GO Classes Test Series 2024 | Mock GATE | Test 12 | Question: 20

Answer 1

One approach is to calculate the execution time for a typical program, having, say, I machinelevel instructions:
$$
\begin{aligned}
\text { Time }_A & =\text { InstructionCount } \times \text { CPI }_A \times \text { CycleTime }_A \\
& =I \times 1.2 \times \frac{1}{2.4 \times 10^9}=\frac{1}{2} I \times 10^{-9} \\
\text { Time }_B & =\text { InstructionCount } \times \text { CPI }_B \times \text { CycleTime }_B \\
& =I \times 2.0 \times \frac{1}{3.0 \times 10^9}=\frac{2}{3} I \times 10^{-9}
\end{aligned}
$$
So, a typical program will execute (slightly) more quickly on Machine A. More precisely, the relative performance would be:
$$
\frac{\text { Time }_B}{\text { Time }_A}=\frac{\frac{2}{3} I \times 10^{-9}}{\frac{1}{2} I \times 10^{-9}}=\frac{4}{3}
$$
So, Machine $A$ is $4 / 3$ times as fast as Machine $B.$

Comments

To say 10 is 2 times of 5 we use 10/5 right , then why cant we use Time A / Time B here

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@Nagendra_voora Lets take $T_{B} = 100s$ and $T_{A} = 2s$. Now will you say $A$ is $\frac{100}{2} = 50$ times faster than $B$ or will you say A is $\frac{2}{100} = 0.02$ times faster than $B$. Clearly, $50$ times faster is the correct conclusion.

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@Suraj Reddy your reasoning questions everybody's understanding of Math & English. @Nagendra_voora even I had the same question according to the vague understanding of the question.

But if this answers was to accepted correct, then u could maybe understand the this statement "Machine A is K-times as fast as Machine B" this way:

Consider that the machine's speed is measured by the no. of instructions that it can executed in a unit time (that's what clock frequency is all about). In that perspective if A is k times faster than B (i.e. A=kB) in terms no. of ins per unit time. But what we have is the "total execution time" of A and B (who's units is in seconds clearly).

So, the mathematical representation would become like this: 
1/(execution time of 'i' no of. ins by A)  =  k * 1/(execution time of 'i' no of. ins by B)
where 'i' is just a constant.

Thus, k = (Exe. Time of B) / (Exe. Time of A)