E p2 1 no independent processing units therefore

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Unformatted text preview: de the block. The answer is B. The reasoning is as follows: P1 will take |K1|*T(P1)=|K1|*1= |K1| ut. P2 will take |K2|*T(P2)=2|K1|*1.5 T(P1) = 2|K1| *1.5 = 3 |K1| ut P3 will take |K3|*T(P3)=3|K1|*0.5 T(P1) = 3|K1| *0.5 = 1.5|K1| ut. So, B. As shown above, A is false, C is also false because P3 takes half the time taken by P2, D is false because the overall time taken is the time taken by the slowest copy, i.e., P2. 1. No independent processing units, therefore achieving the same speed up is harder. 2. The interconnect is shorter (typically, custom ­built) therefore latency tends to be smaller. 3. Locality effects are someFmes a feature of the physical world. For example, in a company that is spread geographically, there’s a performance benefit in keeping local the data about locals, say, keeping data about Manchester employees in Manchester. This avoids paying the transport cost into the remote parallel processor. MCQ answer is B. In SISD, there is only one M ­element, so we fetch exactly one element of either (or both) at a Fme and firing is always of one P ­element on one D ­ element. In SIMD, there are more than one M ­elements, say k, so we f...
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This document was uploaded on 02/10/2014.

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