%		CS Final Exam Solutions	- Sunday May 10, 1998
>> %%%%%%%%%%%%%%%%%%%%%%%%%%%%%  Problem 1 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%
>> Q=[1 -1;-1 1];p=[-1;1];A=[1 -1];b=[-1];
>> M=[Q -A';A zeros(1,1)];q=[p;-b];
>> H=lemketbl(M,q);
              z1=x1      z2=x2      z3=u       1    
       --------------------------------------------
  w1= |     1.0000    -1.0000    -1.0000    -1.0000 
  w2= |    -1.0000     1.0000     1.0000     1.0000 
  w3= |     1.0000    -1.0000     0.0000     1.0000 
>> H=addcol(H,ones(3,1),'z0',4);
              z1         z2         z3         z0         1    
       -------------------------------------------------------
  w1= |     1.0000    -1.0000    -1.0000     1.0000    -1.0000 
  w2= |    -1.0000     1.0000     1.0000     1.0000     1.0000 
  w3= |     1.0000    -1.0000     0.0000     1.0000     1.0000 
>> H=ljx(H,1,4);
              z1         z2         z3         w1         1    
       -------------------------------------------------------
  z0= |    -1.0000     1.0000     1.0000     1.0000     1.0000 
  w2= |    -2.0000     2.0000     2.0000     1.0000     2.0000 
  w3= |     0.0000     0.0000     1.0000     1.0000     2.0000 
>> H=ljx(H,1,1);
              z0         z2         z3         w1         1    
       -------------------------------------------------------
  z1= |    -1.0000     1.0000     1.0000     1.0000     1.0000 
  w2= |     2.0000     0.0000     0.0000    -1.0000     0.0000 
  w3= |    -0.0000     0.0000     1.0000     1.0000     2.0000 
>> 
>> %Complementary tableau! x1=z1=1; x2=z2=0; u=z3=0
>> %Minimum value= -0.5
>> %In fact quadratic function has an unconstrained min on: {x | x1-x2-1=0}
>> %So solution set of QP is: {x | x1-x2-1=0, x1 >= 0}
>> %%%%%%%%%%%%%%%%%%%%%%%%%%%%%  Problem 2 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%
>> % min x3 subject to: -x3=<x1+x2-0=<x3
>> %                    -x3=<x1-x2-2=<x3, (x1,x2,x3)>=0
>> A=[1 1 1;1 -1 1;-1 -1 1;-1 1 1];b=[0 2 0 -2]';c=[0 0 1]';
>> H=totbl(A,b,c);
              x1         x2         x3         1    
       --------------------------------------------
  x4= |     1.0000     1.0000     1.0000    -0.0000 
  x5= |     1.0000    -1.0000     1.0000    -2.0000 
  x6= |    -1.0000    -1.0000     1.0000    -0.0000 
  x7= |    -1.0000     1.0000     1.0000     2.0000 
       --------------------------------------------
   z= |     0.0000     0.0000     1.0000     0.0000 
>> %Dual feasible so use dual simplex!
>> H=ljx(H,2,1);
              x5         x2         x3         1    
       --------------------------------------------
  x4= |     1.0000     2.0000     0.0000     2.0000 
  x1= |     1.0000     1.0000    -1.0000     2.0000 
  x6= |    -1.0000    -2.0000     2.0000    -2.0000 
  x7= |    -1.0000     0.0000     2.0000     0.0000 
       --------------------------------------------
   z= |     0.0000     0.0000     1.0000     0.0000 
>> H=ljx(H,3,3);
              x5         x2         x6         1    
       --------------------------------------------
  x4= |     1.0000     2.0000     0.0000     2.0000 
  x1= |     0.5000     0.0000    -0.5000     1.0000 
  x3= |     0.5000     1.0000     0.5000     1.0000 
  x7= |     0.0000     2.0000     1.0000     2.0000 
       --------------------------------------------
   z= |     0.5000     1.0000     0.5000     1.0000 
>> %Optimal tableau! Solution: x1=1, x2=0, min=z=x3=1
>> %%%%%%%%%%%%%%%%%%%%%%%%%%%%%  Problem 3 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%
>> %(a) False. Because x-bar is feasible for the second LP,
>> %since: Bx >= Ax >= b >= d for Ax >= b,  x>=0. Hence: beta =< alpha
>> %(b) False. Because minimum value could be positive. Counterexample:
>> %M=[0 1;0 1];q=[-1;0];
>> %min {z1*(z2-1)+z2^2 s.t. z2-1>=0, z2>=0,z1>=0,z2>=0}>0
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% End%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%