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Consider two processors P1 and P2 executing the same instruction set. Assume that under identical conditions, for the same input, a program running on P2 takes 25% less time but incurs 20% more CPI (clock cycles per instruction) as compared to the program running on P1. If the clock frequency of P1 is 1GHz, then the clock frequency of P2 (in GHz) is _________.
  • a)
    1.6
  • b)
    3.2
  • c)
    1.2
  • d)
    0.8
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
Consider two processors P1 and P2 executing the same instruction set. ...
For P1 clock period = 1ns 
Let clock period for P2 be t.
Now consider following equation based on specification
7.5 ns = 12*t ns
We get t and inverse of t will be 1.6GHz 
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Most Upvoted Answer
Consider two processors P1 and P2 executing the same instruction set. ...
Given:
- Two processors P1 and P2 executing the same instruction set.
- Program running on P2 takes 25% less time but incurs 20% more CPI compared to P1.
- Clock frequency of P1 is 1GHz.

We need to find the clock frequency of P2.

Let's assume that the program running on P1 takes T1 time, and the CPI for P1 is C1.
Similarly, the program running on P2 takes T2 time, and the CPI for P2 is C2.

We are given that the program running on P2 takes 25% less time compared to P1, so we can write:
T2 = T1 - 0.25T1 = 0.75T1

We are also given that the program running on P2 incurs 20% more CPI compared to P1, so we can write:
C2 = C1 + 0.20C1 = 1.20C1

To calculate the clock frequency of P2, we can use the formula:
Clock frequency = Instructions per program / (CPI * Time)

Since the instruction set and the program are the same for both processors, we can assume the number of instructions per program is constant.
So, the clock frequency of P2 can be calculated as:
Clock frequency of P2 = Instructions per program / (C2 * T2)

Now, substituting the values:
Clock frequency of P2 = Instructions per program / (1.20C1 * 0.75T1)

Since the instruction set and the program are the same for both processors, the number of instructions per program cancels out:
Clock frequency of P2 = 1 / (1.20C1 * 0.75T1)

We know that the clock frequency of P1 is 1GHz, which is equivalent to 1/1GHz = 1/1000 seconds:
Clock frequency of P1 = 1 / (C1 * T1)

Substituting the value of the clock frequency of P1:
1GHz = 1 / (C1 * T1)

Simplifying, we get:
C1 * T1 = 1 / 1GHz

Now, substituting this value in the equation for the clock frequency of P2:
Clock frequency of P2 = 1 / (1.20C1 * 0.75T1)
= 1 / (1.20 * 0.75 * (1 / 1GHz))
= 1 / (0.9 * 1 / 1GHz)
= 1GHz / 0.9
= 1.11111... GHz (approximately)

Therefore, the clock frequency of P2 is approximately 1.11111... GHz, which can be rounded to 1.1 GHz.

However, none of the given options match this value. So, it seems there might be an error in the options provided.
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Consider two processors P1 and P2 executing the same instruction set. Assume that under identical conditions, for the same input, a program running on P2 takes 25% less time but incurs 20% more CPI (clock cycles per instruction) as compared to the program running on P1. If the clock frequency of P1 is 1GHz, then the clock frequency of P2 (in GHz) is _________.a)1.6b)3.2c)1.2d)0.8Correct answer is option 'A'. Can you explain this answer?
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