the dual in linear programming problem
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The dual in linear programming problem
Every linear programming problem has a related problem called the dual problem or simply, dual. Given an original LP problem, referred as the primal problem, or simply primal, the dual can be formulated from the information contained in the primal.
The dual contains economic information useful to management, and it may also be easier to solve, in terms of less computation, than the primal problem. Generally, if the LP primal involves maximizing a profit function subject to less-or-equal-to resource constraints, the dual will involve minimizing total opportunity costs subject to greater-than-or-equal-to product profit constraints. Formulating the dual problem for a given primal is not complex, and once it is formulated, the solution procedure is exactly the same as for any LP problem.
Minimization problem with problem constraints
As with the maximization problem we first restrict our discussion of minimization problem to only those in which the constraints satisfy
a) all variables are nonnegativeb) all the problem constraints are of the form
with , for .
One of the methods to solve such type of minimization problem associated with problem
constraints can be handled by using the method known as the dual problem. For a given
minimization problem with problem constraints we can form the corresponding maximization
problem in the standard form as follows.
In the formation of dual problem from the primal problem we have the following procedures.
1) The sense of optimization is always opposite for corresponding primal and dual problems. (minimization to maximization on the other hand maximization to minimization)
2) The number of variables in the primal always equals the number of constraints in the dual. On the other hand the number of constraints in the primal equals the number of variables in the dual.
3) The objective function coefficient for primal variable equals the right-hand-side
constant for the dual constraint.
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4) The right-hand-side constant for the primal problem constant equals the objective
function coefficient for the dual variable.
5) The coefficients in the primal are the transpose of those in the dual.
Example 1.4 A patient in a hospital is required to have at least 84 units of drug and 120 units
of drug each day. Two substances M and N contain each of these drugs; however, in
addition, suppose both M and N contain an undesirable drug . The relevant
information is contained in the following table.
Amount o f drug per gram ofSubstance M Substance N
Minimum daily requirement
Drug
Drug
10 units 2 units 8 units 4 units
84 units120 units
Drug 3 units 1 units
How many grams of each substance M and N should be mixed to meet the minimum
daily requirement and at the same time minimize the intake of drug . How many units
of the undesirable drug will be in this mixture?
Solution Let the decision variables be the number of grams of substance M used and the
number of grams of substance N used. The linear mathematical model is
Minimize
Subject to
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To form the dual problem:
Step-1 First use the coefficients of the problem constraints and the objective function to form the matrix A with the coefficients of the objective function in the last row.
Then the matrix A given by
Step-2 Now form a second matrix B by using rows of A as columns of B, that is
Step-3 Use the rows of B to form the dual problem of the minimization problem; that is
Maximize
Subject to
In this step to avoid confusion, we shall use different variables in the dual problem.
Theorem 6.1 A minimization LP problem has a solution of and only if its dual problem has a solution. If a solution exists, then the optimal value of the minimization problem is the same as the optimal value of the dual problem.
A minimization LP problem problem constraint whose objective function has nonnegative
coefficients can be solved by applying the simplex method to the dual.
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To solve the above example 1.4, introducing slack variables in which these variables
are the decision variables in the primal LP problem where the reason will become clear later on when we reach to the optimal solution. Hence we have,
Maximize
The initial simplex tableau is
Basic variables
3
1
0
The second simplex tableau is
Basic variables
1 *
30
The third simplex tableau is
Basic variables
4
PC
PC
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Since the rows have no positive entries the optimal solution is attained. The solution to
the dual problem is with the maximum value of the dual
. From theorem 6.1 the primal and dual lead to the same solution even though they are
formulated differently, hence the minimum daily requirement will be units. The entire
solution to the minimization problem can always obtained from the final simplex tableau, the
absolute value of the numbers in the values of the slack variables in the optimal dual
solution represent the optimal values of the primal variables. Hence the values of
can be read from the rows in their column, that is
Example 6.5 A food processing company produces regular and deluxe ice cream in three plants. Per hour of operation, the plant A produces 20 gallons of regular and 10 gallons of deluxe, the plant B produces 10 gallons of regular and 20 gallons of deluxe, and the plant C produces 20 gallons of regular and 20 gallons of deluxe, It costs $70 per hour to operate plant , $75 per hour to operate plant B, and $90 per hour to operate plant C. The company needs at least 300 gallons of regular ice cream and at least 200 gallons of deluxe ice cream each day. How many hours per day should each plant operate in order to produce the required amount of ice cream and minimize the cost of production. Want is the minimum production cost?
Solution Let be the number of hours of plant A operates per day, the number of hours of
plant B operates per day, and be the number of hours of plant C operates per day. The given
information in the problem can be expressed as shown in table 6.3.
