Question

Please be as detailed as possible. This is a multiple part question. Please label the question you are answering.

Consider the CF LPP maximize z 1+ 4x2 - 2x3 subject to the constraints "X1 1 1 1 0 0 -1 10 2-3 2 0 1 0 3 -2 10 0 1| X6. 1,2 2 0. Complete the tableau below, after answering the following questions (a) What is the matrix B for the tableau below? (b) Find B1, using the tableau below (without directly computing the inverse of B) (c) Compute t, t2, t3,ts the columns of the tableau below (without OR) (d) Compute XB. (e) Compute z (the objective function value) (f) Find CB (g) Compute OR\, OR2, .. ,OR6, the objective row entries. You may use the columns (t's) of the tableau below that you already found in part (c)

Answer 1

1. 
   Max subject Z=1, +422 - 283 to * WNL 1007 2 -3 2010 3 -2 10 From the tablen below, it is clear that B = (a o crusung tg 1 -3 1+ 30 -2 , as, 2 variables.) Firstly, we complete the tableu Iteration I clearly the Basic variables initially are 185 1 36 со ТВ | Хе o În - - - o o Zj-cj - 1-472 o o o & Most negative x enters the basis. We choose the pivot element. In the whole column entry corresponding to ty in positive and minimum amongst the positive values This the pivot element is 1 y leaves the basis
   
2. The answers for part d, e, f, g are on this page.
3. 
   Heration In Apply now operations R₂ R2 + 3R & R3 R3 + 2R, 1 3 na ing no C 18 EFFE lalo 312 10 zj – 4 4 4 4 o o zj-G [-5 0 6 4 o o & Most negative enters the basis. Again min & 267 = 26 & Pivot element=1 Thus, de leaves the basis. teration III :- Apply row operations Ry Rut R3 R₂ R₂ + R3 G (1) (4) (-2) (0) (0) (0) CBIB XB 1 x 122 | a3 | 4 | 5 | 6 - - - As cololol85 - - Z I 4 19 14 0 1 zji o o 21 14 o I Thush, Zj-G 30 & aj & B This is the required table. Z (the obj. fanction value) = 4(36) + 0(60) + 1(26) - 170.
   
4. 
   This corresponding to B is the mateix vectors dey, as, do 3 0 1 Atloo *) C = (4,0,1)