InClassActivity03-sol - 1/24/03 5:28 PM Activity #03...

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1/24/03 5:28 PM ECSE-2660 Computer Architecture, Networks, & Operating Systems, Spring 2003 Page 1 of 4 Activity #03 (solutions) Last Name First Name Student ID Number Email Address 3.1 3.2 Total Grader Initials 50 points 50 points 100 points Activity 3.1 (First hour) A. The current machine, codenamed “Jaguar”, has a clock rate of 500MHz. The hardware team claims that it can improve the clock rate to 750MHz. The codename for this improved machine is “Jaguar-EH”. Specs for both the machines are shown in Table 1 below. A standard practice in the computer industry is to perform extensive simulations of new designs before producing them. For instance, we can run a series of benchmark programs on a simulator for a new processor, and make valuable measurements. Such a study yielded the data shown in the last two columns for Jaguar-EH: Table 1. Specs for Jaguar and Jaguar-EH. Instruction Class CPI for Jaguar CPI for Jaguar-EH Frequency A 2 2 35% B 4 3 20% C 3 2 20% D 7 5 15% E 1 1 10% a. (10 points) Calculate the average CPI and MIPS ratings for Jaguar. Answer: Average CPI = 2*0.35 + 4*0.20 + 3*0.20 + 7*0.15 + 1*0.10 = 3.25 cycles/instruction MIPS = MIPS instr instr cycles cycles 154 sec / 10 85 . 153 / 25 . 3 sec / 10 500 6 6 = × = × Grading policy: 5 pts each part. 5 pts for correct answer, 2 pts partial, 0 pts no answer. b. (10 points) Calculate the average CPI and MIPS ratings for Jaguar-EH. Answer: Average CPI = 2*0.35 + 3*0.20 + 2*0.20 + 5*0.15 + 1*0.10 = 2.55 cycles/instruction MIPS = MIPS instr instr cycles cycles 294 sec / 10 12 . 294 / 55 . 2 sec / 10 750 6 6 = × = × Grading policy: 5 pts each part. 5 pts for correct answer, 2 pts partial, 0 pts no answer. c. (10 points) How much faster is Jaguar-EH compared to Jaguar? Answer: The improvement is the inverse ratio of the execution times. For the processor, it is the ratio of the MIPS. Therefore, performance improvement = MIPS JaguarEH / MIPS Jaguar = 294.1/153.9 = 1.91, (EH is 91% faster). Note that ratio of execution times is equivalent to inverse ratio of MIPS in here, because the frequencies of usage of the instructions are the same. Grading Policy: 10 pts for correct answer, 4 pts partial, 0 pts no answer.
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1/24/03 5:28 PM ECSE-2660 Computer Architecture, Networks, & Operating Systems, Spring 2003 Page 2 of 4 Meanwhile, the compiler team comes up with enhancements that reduce the number of instructions executed. They have
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InClassActivity03-sol - 1/24/03 5:28 PM Activity #03...

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