Simplex method:
Standard Minimization Problems
In the last section we learned how to maximize an objective function such as revenue or profit.
In this section we learn how to minimize an objective function such as cost. We study two types
of minimization problems.
- Minimization with £ constraints.
These problems have nearly the same structure as the
standard maximization problems and may be solved by
maximizing the negative of the objective function.
- Standard minimization.
These problems are different from the standard maximization problems in two ways.
(i) The objective function is to be minimized.
(ii) The constraints are of the ³ form.
The standard minimization problem is solved by setting up and solving
a dual problem.
Minimization with £ constraints
This problem is the same as the standard maximization problem except the objective
function is to be minimized. This problem is equivalent to maximizing the negative
of the given objective function.
- The objective function is to be minimized.
- All the variables in the problem are nonnegative.
- Each linear constraint is written as an expression involving the variables set less than
or equal to a nonnegative constant.
Standard minimization problem
- The objective function is to be minimized.
- All the variables in the problem are nonnegative.
- Each linear constraint is written as an expression involving the variables set greater
than or equal to a nonnegative constant.
Dual Problem
Each minimization linear programming problem is associated with a maximization linear
programming problem. The given problem is called the primal problem and the related
problem is called the dual problem. To construct the dual problem write down the a
table for the given primal problem :: constraints first :: objective function second ::
variables then constants. Interchange the rows and columns of this table. Interpreting the
table as a standard maximization problem yields the dual problem.
Solution to standard minimization problem
Solve the standard minimization by solving its dual problem ( a standard maximization
problem). The connection between the solution of the primal problem and the dual problem
is given in the following theorem.
Fundamental Theorem of Duality
A primal problem has a solution if and only if the corresponding dual problem has a
solution. If a solution exists then
- The objective function of both the primal and dual functions attain the same optimal
value.
- The optimal solution of the primal problem appears under the slack variables in the last
row of the final simplex table associated with the dual problem.
Example:     Minimization with £ constraints.
Minimize C = -2x + y subject to the constraints
æ x + 2y £ 6 ö
ç 3x + 2y £ 12 ÷
ç x ³ 0 ÷
è y ³ 0 ø
|
(1) CONVERT TO A MAXIMIZATION PROBLEM
The problem of minimizing C is equivalent to the problem of maximizing P = -C.
Maximize P = 2x - y subject to the constraints
æ x + 2y £ 6 ö
ç 3x + 2y £ 12 ÷
ç x ³ 0 ÷
è y ³ 0 ø
|
(2) SET UP AN INITIAL TABLE
Introduce slack variables u and v to replace the inequalities with equations.
x + 2y + u = 6
3x + 2y + v = 12
Rewrite the objective function.
-2x + y + P = 0
Form augmented matrix with the constraints first and the objective function last.
x y u v P RHS
é 1 2 1 0 0 | 6 ù
ê 3 2 0 1 0 | 12 ú
ë -2 1 0 0 1 | 0 û
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(3) PERFORM A PIVOT OPERATION
We need to perform a pivot operation since there are
negative entries in the last row. Column 1 is the pivot
column since -2 is the most negative entry in the last
row. Form ratios 6/1 = 6 and 12/3 = 4. Row 2 is the
pivot row since it has the smallest ratio. The pivot
element is colored red.
é 1 2 1 0 0 | 6 ù
ê 3 2 0 1 0 | 12 ú
ë -2 1 0 0 1 | 0 û
|
R1 > R1/3
|
We make the pivot element into a 1 by dividing row 2 through by 3.
é 1 2 1 0 0 | 6 ù
ê 1 2/3 0 1/3 0 | 4 ú
ë -2 1 0 0 1 | 0 û
|
R2 > R2 - 2 R1
R3 > R3 + 12 R1
|
Eliminate the entries directly below pivot element.
é 0 4/3 1 -1/3 0 | 2 ù
ê 1 2/3 0 1/3 0 | 4 ú
ë 0 7/3 0 2/3 1 | 8 û
|
There are no negative entries in the last row so we are finished with the pivoting process.
(4) READ THE OPTIMAL SOLUTION
Recall that variables heading columns not in unit form
are assigned a value of 0. The optimal solution for the
maximization problem is
æ x = 4 ö
ç y = 0 ÷
è P = 8 ø
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and hence the optimal solution for the original minimization problem is
æ x = 4 ö
ç y = 0 ÷
è C = -8 ø
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Example:     Standard minimization.
