Assignment 1

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School

Illinois Institute Of Technology *

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350

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Electrical Engineering

Date

Apr 3, 2024

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Lab 1: Performance Due date: Wednesday 1/30/2022 at noon Written Assignment (110 points in total) Show your solution steps of each problems. Please do not just show the final answer. 1. A processor running at 2.5 GHz consumes 60 W of dynamic power and 15 W of leakage power. It briefly enters Turbo-boost mode and operates at a frequency of 3.0 GHz. How much dynamic power and leakage power does the processor consume in Turbo-boost mode? (10 points) f = 2.5 GHz P dynamic = 60 W P leakage = 15 W f turbo = 3.0 GHz P dynamic = ½ * C * V DD 2 * f P dynamic_turbo = ½ * C * V DD 2 * f turbo P dynamic / P dyanmic_turbo = f / f turbo  P dynamic_turbo = P dynamic * f turbo / f = 60W * 1.2 = 72W P leakage_turbo = P leakage = 15W 2. A program executes 100 billion instructions. It executes on an IBM processor that has an average CPI of 1.2 and a clock frequency of 4.0 GHz. How many seconds does the program take to execute? What is the cycle time of this IBM processor? Assume that an ARM processor takes 30 seconds to execute the program. What is the speedup provided by the IBM processor, relative to the ARM processor? (20 points) IC = 100 * 10 9 CPI IBM = 1.2 f IBM = 4.0 GHz  CPU Time IBM = IC * CPI IBM / f IBM = ((100 * 10 9 * 1.2) / (4 * 10 9 )) (s) = (100 * 1.2 / 4) s = 30 s  Cycle Time IBM = 1 / f IBM = (1 / (4.0 * 10 9 )) (s) = 0.25 * 10 -9 s There is no speedup provided by the IBM processor as both the IBM and ARM processor take 30 seconds to execute the program.

3. Alex builds a 1 GHz processor where two important programs, A and B, take one second each to execute. Each program has a CPI of 2. Elaine is tasked with designing the company's next-generation processor. She comes up with an idea that improves the CPI of A to 1.5 and the CPI of B to 1.8. But the idea is so complex that the processor can only be implemented with a cycle time of 1.2 ns. Does Elaine's new processor out-perform Alex’s processor on program A? How about on program B? (20 points) f 1 = 1 GHz CPI A1 = CPI B1 = 2 = CPI 1 Time A 1 = Time B 1 = 1s = Time 1  IC A = IC B = Time 1 * f 1 / CPI 1 = 0.5 * 10 9 f 2 = (1 / 1.2) GHz CPI A2 = 1.5 CPI B2 = 1.8 Time A 2 = IC A * CPI A2 / f 2 = 0.9 s < Time A 1  The new processor out-performs the old one on A Time B 2 = IC B * CPI B2 / f 2 = 1.08 s > Time B 1  The new processor doesn’t out-perform the old one on B 4. 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. (total 60 points, 10 points each) IC = 2.389 * 10 12 Time = 750 s a. Find the CPI if the clock cycle time is 0.333 ns. f = (1 / 0.333) GHz CPI = Time * f / IC = 0.94 b. Find the increase in CPU time if the number of instructions of the benchmark is increased by 10% without affecting the CPI.

CPI = 0.94 f = (1 / 0.333) GHz IC b = 1.1 * IC Time = IC * CPI / f Time b = IC b * CPI / f = 1.1 * IC * CPI / f = Time * 1.1 = 825 s  Time b / Time = 1.1  Increase by a factor of 1.1  Time b – Time = 75 s  Increase of 75 seconds c. 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%. IC c = 1.1 * IC CPI c = 1.05 * CPI Time = IC * CPI / f Time c = IC c * CPI c / f = 1.1 * 1.05 * IC * CPI / f = 1.155 * Time = 866.25  Time c / Time = 1.155  Increase by a factor of 1.155  Time c – Time = 116 s  Increase of 116 seconds d. 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 reduced by 15%. The execution time is reduced to 700 s. Find the new CPI. f = 4 GHz IC d = 0.85 * IC Time d = 700 s

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CPI d = Time d * f / IC d = 1.38 e. This CPI value is larger than obtained in 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? f d / f = 4 / 3 = 1.333 CPI d / CPI = 1.38 / 0.94 = 1.468 The ratios are different because the number of instructions was also reduced. f. By how much has the CPU time been reduced? Time d / Time = 700 / 750 = 0.933 The time was reduced by a factor of 0.933 (6.7%)