Math 187 Solutions to Homework 2: Part 2 Radunskaya 4. Give an ...

which occurs in the entry corresponding to x 4 . So x 4 leaves B, which gives us the desired example, sincex 4 had just entered B on the previous iteration.Geometrically, the algorithm started at the vertex of the feasible region at the origin, moved along thehorizontal axis to the vertex at (.0556, 0), and then moved up to the (optimal) vertex at (0, .5).5. Show that the variable that becomes nonbasic in one iteration of the simplex method cannot become basicin the next iteration.Example: We start with a (hopefully) illuminating example, where we consider an LP in standard formusing the dictionary notation:z = c 1 x 1 + c 2 x 2 + · · · + c k x k− − −− −− − − − − − − − − − − − −w 1 = b 1 − a 11 x 1 − a 12 x 2 − . . .w 2 = b 2 − a 21 x 1 − a 22 x 2 − . . ..w m = b m − a m1 x 1 − a m2 x 2 − . . .Since we can always rearrange the variables, we assume that c 1 is the largest positive coefficient in theobjective function, so that x 1 becomes basic. Also, we assume that a 11 > 0 andb 1a 11≤ b ja j1, 2 ≤ j ≤ m, for all j such that a j1 > 0i.e. we assume that the basic variables are ordered so that w 1 now becomes non-basic.The dictionary then becomes:z =b 1c 1 − c 1w 1 + (c 2 − c 1)x 2 + . . .a 11 a 11 a 12− − −− −− − − − − − − − − − − − −x 1 =b 1− 1 w 1 − a 12x 2 − . . .a 11 a 11 a 11w 2 = b 2 − a 21 ( b 1a 11− 1a 11w 1 . . . ) − a 22 x 2 − . . ..w m = b m − a m1 ( b 1a 11− 1a 11w 1 . . . ) − a m2 x 2 − . . .Since we assumed that both c 1 and a 11 were positive, the coefficient of w 1 in the objective function isnegative, and w 1 cannot become basic in the next iteration.What made this example easy was the fact that we were looking at the first step of the algorithm, so that itwas clear that the coefficient of w 1 in the reduced cost vector was negative. It is clear, also, that we wouldget the same result if we used the matrix notation rather than the dictionary notation. Fortunately, linearalgebra allows us to transform the argument that works for the first iteration of the algorithm so that itapplies to an arbitrary iteration.Solution: Given the standard LP problem: maximize z = c T x, subject to: Ax = b, where A is a m × n realmatrix of rank m, and the vectors c and x are in R n , while b ∈ R m . We claim that a variable that becomesnon-basic in a given iteration cannot become basic in the next iteration.Suppose we are about to start the kth iteration of the simplex method, so that the objective function is:z = c T BB −1 b + (c T N − c T BB −1 N)x N = c T BB −1 b + ¯cx N (1)2