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2 months ago

5

Consider the two computers A and B with the clock cycle times 100 ps and 150 ps respectively for some program. The number of cyc

les per instruction (CPI) for A and B are 2.0 and 1.0 respectively for the same program. Which computer is faster and how much? a) A is 1.33 times faster than B b) Bis 1.22 times faster than A c) A is 1.23 times faster than B d) B is 1.33 times faster than A

Computers and Technology

2 answers:

ivann1987 [1K]2 months ago

7 0

Answer:

Option d): Computer B is 1.33 times faster than A

Given:

Clock cycle times: $t_{A} = 100 ps$

$t_{A} = 150 ps$

Cycles per instruction (CPI): $n_{A} = 2.0$

$n_{B} = 1.0$

Solution:

Let the number of instructions be I.

CPU cycles for A: $f_{A}$ = 2.0 × I

CPU cycles for B: $f_{B}$ = 1.0 × I

The CPU time formula is:

CPU time, T = $CPU clock cycle\times clock time$

$T_{A} = f_{A}\times t_{A} = 2.0 I\times 100 = 200 I ps$

$T_{B} = f_{B}\times t_{B} = 1.0 I\times 100 = 150 I ps$

This shows B runs faster than A.

Calculating performance ratio:

$\frac{Performance of A}{Performance of B} = \frac{T_{A}}{T_{B}}$

$\frac{Performance of A}{Performance of B} = \frac{200}{150}$

B's performance is 1.33 times greater than A's.

oksian1 [950]2 months ago

3 0

The clock cycle duration for Computer A is 100ps

Each instruction requires 2 cycles (CPI = 2)

So, the time to execute one instruction is Clock cycle time × CPI = 100 × 2 = 200 ps

Computer B's clock cycle time equals 150ps

CPI for Computer B is 1

Therefore, the time per instruction is 150 × 1 = 150 ps

Since Computer B takes less time per instruction compared to Computer A, it is the faster machine.

The difference in speed per instruction is 200 ps - 150 ps = 50 ps advantage for Computer B

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