B to make tiny reentrant we must replace all calls to

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Unformatted text preview: this problem is a straightforward application of the formula for disk access time. The average rotational latency (in ms) is surfaces ¢ 10,000 tracks ¢ 2 platter ¢ 2 platters surface disk Ì Ú ÖÓØ Ø ÓÒ 1/2 ¢ ÌÑ 2 ms 1/2 ¢ ´60 secs / 15,000 RPMµ ¢ 1000 ms/sec Ü ÖÓØ Ø ÓÒ The average transfer time is Ì Ú ØÖ Ò× Ö ´ 60 secs / 15,000 RPMµ ¢ 1/500 sectors/track ¢ 1000 ms/sec 0.008 ms Putting it all together, the total estimated access time is Ì ×× Ì Ú × · Ì Ú ÖÓØ Ø ÓÒ · Ì Ú ØÖ Ò× Ö 8 ms · 2 ms · 0.008 ms 10 ms Problem 6.4 Solution: [Pg. 298] To create a stride-1 reference pattern, the loops must be permuted so that the rightmost indices change most rapidly. 1 2 3 4 5 6 7 8 9 10 11 12 13 int sumarray3d(int a[N][N][N]) { int i, j, k, sum = 0; for (k = 0; k < N; k++) { for (i = 0; i < N; i++) { for (j = 0; j < N; j++) { sum += a[k][i][j]; } } } return sum; } B.6. THE MEMORY HIERARCHY 719 This is an important idea. Make sure you understand why this particular loop permutation results...
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This note was uploaded on 09/02/2010 for the course ELECTRICAL 360 taught by Professor Schultz during the Spring '10 term at BYU.

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