Q3. (14 Marks) Consider three different processors P1, P2, and P3 executing the
ID: 3912468 • Letter: Q
Question
Q3. (14 Marks) Consider three different processors P1, P2, and P3 executing the same Instruction set with following clock rate and CPI. CPI 1.5 Processors P1 P2 P3 Clock rate 3 GHz 2.5 GHz 4 GHz 2.2 a. Which processor has the highest performance expressed in instructions per second? b. If the processors each execute a program in 10 seconds, find the number of cycles c, we are trying to reduce the execution time by 30% but this leads to an increase (5 marks) and the number of instructions. (4 marks) of 20% in the CPI, what clock rate should we have to get this time reduction"(5 marks)Explanation / Answer
a) P1 has CPI -1.5 freq = 3 Ghz
Number of instructions per sec = 3 * 10^9/1.5 = 0.2 * 10^9
P2 has CPI -1.5 freq = 2.5 Ghz
Number of instructions per sec = 2.5 * 10^9/1 = 2.5 * 10^9
P3 has CPI -2.2 freq = 4 Ghz
Number of instructions per sec = 4 * 10^9/2.2 = 1.8 * 10^9
P2 has the highest performance
b) P1
t = 10sec
Number of instructions = 0.2 * 10^9 * 10 = 2.0 * 10^9
Number of clocks = 2.0 * 10^9 * 1.5 = 3.0 * 10^9
P2
t = 10sec
Number of instructions = 2.5 * 10^9 * 10 = 2.5 * 10^10
Number of clocks = 2.5 * 10^10 * 1 = 2.5 * 10^10
P3
t = 10sec
Number of instructions = 1.8 * 10^9 * 10 = 1.8 * 10^10
Number of clocks = 1.8 * 10^10 * 2.2 = 4.0 * 10^10
c) Execution time to be reduced by 30%
So 10 sec is now 7 sec
But the CPI increases by 20%
P1
CP1 = 1.5 + 0.3 * 1.5 = 1.5 * 1.3 = 1.95
Freq = x
Number of instructions = 2.0 * 10^9
Number of clocks = 2.0 * 10^9 * 1.95
Total time = 2.0 * 10^9 * 1.95 * (1/x) = 7
= (2.0 * 10^9*1.95)/7 = 0.55 GHz
P2
CP1 = 1 + 0.3 * 1 = 1.3
Freq = x
Number of instructions = 2.5 * 10^10
Number of clocks = 2.5 * 10^10 * 1.3
Total time = 2.5 * 10^10 * 1.3 * (1/x) = 7
= (2.5 * 10^10 * 1.3)/7 = 4.64 GHz
P3
CP1 = 2.2 + 0.3 * 2.2 = 1.3 * 2.2 = 2.86
Freq = x
Number of instructions = 4.0 * 10^10
Number of clocks = 4.0 * 10^10 * 2.86
Total time = 4.0 * 10^10 * 2.86 * (1/x) = 7
= (4.0 * 10^10 * 2.86)/7 = 16.34 GHz
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