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Cls Notes 10-3 Integer Sol

Cls Notes 10-3 Integer Sol - L W9 m 1/51{fly-mun,i.v:gm...

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Unformatted text preview: L W9 m 1/51} {fly-mun,i%:.v:gm Wyfim&fi X2 Duuu»»manmmmmwwA.n.nA W, . Wen»-‘Wwan—mm”H‘FWFWA'Mm-we:m<mflwwwq.hasrv—-u¢»¢v.l.£r¢nh¢v' , , "$4,“ _ __M was ‘ -»7w-’=.-r.-.-m-J—,-,-a..w.u=m-,__ ,7 , mmw‘ ,, Mxmmm wgwwxhgm‘mwwmmw mam-kc “Wu,“ '5 2‘3», ifffi / X) J (€325 x ; {”1556 “P” Vanwf’Hng 3?; 2/? {9} Axwafifijwfififl , h “OW I),Inufypwp‘:L-I:iz-Sfivarw;wmvw".VW'nzmyvnwrnmJ/hxmfymuwnfiw; W ”M,“ u .. - m ‘ww-q“ ”MW“, Jwr-WWMWMWW .‘W , MGR/LP {Revised 1/93) Date: 10/ 3/11 Time: 09:18:08 H:\MORLPS~l\grinteg.lpe MODEL DESCRIPTION ‘ki*1?*fic*i’**********ivki-uk-ki—iwk*********************~kivkfi”): max 2 = 7x1 + 9x2 st —xl +3x2 <= 6 7x1 + x2 <= 35 integer(xl,x2) graphics(0,8,0,5.3) "must have selected linear system and Gomory algorithm" ******‘k*‘k*‘k*‘k**‘k**‘k‘k‘k*******‘k‘k********************** Gomory Cut — Mixed Integer Linear Programming Solving the initial linear program. “A“ Optimal Solution “A“ selected constraint for out 1 Tableau before pivot Tableau: 3 BASES Z X1 X2 Slaokl Slack2 81 SOL Z 1 00 0 00 0 00 2.55 l 36 0 00 63 00 X2 0.00 0 00 l 00 0.32 0 05 0 00 3.50 X1 0.00 l 00 0 00 —0.05 0 14 0 00 4.50 81 0.00 0 00 0 00 -0.32 —0 05 1 00 -0.50 Pivot row is 3 Pivot variable is Slackl Tableau after pivot Tableau: 3 E BASIS 2 x1 x2 Slackl SlackZ Sl SOL E z 1 00 0 00 0 00 0.00 1 00 8 00 59 00 ; X2 0.00 0.00 1.00 0 00 O 00 1 00 3 00 X1 0 00 1.00 0.00 0 00 0.14 —0 l4 4 57 Slaokl 0 00 0.00 0.00 l 00 0.14 —3 14 l 57 selected constraint for out 2 Tableau before pivot BASIS Z X1 X2 Slackl Slack2 51 $2 X2 0.00 0.00 1.00 0.00 0.00 1.00 0.00 3.00 X1 0.00 1.00 0.00 0.00 0.14 —0.14 0.00 4.57 Slackl 0.00 0.00 0.00 1.00 0.14 —3.14 0.00 1.57 82 0.00 0.00 0.00 0.00 —0.l4 —0.86 1.00 —0.57 Pivot row is 4 Pivot variable is SlackZ Tableau after pivot Tableau: 4 BASIS Z X1 X2 Slackl Slack2 81 82 SOL. Z 1.00 0 00 0 00 0.00 0 00 2 00 7 00 55.00 X2 0.00 0.00 1.00 0.00 0.00 1.00 0.00 3.00 X1 0.00 1.00 0.00 0.00 0.00 -1.00 1.00 4.00 Slackl 0.00 0.00 0.00 1.00 0.00 —4.00 1.00 1.00 SlackZ 0.00 0.00 0.00 0.00 1.00 6.00 w7.00 4.00 Integer solution has been obtained Z = 55.0000 X1 = 4.0000 X2 = 3.0000 Slackl : 1.0000 Slaok2 4.0000 m #50 How uflfimnpmcoo Umpomamm -----iiwflm:Sagitmma-:Jmm ..... my ..... mmmiiiimw--- N0 0: 0m 0 00.H 00.0 mm.0 00 0 mx -----:---m-m ..... mm-m-----mm..,m ggggg m3 ..... ma ..... 3m ......... ma: HHE““Humflmméwmmmfin”an"HHMWHHHMHE”Humflfinmmmu m Hammapme Hauooumo msoscfiwcoo mm Uwpmwnu mwaflmflnmp Ham u <<< mafipsaom amaflymo <<< EOHpSHom mmflaamnmonm mecfiq .EMHWOHQ Hmmdfla Hafiuflflfi map mcwbaom mEHEEmeOHm Hmmgflq wamucw vmxflz I #30 >HoEow **%*¥¥¥*¥¥¥¥¥¥«*i******¥**¥¥¥****%**%**¥****%**¥***¥ fimx kmx .nguwwwpcfi HV NNm + wa + HNN Hv mNN + NM + fix me + mxw + axm pm N mNHEHNmE ¥*%*¥*******+*«*********¢¥*¥¥¥*t*%**¥*¥i*¥t*¥******¥ ZO HHmHmUmMQ AMDOE mg .3259 H emmqmoz/ ”m "mEHH fimm\ H Ummflbmmv md\moz HH\m \OH “mumo NMUMHm 00.0 00.0 00.0 mm 00.0 00.H 00.0 HM 00.H 00.0 00.0 NM 00.0 00-0 00.0 MM 00.0 00.0 00.H N NN ax N mHmmm v ”mmmfiQMH pebflm muowmfl Swmaflme H “30 How uflflwupmmou umpumamm qumfim fixomam 00.0 00.H 00.0 HM 00.H 00.0 00.0 mx 00.0 00.0 00.0 mx 00.0 00.0 00.H N NN HM N mHmmm m “Smwaflma pebfla Hmpmm 5mmanme qumam qumam mH mHnMHHmb uo>fim m mfl 30H uo>flm 00.0 mN.0| 00.0 Hm 00.H mm.0 00.0 Nx 00.0 wm.0 00.0 mx 00.0 mN.0 00.fi N NX HM N mHmdm m ”fiwmaflmH uobag mnowwn SMmMQMH 0000.H n Nxomam 0000.N H mx 0000.H H NX 0000.H H HM 0000.0H H N Umcflmpflo mmmn mmg moausaom wamuqH 00.H mm.H: mm.0 00.M 00.0 00.0 00.0 00.0 00.0 NMUMHm 00.H >0.0 hm.MI 00.0 mm.0: 00.0 00.0 00.H 00.0 ax 00.H 00.0 00.H 00.0 00.0 00.0 oo.a 00.0 00.0 NM 00.N mm.0l mm.H 00.0 50.0 00.H 00.0 00.0 00.0 mx 00.0H 00.0 00.H 00.0 00.0 00.0 00.0 00-0 00.H N .AOm mm Hm NMUmHm axumam mx NX HM N mHmflm w "5mmflflme nobm Hmuww 5mmHQMH mxumfim mfi mafimanmb popfim w mH 30H wo>flm ...
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