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The last four requests in the sequence take 15 45 45

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Unformatted text preview: block was invalidated by processor 1. It will thus take 5 + 40 cycles = 45 cycles. R3/Y will hit in the private cache of processor 3 but because the tag directory is not duplicated there will be a delay of 5 cycles. This does not affect the total time though as it is done in parallel with the requests from the other processors. Finally R4/X will miss because of the invalidation by processor 1 and will add another 45 cycles. The last four requests in the sequence take 15 + 45 + 45 cycles = 105 cycles. In total, it takes 180 + 105 cycles = 285 cycles until the entire sequence of requests is completed. b) Processor 1 will be done after W1/X is completed: 180 + 15 cycles = 195 cycles. Processor 2 will be done after the second R2/X is completed: 180 + 15 + 45 cycles = 240 cycles. Processor 3 will be done after the second R3/Y is completed. This request can be carried out right after the first R3/Y and will only take a single cycle as the tag directory is duplicated. Hence, 45 + 45 + 45 + 1 cycles = 136 cycles. Processor 4 will be done 45 cycles after processor 2. Hence, it is done after 285 cycles. Problem 5.11: Processor 1 Processor 2 Processor 3 Miss type 1 RA cold 2 RB cold 3 RC cold 4 WA 5 RC false- sharing 6 RA true- sharing 7 WB 8 RA true- sharing 9 RB a) It is obvious...
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This document was uploaded on 01/18/2014.

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