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

Answer:

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

Explanation:

Processor clock rate = 2 GHz

Execution Time =   ∑  $(\frac{Clock cyles}{Clock rate})$

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,

    = $\frac{512(10^{6}) }{2(10^{9}) }$

 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

($\frac{Clockcycles}{2}$ )=   ($CPI_{FP}$ 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  =  ($\frac{Clockcycles}{2}$ ) -[ ( $CPI_{INT}$ x No. INT instructions) + ( $CPI_{L/S}$  x No. L/S instructions ) + ( $CPI_{branch}$ x No. branch instructions)]

$CPI_{FP improved}$ x 50 x $10^{6}$  = ( $\frac{512(10)^{6} }{2}$ ) - [ (  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