2B-Parallel Min on a Hypercube

2B-Parallel Min on a Hypercube - Processor P 1 sends its...

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Parallel Min on a Hypercube CDA 5110/COP 4520
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The problem • Given one million integers, find the smallest. • Do this in parallel on a hypercube with 8 processors.
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The specifications • Assume each processor has 125,000 integers. • Clock: 2 Ghz • Integer compare and exchange: 8 cycles • Send Latency: 4 μsec • Receive Latency: 1 μsec
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Solution: Step 1 • Each processor finds its local min – 500 μsec
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Solution: Step 2 • Processors P 4 , P 5 , P 6 , P 7 send their smallest to P 0 , P 1 , P 2 , P 3 , respectively. – 5 μsec • P 0 , P 1 , P 2 , P 3 compare against their values and retain the smaller. – 4 nsec
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Solution: Step 3 • Processors P 2 , P 3 send their smallest to P 0 , P 1 respectively. – 5 μsec • P 0 , P 1 compare against their values and retain the smaller. – 4 nsec
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Solution: Step 4
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Unformatted text preview: Processor P 1 sends its value to P . 5 sec P compares against its min value and retains the smaller. 4 nsec Analysis Step 1: 500 sec Steps 2-4: 3 * 5.004 sec = 15.012 sec Total time, T 8 = 515.012 sec Sequential Time, T 1 = 4000 sec Speedup = T 1 / T 8 7.8 Analysis with 10,000 integers Step 1: 5 sec Steps 2-4: 3 * 5.004 sec = 15.012 sec Total time, T 8 = 20.012 sec Sequential Time, T 1 = 40 sec Speedup = T 1 / T 8 2 not too amazing Conclusion Communication overhead is important, knowledge of the architecture and matching algorithmic granularity with machine parameters are important....
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This note was uploaded on 05/12/2010 for the course CS CDA 5110- taught by Professor Deo during the Spring '10 term at University of Central Florida.

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2B-Parallel Min on a Hypercube - Processor P 1 sends its...

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