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# Graphical Method Assignment Help

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Solution of Linear Programming Problems - Graphical Method
**Graphical Method**

The graphical method explains the process of obtaining a solution of a linear programming problem in a simple way. The procedure can be explained in the following steps:

Step 1: Formulate the linear programming problem by identifying the decision variables, the objective function and the constraints.

Step 2: Convert the inequality constraints to their equalities and plot them on a graph (in linear form).

**Figure: Optimum Solution (Graphical method) **

Step 3: Using the inequalities in each constraint, determine the feasible region.

Step 4: Write down the corner points of the solution area. Substitute the values in the objective function. The optimum solution is obtained at any of these points.

The feasible solution area for the problem formulated in the Table 4.3 is shown as OABC in the Figure 4.1. From the figure, we can understand that the optimum solution is at the point (2000/3, 5000/3).

Therefore, the number of circular saws to be manufactured per month is 2000/3 = 667 and the number of table saws to be manufactured is 5000/3 = 1667 per month.

Once the problem is graphically represented, the operations manager should ensure that all the points in the feasible region satisfy all the linear programming constraints.

The point, at which the solution is optimum, can also be found by moving the objective linear equation on the feasible region of the same graph. Starting at the origin or from any point, the objective function is moved parallel to itself in the direction away from the origin until the last point in the feasible region is reached. This is the point where the value of the objective function is optimum.

But the graphical method is applicable only to those problems in which a maximum of two decision variables are involved. If the number of constraints in a problem is more than two, the graph becomes difficult to represent and more complex and it becomes harder to identify the feasible region. Since most practical problems include more than two decision variables, the graphical approach is not very useful in business situations.

In some cases, the objective function line may be parallel to one of the sides of a feasible region. In such cases, the two corner points, also all the points in between them, will result in the same value for the objective function at different values of the decision variables.

But solving a problem using the graphical method is tedious, as the decision-maker has to identify the coordinates of all the extreme points in the feasible region, and then evaluate the objective function at each of them. This method is impractical as real problems have hundreds of variables and constraints. For these reasons, the simplex method is preferred to the graphical method.

**Problem **

The plant manager of a plastic pipe manufacturer has the opportunity of using two different routings for a particular type of plastic pipe. Routing 1 uses Extruder A and routing 2 uses Extruder B. Both the routings require the same melting process. The following table shows the time requirements and capacities of these processes.

Time Requirements (hr/100ft)

**Process **

**Routing 1 **

**Routing 2 **

**Capacity (hr) **

Melting

1

1

45

Extruder A

1

0

80

Extruder B

0

1

120

Each 100ft of pipe processed on routing 1 uses 3kg of raw material, whereas that on routing 2 uses 2kg of raw material. The difference results from differing scrap rates of the extruding machines. Consequently, the profit per 100ft of pipe processed on routing 1 is Rs.300 and on routing 2 is Rs.400. A total of 180 kg of raw material is available.

a. Create a set of linear equations to describe the objective function and the constraints.

b. Use graphic analysis to find the visual solution

c. What is the maximum profit?

**Solution **

a) As the objective is to maximize profit, the objective function for this problem would be

Max Z = 300x + 400y

where, x and y are decision variables representing the length of pipe produced using routing 1 and routing 2 respectively.

In the melting process, a maximum of 45 hrs is available for both the routings. Hence,

x(1) + y(1) ≤ 45

x + y ≤ 45

In routing 1, Extruder A can use a maximum of 80 hrs, which means

x(1) + y(0) ≤ 80

x ≤ 80

Similarly, in routing 2, Extruder B can use a maximum of 120 hrs leading to

x(0) + y(1) ≤ 120

y ≤ 120

The available raw material is 200kg and routing 1 consumes 3kg while routing 2 consumes 2 kg for every 100ft. Thus, 3x + 2y ≤ 180

Hence, the objective function is

Max Z = 300x + 400y

The constraints are

x + y ≤ 45

x ≤ 80

y ≤ 120

3x + 2y ≤ 180

where, x ≥ 0 and y ≥ 0

b) To get the visual solution, the data has to be plotted on a graph. To plot the graph, the inequalities in the constraints have to be converted into equalities.

Thus,

x + y = 45

x = 80

y = 120

3x + 2y = 180

In x + y = 45, when x = 0, y = 45 and when y = 0, x = 45

In 3x + 2y = 180, when x = 0, y = 90, and when y = 0, x = 60

**Figure: Graphical Method **

c) To find the maximum profit, the coordinates of the corner points of the feasible region have to be substituted in the objective function.

Corner point A (0,100) -> 300 x 0 + 400 x100 = 40,000

Corner point B (60,0) -> 300 x 60 + 400 x 0 = 18,000

Corner point C (80,0) -> 300 x 80 + 400 x 0 = 24,000

Corner point S (80,120) -> 300 x 80 + 400 x 120 = 24,000 + 48,000 = 72,000

Corner point D (0,100) -> 300 x 0 + 400 x 100 = 40,000

Hence, Corner point S gives the maximum profit of Rs.72,000.

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**Graphical Method**

**Figure: Optimum Solution (Graphical method)**

**Problem**

**Process **

**Routing 1 **

**Routing 2 **

**Capacity (hr) **

Melting

1

1

45

Extruder A

1

0

80

Extruder B

0

1

120

**Solution**

**Figure: Graphical Method**

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