>> %%%%%%%%%%%%%%  Solutions to CS525 May 11, 1999 Final %%%%%%%%%%%%%
>> %%%%%%%%%%%%%%%%%%%%%%%%% Problem 1 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
>> Q=[4 2;2 1];p=[-1;-2];A=[2 2;1 -1];b=[1;-3];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= |     4.0000     2.0000    -2.0000    -1.0000     1.0000    -1.0000 
  w2= |     2.0000     1.0000    -2.0000     1.0000     1.0000    -2.0000 
  w3= |     2.0000     2.0000     0.0000     0.0000     1.0000    -1.0000 
  w4= |     1.0000    -1.0000     0.0000     0.0000     1.0000     3.0000 
>> H=ljx(H,2,5);
              z1         z2         z3         z4         w2         1    
       ------------------------------------------------------------------
  w1= |     2.0000     1.0000     0.0000    -2.0000     1.0000     1.0000 
  z0= |    -2.0000    -1.0000     2.0000    -1.0000     1.0000     2.0000 
  w3= |     0.0000     1.0000     2.0000    -1.0000     1.0000     1.0000 
  w4= |    -1.0000    -2.0000     2.0000    -1.0000     1.0000     5.0000 
>> H=ljx(H,2,2);
              z1         z0         z3         z4         w2         1    
       ------------------------------------------------------------------
  w1= |                                                            3.0000 
  z2= |                                                            2.0000 
  w3= |                                                            3.0000 
  w4= |                                                            1.0000 
>> %Complementary tableau at x1=z1=0;x2=z2=2;u1=z3=0;u2=z4=0.Minimum= -2.
>> %%%%%%%%%%%%%%%%%%%%%%%%%% Problem 2 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
>> A=[1 -1 1;1 -2 1;-1 1 1;-1 2 1];b=[2 4 -2 -4]';c=[0 0 1]';
>> H=totbl(A,b,c);
              x1         x2         x3         1    
       --------------------------------------------
  x4= |     1.0000    -1.0000     1.0000    -2.0000 
  x5= |     1.0000    -2.0000     1.0000    -4.0000 
  x6= |    -1.0000     1.0000     1.0000     2.0000 
  x7= |    -1.0000     2.0000     1.0000     4.0000 
       --------------------------------------------
   z= |     0.0000     0.0000     1.0000     0.0000 
>> %Dual feasible. Hence use dual simplex.
>> H=ljx(H,2,1);
              x5         x2         x3         1    
       --------------------------------------------
  x4= |     1.0000     1.0000     0.0000     2.0000 
  x1= |     1.0000     2.0000    -1.0000     4.0000 
  x6= |    -1.0000    -1.0000     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= |                                      2.0000 
  x1= |                                      3.0000 
  x3= |                                      1.0000 
  x7= |                                      2.0000 
       --------------------------------------------
   z= |     0.5000     0.5000     0.5000     1.0000 
>> %Optimal tableau. Solution x1=3;x2=0, min z=x3=1
>> %%%%%%%%%%%%%%%%%%%%%% Problem 3  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%
>> A=[1 -1;-1 2];b=[0 -3]';p=[1 1]';q=[-1 0]'; 
>> H=totbl(A,b,p);H=addrow(H,[q' 0],'z0',3); 
              x1         x2         1    
       ---------------------------------
  x3= |     1.0000    -1.0000    -0.0000 
  x4= |    -1.0000     2.0000     3.0000 
  z0= |    -1.0000     0.0000     0.0000 
       ---------------------------------
   z= |     1.0000     1.0000     0.0000 
>> %1-t>=0 implies optimal tableau for -infinity<t=< 1 
>> H=ljx(H,2,1);
              x4         x2         1    
       ---------------------------------
  x3= |    -1.0000     1.0000     3.0000 
  x1= |    -1.0000     2.0000     3.0000 
  z0= |     1.0000    -2.0000    -3.0000 
       ---------------------------------
   z= |    -1.0000     3.0000     3.0000 
>> %-1+t>=0 and 3-2t>=0 imply optimal tableau for 1=<t=<1.5
>> %Unbounded tableau for t>1.5 because column x2 is nonnegative in rows 1 & 2
>> %Summary
>> %-infinity<t=< 1: z(t)=0 at x1=0,x2=0
>> %1=<t=<1.5: z(t)=3-3t at x1=3, x2=0
>> %1.5<t: z(t) unbounded below
>> %%%%%%%%%%%%%%%%%%%%%%%%%% Problem 4 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%  Min = 0. Becuase: (1/2)x'Qx+p'x=-(1/2)x'Qx+b'u=-(1/2)x'Qx-p'x,
%  where the first equality follows from strong duality and the second from
%  b'u=-p'x. Hence: x'Qx+2p'x=0.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%