Question

Example #04 Solve the following problem using the Big M method. max:z 4x1 +5x2-3x3 Subject to x12x2x3= 10 x1-x2+2 6 x13x2+x314

Answer 1

Max: z-4가,t.S경2-예) SubJect te> 에+2개2+기2-10 tax z 4+S2-3+0 S+os-MAI-MAL Soboect bo ,2L+, +A fo ,-2S+ Az -6 <14 A3 ,, 2 o ord ,, ,, , S,s,, AA, o 4 S -3 Min Raho Ye/M AI AL S1 to o 2 A1 1 . -M -1 A2 -M 1 O 14 O M 216 -21 O -M C-N21+4 4+S M-3 -M Colorin index 1s , So Me posibve Naximum Czyis 21+4 ord euting vaniable 1s Minimum Raio ts e ond is loo irdex is 2,sa he leaving basis venable is A The Pivot elenet is DePars hgAKey Elemet 1 R2Ines)- Rlold) Ri(ne)Rilod)-Ratne) 2 loc Rslold) -R2lne) A Enti 7 CJ -M M Mm Rotro S1 2 A2 A A1 1. 응 =·333 M 4. 13) 1 -1 1 -1 .వ O 4 M+4 NI-4M +24 M-4 -301-4 4 -21-4 M44 o3M 9 Posisve Maxi mom C-z is a+9ad its coluMn Index 18 2, $o Me ealeine vaiable is Roo Inddex 1s1, So leavirg Gasis nte ia Ai inirp Rabo 1s .333 ord i me Pivot elemet 1s 3 Enleng , Depanhrg A, , Key elemeit 3, Ritne)R,(dd)3 22ine)Rlold) ,Ine) RsIne) Ralold)-4Rna) 의-| 이-|과- TO Ol T J

stessp S 3 M M nin Rato 2 AL A1 O.332-o- 333 1-333 4 -333 3333 o-3333o.333) 구.3333 o G667 06661 0 3333 O-333 -0-3333o-)333 2-666E S2 -1-332 o.3333 4 3 Z 36 -6 -M-3|-M-1 Aogibve Naximu CJ -2y ia 'ondst colon ndex i 'u 80, fhe eutrnra vaiable si inimuM Raho is ^ard Ita Roo index is 'j, So leaving Sasis rodable is he Pvot elemait is o 333 , Ke elemestO.333 EnleirgS, , Defas brg R,(ne)2lold) o-3233 Rilne) R2 lold)o.G667RI (ne) R3lold) o-3333 e, na) R3ne) areton-4 eJ -3 S M M S 2 AL A L Min Raho S1 4 lo 2 S2 Z=40 8 4 4 O -N-4 Snce all Cs-Zjo heuce oPhr Solohm is 4«dved wi valve 2 vagiable Max z40 O O 이M O T -O O