CS525 Final Exam Solutions Sunday December 20, 1998 >> %%%%%%%%%%%%%%%%%%%%%%%%%%%% Problem 1 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% >> Q=[1 1;1 3];p=[-3;-2];A=[1 2;-1 1];b=[1;-5]; >> M=[Q -A';A zeros(2,2)];q=[p;-b]; >> H=lemketbl(M,q); >> H=addcol(H,ones(4,1),'z0',5); z1 z2 z3 z4 z0 1 ------------------------------------------------------------------ w1= | 1.0000 1.0000 -1.0000 1.0000 1.0000 -3.0000 w2= | 1.0000 3.0000 -2.0000 -1.0000 1.0000 -2.0000 w3= | 1.0000 2.0000 0.0000 0.0000 1.0000 -1.0000 w4= | -1.0000 1.0000 0.0000 0.0000 1.0000 5.0000 >> H=ljx(H,1,5); z1 z2 z3 z4 w1 1 ------------------------------------------------------------------ z0= | -1.0000 -1.0000 1.0000 -1.0000 1.0000 3.0000 w2= | 0.0000 2.0000 -1.0000 -2.0000 1.0000 1.0000 w3= | 0.0000 1.0000 1.0000 -1.0000 1.0000 2.0000 w4= | -2.0000 0.0000 1.0000 -1.0000 1.0000 8.0000 >> H=ljx(H,1,1); z0 z2 z3 z4 w1 1 ------------------------------------------------------------------ z1= | 3.0000 w2= | 1.0000 w3= | 2.0000 w4= | 2.0000 >> %Complementary tableau! x1=z1=3;x2=z2=0;u1=z3=0;u2=z4=0; >> %Minimum value= -9/2 >> %%%%%%%%%%%%%%%%%%%%%%%%%%%%% Problem 2 %%%%%%%%%%%%%%%%%%%%%%%%%%%%% >> % min x3+x4 subject to: -x3=> % -x4==0 >> A=[1 -1 1 0;1 1 0 1;-1 1 1 0;-1 -1 0 1];b=[0 -5 0 5]';c=[0 0 1 1]'; >> H=totbl(A,b,c); x1 x2 x3 x4 1 ------------------------------------------------------- x5= | 1.0000 -1.0000 1.0000 0.0000 -0.0000 x6= | 1.0000 1.0000 0.0000 1.0000 5.0000 x7= | -1.0000 1.0000 1.0000 0.0000 -0.0000 x8= | -1.0000 -1.0000 0.0000 1.0000 -5.0000 ------------------------------------------------------- z= | 0.0000 0.0000 1.0000 1.0000 0.0000 >> %Dual feasible so use dual simplex! >> H=ljx(H,4,4); x1 x2 x3 x8 1 ------------------------------------------------------- x5= | 0.0000 x6= | 10.0000 x7= | 0.0000 x4= | 5.0000 ------------------------------------------------------- z= | 1.0000 1.0000 1.0000 1.0000 5.0000 >> %Optimal tableau! Solution: x1=0, x2=0, min=z=x3+x4=0+5=5 >> %%%%%%%%%%%%%%%%%%%%%Problem 3 Solution%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% >> A=[1 -1];b=[-2];p=[1;1];q=[-1;-2]; >> H=totbl(A,b,p); x1 x2 1 --------------------------------- x3= | 1.0000 -1.0000 2.0000 --------------------------------- z= | 1.0000 1.0000 0.0000 >> H=addrow(H,[q' 0],'z0',2); x1 x2 1 --------------------------------- x3= | 1.0000 -1.0000 2.0000 z0= | -1.0000 -2.0000 0.0000 --------------------------------- z= | 1.0000 1.0000 0.0000 >> %1-t>=0, 1-2t>=0 imply optimal tableau for 1/2>=t>-infinity >> H=ljx(H,1,2); x1 x3 1 --------------------------------- x2= | 1.0000 -1.0000 2.0000 z0= | -3.0000 2.0000 -4.0000 --------------------------------- z= | 2.0000 -1.0000 2.0000 >> %2-3t>=0, -1+2t>=0 imply optimal tableau for 2/3>=t>=1/2 >> %Unbounded below for t>2/3 because column x1 is positive in row 1 >> % Summary >> % 1/2>=t>-infinity: z(t)=0 at x1=0,x2=0 >> % 2/3>=t>=1/2: z(t)=2-4t at x1=0, x2=2 >> % t>2/3: z(t) unbounded below >> %%%%%%%%%%%%%%%%%%%%%%%%%Problem 4 Solution %%%%%%%%%%%%%%%%%%%%%%%%%%%%%% A solution x of the primal also solves the dual (because both problems are the same): max -c'u s.t. -Au = A'u =< c, u >= 0 Hence, c'x = -c'u = -c'x, and consequently 2c'x=0. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%