Question

2. Consider a 2 GHz processor where two important programs, A and B, take one second each to execute. Each program has a CPI of 1.5. In the next-generation design, the company architects came up with the idea that improves the CPI of A to 1.1 and the CPI of B to 1.4. But the idea is so complex that the processor can only be implemented with a cycle time of 0.6 ns. Does the new processor out-perform the original processor on program A? How about on program B? [20%]

Answer 1

Answer:

program A runs in 1 sec in the original processor and 0.88 sec in the new processor.

So, the new processor out-perform the original processor on program A.

Program B runs in 1 sec in the original processor and 1.12 sec in the new processor.

So, the original processor is better then the new processor for program B.

Explanation:

Finding number of instructions in A and B using time taken by the original processor :

The clock speed of the original processor is 2 GHz.

which means each clock takes, 1/clockspeed

= 1 / 2GH = 0.5ns

Now, the CPI for this processor is 1.5 for both programs A and B. therefore each instruction takes 1.5 clock cycles.

Let's say there are n instructions in each program.

therefore time taken to execute n instructions

= n * CPI * cycletime = n * 1.5 * 0.5ns

from the question, each program takes 1 sec to execute in the original processor.

therefore

n * 1.5 * 0.5ns = 1sec

n = 1.3333 * 10^9

So, number of instructions in each program is 1.3333 * 10^9

the new processor :

The cycle time for the new processor is 0.6 ns.

Time taken by program A = time taken to execute n instructions

=  n * CPI * cycletime

= 1.3333 * 10^9 * 1.1 * 0.6ns

= 0.88 sec

Time taken by program B = time taken to execute n instructions

= n * CPI * cycletime

= 1.3333 * 10^9 * 1.4 * 0.6

= 1.12 sec

Now, program A runs in 1 sec in the original processor and 0.88 sec in the new processor.

So, the new processor out-perform the original processor on program A.

Program B runs in 1 sec in the original processor and 1.12 sec in the new processor.

So, the original processor is better then the new processor for program B.