Relying on a subset of the performance equation (CPU execution time is the only
ID: 3885443 • Letter: R
Question
Relying on a subset of the performance equation (CPU execution time is the only valid comparison data), one could select the lower performing processor while they are requiring the higher performing design. For this question, here are the two processors. P1 has a clock rate of 4.3 GHz, average CPI of 1.75, and requires execution of 4.5 times 10^9 instructions. P2 has a clock rate of 3.0 GHz, and average CPI of 1.25 and requires 3.0 times 10^9 instructions. (Question derived from 1.12 in "Computer Organization and Design") Based on CPU execution time, the only valid comparison, which processor is faster? (Answer format is P1 or P2) __________ One fallacy in comparing processors is to consider the computer with the highest clock frequency as having the greatest performance. For this question, does the higher clock rate equate to higher performance/less execution time? (Answer format is true or false) ________ Another fallacy in comparing processors is to use MIPs to compare the performance of two different processors and consider the processor with the higher MIPs rating as the higher performing processor. What is the effective MIPs rating of P1 based on the above? (Answer format is round to an integer) ___________ What is the effective MIPs rating of P2 based on the above? (Answer format is round to an integer) __________ For these processors, does the higher MIPs rating equate to higher performance? (Answer format is true or false) _________ Another fallacy is to consider that the processor executing the largest number of instructions will require greater CPU time. Consider that processor P1 is executing 2 times 10^9 instructions and that the CPI and clock rate of P1 and P2 remain the same specified in this question. Determine the number of instructions that P2 can execute in the time that P1 executes 2 times 10^9 instructions. (Answer format is X.XX times 10^y) __________Explanation / Answer
For processor P1:
The clock rate(R) = 4.3GHz
Instruction count(I)= 4.5 * 109
The average clock per Instruction (CPI) = 1.75
So, execution time = (1/R)*I*CPI= =(1/4.3) * 10-9 * 4.5 * 109 * 1.75 sec = 1.83 sec
MIPs = Clock Rate / (average CPI * 106) = (4.3 * 109 / 1.75) * 10-6 = 2457
For processor P2:
The clock rate(R) = 3 GHz
Instruction count(I)= 3 * 109
The average clock per Instruction (CPI) = 1.25
So, execution time = (1/R)*I*CPI= =(1/3) * 10-9 * 3 * 109 * 1.25 sec = 1.25 sec
MIPs = Clock Rate / (average CPI * 106) = (3 * 109 / 1.25) * 10-6 = 2400
To compare the execution time for two processors the Instruction count should be same; If the instruction count is same for both processor P1 and P2 say it is I; then
Execution time for P1 = (1/4.3) * 10-9 * I * 1.75 sec = 0.406 * 10-9 * I sec
Execution time for P2 = (1/3) * 10-9 * I * 1.25 sec = 0.416 * 10-9 * I sec
Hence P1 is faster than P2.
Based on CPU execution time, the only valid comparison, which processor is faster?
P1
Explanation: Execution time of P1 is less than that of P2.
For this question, does the higher clock rate equate to higher performance / less execution time?
True
Explanation: Execution time of P1 is less than that of P2 and clock rate of P1 is higher than P2.
What is the effective MIPs rating of P1 based on above?
2457
Explanation: MIPs = Clock rate/(CPI * 106).
What is the effective MIPs rating of P2 based on above?
2400
Explanation: MIPs = Clock rate/(CPI * 106).
For these processors, does the higher MIPs rating equate to higher performance?
True
Explanation: Execution time of P1 is less than that of P2 so performance of P1 is higher and MIPs of P1 is higher than P2.
New Instruction count I = 2 * 109
For processor P1:
The clock rate(R) = 4.3GHz
The average clock per Instruction (CPI) = 1.75
So, execution time = (1/R)*I*CPI= =(1/4.3) * 10-9 * 2 * 109 * 1.75 sec = 0.8139 sec
For processor P2:
The clock rate(R) = 3 GHz
The average clock per Instruction (CPI) = 1.25
So, execution time =0.8139 sec
Instruction count =R *Execution time * (1/CPI) = 3 * 109 * 0.8139 * (1/1.25) = 1.95 * 109
The number of instruction that P2 can execute in the time that P1 executes 2 * 109 instructions
1.95 * 109
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P1