Computer Organization: 1.5 4] Consider three different processors P1, P2, and P3
ID: 3748873 • Letter: C
Question
Computer Organization:
1.5 4] Consider three different processors P1, P2, and P3 executing the same instruction set. P1 has a 3 GHz clock rate and a CPI of 1.5. P2 has a 2.5 GHz clock rate and a CPI of 1.0. P3 has a 4.0 GHz clock rate and has a CPI of 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 and the number of instructions c. We are trying to reduce the execution time by 30% but this leads to an increase of 20% in the CPI, what clock rate should we have to get this time reduction?Explanation / Answer
1.5) consider three different processors P1,P2, and P3 executing the same instruction set. P1 has a 3 GHz clock rate and a CPI of 1.5. P2 has a 2.5 GHz clock rate and a CPI of 1.0. P3 has a 4.0 GHz clock rate and has a CPI of 2.2.
a) Which processor has the highest performance expressed in instructions per second?
Solution)
Actual Formula for Performance is:
Performance = Instructions / sec
Process
Clock
CPI
Performance
P1
3 x 10^9
1.5
3 x 10^9 / 1.5 = 2 x 10^9
P2
2.5 x 10^9
1.0
2.5 x 10^9 / 1.0 = 2.5 x 10^9
P3
4 x 10^9
2.2
4 x 10^9 / 2.2 = 1.8 x 10^9
Therefore, P2 has the highest performance
b) If the processors each execute a program in 10 seconds, find the number of cycles and the number of instructions.
Solution)
Process
Clock
Seconds
Cycles
P1
3 x 10^9
10
3 x 10^9 x 10 = 30 x 10^9
P2
2.5 x 10^9
10
2.5 x 10^9 x 10 = 25 x 10^9
P3
4 x 10^9
10
4 x 10^9 x 10 = 40 x 10^9
Process
Clock
Seconds
CPI
Instructions
P1
3 x 10^9
10
1.5
2 x 10^9 * 10 / 1.5 = 2 x 10^10
P2
2.5 x 10^9
10
1.0
2.5 x 10^9 * 10 / 2.5 = 2.5 x 10^10
P3
4 x 10^9
10
2.2
4 x 10^9 * 10 / 2.2 = 1.82 x 10^10
c) We are trying to reduce the execution time by 30% but this leads to an increase of 20% in the CPI. What clock rate should we have to get this time reduction.
Solution)
Formula for Execution Time = (No. of Instructions * CPI) / Clock Rage
So, If we want to reduce the execution time by 30% and CPI increases by 20% then,
Execution Time * 0.7 = (No. of Instructions * CPI * 1.2) / new clock rate
New clock rate = Clock Rate * 1.2 / 0.7 = 1.71 * Clock Rate
Process
Clock Rate
P1
3 GHz * 1.71 = 5.13 GHz
P2
2.5 GHz * 1.71 = 4.27 GHz
P3
4 GHz * 1.71 = 6.84 GHz
Process
Clock
CPI
Performance
P1
3 x 10^9
1.5
3 x 10^9 / 1.5 = 2 x 10^9
P2
2.5 x 10^9
1.0
2.5 x 10^9 / 1.0 = 2.5 x 10^9
P3
4 x 10^9
2.2
4 x 10^9 / 2.2 = 1.8 x 10^9
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