>> %Solutions to CS525 Midterm of March 23, 1993 -------------------------------------------- >> %Note: Tableau count is no more than 4 per problem if done by hand, >> %or if we don't count row/column addition/suppression tableaus, >> %as separate tableaus in MATLAB. >> %Answer 1 >> A=[1 2 -1;1 1 0;0 2 -1;-1 0 1];b=[1 2 -1 -1]'; >> H=totbl(A,b); x1 x2 x3 1 -------------------------------------------- y1= | 1.0000 2.0000 -1.0000 -1.0000 y2= | 1.0000 1.0000 0.0000 -2.0000 y3= | 0.0000 2.0000 -1.0000 1.0000 y4= | -1.0000 0.0000 1.0000 1.0000 >> H=ljx(H,1,1); y1 x2 x3 1 -------------------------------------------- x1= | 1.0000 -2.0000 1.0000 1.0000 y2= | 1.0000 -1.0000 1.0000 -1.0000 y3= | 0.0000 2.0000 -1.0000 1.0000 y4= | -1.0000 2.0000 0.0000 0.0000 >> H=ljx(H,2,2); y1 y2 x3 1 -------------------------------------------- x1= | -1.0000 2.0000 -1.0000 3.0000 x2= | 1.0000 -1.0000 1.0000 -1.0000 y3= | 2.0000 -2.0000 1.0000 -1.0000 y4= | 1.0000 -2.0000 2.0000 -2.0000 >> H=ljx(H,3,3); y1 y2 y3 1 -------------------------------------------- x1= | 1.0000 0.0000 -1.0000 2.0000 x2= | -1.0000 1.0000 1.0000 0.0000 x3= | -2.0000 2.0000 1.0000 1.0000 y4= | -3.0000 2.0000 2.0000 0.0000 >> %Solution: x1=2;x2=0;x3=1.(Note: y4=0 is implied by y1=y2=y3=0.) ___________________________________________________________________ >> %Answer 2 >> A=[1 -2 1;1 -1 2;1 -1 0];b=[1 -1 -2]';p=[1 -1 -1]'; >> H=totbl(-A,-b,-p); x1 x2 x3 1 -------------------------------------------- x4= | -1.0000 2.0000 -1.0000 1.0000 x5= | -1.0000 1.0000 -2.0000 -1.0000 x6= | -1.0000 1.0000 -0.0000 -2.0000 -------------------------------------------- -z= | -1.0000 1.0000 1.0000 0.0000 >> H=addcol(H,[0 1 1 0]','x0',4); x1 x2 x3 x0 1 ------------------------------------------------------- x4= | -1.0000 2.0000 -1.0000 0.0000 1.0000 x5= | -1.0000 1.0000 -2.0000 1.0000 -1.0000 x6= | -1.0000 1.0000 -0.0000 1.0000 -2.0000 ------------------------------------------------------- -z= | -1.0000 1.0000 1.0000 0.0000 0.0000 >> H=addrow(H,[0 0 0 1 0],'z0',5); x1 x2 x3 x0 1 ------------------------------------------------------- x4= | -1.0000 2.0000 -1.0000 0.0000 1.0000 x5= | -1.0000 1.0000 -2.0000 1.0000 -1.0000 x6= | -1.0000 1.0000 -0.0000 1.0000 -2.0000 ------------------------------------------------------- -z= | -1.0000 1.0000 1.0000 0.0000 0.0000 z0= | 0.0000 0.0000 0.0000 1.0000 0.0000 >> H=ljx(H,3,4); x1 x2 x3 x6 1 ------------------------------------------------------- x4= | -1.0000 2.0000 -1.0000 0.0000 1.0000 x5= | 0.0000 0.0000 -2.0000 1.0000 1.0000 x0= | 1.0000 -1.0000 0.0000 1.0000 2.0000 ------------------------------------------------------- -z= | -1.0000 1.0000 1.0000 0.0000 0.0000 z0= | 1.0000 -1.0000 0.0000 1.0000 2.0000 >> H=ljx(H,3,2); x1 x0 x3 x6 1 ------------------------------------------------------- x4= | 1.0000 -2.0000 -1.0000 2.0000 5.0000 x5= | 0.0000 -0.0000 -2.0000 1.0000 1.0000 x2= | 1.0000 -1.0000 0.0000 1.0000 2.0000 ------------------------------------------------------- -z= | 0.0000 -1.0000 1.0000 1.0000 2.0000 z0= | 0.0000 1.0000 0.0000 0.0000 0.0000 >> H(5,:)=[];H(:,2)=[];tbl(H); x1 x3 x6 1 -------------------------------------------- x4= | 1.0000 -1.0000 2.0000 5.0000 x5= | 0.0000 -2.0000 1.0000 1.0000 x2= | 1.0000 0.0000 1.0000 2.0000 -------------------------------------------- -z= | 0.0000 1.0000 1.0000 2.0000 >> %Optimal Phase II tableau! z(min)=-2 at x1=0, x2=2, x3=0. ___________________________________________________________________ >> %Answer 3 >> A=[1 -1 2;2 3 1;-3 -1 1];b=[1 -2 1]';p=[2 1 -2]'; >> H=totbl(-A,-b,-p); x1 x2 x3 1 -------------------------------------------- x4= | -1.0000 1.0000 -2.0000 1.0000 x5= | -2.0000 -3.0000 -1.0000 -2.0000 x6= | 3.0000 1.0000 -1.0000 1.0000 -------------------------------------------- -z= | -2.0000 -1.0000 2.0000 0.0000 >> H=ljx(H,3,2); x1 x6 x3 1 -------------------------------------------- x4= | -4.0000 1.0000 -1.0000 0.0000 x5= | 7.0000 -3.0000 -4.0000 1.0000 x2= | -3.0000 1.0000 1.0000 -1.0000 -------------------------------------------- -z= | 1.0000 -1.0000 1.0000 1.0000 >> H(3,:)=[];H(:,2)=[];tbl(H); x1 x3 1 --------------------------------- x4= | -4.0000 -1.0000 0.0000 x5= | 7.0000 -4.0000 1.0000 --------------------------------- -z= | 1.0000 1.0000 1.0000 >> %Optimal tableau. z(min)=-1 at x1=0, x2=-1, x3=0. _____________________________________________________________________ >> %Answer 4 >> %Dual LP: max w:= u1 - 11u2 + 2u3 >> %s.t. -u1 + u2 + 5u3 <= 9 >> % u1 - 3u2 - u3 <= -1 >> % -u1 + u2 - u3 <= -1 >> % u1, u2, u3 >= 0 >> % u1=1, u2=0, u3=2 is dual feasible with w=5 >> % Since x1=3/4, x2=7/4, x3=0, gives z=5, this a solution. ____________________________________________________________________