Consider two different implementations of the same instruction set architecture (ISA). The instructions can be divided into four
classes according to their CPI (class A, B, C, and D). P1 with a clock rate of 2.5 GHz have CPIs of 1, 2, 3, and 3 for each class, respectively. P2 with a clock rate of 3 GHz and CPIs of 2, 2, 2, and 2 for each class, respectively. Given a program with a dynamic instruction count of 1,000,000 instructions divided into classes as follows: 10% class A, 20% class B, 50% class C, and 20% class D.which implementation is faster?
a. What is the global CPI for each implementation?
b. Find the clock cycles required in both cases.
a. What is the global CPI for each implementation?
b. Find the clock cycles required in both cases.
1 answer:
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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:
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:
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:
Speedup gain is 1.81 times.