The results of the SPEC CPU2006 bzip2 benchmark running on a

The results of the SPEC CPU2006 bzip2 benchmark running on an AMD Barcelona has an instruction count of 2.389E12, an execution time of 750 s, and a reference time of 9650 s.

1) Find the CPI if the clock cycle time is 0.333 ns.

2) Find the increase in CPU time if the number of instructions of the benchmark is increase by 10% without affecting the CPI.

3) Find the change in the SPECratio for this change.

Please show all the work. Thank you!

Solution

Given that, Instruction count = 2.389E12,

Execution time = 750 s,

Reference time = 9650 s,

1) If the clock cycle time is 0.333 ns, Then we have to calculate CPI by the following formula,

CPI = (execution time / (instruction count * clock cycle time))

CPI = ( 750 s / ( 2.389E12 * 0.333 ns))

CPI = 0.94

2) If the number of instructions of the benchmark is increased by 10% without affecting the CPI. Then, the increase in CPU time is,

Increase in CPU time = (1 - (increase of 10% of number of instructions of the benchmark / Execution Time))

Increase of 10% of number of instructions of the benchmark = (750 + ((750 * 10) /100)) = 825 s

Increase in CPU time = (1-(825/750)) = 0.1 (10%)

3) The change in SPECratio for this change is,

SPECratio = (reference time / execution time)

SPECratio = (9650 s / 825 s)

SPECratio = 11.7

instruction count * 1.1 = 2.628, CPU time = 823s