CS 525 Linear Programming
                     Midterm March 7, 1996 Solutions
		     -------------------------------
>> %Solution 1
>> A=[-1 0 1;1 1 1;3 2 1];b=[1 -1 -4]';
>> H=totbl(A,b);
              x1         x2         x3         1    
       --------------------------------------------
  y1= |    -1.0000     0.0000     1.0000    -1.0000 
  y2= |     1.0000     1.0000     1.0000     1.0000 
  y3= |     3.0000     2.0000     1.0000     4.0000 
>> H=ljx(H,1,1);
              y1         x2         x3         1    
       --------------------------------------------
  x1= |    -1.0000     0.0000     1.0000    -1.0000 
  y2= |    -1.0000     1.0000     2.0000     0.0000 
  y3= |    -3.0000     2.0000     4.0000     1.0000 
>> H=ljx(H,2,2);
              y1         y2         x3         1    
       --------------------------------------------
  x1= |    -1.0000     0.0000     1.0000    -1.0000 
  x2= |     1.0000     1.0000    -2.0000    -0.0000 
  y3= |    -1.0000     2.0000     0.0000     1.0000 
>> %System has no solution, because y1(x)=y2(x)=0 ===> y3(x)=1.
>> 
>> 
>> %Solution 2
>> A=[1 -1 1;2 -1 1;-1 1 2];b=[-1 -2 0]';p=[1 -1 2]';
>> H=totbl(A,b,p);
              x1         x2         x3         1    
       --------------------------------------------
  x4= |     1.0000    -1.0000     1.0000     1.0000 
  x5= |     2.0000    -1.0000     1.0000     2.0000 
  x6= |    -1.0000     1.0000     2.0000    -0.0000 
       --------------------------------------------
   z= |     1.0000    -1.0000     2.0000     0.0000 
>> H=ljx(H,1,2);
              x1         x4         x3         1    
       --------------------------------------------
  x2= |     1.0000    -1.0000     1.0000     1.0000 
  x5= |     1.0000     1.0000     0.0000     1.0000 
  x6= |     0.0000    -1.0000     3.0000     1.0000 
       --------------------------------------------
   z= |     0.0000     1.0000     1.0000    -1.0000 
>>
>> %Optimal tableau! z_min=-1 at x1=0, x2=1, x3=0.
>> 
>> 
>> %Solution 3
>> A=[2 1 -1;3 5 1;-1 3 -2];b=[-2 -1 1]';p=[-1 -2 -1]';
>> H=totbl(A,b,p);
              x1         x2         x3=unres.  1    
       --------------------------------------------
  x4= |     2.0000     1.0000    -1.0000     2.0000 
0=x5= |     3.0000     5.0000     1.0000     1.0000 
  x6= |    -1.0000     3.0000    -2.0000    -1.0000 
       --------------------------------------------
   z= |    -1.0000    -2.0000    -1.0000     0.0000 
>> H=ljx(H,2,3);
              x1         x2         x5=0       1    
       --------------------------------------------
  x4= |     5.0000     6.0000    -1.0000     3.0000 
unx3= |    -3.0000    -5.0000     1.0000    -1.0000 
  x6= |     5.0000    13.0000    -2.0000     1.0000 
       --------------------------------------------
   z= |     2.0000     3.0000    -1.0000     1.0000 
>>
>> %Optimal tableau if we ignore column x5 (x5=0 never becomes
>> %basic again) and ignore row x3 (x3 is unrestricted).
>> %Solution: z_min=1 at x1=0, x2=0, x3=-1.
>> 
>> 
>> Answer 4
>> A=[4 1 17;1 -1 1;6 2 13];b=[2 1 5]';p=[3 2 1]';
>> H=totbl(A,b,p);
              x1         x2         x3         1    
       --------------------------------------------
  x4= |     4.0000     1.0000    17.0000    -2.0000 
  x5= |     1.0000    -1.0000     1.0000    -1.0000 
  x6= |     6.0000     2.0000    13.0000    -5.0000 
       --------------------------------------------
   z= |     3.0000     2.0000     1.0000     0.0000 
>> %Nonnegative bottom row. Hence dual feasible.
>> %Pick a pivot row that results in an easy pivot.
>> %Row x4 gives pivot=17. Row x6 gives pivot=13. So.....
>> H=ljx(H,2,3);
              x1         x2         x5         1    
       --------------------------------------------
  x4= |   -13.0000    18.0000    17.0000    15.0000 
  x3= |    -1.0000     1.0000     1.0000     1.0000 
  x6= |    -7.0000    15.0000    13.0000     8.0000 
       --------------------------------------------
   z= |     2.0000     3.0000     1.0000     1.0000 
>>
>> %Optimal tableau! z_min=1 at x1=0, x2=0, x3=1.               
>> diary off