Question

Consider the multicore processor with 6 heterogeneous cores labelled C1, C2, C3, C4, C5, and C6. Assume cores C1 and C6 have the same speed. C2 runs twice as fast as C1 and core C3 runs 3 times faster than C1, and C4 and C5 are four times faster than C1. Assume all six cores start executing the application CUBE-X at the time and no cache misses are encountered in all core operations. Suppose the CUBE-X application computes the cube of each element in an array of 1024 elements. Assume 1 unit of time for C1 (or C6) to compute the cube of an element. Given the following division of labour in six cores:

C1 64 elements

C2 128 elements

C3 64 elements

C4 512 elements

C5 128 elements

C6 128 elements

Compute the total execution time (in time units) for the six-core processor to compute the cube of 1024 elements in parallel. Remember elements may finish before others, but the time to complete should be defined by the last core to complete.

Answer 1

As given c1 take 1unit of time to compute the cube of an element

Total time taken by to compute cube of 64 element =64

Speed of c1=c6

Total time takenyby c6 to compute cube of 128 element =128

C2 run twice of c1

Mean c2 take 1/2 unit of time to compute cubic of an element

Total time required for c2=1/2*128=64

C3 runs thrice of c1

C3 required time=(1/3)*64=64/3

C4 and c5 run 4times of c1

So c4 and c5 take 1/4 unit of time to compute cubic of an element

Total time required by c4=(1/4)*512=128

Total time required by c5=(1/4)*128=32

So total execution time=128