Question

Week 7, Thu 11/7/19 Introduction to Optimization Theory (math305), Fall 19 4. (16pts) Given the following LP problem Max Z=3x, + 2x - 4x, st: -1, + 2x - 4x = -10 2x, + 2x, – Xj28 x, + xy + x = 15 X, X, X, 20 (A) Set up a valid initial Simplex Tableau for the auxiliary problem of the Phase 1 of two phase method that will give an initial basic feasible solution for phase I DO NOT SOLVE FOR PHASE 1. (B) Find an optimal solution (must be feasible, not necessarily basic) for the auxiliary problem without doing any pivoting steps Answer: Phase I Optimal Solution (X1, X2, X3, X4, Xs, Yı, y2) = Show your work:

Answer 1

given Max Z=301, +2302 -4013 &t: -7,+212 -403 = -10 22, +2012 - 33 38 ? + a2 + N = 15 M, 12, 013 20 Tableaul ololololo. Base e Pol P P2 P3 Put Ps. 1 - 10 -2 1 4 0 P7 8 22 – olo i Ps 10.15 1. 0 1 18 1-3 10l-31 1 olo To The leaving rasiable is Pf and entring rasaible is ? Tableauz 0 0100 Jo --, Base colpo PP BP PSP6 P7 Po -16 To 1-3/4.5 0.5 0 I 1-0.5 P Tou I -0.5-0-5 o lo 0.5 5 1o l olol.5 0.5 II 0 1-0.5 z T61 o 31-451-0.5) o lo 11.5 the leaving variable is 'Ps and entring variable is P3

Tableaus olololo 10 - Tal Base a lo IP P P P PSPG I P2 P3 10 1.3333] 0-0.666671 0: 0 .22222 -0.11) | |ou. CG-61 | |--GCG 67 | 0 -0.449 Moo-111) | Ծ• ԿԿԿԿԿ P5 Lola lol lol 0.33333|1|-0-33333 -0.33333 Z There is any possible solution for the problem, so we can controue to phase II to calcolate It Tableaul Base o Po P P P P I P3 -4 1.33330 -0.66667 1 0.0 10 P 34 .66667) 1 .66666) 0 -0.44444 | 0 PS lol 9 lo IIo 10.33333 Z 18.66667 0 2.66667 0 -1.77778 The leaving variable is P3 and entring variable is Py.

Tableaua Base co o P P o P P5 R4 To 120-69 P31101 1-2 4 lolo Tolslo 3.- 30 300 -8 | 1610 the leaving Variable is Pss and en hering variable is P2 32-4 ol dose | с. Т о е е, | rs | Fs Pylo 22. lolol. 311 3.13.33333 1J 0 0.666667 2 1.6666667 0 0 0.333333 43:33333 0 .0 8 0 2-66667 The optimal solotion rowe is Z=43.3333 Tableau? 2 040 - ll, = 13.33333 92= 1.66667 83=0