Finite Mathematics and Calculus with Applications
(6th Edition)
by Lial, Greenwell, and Ritchey
Section 4.3- Minimization Problems; Duality
7.
State the dual problem for the problem:
Minimize          subject to
        
        
with             
Consider the maxtirx
      
      don't put negative numbers in last row
   
 
Find the transpose of this matrix. The transpose of this 3  5 matrix
is a 5  3 matrix where the rows and columns are interchanged.
The transpose is


 
 






 





    
The dual problem is a maximization problem where we have 4
inequalities obtained from the 1st 4 rows of the matrix, and the last
row gives the function to maximize. So the dual problem is
Maximize      subject to
    
    
    
    
where all the variables are non-negative.
9.
Use the simplex method to find    and    such that
    
    
and      is minimized.
  
Consider the matrix    . The 1st and 2nd rows are the
  
coefficients from the 2 inequalities. The 1st 2 numbers in the
last row come from the equation for . Don't put negatives
in the last row.
  
The transpose of this matirx is    which is obtained by
  
interchanging the rows and columns.
The solution to the minimization problem stated above can be
found by solving the following maximization problem called the
dual problem:
Maximize      subject to
    
    
The initial matrix for this maximization problem is
x1
x2
s1
s2
2
3
2
1
1
0
0
1
5
2
-6
-7
0
0
0
It is probably best to pivot on a number in the 2nd column.
Using the division technique, we see we must pivot on the 1
in the 2nd row.
x1
x2
s1
s2
2
3
2
1
1
0
0
1
5
2
-6
-7
0
0
0
5/2 = 2.5
1/2 = 0.5
Adding -2 time row 2 to row 1 and adding 7 times row 2 to
row 3 we get the matrix
x1
x2
s1
s2
-4
3
0
1
1
0
15
0
0
-2
1
7
1
2
14
Since there are no negative numbers in the last row we
are done. The maximum value of  is 14 which is also
the minimum value of . The values of  and 
are the values in the last row underneath and  
They are 0 and 7.
Ans: The minimum value of  is 14 which occurs when
   and   
13.
Minimize        subject to
      
  
 
with          
     
Consider the matrix     
  
 
The transpose of this matrix is
 
 
 
 

  this is like a maximization problem with 3

 
    
inequalities and 2 variables. The maximization problem is
maximize      subject to
    
    
  
This means we need 3 slack variables  So solve
the maximization problem where the initial matrix is
x1 x2 s 1
s2
s3
1
1
1
2
1
0
1
0
0
0
1
0
0
0
1
2
1
3
-100 -50
0
0
0
0
Pivot on the 1 in the 1st column, 2nd row. Add -1 times row 2
to row 1, add -1 times row 2 to row 3, and add 100 times row 2
to the bottom row. This results in the matrix
0
1
0
1
1
-1
1
0
0
-1
1
-1
0
0
1
1
1
2
0
50
0
100
0
100
y1
y2
y3
Minimum value of w
Ans: The minimum value of  is 100 when      
and    