Architecture -1

1. A program runs in 10 seconds on Computer A which runs at 2 GHz. The plan is to build a computer B which will run this program in 6 seconds. The designer feels that it is possible to raise the clock frequency but the effect will be to affect the number of clock cycles required adversely (due to the decrease in clock cycle duration). B will require 1.2 times as many clock cycles as computer A. What should be the clock rate for B to run the program in 6 seconds?

Answer:

-According to performance equation,

Execution time= Instruction count X cycles per instruction X clock cycle time

-For computer A,

Execution time = 10 seconds, clock cycle time= 0.5 ns

Let us assume that for one instruction count computer A takes ‘y’ cycles per instruction. –(1

According to performance equation,

10= 1 X y X 0.5

So, y=20 __(2)

-For computer B,

Execution time= 6 seconds, clock cycle time=? ,

As computer B will require 1.2 times as many clock cycles as computer A.

So, as per (1) for one instruction count computer B takes ‘1.2 y’ cycles per instruction.

As per (2) cycles per instruction = 1.2 X 20= 24

According to performance equation,

6= 1 X 24 X clock cycle time

So, clock cycle time= 0.25 ns

* Computer B should run at 4 GHz to run the program in 6 seconds.

2. Computer A has a clock cycle time of 250 ps and a CPI of 2.0 for a program. Computer B has a clock cycle time of 500 ps and a CPI of 1.2 for the same program. Which computer is faster and by how much?

* For computer A,

Clock cycle time= 250 ps , CPI= 2.0

According to performance equation,

Execution time= instruction count X Cycles per instruction X clock...