Consider the following problem.
Minimize Z = 2x1 + 3x2 + 2x3,
Subject to

And x1 ‰¥ 0, x2 ‰¥ 0, x3 ‰¥ 0.
Let x4 and x6 be the surplus variables for the first and second constraints, respectively. Let x-bar5 and x-bar7 be the corresponding artificial variables. After you make the adjustments described in Sec. 4.6 for this model form when using the Big M method, the initial simplex tableau ready to apply the simplex method is as follows:

After you apply the simplex method, a portion of the final simplex tableau is as follows:

(a) Based on the above tableaux, use the fundamental insight presented in Sec. 5.3 to identify the missing numbers in the final simplex tableau. Show your calculations.
(b) Examine the mathematical logic presented in Sec. 5.3 to validate the fundamental insight (see the T* = MT and t* = t + vT equations and the subsequent derivations of M and v). This logic assumes that the original model fits our standard form, whereas the current problem does not fit this form. Show how, with minor adjustments, this same logic applies to the current problem when t is row 0 and T is rows 1 and 2 in the initial simplex tableau given above. Derive M and v for this problem.
(c) When you apply the t* = t + vT equation, another option is to use t = [2, 3, 2, 0, M, 0, M, 0], which is the preliminary row 0 before the algebraic elimination of the nonzero coefficients of the initial basic variables x-bar5 and x-bar7. Repeat part (b) for this equation with this new t. After you derive the new v, show that this equation yields the same final row 0 for this problem as the equation derived in part (b).

Transcribed Image Text:

x42 2x38 3xi 212 26 Coefficient of: Basic Variable! Eq.|Z| Right x4XsKgXyI side x1 x2 x3 Z (0)-1-4M2 -6M3 -2M+2 M 0 M 014M 01 21100 8 0 0 06 4 Coefficient of: Right Variable Eq Zx1 2 Xa X4 XSide Basic 2(0)-1 M- 0.5 M 0.5 x2 (1) 0 X1(2 0 0.3 -0.2 -0.1 0.4