Definition a basic feasible solution is a basic

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Definition: A basic feasible solution is a basic solution which also satisfies the non-negativity constraints. Thus optimal solution can be found by performing a search of the basic feasible solution. The Simplex Method Before we begin our discussions of the simplex method, let’s provide a generalized statement of an LP model. Given the definitions x j = j- th decision variable c j = coefficient on j- th decision variable in the objective function. a ij = coefficient in the i- th constraint for the j- th variable. b i = right-hand-side constant for the i- th constraint The generalized LP model can be stated as follows: Optimize (maximize or minimize) z = c 1 x 1 + c 2 x 2 + ... + c n x n subject to a 11 x 1 + a 12 x 2 + ... + a 1n x n (≤ , ≥ , =) b1 (1) a 21 x 1 + a 22 x 2 + ... + a 2n x n (≤ , ≥ , =) b2 (2) ... a m1 x 1 + a m2 x 2 + ... + a mn x n (≤ , ≥ , =)b m (m) x 1 0 x 2 0 ... x n 0 Solution by Enumeration Consider a problem having m (≤) constraints and n variables. Prior to solving by the simplex method, the m constraints would be changed into equations by adding m slack variables. This
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74 Business Mathematics MTH-367 restatement results in a constraint set consisting of m equations and m + n variables. Example Solve the following LP problem. Maximize z = 5x 1 + 6x 2 subject to 3x 1 + 2x 2 ≤ 120 (1) 4x 1 + 6x 2 ≤ 260 (2) x 1 , x 2 ≥ 0 Solution: The constraint set must be transformed into the equivalent set. 3x 1 + 2x 2 + S 1 = 120 4x 1 + 6x 2 + S 2 = 260 x 1 , x 2 , S 1 , S 2 ≥ 0 The constraint set involves two equations and four variables. Of all the possible solutions to the constraint set, an optimal solution occurs when two of the four variables in this problem are set equal to zero and the system is solved for the other two variables. The question is, which two variables should be set equal to 0 (should be non-basic variables)? Let’s enumerate the different possibilities. 1. If S 1 and S 2 are set equal to 0, the constraint equations become 3x 1 + 2x 2 = 120 4x 1 + 6x 2 = 260 Solving for the corresponding basic variable x 1 and x 2 results in x 1 = 20 and x 2 = 30 2. If S 1 and x 1 are set equal to 0, the system becomes 2x 2 = 120 6x 2 + S 2 = 260 Solving for the corresponding basic variables x 2 and S 2 results in x 2 = 60 and S 2 = – 100 Following table summarizes the basic solutions, that is, all the
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75 Business Mathematics MTH-367 solution possibilities given that two of the four variables are assigned 0 values. Notice that solutions 2 and 5 are not feasible. They each contain a variable which has a negative value, violating the non-negativity restriction. However, solutions 1, 3, 4 and 6 are basic feasible solutions to the linear programming problem and are candidates for the optimal solution. The following figure is the graphical representation of the set of constraints.
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76 Business Mathematics MTH-367 Specifically, solution 1 corresponds to corner point C, solution 3 corresponds to corner point B, solution 4 corresponds to corner point D, and solution 6 corresponds to corner point A. Solution 2 and 5, which are not feasible, correspond to the points E and F shown in the figure.
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  • Spring '16
  • Business mathematics, Quadratic equation

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