Answer:
a) Speedup gain is 1.428 times.
b) Speedup gain is 1.81 times.
Explanation:
in order to calculate the speedup again of an application that has a 60 percent parallel component using Anklahls Law is speedup which state that:
[tex]t=\frac{1 }{(S + (1- S)/N) }[/tex]
Where S is the portion of the application that must be performed serially, and N is the number of processing cores.
(a) For N = 2 processing cores, and a 60%, then S = 40% or 0.4
Thus, the speedup is:
[tex]t = \frac{1}{(0.4 + (1-0.4)/2)} =1428671[/tex]
Speedup gain is 1.428 times.
(b) For N = 4 processing cores and a 60%, then S = 40% or 0.4
Thus, the speedup is:
[tex]t=\frac{1}{(0.4 + (1-0.4)/4)} = 1.8181[/tex]
Speedup gain is 1.81 times.
The speedup gain of an application that has a 60 percent parallel component for two processing cores is 1.428 times.
The speed up gain will be calculated thus for the application that has a 60 percent parallel component for two processing cores.
= 1/[S + (1 - S)/N]
= 1[(0.4 + (1-0.4)/2]
= 1.428
The speed up gain for four processing cores will be:
= 1/[(0.4 + (1 - 0.4)/4]
= 1.81 times
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