DRDO CSE 2022 Paper 2 | Question: 3

Question

The execution of a program occurs on a $250 \mathrm{~GHz}$ processor that executes millions of instructions. Type, $\text{CPI}$ (cycles per instruction) and $\%$ of four instructions are provided in the table.
\[\begin{array}{|c|c|c|}
\hline \text{Instruction Type} & \% & \text{CPI} \\
\hline \hline \text{Data Manipulation} & 45 & 1 \\
\text{Data Transfer} & 19 & 3 \\
\text{Sequence } \& \text{ Control} & 21 & 5 \\
\text{Input and Output} & 15 & 6 \\
\hline \end{array}\]
Compute the average $\text{CPI}$ and $\text{MIPS}$ (millions of instructions per second) rate of the processor.

Answer 1

Average CPI = $1*0.45$ + $3*0.19$ + $5*0.21$ + $0.15*6$ = $2.97$

Clock rate = $250$ $GHz$ , hence Clock cycle time = $1/250$ $GHz$ = $0.004$ $ns$

Average instruction execution time = $0.004 * 2.97$ = $0.01188$ $ns$

This signifies the number of instructions executed in 1ns,

Now by unitary method

in $0.01188$ $*$ $10^{-9}$ second number of instructions executed  = $1$

hence in $1s$ number of instructions executed  = $\frac{1}{0.01188*10^{-9}}$ = $8.417508*10^{10}$ = $84175.08*10^{6}$

hence MIPS rate = $84175.08$