QUESTION 1: A = [-3 2; 0.25 0.5]; b = [0; 0.5]; p = [-1 3]'; T = totbl(A,b,p); x1 x2 1 --------------------------------- x3 = | -3.0000 2.0000 -0.0000 x4 = | 0.2500 0.5000 -0.5000 --------------------------------- z = | -1.0000 3.0000 0.0000 T = addcol(T,[0 1 0]','x0',3); x1 x2 x0 1 -------------------------------------------- x3 = | -3.0000 2.0000 0.0000 -0.0000 x4 = | 0.2500 0.5000 1.0000 -0.5000 -------------------------------------------- z = | -1.0000 3.0000 0.0000 0.0000 T = addrow(T,[0 0 1 0],'z0',4); x1 x2 x0 1 -------------------------------------------- x3 = | -3.0000 2.0000 0.0000 -0.0000 x4 = | 0.2500 0.5000 1.0000 -0.5000 -------------------------------------------- z = | -1.0000 3.0000 0.0000 0.0000 z0 = | 0.0000 0.0000 1.0000 0.0000 T = ljx(T,2,3); x1 x2 x4 1 -------------------------------------------- x3 = | -3.0000 2.0000 0.0000 -0.0000 x0 = | -0.2500 -0.5000 1.0000 0.5000 -------------------------------------------- z = | -1.0000 3.0000 0.0000 0.0000 z0 = | -0.2500 -0.5000 1.0000 0.5000 T = ljx(T,2,2); x1 x0 x4 1 -------------------------------------------- x3 = | -4.0000 -4.0000 4.0000 2.0000 x2 = | -0.5000 -2.0000 2.0000 1.0000 -------------------------------------------- z = | -2.5000 -6.0000 6.0000 3.0000 z0 = | 0.0000 1.0000 0.0000 0.0000 T = delcol(T,'x0'); x1 x4 1 --------------------------------- x3 = | -4.0000 4.0000 2.0000 x2 = | -0.5000 2.0000 1.0000 --------------------------------- z = | -2.5000 6.0000 3.0000 z0 = | 0.0000 0.0000 0.0000 T = delrow(T,'z0'); x1 x4 1 --------------------------------- x3 = | -4.0000 4.0000 2.0000 x2 = | -0.5000 2.0000 1.0000 --------------------------------- z = | -2.5000 6.0000 3.0000 T = ljx(T,1,1); x3 x4 1 --------------------------------- x1 = | -0.2500 1.0000 0.5000 x2 = | 0.1250 1.5000 0.7500 --------------------------------- z = | 0.6250 3.5000 1.7500 % x1 = 0.5, x2 = 0.75, z = -1.75 (max) QUESTION 2: A = [1 -2 1; 3 1 0; 0 0 1]; b = [1; -5; 2]; p = [1 1 1]'; T = totbl(A,b,p,3); x1 x2 x3 1 -------------------------------------------- x4 = | 1.0000 -2.0000 1.0000 -1.0000 x5 = | 3.0000 1.0000 0.0000 5.0000 x6 = | 0.0000 0.0000 1.0000 -2.0000 -------------------------------------------- z = | 1.0000 1.0000 1.0000 3.0000 T = ljx(T,1,1); x4 x2 x3 1 -------------------------------------------- x1 = | 1.0000 2.0000 -1.0000 1.0000 x5 = | 3.0000 7.0000 -3.0000 8.0000 x6 = | 0.0000 0.0000 1.0000 -2.0000 -------------------------------------------- z = | 1.0000 3.0000 0.0000 4.0000 T = delcol(T,'x4'); x2 x3 1 --------------------------------- x1 = | 2.0000 -1.0000 1.0000 x5 = | 7.0000 -3.0000 8.0000 x6 = | 0.0000 1.0000 -2.0000 --------------------------------- z = | 3.0000 0.0000 4.0000 T = ljx(T,3,2); x2 x6 1 --------------------------------- x1 = | 2.0000 -1.0000 -1.0000 x5 = | 7.0000 -3.0000 2.0000 x3 = | -0.0000 1.0000 2.0000 --------------------------------- z = | 3.0000 0.0000 4.0000 T = permrows(T,[2 4 1 3]); x2 x6 1 --------------------------------- x5 = | 7.0000 -3.0000 2.0000 --------------------------------- z = | 3.0000 0.0000 4.0000 x1 = | 2.0000 -1.0000 -1.0000 x3 = | -0.0000 1.0000 2.0000 % x1 = -1; x2 = 0; x3 = 2; z = 4 QUESTION 3: A = [-1 2 4; -1 -2 -5]; b = [1; -2]; p = [1 0 1]'; T = totbl(-A,-b,p); x1 x2 x3 1 -------------------------------------------- x4 = | 1.0000 -2.0000 -4.0000 1.0000 x5 = | 1.0000 2.0000 5.0000 -2.0000 -------------------------------------------- z = | 1.0000 0.0000 1.0000 0.0000 T = ljx(T,2,2); x1 x5 x3 1 -------------------------------------------- x4 = | 2.0000 -1.0000 1.0000 -1.0000 x2 = | -0.5000 0.5000 -2.5000 1.0000 -------------------------------------------- z = | 1.0000 0.0000 1.0000 0.0000 T = ljx(T,1,1); x4 x5 x3 1 -------------------------------------------- x1 = | 0.5000 0.5000 -0.5000 0.5000 x2 = | -0.2500 0.2500 -2.2500 0.7500 -------------------------------------------- z = | 0.5000 0.5000 0.5000 0.5000 % x1 = 0.5 x2 = 0.75 x3 = 0; z = 0.5 QUESTION 4: a) max u1 - u2 + 3 u3 st u1 + u2 + u3 <= 1 -3u1 + u3 <= -1 5u1 + 4u2 + u3 <=2 2u1 + 3u2 + u3 <= 1 u1,u2,u3 >= 0 b) Point is feasible for dual. From KKT(d), since slack in first and fourth dual constraints is positive, x1 = x4 = 0 Since u1 and u3 are positive, first and third primal constraints must be equations, hence -3 x2 + 5 x3 = 1 x2 + x3 = 3 Solving gives: x2 = 1.75, x3 = 1.25 For this feasible x, and the given u, primal obj = dual obj = 0.75, so both are optimal. c) Opt soln = (0, 1.75, 1.25, 0); d) Opt value = 0.75