DRDO CSE 2022 Paper 2 | Question: 4

Question

A system $\mathrm{X}$ with $2 \mathrm{~GHz}$ clock speed runs a program in $10$ seconds. We want to build a system $\mathrm{Y}$ to run the same program in $6$ seconds. For this, system $\mathrm{Y}$ needs $1.2$ times as many clock cycles as system $\mathrm{X}$. What should be the clock speed of the system $\mathrm{Y}?$

Answer 1

We Know that Total time for Instruction Execution = No. of Clock Cycles * Cycle Time and Time = 1/frequency.

For System X:

1. Time for 1 clock cycle = 1/2GHz = 1/(2 * 10^9) = 0.5nsec
2. Total time taken by System X to complete Instruction Execution = 6sec, Hence no. of clock cycles taken = (6 * 10^9)nsec / 0.5nsec = 2 * 10^10 clocks.

For System Y:

1. It needs 1.2 times the no. of clock cycles as X. Hence total clock cycles = 1.2 * 2 * 10^10 = 2.4 * 10^10
2. Total time System Y takes to complete is 6 seconds. Hence time for 1 clock cycle for System Y ==> 6/(2.4 * 10^10) = 0.25nsec
3. Clock Speed/ Frequency for System Y = 1/cycle time = 1/0.25nsec = 4GHz

Hence Clock Speed for System Y = 4GHz