FoatingPointNotes

FoatingPointNotes - Silvana Ilie - MTH510 Lecture Notes...

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Unformatted text preview: Silvana Ilie - MTH510 Lecture Notes Binary Floats Machine numbers in floating point representation are of the form: x = (1 + f )2 e where f is mantissa, f = (0 .d 1 d 2 ...d t- 1 d t ) 2 (in base 2) t = precision (number of digits, t > 0) e min = lower bound on exponent e max = upper bound on exponent d k = 0 or 1 are binary digits, k = 1 ,...,t e = exponent, e min e e max in normalized form: leading implicit bit 1, not stored Toy floating point system Assume t = 2 ,e min =- 1 ,e max = 1. Then [1 . 00] 2 2- 1 = 1 2 [1 . 00] 2 2 = 1 [1 . 00] 2 2 1 = 2 [1 . 01] 2 2- 1 = 1 2 (1 + 1 4 ) = 5 8 [1 . 01] 2 2 = 5 4 [1 . 01] 2 2 1 = 5 2 [1 . 10] 2 2- 1 = 1 2 (1 + 1 2 ) = 3 4 [1 . 10] 2 2 = 3 2 [1 . 10] 2 2 1 = 3 [1 . 11] 2 2- 1 = 7 8 [1 . 11] 2 2 = 7 4 [1 . 11] 2 2 1 = 7 2 3 possible exponents, 4 mantissas, 2 signs give 24 possible machine numbers (excluding 0) Non-uniform spacing of machine numbers: [2- 1 , 2 ] | {z } spacing 2...
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This note was uploaded on 03/07/2012 for the course MTH MTH510 taught by Professor Dr.silvanailie during the Winter '12 term at Ryerson.

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FoatingPointNotes - Silvana Ilie - MTH510 Lecture Notes...

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