Question

The following table shows results for SPEC CPU2006 benchmark program bzip2 running on an AMD Barcelona Name | Intr. Count x 10° | Execution Time (seconds) | Reference Time (seconds) 820 3120 9650 bzip2 1.111 [5] <1.7> Find the CPI if the clock cycle time is 0.250 ns. 1.11.2 [5] <1.7> Find the SPECratio. 1.11.3 [5] <1.7> Find the increase in CPU time if the number of instructions of the benchmark is increased by 15% without affecting the CPI. 1.114 [5] <1.7> Find the increase in CPU time if the number of instructions of the benchmark is increased by 7% and the CPI is increased by 12%. 1.11.5 [5] <1.7> Find the change in the SPECratio for this change

Answer 1

1.11.1

execution time = CPI x cycle time x instruction count

          820        = CPI x 0.250 x 10 ^ -9 x 3120 x 10 ^ 9

CPI = 1.051

1.11.2

SPECratio = time on reference computer / time on computer being rated

                 = 9650 / 820 = 11.76

1.11.3

Number of instruction increased by 15 %

so total number of onstruction = 3120 + 0.15 * 3120 = 3588

CPI = 1.051

cycle time = 0.250ns

execution time = CPI x cycle time x instruction count

                        = 1.051 x 0.250 x 10 ^ -9 x 3588 x 10 ^9

                        = 942.7 sec

1.11.4

instruction increased by 7%

so total number of instruction = 3120 + 0.07 * 3120 = 3339

new cpi = old cpi + 0.12 * old cpi

             = 1.051 + 0.12 * 1.051 = 1.177

so execution time = 1.177 x 3339 x 10^9 x 0.250 x 10^-9 = 982.6

1.11.5

SPECratio = 9650 / 982.6 = 9.82

so change in SPECratio = 11.76 - 9.82 = 1.94