Suppose we have two implementations of the same instruction set architecture. Computer A has a clock cycle time of 250 ps and a CPI of 2.0 for some program, and computer B has a clock cycle time of 500 ps and a CPI of 1.2 for the same program. Which computer is faster for this program, and by how much? We know that each computer executes the same number of instructions for the program; let's call this number I. First, find the number of processor clock cycles for each computer:
CPU clock cycles A - — 172 2.0
CPU clock cyclesB = E 1.2
Now we can compute the CPU time for each computer:
CPU timeA = CPU clock cyclesA Clock cycle timeA
= 122.0 z 250 ps = 500 ps Likewise, for B:
CPU timeB = 12 1.2 Z 500 ps = 600
Clearly, computer A is faster. The amount faster is given by the ratio of the execution times:
CPU performanceA Execution timeB 600 1 ps = 1.2 CPU performanceB Execution time 500 1 ps
We can conclude that computer A is 1.2 times as fast as computer B for this program.