>> %%%%%%%%%%% CS525 Midterm March 9, 2000 Answers %%%%%%%%%%% >> %%%%%%%%%%%%%%%%%%%%% Question 1 Answer %%%%%%%%%%%%%%%%%%% >> A=[1 -1 -1;-1 2 0;1 1 -3];b=[-1 2 -1]';H=totbl(A,b); x1 x2 x3 1 -------------------------------------------- 0=y1= | 1.0000 -1.0000 -1.0000 1.0000 0=y2= | -1.0000 2.0000 0.0000 -2.0000 0=y3= | 1.0000 1.0000 -3.0000 1.0000 >> H=ljx(H,1,1); 0=y1 x2 x3 1 -------------------------------------------- x1= | 1.0000 1.0000 1.0000 -1.0000 0=y2= | -1.0000 1.0000 -1.0000 -1.0000 0=y3= | 1.0000 2.0000 -2.0000 0.0000 >> H=ljx(H,2,2); 0=y1 0=y2 x3 1 -------------------------------------------- x1= | 2.0000 1.0000 2.0000 0.0000 x2= | 1.0000 1.0000 1.0000 1.0000 0=y3= | 3.0000 2.0000 0.0000 2.0000 >> %Blocked! No solution because: y1=0 & y2=0 imply y3=2. >> %%%%%%%%%%%%%%%%%%%%% Question 2 Answer %%%%%%%%%%%%%%%%%%% >> A=[1 3 -2;1 1 -1;0 2 -1];b=[1 1 -1]';p=[-1 -1 2]';H=totbl(A,b,p); x1 x2 free x3= 1 -------------------------------------------- x4= | 1.0000 3.0000 -2.0000 -1.0000 0=x5= | 1.0000 1.0000 -1.0000 -1.0000 x6= | 0.0000 2.0000 -1.0000 1.0000 -------------------------------------------- z = | -1.0000 -1.0000 2.0000 0.0000 >> %Exchange 0=x5 with free x3 >> H=ljx(H,2,3); x1 x2 x5=0 1 -------------------------------------------- x4= | -1.0000 1.0000 2.0000 1.0000 frx3= | 1.0000 1.0000 -1.0000 -1.0000 x6= | -1.0000 1.0000 1.0000 2.0000 -------------------------------------------- z = | 1.0000 1.0000 -2.0000 -2.0000 >> %Optimal tableau (ignoring col x5 row x3)! z_min=-2 at x1=0, x2=0, x3=-1. >> %%%%%%%%%%%%%%%%%%%%% Question 3 Answer %%%%%%%%%%%%%%%%%%% >> A=[-13 1 -2;1 0 1;2 1 3];b=[-1 -2 -1]';p=[-3 -1 1]';H=totbl(A,b,p); x1 x2 x3 1 -------------------------------------------- x4 = | -13.0000 1.0000 -2.0000 1.0000 x5 = | 1.0000 0.0000 1.0000 2.0000 x6 = | 2.0000 1.0000 3.0000 1.0000 -------------------------------------------- z = | -3.0000 -1.0000 1.0000 0.0000 >> %Primal feasible, BUT unbounded! z-->-infinity for x1=x3=0 & x2-->+infinity >> %%%%%%%%%%%%%%%%%%%%% Question 4 Answer %%%%%%%%%%%%%%%%%%% >> A=[1 -2 -1;-1 1 1;-1 4 2];b=[3 -1 -8]';p=[1 2 1];H=totbl(A,b,p); x1 x2 x3 1 -------------------------------------------- x4 = | 1.0000 -2.0000 -1.0000 -3.0000 x5 = | -1.0000 1.0000 1.0000 1.0000 x6 = | -1.0000 4.0000 2.0000 8.0000 -------------------------------------------- z = | 1.0000 2.0000 1.0000 0.0000 >> %Dual feasible! Apply dual simplex phase II. >> H=dualbl(H);H=ljx(H,1,1); u4 = u5 = u6 = w = x1 x2 x3 1 -------------------------------------------- -u1 x4 = | 1.0000 -2.0000 -1.0000 -3.0000 -u2 x5 = | -1.0000 1.0000 1.0000 1.0000 -u3 x6 = | -1.0000 4.0000 2.0000 8.0000 -------------------------------------------- -1 z = | 1.0000 2.0000 1.0000 0.0000 u1 = u5 = u6 = w = x4 x2 x3 1 -------------------------------------------- -u4 x1 = | 1.0000 2.0000 1.0000 3.0000 -u2 x5 = | -1.0000 -1.0000 0.0000 -2.0000 -u3 x6 = | -1.0000 2.0000 1.0000 5.0000 -------------------------------------------- -1 z = | 1.0000 4.0000 2.0000 3.0000 >> %Dual unbounded! w-->+infinity for u4=u3=0 and u2-->+infinity. >> %Equivalently by Strong Duality or by above tableau primal is infeasible, >> %because: x4>=0, x2>=0, x3>=0 imply x5=<-2. >> diary off