...
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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.
 Spring '12
 kavi
 Computer Architecture

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