Chapter Floating Point Representation

Transcription

1 Chaper Floaing Poin Represenaion Afer reading his chaper, you should be able o: 1. conver a base- number o a binary floaing poin represenaion,. conver a binary floaing poin number o is equivalen base- number, 3. undersand he IEEE-754 specificaions of a floaing poin represenaion in a ypical compuer, 4. calculae he machine epsilon of a represenaion. Consider an old ime cash regiser ha would ring any purchase beween 0 and unis of money. Noe ha here are five (no six) working spaces in he cash regiser (he decimal number is shown jus for clarificaion). Q: How will he smalles number 0 be represened? A: The number 0 will be represened as Q: How will he larges number be represened? A: The number will be represened as Q: Now look a any ypical number beween 0 and , such as How would i be represened? A: The number will be represened as Q: Wha is he smalles change beween consecuive numbers? A: I is 0.01, like beween he numbers and Q: Wha amoun would one pay for an iem, if i coss ? A: The amoun one would pay would be rounded off o or chopped o In eiher case, he maximum error in he paymen would be less han Q: Wha magniude of relaive errors would occur in a ransacion? A: Relaive error for represening small numbers is going o be high, while for large numbers he relaive error is going o be small

2 Chaper For example, for , rounding i off o accouns for a round-off error of = The relaive error in his case is = = %. For anoher number, 3.546, rounding i off o 3.55 accouns for he same round-off error of = The relaive error in his case is = = %. Q: If I am ineresed in keeping relaive errors of similar magniude for he range of numbers, wha alernaives do I have? A: To keep he relaive error of similar order for all numbers, one may use a floaing-poin represenaion of he number. For example, in floaing-poin represenaion, a number is wrien as , is wrien as , and is wrien as The general represenaion of a number in base- forma is given as exponen sign manissa or for a number y, e y = σ m Where σ = sign of he number, + 1or -1 m = manissa, 1 m < e = ineger exponen (also called ficand) Le us go back o he example where we have five spaces available for a number. Le us also limi ourselves o posiive numbers wih posiive exponens for his example. If we use he same five spaces, hen le us use four for he manissa and he las one for he exponen. So 9 he smalles number ha can be represened is 1 bu he larges number would be By using he floaing-poin represenaion, wha we lose in accuracy, we gain in he range of numbers ha can be represened. For our example, he maximum number represened 9 changed from o Wha is he error in represening numbers in he scienific forma? Take he previous example of I would be represened as.568 and in he five spaces as Anoher example, he number would be represened as and in five spaces as So, how much error is caused by such represenaion. In represening 56.78, he round off error creaed is = 0. 0, and he relaive error is

3 Floaing Poin Represenaion = 0 = %, In represening , he round off error creaed is = 9. 78, and he relaive error is 9.78 = 0 = % Wha you are seeing now is ha alhough he errors are large for large numbers, bu he relaive errors are of he same order for boh large and small numbers. Q: How does his floaing-poin forma relae o binary forma? A: A number y would be wrien as e y = σ m Where σ = sign of number (negaive or posiive use 0 for posiive and 1 for negaive), m = manissa, () 1 m < ( ), ha is, ( 1 ) m < ( ), and e = ineger exponen. Example 1 Represen ( 54.75) in floaing poin binary forma. Assuming ha he number is wrien o a hypoheical word ha is 9 bis long where he firs bi is used for he sign of he number, he second bi for he sign of he exponen, he nex four bis for he manissa, and he nex hree bis for he exponen, Soluion (5) (.75) = (11.11) = ( 1.111) 54 The exponen 5 is equivalen in binary forma as () 5 = ( 1) Hence ( 54.75) ( 1.111) (1) = The sign of he number is posiive, so he bi for he sign of he number will have zero in i. σ = 0 The sign of he exponen is posiive. So he bi for he sign of he exponen will have zero in i. The manissa m = 11 (There are only 4 places for he manissa, and he leading 1 is no sored as i is always expeced o be here), and he exponen e = 1. we have he represenaion as

4 Chaper Example Wha number does he below given floaing poin forma represen in base- forma. Assume a hypoheical 9-bi word, where he firs bi is used for he sign of he number, second bi for he sign of he exponen, nex four bis for he manissa and nex hree for he exponen. Soluion Given Bi Represenaion Par of Floaing poin number 0 Sign of number 1 Sign of exponen 11 Magniude of manissa 1 Magniude of exponen The firs bi is 0, so he number is posiive. The second bi is 1, so he exponen is negaive. The nex four bis, 11, are he magniude of he manissa, so m = 1.11 = = The las hree bis, 1, are he magniude of he exponen, so 1 0 e = ( 1 ) = ( ) = ( 6) The number in binary forma hen is ( 1) ( 1.11) The number in base- forma is 6 = = ( ) ( ) ( ) Example 3 A machine sores floaing-poin numbers in a hypoheical -bi binary word. I employs he firs bi for he sign of he number, he second one for he sign of he exponen, he nex four for he exponen, and he las four for he magniude of he manissa. a) Find how will be represened in he floaing-poin -bi word. b) Wha is he decimal equivalen of he -bi word represenaion of par (a)? Soluion a) For he number, we have he ineger par as 0 and he fracional par as Le us firs find he binary equivalen of he ineger par Ineger par ( 0 ) = ( 0) Now we find he binary equivalen of he fracional par Fracional par:

5 Floaing Poin Represenaion Hence ( 0.083) ( ) 6 = ( 1.101) 6 ( 1.10) The binary equivalen of exponen is found as follows Quoien Remainder 6/ 3 0 = a0 3/ 1 1= a1 1/ 0 1 = a So So ( 6 ) = ( 1) ( 1) ( 0.083) ( ) = 1.10 ( 01) = ( 1.10) Par of Floaing poin number Bi Represenaion Sign of number is posiive 0 Sign of exponen is negaive 1 Magniude of he exponen 01 Magniude of manissa 10 The en-bi represenaion bi by bi is b) Convering he above floaing poin represenaion from par (a) o base by following Example gives ( 01) ( 1.10) = ( ) ( ) ( 6) = ( 1.75) = Q: How do you deermine he accuracy of a floaing-poin represenaion of a number?

