Compute Load Store Branch Total a. Program1 1000 400 100 50 1550 b. Program4 150
ID: 3616674 • Letter: C
Question
Compute Load Store Branch Total a. Program1 1000 400 100 50 1550 b. Program4 1500 300 100 100 1750 a) Assuming that computes take 1 cycle, loads and storeinstructions take 10 cycles, and branches take 3 cycles, find theexecution time of each program on a 3 GHz MIPS processor. b) Assuming that computes take 1 cycle, loads and storeinstructions take 2 cycles, and branches take 3 cycles, find theexecution time of each program on a 3 GHz MIPS processor. c) Assuming that computes take 1 cycle, loads and storeinstructions take 2 cycles, and branches take 3 cycles, what is thespeed-up of a program if the number of compute instruction can bereduced byone-half? Compute Load Store Branch Total a. Program1 1000 400 100 50 1550 b. Program4 1500 300 100 100 1750 a) Assuming that computes take 1 cycle, loads and storeinstructions take 10 cycles, and branches take 3 cycles, find theexecution time of each program on a 3 GHz MIPS processor. b) Assuming that computes take 1 cycle, loads and storeinstructions take 2 cycles, and branches take 3 cycles, find theexecution time of each program on a 3 GHz MIPS processor. c) Assuming that computes take 1 cycle, loads and storeinstructions take 2 cycles, and branches take 3 cycles, what is thespeed-up of a program if the number of compute instruction can bereduced byone-half?Explanation / Answer
(a)
Execution time of the program = Clock cycles for a program /clock rate
Clock cycles for the program = (1000 x 1) + (400 x 10) + (100 x10) + (50 x 3)
= 1000 + 4000 + 1000 + 150
= 6150 cycles
Clock rate = 3GHz
Therefore, Execution time = 6150 / 3
= 2050 ns
(b)
Execution time of the program = Clock cycles for a program /clock rate
Clock cycles for the program = (1000 x 1) + (400 x 2) + (100 x2) + (50 x 3)
= 1000 + 800 + 200 + 150
= 2150 cycles
Clock rate = 3GHz
Therefore, Execution time = 2150 / 3
= 716.6 ns
(c)
Number of compute instructions after reduced to half = 500
Execution time of the program = Clock cycles for a program /clock rate
Clock cycles for the program = (500 x 1) + (400 x 2) + (100 x 2)+ (50 x 3)
= 500 + 800 + 200 + 150
= 1650 cycles
Clock rate = 3GHz
Therefore, Execution time = 1650 / 3
= 550 ns
Execution time before reducing the compute instructions = 716.6ns
Execution time after reduction = 550 ns
Hence, speed up = 716 / 550
= 1.3
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