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DUAL LINEAR PROGRAMMING PROBLEMS
4. 1 Introduction For every linear programming problem there is a corresponding linear programming problem called the dual. If the original problem is a maximization problem then the dual problem is minimization problem and if the original problem is a minimization problem then the dual problem is maximization problem. In either case the final table of the dual problem will contain both the solution to the dual problem and the solution to the original problem. The solution of the dual problem is readily obtained from the original problem solution if the simplex method is used. The formulation of the dual problem also sometimes referred as the concept of duality is helpful for the understanding of the linear programming. The variable of the dual problem is known as the dual variables or shadow price of the various resources. The dual problem is easier to solve than the original problem. The dual problem solution leads to the solution of the original problem and thus efficient computational techniques can be developed through the concept of duality. Finally, in the competitive strategy problem solution of both the original and dual problem is necessary to understand the complete problem. 4. 2 Dual Problem Formulation If the original problem is in the standard form then the dual problem can be formulated using the following rules: The number of constraints in the original problem is equal to the number of dual variables. The number of constraints in the dual problem is equal to the number of variables in the original problem. The original problem profit coefficients appear on the right hand side of the dual problem constraints. If the original problem is a maximization problem then the dual problem is a minimization problem. Similarly, if the original problem is a minimization problem then the dual problem is a maximization problem. The original problem has less than or equal to (≤) type of constraints while the dual problem has greater than or equal to (≥) type constraints. The coefficients of the constraints of the original problem which appear from left to right are placed from top to bottom in the constraints of the dual problem and vice versa. The Dual Linear Programming Problem is explained with the help of the following Example 4 .1. Example 4. 1 Consider the following product mix problem:
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MBA-H 2040 Quantitative Techniques for Managers 0 0 s 5 s 6
zj-cj - 12 - 3 - 1 0 0 0 Table 1 Note that the basic variables are s 4 , s 5 and s 6. Therefore CB 1 = 0 , CB 2 = 0, CB 3 =
The smallest negative element in the above table of z 1 – c 1 is - 12. Hence, x 1 becomes a basic variable in the next iteration.
Determine the minimum ratios Min 100 , 72 , 80 = 10 10 7 2 Here the minimum value is s 4 , which is made as a non-basic variable.
The next Table 2 is calculated using the following rules: (i) The revised basic variables are x 1 , s 5 , s 6. Accordingly we make CB 1 =22, CB 2 =0 and CB 3 =0. (ii) Since x 1 is the incoming variable we make x 1 coefficient one by dividing each element of row 1 by 10. Thus the numerical value of the element corresponding to x 2 is 2/10, corresponding to x 3 is 1/10, corresponding to s 4 is 1/10, corresponding to s 5 is 0/10 and corresponding to s 6 is 0/10 in Table 2. (iii) The incoming basic variable should only appear in the first row. So we multiply first row of Table 2 by 7 and subtract if from the second row of Table 1 element by element. Thus, The element corresponding to x 1 in the second row of Table 2 is zero The element corresponding to x 2 is 3 – 7 * 2 = 16 10 10 By using this way we get the elements of the second and the third row in Table 2. Similarly, the calculation of numerical values of basic variables in Table 2 is done. CB 12 0 0 Basic Variable x 1 s 5 s 6 Cj XB 10 7 60
x 1 1 0 0
x 2 2/ 16/ 18/
x 3 1/ 13/ 4/
s 4 1/
s 5 0 1 0
s 6 0 0 1 zj-cj 0 - 3/5 1/5 6/5 0 0
Table 2 64
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MBA-H 2040 Quantitative Techniques for Managers
- The total amount offer by the investor to the three resources viz. A, B and C required to produce one unit of each product must be at least as high as the profit gained by the producer per unit. Since, these resources enable the producer to earn the specified profit corresponding to the product he would not like to sell it for anything less assuming he is behaving rationally. This leads to the following conditions: 10w 1 + 7w 2 + 2w 3 ≥ 12 2w 1 + 3w 2 + 4w 3 ≥ 3 w 1 + 2w 2 + w 3 ≥ 1 Thus, in this case we have a linear problem to ascertain the values of the variable w 1 , w 2 , w 3. The variables w 1 , w 2 and w 3 are called as dual variables. Note: The original (primal) problem illustrated in this example a. considers the objective function maximization b. contains ≤ type constraints c. has non-negative constraints This original problem is called as primal problem in the standard form. 4. 3 Dual Problem Properties The following are the different properties of dual programming problem: i. If the original problem is in the standard form, then the dual problem solution is obtained from the zj – cj values of slack variables. For example: In the Example 4 .1, the variables s4, s 4 and s 6 are the slack variables. Hence the dual problem solution is w 1 = z 4 – c 4 = 15/16, w 2 = z 5 – c 5 = 3/8 and w 3 = z 6 – c 6 = 0. ii. The original problem objective function maximum value is the minimum value of the dual problem objective function. For example: From the above Example 4. 1 we know that the original problem maximum values is 981/8 = 122 .625. So that the minimum value of the dual problem objective function is 10015/16 + 773/8 + 80*0 = 981/ Here the result has an important practical implication. If both producer and investor analyzed the problem then neither of the two can outmaneuver the other. iii. Shadow Price: A resource shadow price is its unit cost, which is equal to the increase in profit to be realized by one additional unit of the resource.
