Floating Point Number System -(1.3)

Transcription

1 Floting Point Numbr Sstm -(.3). Floting Point Numbr Sstm: Comutrs rrsnt rl numbrs in loting oint numbr sstm: F,k,m,M k ;0, 0 i, i,...,k, m M. Nottions: th bs 0, k th numbr o igts in th bs xnsion or k 8, 0 m th minimum xonnt 0 M th mximum xonnt,0 Exml: F0,,0, 0, 0., 0. 0, 0. 0 ;,,...,9, n 0,,,...9 () F0,,0, contins numbrs. () Thsmllst mgnitut is 0 n th lrgst mgnitut is This sstm cnnot rrsnt n numbr tht is lrgr thn 99. An comuttion xcs th lrgst numbr will cus n ovrlow xction (comutr stos) n th smllr nonzro numbr will cus unrlow (unxct 0). Unxct zros will lso cus n ovrlow roblm. For xml, F0,,0, (3) Th numbr nxt to 0 (rom th right) in F0,,0, is 0.0. So, rl numbr in 0, 0. is rrsnt b ithr 0 or 0.. In othr wors, 0 or 0. rrsnts mn rl numbrs.. Roun-o Errors: Roun-o rrors is rouc whn clcultor or comutr is us to rorm rl numbr clcultions. Tht is bcus th rithmtic rorm in mchin involvs numbrs with onl init numbr o igits n th clcult rsults r onl roximtions o th ctul numbrs. Lt brl numbr n l b lting oint numbr in F,k,m,M rrsnting. Suos tht k k...0 n ( 0). Choing mtho: l c k 0 n Rouning mtho: l r k 0 n i k k 0 n i k 5 Exml Giv th loting -oint orm o using 5-igit choing; n b 5-igit rouning l c b l r Exml Us 5-igit rouning rithmtic to rorm th clcultion 3. x x x x x x x x

2 Absolut Error n Rltiv Error: Lt b n roximtion to. Thn th bsolut rror is in s n th rltiv rror is in s rovi tht 0. Rltiv rrors or choing roximtions: k k...0 n, l c k l c 0. k k...0 n k.0 0 n k 0 n k k...0 n n 0 n l c 0n k 0 k 0 n Rltiv rrors or rouning roximtions: k k...0 n, l r k i k k i k 0.5 l r 0. k k...0 n k i k k k k i k n k i k k k i k n k l r 0n k n l r 0 k From th ur bouns or choing n rouning roximtions, w lrn rouning roximtion is bttr in gnrl. Th Mchin Prcision: Th mchin rcision u is givn b u Signiicnt bs igits: Suos tht t l t 0 k choing 0 k rouning. or som ositiv intgr t. Thn w s tht l n gr to t lst t n t most t signiicnt bs igits. Exml Lt n l r n l c Fin th bsolut rrors n rltiv rrors o ths roximtions n trmin th numbrs o signiicnt ciml igits.

3 Hnc, th numbr o signiicnt ciml igits or is roximt b l r is Hnc, th numbr o signiicnt ciml igits or is roximt b l c is 4. Exml Lt , n ; n lt , n Comut th bsolut rror n rltiv rror or ch roximtion Th bsolut rror o is much lrg thn th on or, th rltiv rrors or both n r th sm. From this xml, w s tht th bsolut rror ns on th mgnitu o, on th othr hn, th rltiv rror os not n on th mgnitu o. So, th rltiv rror is usull us to vlut th closnss o th roximtion. Exml Solv x n Rcll tht th sstm solution is x b c c x b hs uniqu solution i bc 0 n th bc c b For our roblm, bc So, th sstm os not hv uniqu solution. 3

4 Is th sstm consistnt? Tht is, is in th rng o c b? How to chck this? Rcll, i th rnk c b n th rnk b c r th sm, thn th sstm hs ininitl mn solutions, othrwis th sstm is inconsistnt (no solution). W know rnk Chck th rnk o Gussin Elimintion.0. R R R Th sstm hs no solution Th IEEE Stnr: Th signrs o th IEEE (Institut or Elctricl n Elctronics Enginrs) binr loting oint rithmtic stnr slct bso(. Th singl n oubl rcision loting oint numbr sstms r F,4, 5,8 n F,53, 0,04, rsctivl. For xml, singl rcision: Th smllr n lrgr numbrs r: s igits l igits Hnc, rrsnts mn numbrs in th intrvl s, l , Th smllst ositiv numbr n th lrgst numbr in th oubl rcision loting oint sstm r smllst igits lrgst igits

5 loting oint sstm mchin rcision (rouning) smllst ostiv numbr lrgst ositiv numbr singl rcision oubl rcision u u Exml Solv th qution x 6.0x 0 using 4-igit rouning rithmtic. W know i b 4c 0 thn th qution x bx c 0hstworlsolutionsnthr x b b 4c, x b 4c Now comut x n x st b st in 4-igit rouning rithmtic: St Exrssion Vlu b b 4c b 4c b b 4c b b 4c x 7 b 4c b 4c x Tru solutions (or solutions comut using k-igit rouning rithmtic whr k 4 : b 4c Rltiv rrors: x x x x x x x x Wh th roximtion o x is so oor? Not tht St 4 involvs subtrction o two clos numbrs. Chck out th rltiv rror or this subtrction: 5

6 roximtion - tru - irnc tru irnc I th subtrction o two numbrs in clos mgnitus cn b voi, thn th ccurc o th comuttion o x cn b imrov. Rwrit th ormul or x : x x x x b b 4c b b 4c b 4c x b b 4c c b 4c b 4c b 4c x