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Here, m=2 and there are 10 subsets of 2 indices, however, not all of them are bases: the set {3,5} is not a basis since columns 3 and 5 are linearly dependent. The set B={2,4} is a basis, since the matrix = is non-singular.
Let S 1 be the selling price of wheat and S 2 be the selling price of barley, per hectare. If we denote the area of land planted with wheat and barley by x 1 and x 2 respectively, then profit can be maximized by choosing optimal values for x 1 and x 2. This problem can be expressed with the following linear programming problem in the standard form:
However, there is a fractional solution in which each set is assigned the weight 1/2, and for which the total value of the objective function is 3/2. Thus, in this example, the linear programming relaxation has a value differing from that of the unrelaxed 0–1 integer program.
where x i is a basic variable and the x j 's are the nonbasic variables (i.e. the basic solution which is an optimal solution to the relaxed linear program is = ¯ and =). We write coefficients b ¯ i {\displaystyle {\bar {b}}_{i}} and a ¯ i , j {\displaystyle {\bar {a}}_{i,j}} with a bar to denote the last tableau produced by the simplex method.
The combined LP has both x and y as variables: Maximize 1. subject to Ax ≤ b, A T y ≥ c, c T x ≥ b T y, x ≥ 0, y ≥ 0. If the combined LP has a feasible solution (x,y), then by weak duality, c T x = b T y. So x must be a maximal solution of the primal LP and y must be a minimal solution of the dual LP. If the combined LP has no ...
Columns 2, 3, and 4 can be selected as pivot columns, for this example column 4 is selected. The values of z resulting from the choice of rows 2 and 3 as pivot rows are 10/1 = 10 and 15/3 = 5 respectively. Of these the minimum is 5, so row 3 must be the pivot row. Performing the pivot produces
Solve the problem using the usual simplex method. For example, x + y ≤ 100 becomes x + y + s 1 = 100, whilst x + y ≥ 100 becomes x + y − s 1 + a 1 = 100. The artificial variables must be shown to be 0. The function to be maximised is rewritten to include the sum of all the artificial variables.
Choose q = 3 as the entering index. Then d = [1 3] T, which means a unit increase in x 3 results in x 4 and x 5 being decreased by 1 and 3, respectively. Therefore, x 3 is increased to 5, at which point x 5 is reduced to zero, and p = 5 becomes the leaving index. After the pivot operation,