Question

Excel

Use Simplex method and Exel To solve the following LPPs. Maximize Maximize P-3x + x2 subject to the constraints x1 + x2 = 2 2x) + 3x2 s 12 3x + = 12 x 20 x220 P = 5x1 + 7x2 subject to the constraints 2xy + 3x2 = 12 3x + x2 = 12 x 20 *2 2 0 Maximize Maximize P = 2x2 + 4x2 + x3 subject to the constraints -*1 + 2x2 + 3x3 5 6 -*, + 4x2 + 5x3 5 *1 + 5x2 + 7x3 = 7 X, 20 X220X320 P = 6x + 3x2 + 2x3 subject to the constraints 2x1 + 2x2 + 3x3 = 30 2x1 + 2x2 + x3 = 12 x 20 x 20 x 20 Maximize P = 4x1 + 2x2 + 5x3 subject to the constraints *1 + 3x2 + 2x3 = 30 2xy + x2 + 3x3 = 12 X 20 X220 x 20

Answer 1

The Given Problem is – Pmax = 3x1+x2 Subject to : X1 + X2 <2, 2X1 + 3x2 < 12 3x1 + x2 < 12. 217,0, X270 we courite the problem in Standard Form - may = 3x + y 2 +0.5 -0.99 +0.83 Subject to: = 2 - 12 +53 =12 X1 + 2 +91 2x4+ 3H2 +52 334 + x2 . x1,x2,5, S2, S37,0 Min Rati 3. 1 BCB XB x % 22 x si o 1 1 12 2 3 S3 o 12 13 1 2 OOO 0 S3 0 0 1. S2 0 1 0 doo lo ce v -31 - 1 Zj-Cj 2, 3 se o So, 0 2 s 6 1 1 1 0 0 o 1 -2 10 0 -2 -3 0 1. o. O Zj-ej 10 2 3 0 0 Sipce au z -ej 70. So optimal Solution arrived and it's x2= 2 x2=0 Pmax = 6 ( Ans)

Given problem: Pmax = 521 + 7X2 Sub.to: 2X1+3x2 < 12 3X1+ X2 <12 x 70 x 70 Ce conite the problem in Standard from - P max = 5X1 + 7%, +0:51+0.Sz Subject to: 2X1+ 3x2 + $i = 12 BX1+ x2 + S2 = 12. and X1, X2, S1, S2 7,0. 7 , Radio C 5 | ® СвХВ x si o 12 2. Sz 0 12 3 0 Si 1 0 0 S2 .o 1 4 ~ 1 |2477 Zj-e; -5 -77 o o x2 7 4 2/3 1 1/3 o 52 0 8. (7/30 -1/3 1 zieli -1/310 7/ 3 0 des 7 1.287 0 1 317 -217 1. 5 24171 o 1/7.. 317. Zj-ej o o 1617 17. Since au 73-217,0 So optimal Sotalton arrived and it is x1- 2417 m2 = 12/7 P max = 204 (aws)

The Gevem problem is - P max = 2X1+ 4%2+3 S.+: - X1 + 2x2 + 3x3 <6 - 24 +42 +5X3 <5 21 +5X2 + 7X3 < 7 X1, X1, X3O . Coe conite the problem in Standard Form - Trox = 2X + 4 x2 + 3 + 0.5 +0.S2 +0.83 Subjeetto: --X1 + 2x2 + 3x3 + SI -X14 4 824 509 +52 = 5 x + 5 X2 + 743 + 5z = 7 X1, X2, X3, S.,S2, S3 > 0. Min Ratio Cj 2 4 0 B CB XBX1 X2 X3 si si o 6 1 2 3 1 52 0 5 -1 4 5 0 S3.0 F. 15 70 ool 0 S2 0 1 o 10 Pools to 514 → 7/5 1/3 → Zj-ej 1-2 - 47 -1 0 0 0 Isi o 722 -2 0 1/2 2 1/2 o x2 4 514 114.1 5/4 . 4. o 53.0 314 14 0 34 0 - 5/4 1. I zili -31 0 4 0 1 o. S. 0 12 0 0 2 2 H 210 x2 4 413 0 1 42 o 1/6 1/9.. de 2 1/3 1 0 1 0 -5/9, 419. ziej Too 50 -2/31 43 si o 13o I 10 1 1 Szo 120 9 120 1 xy2 7 1 5 7 0 0 Zj-ej. To 6 13 0 og The obl mal solution is Xi = 7 *2=0 X =0 P max = 14 ( Ans )

The Givem Problem is Pmax = 6x1t3x2 +243 Sub. to! 22, + 2x2 + 3x3 <30. 2x4 + 2X2 + x3 <12 . 201, X2, X37,0 We write the broblem in standard foom - mone 6X1 + 3x2 +233 +0.5, +0°S 2 Sub.to.: 231 +239 + 3x3 + SI = 30. 2xy +222 + 3 + S2 = 12. X1, X2, 43, S1, S2 »,0... 0 Si Min Ratio 0 S2 0. 1 15 C 6 3 2 в СвХp | 2, x, x, s, o 302 2 3 s2 0 122 2 1 Z-cj -6 1 -3 -2 SO 18 0 0 2 a. 6 6 1 1. 2/2 0 0 1 O 0 - 2 1/2 Zj-ej To 3 1. 3 Sinee, au 7-ey 7,0 the optimal solation arrived and it is x, = 6 x=0, x=0 Pmax=36 (Ans)

The Giver Problem is - Pmax = 4X1 + 2X2 +5%3 Subjec to: % +3% + 2x3 < 30 2x4+x2 + 3x3 <12. X1, X2, X3 >, O. We comorite the problem in Standard Form -- Pmax = 4x2+2x2 +503 +0.51+0.82. 10: = 12. X1 +3x2+2x3 + Si 2x 1 + x2 + 3x3 + S2 X1, X2, X3, S1, S2 > 0. 5 To Ratio si 15 4. I Cj 14 2 B CB XB.. x X2 xz si o 30 2 3 2 52 O 12 2 1 3 I zirei - 4 -2 -5 si o 22 | -1/3 7/3 o x3 5 4.2/3 23 1 I Zj-lj - 2/39 -1/3 0 Isi 0 24 10 5/2 1/2 x 4 6 1 1/23/2 0 ↑ 0 1 o 0 2 O o o S2 0 1 0 – 2/3 1/3 5/3 1/2 1/2. Zi-ci Too 1 o e As au zj-ej >0 So, optimal solution arrived, The optimal Solution is - x1 = 6,2 = 0 , X3 =D max = 24 (AND)