Minimize C = 8x + 12y subject to the constraints
æ x + 3y ³ 2 ö
ç 2x + 2y ³ 3 ÷
ç x ³ 0 ÷
è y ³ 0 ø
|
(1) SET UP THE DUAL PROBLEM
Write down the following table for the primal problem
x | y | RHS
--------------
1 | 3 | 2
2 | 2 | 3
8 | 12 |
Interchange the rows and columns to obtain
u | v | RHS
--------------
1 | 2 | 8
3 | 2 | 12
2 | 3 |
Interpret the table as if it were the initial table
for a standard maximization problem with the exception
that the signs of the coefficients of the objective
function are not reversed.
Maximize P = 2u + 3v subject to the constraints
æ u + 2v £ 8 ö
ç 3u + 2v £ 12 ÷
ç u ³ 0 ÷
è v ³ 0 ø
|
(2) SOLVE STANDARD MAXIMIZATION PROBLEM:
SET UP AN INITIAL TABLE
Introduce slack variables x and y to replace the inequalities with equations.
u + 2v + x = 8
3u + 2v + y = 12
Rewrite the objective function.
-2u - 3v + P = 0
Form augmented matrix with the constraints first and the objective function last.
u v x y P RHS
é 1 2 1 0 0 | 8 ù
ê 3 2 0 1 0 | 12 ú
ë -2 -3 0 0 1 | 0 û
|
(3) PERFORM A PIVOT OPERATION
We need to perform a pivot operation since there are
negative entries in the last row. Column 2 is the pivot
column since -3 is the most negative entry in the last
row. Form ratios 8/1 = 8 and 12/3 = 4. Row 1 is the
pivot row since it has the smallest ratio. The pivot
element is colored red.
é 1 2 1 0 0 | 8 ù
ê 3 2 0 1 0 | 12 ú
ë -2 -3 0 0 1 | 0 û
|
R1 > R1/2
|
We make the pivot element into a 1 by dividing row 1 through by 2.
é 1/2 1 1/2 0 0 | 4 ù
ê 3 2 0 1 0 | 12 ú
ë -2 -3 0 0 1 | 0 û
|
R2 > R2 - 2 R1
R3 > R3 + 3 R1
|
Eliminate the entries directly below pivot element.
é 1/2 1 1/2 0 0 | 4 ù
ê 2 0 -1 1 0 | 4 ú
ë -1/2 0 3/2 0 1 | 12 û
|
There is a negative entry in the last row. Hence, a second pivot operation is required.
(4) PERFORM A SECOND PIVOT OPERATION
We need to perform a pivot operation since there are
negative entries in the last row. Column 1 is the pivot
column since -1/2 is the most negative entry in the last
row. Form ratios 4/(1/2) = 8 and 4/2 = 2. Row 2 is the
pivot row since it has the smallest ratio. The pivot
element is colored red.
é 1/2 1 1/2 0 0 | 4 ù
ê 2 0 -1 1 0 | 4 ú
ë -1/2 0 3/2 0 1 | 12 û
|
R2 > R2/2
|
We make the pivot element into a 1 by dividing row 2 through by 2.
é 1/2 1 1/2 0 0 | 4 ù
ê 1 0 -1/2 1/2 0 | 2 ú
ë -1/2 0 3/2 0 1 | 12 û
|
R1 > R1 - 1/2 R2
R3 > R3 + 1/2 R2
|
Eliminate the entries directly above and below the pivot element.
u v x y P RHS
é 0 1 3/4 -1/4 0 | 3 ù
ê 1 0 -1/2 1/2 0 | 2 ú
ë 0 0 5/4 1/4 1 | 13 û
|
There are no negative entries in the last row so we are finished with the pivoting process.
(5) READ THE OPTIMAL SOLUTION
The optimal solution to the dual (maximization) problem
is read from the final table in the usual manner as
æ u = 2 ö
ç v = 3 ÷
è P = 13 ø
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Recall the optimal solution of the primal problem appears
under the slack variables in the last row of the final
simplex table associated with the dual problem. The
optimal solution is colored blue.
æ x = 5/4 ö
ç y = 1/4 ÷
è C = 13 ø
|
Note that the maximum value of P is equal to the minimum
value of C as promised by the fundamental theorem.
Exercises
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