Question

Problem #5 -- Consider the following linear programming problem: Maximize Z = 2x1 + 4x2 + 3x3 subject to: X1 + 3x2 + 2x3 S 30 best to X1 + x2 + x3 S 24 3x1 + 5x2 + 3x3 5 60 and X120, X220, X3 2 0. You are given the information that x > 0, X2 = 0, and x3 >O in the optimal solution. Using the given information and the theory of the simplex method, analyze the constraints of the problem in order to identify a system of three constraint boundary equations (defining equations) whose simultaneous solution must be the optimal solution (not augmented). Then solve this system of equations to obtain this solution. (15 points)

Answer 1

> mon-2X1 +11%2+373 Ry+872 +243 $30, +72+73 (24, 37, 4577 +375 60. 2,192,7320 :. 3 fandand form, ma andud gorm, max 2: 271 +11%9+372 +0.51 +0.52+0.5m. 5.1. 2, +379 +243 + 51:30. X1 + X2 + x3 + 32-24 321 +5x2 + 3x3=60 t6 3=60 1,182,3870, 51152152206 slack vad. ist It. 0 0 0 Go 2 4 3 0 0 0 B. Basis x X2 X3 B, 52 sa b. a = 6/22 0 si 1 131 2 1 0 0 30 30%-10% lo 52 1 0 1 0 24 24,:24 O. 53 3 5 0 0 1 60 6015 = 12 z 0 0 0 0 0 0 - G-tj 2 up ↑ 3 o o o. .. Here all G-j ko, 'B' is pinot elt. R2 is incoming vad. and oi is outgoing vad. :. R1/3, R2-R1, Rz -5R1. 2nd lt. G 2 4 3 o o o. CB Basis ni x2 xz si 62 63 b o.. 4 82 113 1 213 43 0 0 10. 30. 0 32 6.213 0. 113 -13 1 0 14 21 0 62 14131.0. 113-513 0 I 10. 7.5<- 413 4 813 413 0 0. 67712137 0. 113 -413 0 0 o I mines now

J at is i All cj . Zj 80, 413 is Pinot elt. x is in coming vat. 53 is outgoing. Vad 2 R3, R2 - 2 R3, R1 - 1/6 RB . gdd It - Go 2 4 3 0 0 0 CB. 1 Basis. 21 53 X2 X3 si 52 b Q. Q. 22 1314) 314 0 -14 1512 106 O s2 0 0 112 42 1 -12 g 18 2. hod 1 0. po 314 7.5= -30 1 | 2 4 512 412 o 42 G-zi o o 429-1/2 0 -12. :. All ci-Zj 80, 314 is piuot elt. - az is incoming vad. and a2 is outgoing vad. .:. Rixų R2-12 R1, R₂ + 1/4 Rp. 4th It o Ca. 8 0 2 12 413 -23 az 1 0 Basis. ai 13 52 0 X 1 Z 2 Gj-23 o. si S2 S3 b. 1 0 -113 D. 0 1 -213 4 - 0 23 10. 1 0 43 50 - 0 -113 y -213 3 0

All cj-; so :. Max. 2 : 50. by calculation , 21770, 22=0, 23=10, 3120, 52= 4153=0. :. Max z = 2(10) + 460) +3(10) +0. +0.+0. Max z = 50 (10,0,10).