MBA-H 2040 Quantitative Techniques for Managers 10015/16 + 773/8 + 80* If the first resource is increased by one unit the maximum profit also increases by 15/16, which is the first dual variable of the optimum solution. Therefore, the dual variables are also referred as the resource shadow price or imputed price. Note that in the previous example the shadow price of the third resource is zero because there is already an unutilized amount, so that profit is not increased by more of it until the current supply is totally exhausted. iv. In the originals problem, if the number of constraints and variables is m and n then the constraint and variables in the dual problem is n and m respectively. Suppose the slack variables in the original problem is represented by y 1 , y 1 , ….., yn and the surplus variables are represented by z 1 , z 2 , …, zn in the dual problem. v. Suppose, the original problem is not in a standard form, then the dual problem structure is unchanged. However, if a constraint is greater than or equal to type, the corresponding dual variable is negative or zero. Similarly, if a constraint in the original problem is equal to type, then the corresponding dual variables is unrestricted in sign. Example 4. 2 Consider the following linear programming problem Maximize 22 x1 + 25 x2 +1 9 x 3 Subject to: 18 x1 + 26 x2 + 22 x 3 ≤ 350 14 x1 + 18 x2 + 20 x 3 ≥ 180 17 x1 + 19 x2 + 18 x3 = 205 x1, x2, x 3 ≥ 0 Note that this is a primal or original problem. The corresponding dual problem for this problem is as follows: Minimize 250w1 + 80w2 +105w Subject to: 18w1 + 4w2 + 7 w 3 ≥ 22 26w1 + 18w2 + 19w3 ≥ 25 22w1 + 20w2 + 18w3 ≥ 19 w1 ≥ 0, w2 ≤, and w3 is unrestricted in sign (+ or
- ). Now, we can solve this using simplex method as usual. 4. 4 Simple Way of Solving Dual Problem Solving of dual problem is simple; this is illustrated with the help of the following Example 4 .3. 67
MBA-H 2040 Quantitative Techniques for Managers Example 4.3: Minimize P = x 1 + 2 x 2 Solution: Subject to: x 1 + x 2 ≥ 8 2 x 1 + y ≥ 12 x 1 ≥ 1 Step 1: Set up the P-matrix and its transpose 1 1 8 2 1 12 P = 1 0 1 1 2 0 1 2 1 1 PT = 1 1 0 2 8 12 1 0 w 1 w 2 w 3 s 1 s 2 g z 1 2 1 1 0 0 1 Step 2: Form the 1 1 0 0 1 0 2 constraints and the objective function for the dual
- 8 - 12 - 1 0 0 1 0 w 1 + 2w 2 + w 3 ≤ 1 w 1 + w 2 ≤ 2 z = 8w 1 = 12w 2 + 2 Step 3: Construct the initial simplex tableau for the dual Since there are no negative entries in the last column above the third row, we have a standard simplex problem. The most negative number in the bottom row to the left of the last column is - 12. This establishes the pivot column. The smallest nonnegative ratio is 1/2. The pivot element is 2 in the w 2 - column.
