Optimisation Lecture 7 - Easter 2017 Michael Tehranchi How to use the simplex algorithm We are still considering the problem to maximise cx subject to Ax = b, x ? 0. Suppose we have a given basic feasible solution xº. Let B C {1, ... , n} the set of basic indices and N the set of non-basic indices, and so that we can rewrite the problem as maximise CBXB + CNIN subject to ABIB + ANIN = b, XB, IN ? 0. Since the given b.f.s. 20 is such that x) = 0 and hence xB = A=1b, we can further rewrite the problem as maximise cTx0 + (CN - ANA)EN subject to AB ANIN + XB = XB, XB, IN > 0, where > = (AB)CB is the Lagrange multiplier matched to the b.f.s. r° by complementary slackness. To avoid all this heavy notation, we will rewrite our problem once more as maximize cxN + k subject to AxN + B=b, xB ? 0, IN ? 0 where b, c and A are not necessarily the same as in the original problem. Note that the initial b.f.s. solution in this new notation is CB = (0). Note that also that k is the value of the objective function at the initial b.f.s. : N. Step(0). The initial simplextableau is ? payoff XB A I b C 0 -k where I is the m x m identity. (1) Testforoptimality. If c ? 0 then STOP! Otherwise go to step (2). (2) Choose thepivotcolumn. Pick a j such that c; > 0. (Rule of thumb: Pick j such that cj is largest) (3) Choose the pivotrow. Look within the pivot column j and find the i which minimizes b? / ai,j over all i = 1, ... , m such that ai,j > 0. (4) Performthepivotoperation. Move to the next b.f.s. as follows: · Replace row i with (old row i) / ai,j. · Replace row k with (old row k) - (old row i) x ak,j / ai,j, for k # i and including the payoff row 1
Now return to step (1). Example. Consider the linear program to P : maximise 3x1 + 2x2 subject to 2x1 + x2 ? 4, x1, X2 > 0 2x1 + 3x2 ? 6. Before using the simplex algorithm, we do some side computations to see what is going on. The dual problem is to D : minimise 411 + 612 subject to 2×1 + 2×2 > 3, 1, 2 > 0 >1 +32 ? 2 We introduce slack variables as usual to both problems, and list all of the basic solutions, paired by complementary slackness: X1V1 = X2V2 = 21\1 = 22\2 = 0. The graph shows the set of feasible solutions of both problems. ?1= 0 F À1 = 0 2x1 +3x2 = 6 E C D M1 + 3^2 = 2 - E A B C -@2 = 0 2x +@2 = 4 D A B - X2 =