ECE314sp08_HW6

# ECE314sp08_HW6 - ECE/CS 314 Spring 2008 Hung Dang(hvd2...

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ECE/CS 314 Spring 2008 Hung Dang (hvd2) Homework 6 Due Tuesday, April 15, 2008 before 10:00pm EST Problem 1. Floating Point Representation (20 points) (a) What IEEE 754 floating point number does this bit pattern represent? 1010 1101 0101 0000 0000 0000 0000 0000 = (-1) 1 x (1+0.5+0.125) x 2 (90-127) = -1 x 1.625 x 2 -37 = -1.625 x 2 -37 (b) Show the IEEE 754 binary representation for the floating point number 30 ten in single and double precision. Single: 30 ten = (-1) 0 x (1+0.5+0.25+.125) x 2 (131-127) hence 0100 0001 1111 0000 0000 0000 0000 0000 Double: 30 ten = (-1) 0 x (1+0.5+0.25+.125) x 2 (1027-1023) hence 0100 0000 0011 1110 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 (c) Show the IEEE 754 binary representation for the floating point number 0.025 ten in single and double precision. Single: 0.025 ten = (-1) 0 x (1+.6) x 2 (121-127) hence 0011 1100 1100 1100 1100 1100 1100 1100 Double: 0000 0111 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 (d) Add 2.85 ten x 10 3 to 9.84 ten x 10 4 , assuming that you have only three significant digits, first with guard and round digits and then without them. With guard and round digits: 2.85 ten x 10 3 + 9.84 ten x 10 4 = 0.2850 x 10 4 + 9.8400 ten x 10 4 = 10.125 ten x 10 4 which equals to 10.1 ten x 10 4 Without guard and round digits: 2.85 ten x 10 3 + 9.84 ten x 10 4 = 0.28 x 10 4 + 9.84 ten x 10 4 = 10.1 ten x 10 4

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Problem 2. Caches (30 points) In this problem we are going to compare the performance of two ISAs, the normal MIPS ISA and an ISA with two enhancements, described below. 1)
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ECE314sp08_HW6 - ECE/CS 314 Spring 2008 Hung Dang(hvd2...

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