By Michael R. Greenberg

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**Example text**

In the following tableau, vectors Xx and X3 are part of the basis and have the values of 2 and 1, respectively. Vector 2 is not part of the basis and has a value of 0. Basis vectors Solution values (b) 1 3 2 1 Vectors 1 2 3 1 3 0 0 2 1 In the second tableau, vector X3 has a value of 4 and vector 2 has a value of 2. Vector 1 has a value of 0. Vectors Basis vectors Solution values (b) 3 4 2 2 2 6 1 2 0 3 1 1 0 The exception to the rule that basis vectors assume nonzero values is the degenerate solution.

If you follow these three guidelines, the maximum number of row transformations that are required equals the number of elements in the matrix. In the matrix Ä case presented above, seven transformations were necessary to convert Ä into /. Two steps below the maximum were possible because in the column 1 operations (between steps 2 and 3), the element a31 was 0 in the original Ä matrix. In the column 3 operations, the element a 3 3 was the required value 1 after step 5. As a summary of matrix addition, subtraction, and multiplication, the following equation will be solved : ÄÄ + AB1 - 2BÄ where *-Gîl· -[-13 To find AA [o Î] [o Η ί 3 To find B~1 [-1 S H.

Maximize: Z = 2XX + X2 + 0X 3 + 0X 4 + 0X5 k Subject tO : X1 + 2X2 + Ί ^ 2XX + 3X2 3X, + X2 (1) variabless l a c ' + X± N = 10 (2) =12 (3) + X5 = 15 (4) Xl >0 (5) X2 >0 (6) Optimal solution: Z = lOf ; Xx =ψ;Χ2= f ; X3 =^;X4 = 0; X5 = 0. Inspection of the graphical solution (Fig. 1) indicates that Eqs. (3) and (4) are the restrictive constraints. If we change the availability of the resource in constraint Eq. (2) from Xx + 2X2 + X3 = 10 to Xt + 2X2 + X3 = 9, the optimal value of the solution will remain at lOf.