Reformulate the following linear program into standard form: minimize -2x_1 + x_

Question

Reformulate the following linear program into standard form: minimize -2x_1 + x_2 subject to x_1 + 2x_2 lessthanorequalto 10 x_1 - x_2 lessthanorequalto -5 x_1greaterthanorequalto 0.x_2greaterthanorequalto 0 Show that the reformulated problem is equivalent to the original problem in the following sense: Given a feasible point for one of the problems, one can construct a feasible point for the other problem that has the same objective function value. Give a precise rule for constructing such vectors. If two linear programs are equivalent in the sense discussed in part (b), and if one problem has a finite optimal value, then why must the other problem have the same optimal value programs are equivalent in the sense discussed in part (b), and if one problem is feasible but has no finite optimal value, then why must this also be true of the other problem? If two linear programs are equivalent in the sense discussed (b), and if in part one problem has no feasible vectors, then why must this be true of the other problem as well?

Explanation / Answer

solution-:

Tableau #1
x      y      s1     s2     s3     s4     -p          
1      2      1      0      0      0      0      10   
1      -1     0      1      0      0      0      -5   
1      0      0      0      -1     0      0      0    
0      1      0      0      0      -1     0      0    
-2     1      0      0      0      0      1      0    

Tableau #2
x      y      s1     s2     s3     s4     -p          
1      2      1      0      0      0      0      10   
1      -1     0      1      0      0      0      -5   
-1     0      0      0      1      0      0      0    
0      1      0      0      0      -1     0      0    
-2     1      0      0      0      0      1      0    

Tableau #3
x      y      s1     s2     s3     s4     -p          
1      2      1      0      0      0      0      10   
1      -1     0      1      0      0      0      -5   
-1     0      0      0      1      0      0      0    
0      -1     0      0      0      1      0      0    
-2     1      0      0      0      0      1      0    

Tableau #4
x      y      s1     s2     s3     s4     -p          
1      2      1      0      0      0      0      10   
0      -3     -1     1      0      0      0      -15  
0      2      1      0      1      0      0      10   
0      -1     0      0      0      1      0      0    
0      5      2      0      0      0      1      20    

Optimal Solution: p = -20; x = 10, y = 0 answer

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