Answer :
Answer:
Computer A is 1.41 times faster than the Computer B
Explanation:
Assume that number of instruction in the program is 1
Clock time of computer A is [tex]CT_{A} =200 ps[/tex]
Clock time of computer B is [tex]CT_{B} =250 ps[/tex]
Effective CPI of computer A is [tex]CPI_{A} =1.5[/tex]
Effective CPI of computer B is[tex]CPI_{B} =1.7[/tex]
CPU time of A is
[tex]CPU_{time}=instructions \times CPA_{A} \times CT_{A}\\CPU_{time}=1 \times 1.5 \times 200=300 sec[/tex]
CPU time of B is
[tex]CPU_{time}=instructions \times CPA_{B} \times CT_{B}\\CPU_{time}=1 \times 1.7 \times 250=425 sec[/tex]
Hence Computer A is Faster by [tex]\frac{425}{300} =1.41[/tex]
Computer A is 1.41 times faster than the Computer B