CS 525 Final Exam Solutions Wednesday December 18, 1996 >> %Problem 1 Solution%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% >> Q=[1 -1;-1 1];p=[8 -2]';A=[1 2;1 -1];b=[1 -7]'; >> M=[Q -A';A zeros(2,2)];q=[p;-b]; >> H=lemketbl(M,q); z1 z2 z3 z4 1 ------------------------------------------------------- w1= | 1.0000 -1.0000 -1.0000 -1.0000 8.0000 w2= | -1.0000 1.0000 -2.0000 1.0000 -2.0000 w3= | 1.0000 2.0000 0.0000 0.0000 -1.0000 w4= | 1.0000 -1.0000 0.0000 0.0000 7.0000 >> H=addcol(H,[1 1 1 1]','z0',5); z1 z2 z3 z4 z0 1 ------------------------------------------------------------------ w1= | 1.0000 -1.0000 -1.0000 -1.0000 1.0000 8.0000 w2= | -1.0000 1.0000 -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 7.0000 >> H=ljx(H,2,5); z1 z2 z3 z4 w2 1 ------------------------------------------------------------------ w1= | 2.0000 -2.0000 1.0000 -2.0000 1.0000 10.0000 z0= | 1.0000 -1.0000 2.0000 -1.0000 1.0000 2.0000 w3= | 2.0000 1.0000 2.0000 -1.0000 1.0000 1.0000 w4= | 2.0000 -2.0000 2.0000 -1.0000 1.0000 9.0000 >> H=ljx(H,2,2); z1 z0 z3 z4 w2 1 ------------------------------------------------------------------ w1= | 0.0000 2.0000 -3.0000 0.0000 -1.0000 6.0000 z2= | 1.0000 -1.0000 2.0000 -1.0000 1.0000 2.0000 w3= | 3.0000 -1.0000 4.0000 -2.0000 2.0000 3.0000 w4= | 0.0000 2.0000 -2.0000 1.0000 -1.0000 5.0000 >> %Complementary tableau! >> %x1=z1=0, x2=z2=2, u1=z3=0, u2=z4=0, z0=0 >> % w1=6, w2=0, w3=3, w4=5. Hence z'w=0. >> % min f(x)=-2. %Problem 2 Solution%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% >> %min x3 subject to: -x3=> % -x3=> A=[1 1 1;-1 -1 1;1 -1 1;-1 1 1];b=[1 -1 -2 2]';c=[0 0 1]'; >> H=totbl(A,b,c); x1 x2 x3 1 -------------------------------------------- x4= | 1.0000 1.0000 1.0000 -1.0000 x5= | -1.0000 -1.0000 1.0000 1.0000 x6= | 1.0000 -1.0000 1.0000 2.0000 x7= | -1.0000 1.0000 1.0000 -2.0000 -------------------------------------------- z= | 0.0000 0.0000 1.0000 0.0000 >> H=ljx(H,4,2); x1 x7 x3 1 -------------------------------------------- x4= | 2.0000 1.0000 0.0000 1.0000 x5= | -2.0000 -1.0000 2.0000 -1.0000 x6= | 0.0000 -1.0000 2.0000 0.0000 x2= | 1.0000 1.0000 -1.0000 2.0000 -------------------------------------------- z= | 0.0000 0.0000 1.0000 0.0000 >> H=ljx(H,2,3); x1 x7 x5 1 -------------------------------------------- x4= | 2.0000 1.0000 0.0000 1.0000 x3= | 1.0000 0.5000 0.5000 0.5000 x6= | 2.0000 0.0000 1.0000 1.0000 x2= | 0.0000 0.5000 -0.5000 1.5000 -------------------------------------------- z= | 1.0000 0.5000 0.5000 0.5000 >> %Optimal tableau! >> %min x3=z=1/2 at x1=0, x2=1.5. >> %Problem 3 Solution%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% >> A=[-1 1;-1 0];b=[-1 -2]';p=[-1 0]';q=[2 1]'; >> H=totbl(A,b,p); x1 x2 1 --------------------------------- x3= | -1.0000 1.0000 1.0000 x4= | -1.0000 0.0000 2.0000 --------------------------------- z= | -1.0000 0.0000 0.0000 >> H=addrow(H,[q' 0],'z0',3); x1 x2 1 --------------------------------- x3= | -1.0000 1.0000 1.0000 x4= | -1.0000 0.0000 2.0000 z0= | 2.0000 1.0000 0.0000 --------------------------------- z= | -1.0000 0.0000 0.0000 >> %-1+2t>=0, 0+t>=0 imply optimal tableau for t>=1/2 >> %-1+2t<0 when t<1/2. So first column is pivot column. >> H=ljx(H,1,1); x3 x2 1 --------------------------------- x1= | -1.0000 1.0000 1.0000 x4= | 1.0000 -1.0000 1.0000 z0= | -2.0000 3.0000 2.0000 --------------------------------- z= | 1.0000 -1.0000 -1.0000 >> %1-2t>=0, -1+3t>=0 imply optimal tableau for 1/2>=t>=1/3 >> %-1+3t<0 when t<1/3 so column 2 is pivot column. >> H=ljx(H,2,2); x3 x4 1 --------------------------------- x1= | 0.0000 -1.0000 2.0000 x2= | 1.0000 -1.0000 1.0000 z0= | 1.0000 -3.0000 5.0000 --------------------------------- z= | 0.0000 1.0000 -2.0000 >> %0+t>=0, 1-3t>=0 imply optimal tableau for 1/3>=t>=0 >> %0+t<0 fot t<0 so column 1 is pivot column. But..... >> %column 1 has no negative elements so z(t) is unbounded below for t<0. >> %Summary%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% >> %For t>=1/2, optimal z(t)=0 at x1=0, x2=0 >> %For 1/2>=t>=1/3, optimal z(t)=-1+2t at x1=1, x2=0 >> %For 1/3>=t>=0, optimal z(t)=-2+5t at x1=2, x2=1 >> %For 0>t, z(t) is unbounded below %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%