GATE CS 2014 Set 1 Q.65 (CS Q.55): ALU, Datapath and Control Unit

Consider two processors $P_1$ and $P_2$ executing the same instruction set. Assume that under identical conditions, for the same input, a program running on $P_2$ takes 25% less time but incurs 20% more CPI (clock cycles per instruction) as compared to the program running on $P_1$ . If the clock frequency of $P_1$ is 1GHz, then the clock frequency of $P_2$ (in GHz) is _________.

Your answer

Official answer and solution

Official answer: 1.6
From: Official GATE 2014 answer key

Solution

Relate CPU execution time to instruction count, clock cycles per instruction (CPI), and clock frequency.

The basic CPU execution time equation is:

T = \frac{I \times \text{CPI}}{f}

where $I$ is instruction count, $\text{CPI}$ is cycles per instruction, and $f$ is clock frequency.

Given conditions for $P_1$ and $P_2$ :

Same instruction set and identical program input $\implies I_1 = I_2 = I$ .
$P_2$ takes 25% less time than $P_1$ : $T_2 = (1 - 0.25) T_1 = 0.75 T_1$
$P_2$ incurs 20% more CPI than $P_1$ : $\text{CPI}_2 = (1 + 0.20) \text{CPI}_1 = 1.20 \text{CPI}_1$
Clock frequency of $P_1$ : $f_1 = 1\text{ GHz}$ .

Taking the ratio of execution times:

\frac{T_2}{T_1} = \frac{\text{CPI}_2}{\text{CPI}_1} \times \frac{f_1}{f_2}

Substitute the known ratios:

0.75 = 1.20 \times \frac{1\text{ GHz}}{f_2}

Solve for $f_2$ :

f_2 = \frac{1.20}{0.75} \times 1\text{ GHz} = \frac{120}{75} = \frac{8}{5} = 1.6\text{ GHz}

Thus, the clock frequency of $P_2$ is $1.6\text{ GHz}$ .

Final answer

1.6

Common trap

Dividing $0.75 / 1.20 = 0.625$ instead of solving for $f_2$ in the denominator.

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2014 Computer Science and Information Technology (CS) Session 1
Question: 65 of 65, printed as CS Q.55
Section: Computer Science and Information Technology

Report an error in this question

GATE questions are the work of the organising institutes. This site is independent and is not affiliated with them.

Question

Solution

Source