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LINEAR PROGRAMMING –GRAPHICAL METHOD
Introduction to Linear Programming Linear Programming is a special and versatile technique which can be applied to a variety of management problems viz. Advertising, Distribution, Investment, Production, Refinery Operations, and Transportation analysis. The linear programming is useful not only in industry and business but also in non-profit sectors such as Education, Government, Hospital, and Libraries. The linear programming method is applicable in problems characterized by the presence of decision variables. The objective function and the constraints can be expressed as linear functions of the decision variables. The decision variables represent quantities that are, in some sense, controllable inputs to the system being modeled. An objective function represents some principal objective criterion or goal that measures the effectiveness of the system such as maximizing profits or productivity, or minimizing cost or consumption. There is always some practical limitation on the availability of resources viz. man, material, machine, or time for the system. These constraints are expressed as linear equations involving the decision variables. Solving a linear programming problem means determining actual values of the decision variables that optimize the objective function subject to the limitation imposed by the constraints. The main important feature of linear programming model is the presence of linearity in the problem. The use of linear programming model arises in a wide variety of applications. Some model may not be strictly linear, but can be made linear by applying appropriate mathematical transformations. Still some applications are not at all linear, but can be effectively approximated by linear models. The ease with which linear programming models can usually be solved makes an attractive means of dealing with otherwise intractable nonlinear models.
2.2 Linear Programming Problem Formulation The linear programming problem formulation is illustrated through a product mix problem. The product mix problem occurs in an industry where it is possible to manufacture a variety of products. A product has a certain margin of profit per unit, and uses a common pool of limited resources. In this case the linear programming technique identifies the products
combination which will maximize the profit subject to the availability of limited resource constraints.
Example 2.1:
Suppose an industry is manufacturing tow types of products P1 and P2. The profits per Kg of the two products are Rs.30 and Rs.40 respectively. These two products require processing in three types of machines. The following table shows the available machine hours per day and the time required on each 18
Subject to: 3x1 + 2x 2 ≤ 600 3x 1 + 5x 2 ≤ 800 5x 1 + 6x 2 ≤ 1100 x 1 ≥ 0, x 2 ≥ 0
2.3 Formulation with Different Types of Constraints
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MBA-H2040 Quantitative Techniques for Managers The constraints in the previous example 2.1 are of “less than or equal to” type. In this section we are going to discuss the linear programming problem with different constraints, which is illustrated in the following Example 2.2. Example 2.2:
A company owns two flour mills viz. A and B, which have different production capacities for high, medium and low quality flour. The company has entered a contract to supply flour to a firm every month with at least 8, 12 and 24 quintals of high, medium and low quality respectively. It costs the company Rs.2000 and Rs.1500 per day to run mill A and B respectively. On a day, Mill A produces 6, 2 and 4 quintals of high, medium and low quality flour, Mill B produces 2, 4 and 12 quintals of high, medium and low quality flour respectively. How many days per month should each mill be operated in order to meet the contract order most economically. Solution: Let us define x1 and x2 are the mills A and B. Here the objective is to minimize the cost of the machine runs and to satisfy the contract order. The linear programming problem is given by Minimize 2000x1 + 1500x 2 Subject to: 6x1 + 2x 2 ≥ 8 2x 1 + 4x 2 ≥ 12 4x 1 + 12x 2 ≥ 24 x 1 ≥ 0, x 2 ≥ 0
2.4 Graphical Analysis of Linear Programming
This section shows how a two-variable linear programming problem is solved graphically, which is illustrated as follows:
Example 2.3:
Consider the product mix problem discussed in section 2.
Maximize
MBA-H2040 Quantitative Techniques for Managers From the first constraints 3x 1 + 2x 2 ≤ 600, draw the line 3x 1 + 2x 2 = 600 which passes through the point (200, 0) and (0, 300). This is shown in the following graph as line 1.
3x1 + 2x 2 = 600(line 1)
X 2
C
100
3x 1 + 5x 2 = 800(line 2)
5x 1 + 6x 2 = 1100(line 3)
100 150 D X
Graph 1 : Three closed half planes and Feasible Region Half Plane - A linear inequality in two variables is called as a half plane.
B
A
1
In the previous example we discussed about the less than or equal to type of linear programming problem, i.e. maximization problem. Now consider a minimization (i.e. greater than or equal to type) linear programming problem formulated in Example 2.2.
Minimize 2000x 1 + 1500x 2
Subject to: 6x 1 + 2x 2 ≥ 8 2x 1 + 4x 2 ≥ 12 4x1 + 12x 2 ≥ 24 x1 ≥ 0, x 2 ≥ 0 The three lines 6x 1 + 2x 2 = 8, 2x 1 + 4x 2 = 12, and 4x 1 + 12x 2 = 24 passes through the point (1.3,0) (0,4), (6,0) (0,3) and (6,0) (0,2). The feasible region for this problem is shown in the following Graph 2. In this problem the constraints are of greater than or equal to type of feasible region, which is bounded on one side only.
