The Simplex Method in Matrix Form: Steps and Example, Exams of Optimization Techniques in Engineering

Download The Simplex Method in Matrix Form: Steps and Example and more Exams Optimization Techniques in Engineering in PDF only on Docsity! Math 164 – Fall 2006 The Simplex method II Stefán Ingi Valdimarsson October 23, 2006 Let us describe the simplex method in matrix form. We are considering the linear programming problem in standard form min cTx s.t. Ax = b x ≥ 0. We assume that we have found a basic feasible solution, corresponding to the basis xB. We split A and x according to this basis, so A = ( B N ) , x = ( xB xN ) , c = ( cB cN ) . 1. The first step is to express the equality constraints Ax = b so that the basic variables appear in as simple form as possible. With our splitting, the equality constraints are BxB + NxN = b. We multiply this from the left by B−1. Note that the fact that B is invertible comes from the definition of xB being basic variables. This gives xB + B −1NxN = b̂ (1) where we define b̂ = B−1b. 1 2. We must now determine if we can decrease the value of the objective function by letting any of the non-basic variables become basic. In order to do this, we represent the objective function using only the non-basic variables. The original form of the objective function is z = cTx = cTBxB + c T NxN . Using (1) this becomes z =cTB(B −1b − B−1NxN ) + c T NxN =cTBB −1b + (cTN − c T BB −1N)xN =ẑ + ĉTNxN . where we define ĉTN = c T N − c T BB −1N and ẑ = cTBB −1b. We refer to the coefficient ĉt multiplying the variable xt of xN as the reduced cost of xt. We now examine the values in the vector ĉ. If a particular ct from this vector is negative then the value of the objective function will decrease if we let the non-basic xt enter the basis. • If no ct has a negative value then we should not choose any of the non-basic variables enter the basis and in fact this means that the current feasible basic solution is an optimal basic feasible solution and we are done. • If on the other hand there is a negative value in the ĉ we choose the variable xt such that ĉt has the largest (in magnitude) to enter the basis. This will decrease the objective function the fastest. 3. Next we must determine how large a value the variable that enters the basis can take. We note that since all the non-basic variables except xt will remain at the value 0 we get from (1) that xB = b̂ − Âtxt where Ât = B −1At and At is the column of A corresponding to xt. We examine this equality, component for component, with respect to the constraint that all of the variables in xB must remain non-negative. Thus (xB)i = b̂i − âi,txt ≤ 0. 2 We calulate B−1 =     1 −1 0 0 0 1 0 0 0 −3 1 0 0 0 0 1     ĉN = cN − N T(B−1)TcB = ( −3 2 ) and since −3 is the only negative number in ĉN we choose x1 to enter the basis. Then xB,old =     x3 x2 x5 x6     , Ât =     1 −1 4 1     , b̂ =     1 4 6 6     . The ratio test gives that x3 should leave the basis. Thus the next basis is xB = (x1, x2, x5, x6), with xN = (x3, x4). Then B =     0 1 0 0 −1 1 0 0 1 3 1 0 1 0 0 1     N =     1 0 0 1 0 0 0 0     cB =     −1 −2 0 0     cN = ( 0 0 ) . We calulate B−1 =     1 −1 0 0 1 0 0 0 −4 1 1 0 −1 1 0 1     ĉN = cN − N T(B−1)TcB = ( 3 −1 ) and since −1 is the only negative number in ĉN we choose x4 to enter the basis. Then xB,old =     x1 x2 x5 x6     , Ât =     −1 0 1 1     , b̂ =     1 5 2 5     . The ratio test gives that x5 should leave the basis. 5 Thus the next basis is xB = (x1, x2, x4, x6), with xN = (x3, x5). Then B =     0 1 0 0 −1 1 1 0 1 3 0 0 1 0 0 1     N =     1 0 0 0 0 1 0 0     cB =     −1 −2 0 0     cN = ( 0 0 ) . We calulate B−1 =     −3 0 1 0 1 0 0 0 −4 1 1 0 3 0 −1 1     ĉN = cN − N T(B−1)TcB = ( −1 1 ) and since −1 is the only negative number in ĉN we choose x3 to enter the basis. Then xB,old =     x1 x2 x4 x6     , Ât =     −3 1 −4 3     , b̂ =     3 5 2 3     . The ratio test gives that x6 should leave the basis. Thus the next basis is xB = (x1, x2, x4, x3), with xN = (x5, x6). Then B =     0 1 0 1 −1 1 1 0 1 3 0 0 1 0 0 0     N =     0 0 0 0 1 0 0 1     cB =     −1 −2 0 0     cN = ( 0 0 ) . We calulate B−1 =     0 0 0 1 0 0 1 3 −1 3 0 1 −1 3 4 3 1 0 −1 3 1 3     ĉN = cN − N T(B−1)TcB = ( 2 3 1 3 ) and since there is no negative number in ĉN we see that we have found a minimum and thus the solution to the problem. 6 Finally, we calculate the value of the basis variables at the optimum solution x̄B =     x1 x2 x4 x3     = B−1b =     6 4 6 1     and the value of the objective function ẑ = cTBx̄B = −14. Note: The part of the above calculation which appears to take the most time is the calculation of the inverse B−1 at each step. However, we can simplify this considerably if we do not start calculating the inverse from scratch in each step. Rather we note that when xt leaves the basis and xi′ enters the basis, the matrix B remains unchanged except that the column At gets replaced by Ai′ . Since B is invertible, this can be achieved by m suitably chosen column operations. Since carrying out column operations on B corresponds to doing row operations on B−1 we see that we can pass from B−1 old to B−1 new by carrying out m row operations. In fact these operations can be easily described as we noted in class when we carried out the simplex algorithm originally. We form the augmented matrix ( Ât | B −1 ) (2) and apply row operations to it until the column Ât has been transformed in to a column of only zeroes except there is a 1 in the place of the pivotal element, the one circled in the example above. 7