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Unformatted text preview: C&O 350 Linear Optimization Winter 2009 Solutions to Assignment 8 Question # Max. marks Part marks 1 16 1 1 2 1 3 2 1 2 3 2 23 1 2 1 5 1 2 5 2 2 2 3 14 5 9 4 12 4 4 4 5 No Credit 6 No Credit Total 65 1. 16 marks = 1+1+2+1+3+2+1+2+3 part (e): 1 for y , 1 each for checking c j ,j N Consider the linear programming problem ( P ) max { c T x : Ax = b, x } , where c T = , 1 , 2 , 1 , 2 , A = 2 1 5 4 1 1 1 1 1 1 2 1 1 1 , and b = 4 2 10 . Given B = { 3 , 4 , 5 } is a feasible basis. (Note: Parts (d) to (i) are independent of each other.) (a) Find the basic solution x * determined by basis B . (b) Find the solution y of A T B y = c B . (c) Compute c 1 and c 2 and show that x * is an optimal solution. (d) Suppose that c 1 in ( P ) is changed from zero to a real number . For what values of is x * still optimal? (e) Suppose that c 3 in ( P ) is changed from 2 to a real number . For what values of is x * still optimal? (f) Suppose that the value of b 1 is changed from 4 to a real number . For what values of does the basis B remain optimal? (g) Suppose that ( P ) is modified by the addition of a variable x 6 with A 6 = [1 , 2 , 3] T and c 6 = . For what values of does the basis B remain optimal? (h) Compute the tableau corresponding to the basis B . (Recall Chapters 6.7 and 9.1. The zrow is given by z j N c j x j = v , where c j = c j A T j y, j , v = c T x * = b T y , where y is the solution to the system A T B y = c B . The other rows are given by x B + A 1 B A N x N = A 1 B b .) (i) Suppose that a new constraint x 1 + x 2 x 3 + x 4 1 is added to ( P ). Does x * remain an optimal solution of the new problem? If not, how would you proceed to solve the new problem? (If pivots are needed, then give the entering and leaving variables for the first pivot, but do NOT go further.) 1 Solution : (a) x * = [0 , , 8 , 6 , 12] T . (b) y = [3 , 8 , 5] T . (c) c 1 = c 1 A T 1 y = 0 [2 , 1 , 1] 3 8 5 = 19 < 0. c 2 = c 2 A T 2 y = 1 [ 1 , 1 , 2] 3 8 5 = 0. Since c j 0 for all j / B , x * is an optimal solution. (d) Since 1 / B , x * remains optimal if and only if 0 c 1 =  19 if and only if 19. (e) Solving 5 1 1 4 1 1 1 1 y 1 y 2 y 3 = 1 2 , we get y 1 = + 1, y 2 = 3  2, y 3 = + 3. Hence, x * remains optimal if and only c 1 = 0 [2 , 1 , 1] + 1 3  2 + 3 = 6  7 and c 2 = 1 [ 1 , 1 , 2] + 1 3  2 + 3 = 2  4 if and only if 7 6 2 . (f) Solving 5 4 1 1 1 1 1 1 x * 3 x * 4 x * 5 = 2 10 , we get x * 3 = + 4, x * 4 = + 2, x * 5 = 12....
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 Spring '07
 S.Furino,B.Guenin

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