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hw3_standard_answer

hw3_standard_answer - CSE721 Homework 3 Due on Tuesday...

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CSE721 Homework 3 Due on Tuesday, 1/18/2011 Thanks to Xin Huo, S M Faisal 1
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Problem 1 (5.3) 1. Compute the degree of concurrency. Degree of concurrency is the maximum number of tasks that can be executed in parallel. a) 2 n 1 b) 2 n 1 c) n d) n 2. Compute the maximum possible speedup if an unlimited number of processing elements is available. a) The sequential time T s = 2 n 1. The Parallel run time T p = n . Speedup S = (2 n 1) /n b) The sequential time T s = 2 n 1. The Parallel run time T p = n . Speedup S = (2 n 1) /n c) The sequential time T s = n 2 . The Parallel run time T p = 2 n 1. Speedup S = n 2 / (2 n 1) c) The sequential time T s = n ( n + 1) / 2. The Parallel run time T p = n . Speedup S = ( n + 1) / 2 3. Compare the values of speedup, e ciency, and overhead function if the number of processing elements is (i) the same as the degree of concurrency and (ii) equal to half of the degree of concurrency. (i) processor equals to the degree of concurrency. 1) Number of processors: P a = 2 n 1 P b = 2 n 1 P c = n P d = n 2) Speedup: S a = (2 n 1) /n S b = (2 n 1) /n S c = n 2 / (2 n 1) S d = ( n + 1) / 2 3) E ciency: E a = S a /P a = (2 n 1) /n 2 n 1 = 2 n 1 n 2 n 1 E b = S b /P b = (2 n 1) /n 2 n 1 = 2 n 1 n 2 n 1 E c = S c /P c = n 2 / (2 n 1) n = n/ (2 n 1) E d = S d /P d = ( n + 1) / 2 n 1
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4) Overhead: T o ( a ) = 2 n 1 × n 2 n + 1 T o ( b ) = 2 n 1 × n 2 n + 1 T o ( c ) = 2 n 2 n n 2 = n 2 n T o ( d ) = n 2 n ( n + 1) / 2 = ( n 2 n ) / 2 (ii) processor equals to half of degree of concurrency. 1) Number of processors: P a = 2 n 2 P b = 2 n 2 P c = n/ 2 P d = n/ 2 2) Speedup: Parallel Time for these four graphs T p ( a ) = n + 1 T p ( b ) = n + 1 For graph (c), the execution order is from the left top matrix to the right bottom. In the first n/ 2 and last n/ 2 times of execution, the degree of concurrency is increased from 1 to n/ 2. However, in the center of the matrix, there are enough data can be parallelized in all the processors. Thus, the run time of two ends is n/ 2 × 2 = n . The left data in the center is n 2 (1+ n/ 2) n/ 2 2 × 2 = 3 n 2 2 n 4 . Thus, T p ( c ) = n + 3 n 2 2 n 4 / ( n 2 ) = 3 n 2 2 + n = 5 n 2 2 The method for computing parallel time in (d) is very similar to graph (c). For the top n/ 2 rows, the run time is n/ 2. The the left data is
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