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Unformatted text preview: m = 16, s = 1. We access each of the 16 elements 16 times, and we have 2 cache misses, one for each block of 8 elements. So the total time is 256 + 2 ∗ 16 ns for the 256 accesses, for an average of 1 . 125 ns. When s is increased to 16, we access only z (1), so the total time drops to 256 + 16 ns. For m = 64, the array no longer ±ts in cache and each block that we use must be reloaded for each cycle. For s = 4, we have a cache miss for every other access to the array, so the average access time is (1 + 16 / 2) = 9 ns. The other entries are similar. 15...
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This note was uploaded on 01/21/2012 for the course MAP 3302 taught by Professor Dr.robin during the Fall '11 term at University of Florida.
- Fall '11