G cmaj9/g chord -Googl Stewart Calculus: Early Calculus I Syllabus Fal xmy.baker
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G cmaj9/g chord -Googl Stewart Calculus: Early Calculus I Syllabus Fal xmy.bakeru.edu x e chegg Study | Guided × /D Homework C file:///C:/Users/Carlos/Downloads/Homework03.pdf O@ 1. The results of the SPEC CPU2600 bzip2 benchmark running on an AMD Barcelona has an instruction count of 2.389x1012, an execution time of 750s, and a reference time of 9650s. Find the CPI if the clock cycle time is 0.333ns. Find the SPECratio. Find the increase in CPU time if the number of instructions of the benchmark is increased by 10% without affecting the CPI. Find the increase in CPU time if the number of instructions of the benchmark is increased by 10% and the CPI is increased by 5%. Find the change in the SPECratio for this change Suppose that we are developing a new version of the AMD Barcelona processor with a 4 GHz clock rate. We have added some additional instructions to the instruction set in such a way that the number of instructions has been reduces by 15%. The execution time is reduced to 700s and the new SPECratio is 13.7. Find the new CPI This CPI value is larger than obtained in part a as the clock rate was increased from 3 GHZ to 4 GHz. Determine whether the increase in the CPI is similar to that of the clock rate. If they are dissimilar, why? By how much has the CPU time been reduced? For a second benchmark, libquantum, assume an execution time of 960 ns, CPI of 1.61, and clock rate of 3 GHz. If the execution time is reduced by an additional 10% without affecting the CPI and with a clock rate of 4 GHz, determine the number of instructions. Determine the clock rate required to give a further 10% reduction in CPU time while maintaining the number of instructions with the CPI unchanged. Determine the clock rate if the CPI is reduced by 15% and the CPU time by 20% while the number of instructions is unchanged. a. b. c. d. e. f. g. h. i. j. k. 7:27 AM 9/6/2017 Type here to searchExplanation / Answer
Given,
Clock rate =1/cycle
Time =3 GHz
CPI =clock rate * CPU time/instr count
a. CPI(bzip2) = 3 *109* 750/(2,389 *109) = 0.94
SPECratio =ref. time/execution time
b. SPECratio(bzip2) = 9,650/750 = 12.86
CPU time = (No. instructions * CPI)/clock rate
c. Hence, If CPI and clock rate do not change, the CPU time increase is equal to the increase in the number of instructions, that is, 10%.
d. If given Instruction increased by 10% then the final number of instuction is
=(110/100) * Initial no. of Instruction
= 1.1 * Initial no. of Instructions.
CPI also increased by 5% then the final CPI will be
=(105/100) * Initial CPI
= 1.05 * Initial CPI
Final CPU Time = (Final no. of Instruction * Final CPI)/Clock Rate
= 1.155 ( Final Ins. * Final CPI) / Clock Rate
= 1.155 * Initial CPU time
CPU Time has been increased by = (1.155 - 1) * 100 = 15.5%.
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