Question

Assume a program requires the execution of 50 x 10^6 FP instructions, 110 x 10^6 INT instructions, 80 x 10^6 Load/Store (L/S) instructions and 16 x 10^6 branch instructions. The CPI for each type of instruction is 1, 1, 4 and 2, respectively. Assume that the processor has a 2 GHz clock rate.By how much must we improve the CPI of FP instructions if we want the program to run two times faster?

Answer 1

We can not improve CPI of FP instructions when we run the program two times faster because it would be negative.

Step-by-step explanation:

Processor clock rate = 2 GHz

Execution Time = ∑
`((Clock cyles)/(Clock rate))`

Clock cycles can be determined using following formula

Clock cycles = (
`CPI_(FP)` x No. FP instructions )+ (
`CPI_(INT)` x No. INT instructions) + (
`CPI_(L/S)` x No. L/S instructions ) + (
`CPI_(branch)` x No. branch instructions)

Clock cycles = ( 50 x
`10^(6)` x 1) + ( 110 x
`10^(6)` x 1) + ( 80 x
`10^(6)` x 4) + ( 16 x
`10^(6)` x 2)

Clock cycles = 512 x 10⁶

So,Initial Execution time for FP instructions is,

=
`(512(10^(6)) )/(2(10^(9)) )`

Initial execution Time = 256 x 10⁻³

For 16 processors ,

clock cycle = 512 x 10⁶

Execution Time = 256 x 10⁻³

To run the program two times faster, half the number of clock cycles

(
`(Clockcycles)/(2)` )= (
`CPI_(FP)` x No. FP instructions )+ (
`CPI_(INT)` x No. INT instructions) + (
`CPI_(L/S)` x No. L/S instructions ) + (
`CPI_(branch)` x No. branch instructions)

`CPI_(FP improved)` x No. FP instructions = (
`(Clockcycles)/(2)` ) -[ (
`CPI_(INT)` x No. INT instructions) + (
`CPI_(L/S)` x No. L/S instructions ) + (
`CPI_(branch)` x No. branch instructions)]

`CPI_(FP improved)` x 50 x
`10^(6)` = (
`(512(10)^(6) )/(2)` ) - [ ( 110 x
`10^(6)` x 1) + ( 80 x
`10^(6)` x 4) + ( 16 x
`10^(6)` x 2)]

`CPI_(FP improved)` x 50 x
`10^(6)` = - 206 x
`10^(6)`

`CPI_(FP improved)` = - 206 x
`10^(6)` / 50 x
`10^(6)`

`CPI_(FP improved)` = - 4.12 < 0