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8
6
X2 A (^4) 6x1 + 2x 2 ≥ 8
B
2 2x1 + 4x 2 ≥ 12
0 C 4x1 + 12x 2 ≥ 24 X 2 4 6 8
Graph 2 : Feasible Region 2.5 Graphical Liner Programming Solution A two variable linear programming problem can be easily solved graphically. The method is simple but the principle of solution is depends on certain analytical concepts, they are: C on ve x Re gi on : A region R is convex if and only if for any two points on the region R the line connecting those points lies entirely in the region R. E xt re m e P oi
nt :
The extreme point E of a convex region R is a point such that it is not possible to locate two distinct points in R, so that the line joining them will include E. The extreme points are also called as corner points or vertices. Thus, the following result provides the solution to the linear programming model: F 0B 7 “If the minimum or maximum value of a linear function defined over a convex region exists, then it must be on one of the extreme points”. 23 MBA-H2040 Quantitative Techniques for Managers
In this section we are going to describe linear programming graphical solution for both the maximization and minimization problems, discussed in Example 2.3 and Example 2.4.
Example 2. 5: Consider the maximization problem described in Example 2.3.
100 150 X 1
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Graph 3 : Graphical Linear Programming Solution In this problem the objective function is 30x 1 + 40x 2. Let be M is a parameter, the graph 30x 1 +
40x 2 = M is a group of parallel lines with slope – 30/40. Some of these lines intersects the feasible region and contains many feasible solutions, whereas the other lines miss and contain no feasible solution. In order to maximize the objective function, we find the line of this family that intersects the feasible region and is farthest out from the origin. Note that the farthest is the line from the origin the greater will be the value of M. Observe that the line 30x 1 + 40x 2 = M passes through the point D, which is the intersection of the lines 3x 1 + 5x 2 = 800 and 5x 1 + 6x 2 = 1100 and has the coordinates x 1 = 170 and x 2 = 40. Since D is the only feasible solution on this line the solution is unique. The value of M is 6700, which is the objective function maximum value. The optimum value variables are x 1 = 170 and X 2 = 40. The following Table 1 shows the calculation of maximum value of the objective function.
Extreme Point Coordinates X 1 X
Objective Function 30x1 + 40x 2 A X 1 = 0 X 2 = 0 0 B X 1 = 0 X 2 = 160 6400 C X 1 = 110 X 2 = 70 6100 D X 1 = 170 X 2 = 40 6700 E X 1 = 200 X 2 = 0 6000
Table 1 : Shows the objective function Maximum value calculation Example 2.6: Consider the minimization problem described in Example 2.4.
Minimize 2000x1 + 1500x 2 Subject to: 6x1 + 2x 2 ≥ 8 2x 1 + 4x 2 ≥ 12 4x 1 + 12x 2 ≥ 24 x 1 ≥ 0, x 2 ≥ 0 The feasible region for this problem is illustrated in the following Graph 4. Here each of the half planes lies above its boundary. In this case the feasible region is infinite. In this case, we are concerned with the minimization; also it is not possible to determine the maximum value. As in the previous
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0 2000x1+ 1500x 2=M
C
X 2 4 6 8
Graph 4 : Graphical Linear Programming Solution
Extreme Point Coordinates X 1 X 2
Objective Function 2000x1 + 1500x 2 A X 1 = 0 X 2 = 4 6000 B X 1 = 0.5 X 2 = 2.75 5125 C X 1 = 6 X 2 = 0 12000
Table 2: Shows the objective function Minimum value computation The minimum value is 5125 at the extreme point B, which is the value of the M (objective function). The optimum values variables are X 1 = 0.5 and X 2 = 2.75.
2.6 Multiple Optimal Solutions
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MBA-H2040 Quantitative Techniques for Managers When the objective function passed through only the extreme point located at the intersection of two half planes, then the linear programming problem possess unique solutions. The previous examples i.e. Example 2.5 and Example 2.6 are of this types (which possessed unique solutions). When the objective function coincides with one of the half planes generated by the constraints in the problem, will possess multiple optimal solutions. In this section we are going to discuss about the multiple optimal solutions of linear programming problem with the help of the following Example 2.7.
Example 2.7:
A company purchasing scrap material has two types of scarp materials available. The first type has 30% of material X, 20% of material Y and 50% of material Z by weight. The second type has 40% of material X, 10% of material Y and 30% of material Z. The costs of the two scraps are Rs.120 and Rs.160 per kg respectively. The company requires at least 240 kg of material X, 100 kg of material Y and 290 kg of material Z. Find the optimum quantities of the two scraps to be purchased so that the company requirements of the three materials are satisfied at a minimum cost. Solution
First we have to formulate the linear programming model. Let us introduce the decision variables x1 and x2 denoting the amount of scrap material to be purchased. Here the objective is to minimize the purchasing cost. So, the objective function here is Minimize 120x1 + 160x 2 Subject to: 0.3x1 + 0.4x 2 ≥ 240 0.2x 1 + 0.1x 2 ≥ 100 0.5x 1 + 0.3x 2 ≥ 290 x 1 ≥ 0; x 2 ≥ 0
Multiply by 10 both sides of the inequalities, then the problem becomes Minimize 120x1 + 160x 2 Subject to: 3x1 + 4x 2 ≥ 2400 2x 1 + x 2 ≥ 1000 5x 1 + 3x 2 ≥ 2900 x 1 ≥ 0; x 2 ≥ 0
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