Question

1. (20 pts.) Consider the following linear program: max 4x4 +xz+5x3 +3x4 s.t. *1 -X2 -X3 +3X, 51 5x +xz+3X3 +8X555 -X2 +2x2+3x3 -5x53 It is claimed that the solution x* = (0,14,0,5) is an optimal solution to the problem. Give a proof of the claim. Do not use the simplex method to solve this problem.

Answer 1

1. Given Linear programing problem Manz: 40, +Mz +5 hz + 3y SE X-Mz - Az + 384 El. 591 +92 +3 X3 +804 555 -dj +2 M2 +3 %3 - 5*4 = 3 This problem can be solved using Two-phase method phase-Ikki okay The canonical form is obtained by adding slack, Surplus and artificial variable as appropiate type, we can add slacke variables All constraints are s Si, s, and san norporio plumnos ant loimolog Manza - T - X,-&g -43.+3x4 + S - 54, +*2 +3×3 + 8X4 + Sg 2 55 -2, +2, +3% -5x4 + 5z = 3, 81, 82, 23, 24, 51, 52, 5320 Iteration - I ci o oo oooo mi CB XB * B S * a *4 Sa Sg S3 Rahib s ol 1 -1 -1 3 100 55 51 38 OO 0 3 -1 2 3 500 z 0 0 0 0 0 0 0 Zj-cj o o o o o do

Since all zj-C520 Hence optimal solution is reached "=0, x=0,030 14:0 Manz=0 Phase 2 Maxz=48,+ 22 +503 + 344 too Ratio 'Iteration 1 ci 4,5 3 00 B C1B x x 22 xs xy S4 3 53 S, O L - L31 00:- S 0 155 513 8 010 S; O 3 - 2 3 -5 00! Zj-Cja), 3-4 -1 -51-3 000 from the table negative min of 2;-; IB So the entering variable is x3 Minimum ratio is, ső leaving variable is Sz. (key) The pivot element is 3 Rz (new) = R3 Cold) :3 R(new) = R, cold) + Rz bew) Rg Cnew) - Recold) – 3R3 (new)

Iteration-a es cj XB 4 a 5 & 3 0 00 ass Sy Sg Parsis uz B e 1301- 59 culto calo wlw Zj-Cj Negative min of 25-5; is -34 So xu is entering vanable minimum Ratio is so s, is leaving Now R, (new) = RICO Rz (new) - Rz6ld) _13 R, (new) R3 (new) - Recold) + 5 R. (new) Iteration 3 ; 4 i 5 3 o o o min. B CB *Bwax nys, Sz Sz Kaho Mo 82 0 5 9 0 0 3 ola Na Z-92 N zz 4 2;_c; o IT 5 3 og o no

negative min 2j-cj is So xz is enleving variable Min Ratio is 14, so do is the leaving variable is the key element Rz (new) - Rz Cold ) x4 Ra Coeli e cold it I Rz (new) 4 R2 Cnew , R old)-9 Rg (new) 0 min Ratio 8 Iteration - 4. c; 0 4 , 5 3 o o B CB xodies & au si sa Se oi-s og op! X2 i 14 do 5 2=29 Since all 2;-C; 20 optimal solution is reached &p=0, xy=5 *I=0, 22=14 Max 2-29 N= (0,14,0,5) :. The optimal solution Hena proved.