MBA-H 2040 Quantitative Techniques for Managers Step 4: Pivoting Pivoting about the 2 we get: w 1 w 2 w 3 s 1 s 2 g z 1/2 1 1/2 1/2 0 0 1/ 1 1 0 0 1 0 2
- 8 - 12 - 1 0 0 1 0 w 1 w 2 w 3 s 1 s 2 g z 1/2 1 1/2 1/2 0 0 1/ 1/2 0 - 1/2 - 1/2 1 0 3/
- 2 0 5 6 0 1 6 The most negative entry in the bottom row to the left of the last column is - 2. The smallest non- negative ratio is the 1/2 in the first row. This is the next pivot element. w 1 w 2 w 3 s 1 s 2 g z 1/2 1 1/2 1/2 0 0 1/ 1/2 0 - 1/2 - 1/2 1 0 3/ Pivoting about the - 2 0 5 6 0 1 6 1/ 2 : 69 w 1 w 2 w 3 s 1 s 2 g z 1 2 1 1 0 0 1 1/2 0 - 1/2 - 1/2 1 0 3/
- 2 0 5 6 0 1 6
MBA-H 2040 Quantitative Techniques for Managers Since there are no negative entries in the bottom row and to the left of the last column, the process is complete. The solutions are at the feet of the slack variable columns. Therefore, The w 1 w 2 w 3 s 1 s 2 g z 1 2 1 1 0 0 1 optimum^ solution provided by x 1 = 8 and x 2 = 0 0 - 1 - 1 - 1 1 0 1 The Minimum Value is: 8 4. 5 Summary
For every linear programming problem there is a dual problem. The variables of the dual problem are called as dual variables. The variables have economic value, which can be used for planning its resources. The dual problem solution is achieved by the simplex method calculation of the original (primal) problem. The dual problem solution has certain properties, which may be very useful for calculation purposes. 4. 6 Key Terms Original Problem : This is the original linear programming problem, also called as primal problem. Dual Problem : A dual problem is a linear programming problem is another linear programming problem formulated from the parameters of the primal problem. Dual Variables : Dual programming problem variables. Optimum Solution : The solution where the objective function is minimized or maximized. Shadow Price : Price of a resource is the change in the optimum value of the objective function per unit increase of the resource. 4. 7 Self Assessment Questions Q1. An organization manufactures three products viz. A, B and c. The required raw material per piece of product A, B and C is 2kg, 1kg, and 2kg. Assume that the total weekly availability is 50 kg. In order to produce the products the raw materials are processed on a machine by the labour force and on a weekly
MBA-H 2040 Quantitative Techniques for Managers the availability of machine hours is 30. Assume that the available total labour hour is 26. The following table illustrates time required per unit of the three products. The profit per unit from the products A, B and C are #25, #30 and #40. Formulate the dual linear programming problem and determine the optimum values of the dual variables. Q2. Consider the following dual problem Minimize 3w 1 + 4w 2 Subject to: 3w 1 + 4w 2 ≥ 24 2w 1 + w 2 ≥ 10 5w 1 + 3w 2 ≥ 29 w 1 , w 2 ≥ 0 4. 8 Key Solutions Q1. Minimize 50w 1 + 30w 2 + 26w 3 Subject to: 2w 1 + 0 .5w 2 + w 3 ≥ 25 w 1 + 3w 2 + 2w 3 ≥ 30 2w 1 + w 2 + w 3 ≥ 40 w 1 = 50/3 = 16. 6 , w 2 = 0 and w 3 = 20 /3 = 6. 6 Q2. Maximize 24 x 1 + 10 x 2 + 29 x 3
Subject to: Value is: 24 3 x 1 + 2 x 2 + 5 x 3 ≤ 3 4 x 1 + x 2 + 3 x 3 ≤ 4 x 1 , x 2 , x 3 ≥ 0 and w 1 = 4 , w 2 = 3 Objective Function Maximum Product Labour Hour Machine Hour A 0. 5 1 B 3 2 C 1 1
