NIELIT 2022 April Scientist B | Section B | Question: 102

Question

Consider a machine with $40 \; \text{MHz}$ processor which has run a benchmark program. The executed program consists of $100,000$ instruction executions, with the following instructions mix and clock cycle count. What will be the effective $\text{CPI, MIPS}$ rate, and execution time.

 

$$\begin{array}{|l|l|r|} \hline \textbf{Instruction Type}& \textbf{Instruction Count}& \textbf{Cycles/Instructions} \\\hline \text{Integer arithmetic}&45000&1 \\\hline \text{Data Transfer}&32000&2 \\\hline \text{Floating point}&15000&2 \\\hline \text{Control transfer}&8000&2 \\\hline \end{array}$$

• A. $\text{CPI} : 3.55; \; \text{MIPS : 30;}$ Execution time $:1.87 \; \text{ms}$
• B. $\text{CPI} : 1.55; \; \text{MIPS : 25.8;}$ Execution time $:3.87 \; \text{ms}$
• C. $\text{CPI} : 5.60; \; \text{MIPS : 45.8;}$ Execution time $:2.87 \; \text{ms}$
• D. $\text{CPI} : 2.55; \; \text{MIPS : 35.8;}$ Execution time $:4.87 \; \text{ms}$

Answer 1

Given,

Frequency =40 MHz

Clock time = $\frac{1}{40*10^{6}} sec$

Instruction type	Instruction count	cycle/instructions	% of instruction type
Integer arithmetic	45000	1	45%
Data transfer	32000	2	32%
Floating point	15000	2	15%
Control Transfer	8000	2	8%
Total	100000		100%

 

$\text{CPI}= \displaystyle  \sum_i (\text{Fraction of instruction type } i \times \text{Cycles needed by instruction type }i)$

$\implies \text{CPI}= 0.45*1 +(0.32+0.08+0.15)*2 = 1.55$

$\text{Execution time} = \text{CPI} * \text{total number of instructions} * \text{Clock time}$ 

$\qquad =1.55*10^{5}*\frac{1}{40*10^{6}}$ sec

$\qquad =3.875 \;ms$

$\text{MIPS} = \frac{\text{Clock Frequency}}{\text{CPI}*10^{6}}$

$\qquad =\frac{40*10^{6}}{1.55*10^{6}}$

$\qquad =25.8$

So correct answer is option (B).

Reference :- https://en.wikipedia.org/wiki/Cycles_per_instruction#:~:text=For%20example%2C%20with%20two%20executions,1%2F2%20%3C%2001).

The question was copied from wikipedia itself.

Comments

Can we get MIPS from execution time?

---

Yes sir we can.

In $3.875 *10^{-3}$ s we execute  $10^{5}$ instruction

1 s                          we execute   $\frac{10^{5}}{3.875*10^{-3}}$ instruction

so now MIPS =$\frac{10^{8}}{3.875*10^{6}}$

                         =25.8

---

Execution time is 3.875 ms.

ie in 3.875 x 10^-3 sec cpu ran 10^5 instructions

Therefore, in 1 sec we cpu can run

= 10^5 / (3.875 x 10^-3)

= (100 x 10^6) / 3.875

=  25.806 mips