Production per hourPlant A Plant B Plant C
Minimum amount required per day
5
Regular (in gallons)Deluxe (in gallons)
20 10 20 10 20 20
300 ( in gallons)200 (in gallons)
Production cost per hour $70 $75 $90Table 6.3 Production information
The LP problem is
Minimize
Subject to
To form the dual problem:
Then the matrix A will be
The transpose of A will be
The dual problem of the minimization problem will be
Maximize
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A =
Subject to
Introduce slack variables all nonnegative we have
The initial simplex tableau is
Basic variables
70
75 7.5
90
0 0
The second simplex tableau is
Basic variables
7
PC
7
8
2 2*
1050
The third simplex tableau is
Basic variables
2 2
1150
Therefore, the optimal dual solution will be with the Z=1150. On the
other hand, the primal problem will have a solution of and with the
minimum cost of production of $1150.
Note that multiplying a problem constraint by a number (usually in order to simplify calculation) changes the units of the slack variables. This requires some special interpretation of the value of the slack variable in the optimal solution, but causes no series problems. However, when using the dual method, multiplying a problem constraint in the dual problem by a number can have
some series consequence; the row of the simplex tableau may no longer give the correct
solution to the primal problem (minimization problem as shown below).
Equivalently expressed as
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PC
After introducing the slack variables the simplex matrix tableau will result
Basic variables
7
15 7.5
9
0 0
The second simplex tableau is
Basic variables
7
8
2 2*
1050
The third simplex tableau is
Basic variables
2 2
1150
9
PC
PC
Therefore, the optimal dual solution will be with the Z=1150. While, the
primal problem will have a solution of and with the minimum cost
of production of $1150 which is a wrong conclusion.
Thus one should never multiply a problem constraint in a maximization problem by a number if the maximization problem is being used to solve a minimization problem. You may still simplify problem constraints in minimization problem before forming the dual problem.
Maximization and Minimization with mixed Problem constraints
The simplex method has been described in the above sections for solving LP problem in which the objective function is maximized or minimized with the special type of constraints less-than-
or equal-to or greater-than-or-equal-to , respectively. To use the simplex method, each of
these must be converted to a special form. If they are not, the simplex technique is unable to set up an initial feasible solution in the initial simplex tableau.
Minimization with mixed problem constraints
For maximization LP problem having a mix of ( problem constraints the simplex
method works similarly as for all problem constraints. The only change is in transforming
constraints to the standard equation form with appropriate supplementary variables. Recall that
for each problem constraint, we have added a slack variable is added and changed the
inequality to equation. If the problem constraints involve the greater-than-or-equal-to , they
involve the subtraction of a surplus variable rather than addition of a slack variable. The surplus variable tells us how much the solution exceeds the constraint resource. For example a problem constraint:
To convert this constraint we subtract a surplus variable , to create an equation as follows:
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Now we have one small problem in trying to use this constraint (as it has just been rewritten) in
setting up an initial simplex solution. Since all the “real” variables are set to 0 in
the initial simplex tableau, then takes a negative value i.e.
, Which implies
All variables in LP problems be they real, slack, or surplus, must be nonnegative at all times. If
, then this important condition is violated. To resolve the situation, we introduce one
last kind of variable, called the artificial variable. We simply add the artificial variable, , to the
constraint as follows:
Now, not only the variables may be set to 0 in the initial simplex solution, but the
surplus variable as well. This will leave us with
If problem constraints involve equality sign , they involve the addition of an artificial variable
to it. For example a problem constraint:
To convert this constraint we add an artificial variable , to it as follows:
The reason for inserting an artificial variable into an equality constraint deals with the usual problem of finding an initial LP solution to the initial simplex tableau.
Artificial variables have no meaning in the physical sense, and are nothing more than computational tools for generating initial LP solutions for the initial simplex tableau. Before the final simplex solution has been reached, all artificial variables must be gone from the solution mix. This matter is handled through the problem’s objective function.
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Finally, whenever an artificial or surplus variable is added to one of the constraints, it must be also included in the other equations and in the problem’s objective function, just as we have done for slack variables. Since artificial variables must be forced out of the solution, we can assign a
very high cost to each. In minimization problems, variable with low costs are the most
desirable ones and the first to enter the solution. Variables with high costs leave the solution quickly, or never enter it at all. Whenever we use an artificial variable I n the objective function we use very large positive number M as a coefficient, however the slack and surplus variables, contribute 0 for the cost, hence the coefficients of the slack and surplus variables are 0 in the objective functions. For example; if our objective function is:
Minimize
Subject to
Then the completed objective functions and constraints would appear as follows:
Minimize
Subject to
Remark: In any LP problem, the initial set of basic variables will consists of all slack variables and all the artificial variables which appear in the problem.
Minimization problems are quite similar to maximization problems tackled earlier in the
previous section. The significant difference involves the row. Since our objective is now
to minimize costs the new variable to enter the solution in each tableau ( the pivot column) will
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be the one with the largest negative number in the row. Thus, we choose the variable that
decrease costs the most.
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