6 Chaper A: The machine epsilon, mach is a measure of he accuracy of a floaing poin represenaion and is found by calculaing he difference beween 1 and he nex number ha can be represened. For example, assume a -bi hypoheical compuer where he firs bi is used for he sign of he number, he second bi for he sign of he exponen, he nex four bis for he exponen and he nex four for he manissa. We represen 1 as and he nex higher number ha can be represened is The difference beween he wo numbers is (0000) (0000) ( ) ( ) = ( ) = (1 4 ) = (0.065). The machine epsilon is mach = The machine epsilon, mach is also simply calculaed as wo o he negaive power of he number of bis used for manissa. As far as deermining accuracy, machine epsilon, mach is an upper bound of he magniude of relaive error ha is creaed by he approximae represenaion of a number (See Example 4). Example 4 A machine sores floaing-poin numbers in a hypoheical -bi binary word. I employs he firs bi for he sign of he number, he second one for he sign of he exponen, he nex four for he exponen, and he las four for he magniude of he manissa. Confirm ha he magniude of he relaive rue error ha resuls from approximae represenaion of in he -bi forma (as found in previous example) is less han he machine epsilon. Soluion From Example, he en-bi represenaion of bi-by-bi is Again from Example, convering he above floaing poin represenaion o base- gives ( 01) ( 1.10) ( 6) = ( 1.75) = ( ) The absolue relaive rue error beween he number and is approximae represenaion is = = which is less han he machine epsilon for a compuer ha uses 4 bis for manissa, ha is,

7 Floaing Poin Represenaion = 4 mach. = Q: How are numbers acually represened in floaing poin in a real compuer? A: In an acual ypical compuer, a real number is sored as per he IEEE-754 (Insiue of Elecrical and Elecronics Engineers) floaing-poin arihmeic forma. To keep he discussion shor and simple, le us poin ou he salien feaures of he single precision forma. A single precision number uses 3 bis. A number y is represened as e (. a a a ) y = σ where σ = sign of he number (posiive or negaive) a i = enries of he manissa, can be only 0 or 1, i = 1,..,3 e =he exponen Noe he 1 before he radix poin. The firs bi represens he sign of he number (0 for posiive number and 1 for a negaive number). The nex eigh bis represen he exponen. Noe ha here is no separae bi for he sign of he exponen. The sign of he exponen is aken care of by normalizing by adding 17 o he acual exponen. For example in he previous example, he exponen was 6. I would be sored as he binary equivalen of = 133. Why is 17 and no some oher number added o he acual exponen? Because in eigh bis he larges ineger ha can be represened is ( ) = 55, and halfway of 55 is 17. This allows negaive and posiive exponens o be represened equally. The normalized (also called biased) exponen has he range from 0 o 55, and hence he exponen e has he range of 17 e 18. If insead of using he biased exponen, le us suppose we sill used eigh bis for he exponen bu used one bi for he sign of he exponen and seven bis for he exponen magniude. In seven bis, he larges ineger ha can be represened is ( ) = 17 in which case he exponen e range would have been smaller, ha is, 17 e 17. By biasing he exponen, he unnecessary represenaion of a negaive zero and posiive zero exponen (which are he same) is also avoided. Acually, he biased exponen range used in he IEEE-754 forma is no 0 o 55, bu 1 o 54. Hence, exponen e has he range of 16 e 17. So wha are e = 17 and e = 18 used for? If e = 18 and all he manissa enries are zeros, he number is ± ( he sign of infiniy is governed by he sign bi), if e = 18 and he manissa enries are no zero, he number being represened is No a Number (NaN). Because of he leading 1 in he floaing poin represenaion, he number zero canno be represened exacly. Tha is why he number zero (0) is represened by e = 17 and all he manissa enries being zero. The nex weny-hree bis are used for he manissa. The larges number by magniude ha is represened by his forma is ( ) 38 = 3.40 The smalles number by magniude ha is represened, oher han zero, is

8 Chaper ( ) Since 3 bis are used for he manissa, he machine epsilon, 3 mach =. 7 = 1.19 = 1.18 Q: How are numbers represened in floaing poin in double precision in a compuer? A: In double precision IEEE-754 forma, a real number is sored in 64 bis. The firs bi is used for he sign, he nex 11 bis are used for he exponen, and he res of he bis, ha is 5, are used for manissa. Can you find in double precision he range of he biased exponen, smalles number ha can be represened, larges number ha can be represened, and machine epsilon? 38 INTRODUCTION TO NUMERICAL METHODS Topic Floaing Poin Represenaion Summary Texbook noes on floaing poin represenaion Major General Engineering Auhors Auar Kaw Dae April 4, 009 Web Sie hp://numericalmehods.eng.usf.edu