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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:
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. Then row reductions are applied to gain a final solution.
In mathematical optimization, the fundamental theorem of linear programming states, in a weak formulation, that the maxima and minima of a linear function over a convex polygonal region occur at the region's corners.
For the rest of the discussion, it is assumed that a linear programming problem has been converted into the following standard form: =, where A ∈ ℝ m×n.Without loss of generality, it is assumed that the constraint matrix A has full row rank and that the problem is feasible, i.e., there is at least one x ≥ 0 such that Ax = b.
Suppose we have the linear program: Maximize c T x subject to Ax ≤ b, x ≥ 0.. We would like to construct an upper bound on the solution. So we create a linear combination of the constraints, with positive coefficients, such that the coefficients of x in the constraints are at least c T.
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.
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.
The optimal answer requires 73 master rolls and has 0.401% waste; it can be shown computationally that in this case the minimum number of patterns with this level of waste is 10. It can also be computed that 19 different such solutions exist, each with 10 patterns and a waste of 0.401%, of which one such solution is shown below and in the picture: