Graphical Method Linear Programming

Graphical Method of Solving Linear Programming Problems

It is not hidden that the simplex method is a well-studied and widely used method for solving linear programming problems. But as far as non-linear programmings are concerned, such a universal method does not exist. With graphical methods, any optimization programming problems consisting of only two variables can easily be solved. These variables can be referred as x₁ and x₂ and with the help of these variables, most of the analysis can be done on a two-dimensional graph. Although we cannot generalize a large number of variables using a graphical approach, the basic concepts of linear programming in a two-variable context can be easily demonstrated. We can always turn to two variable problems if the problems seem to be complicated and we find ourselves in a pool of questions. And we can always search for answers in a two variable case using graphs, that is solving linear programming problems graphically. 

The graphical approach wraps itself with another advantage and that is its visual nature. It provides us with a picture to get along with the algebra of linear programming. This picture can quench our thirst for understanding the basic definitions and possibilities. These reasons are proof that the graphical approach works smoothly with linear programming concepts.

Now, for solving linear programming problems graphically, we must two things:

  1. Inequality constraints

  2. And the objective function.

The graphical method of solving linear programming problems is based on a well-defined set of logical steps. With the help of these steps, we can master the graphical solution of linear programming problems.

Linear Programming Graphical Method

To find the graphical solution of linear programming problems, we have to follow a few steps. 

Step 1) Formulate the problem using the objective and the constraints.

Step 2) Frame the graph by plotting the constraints lines.

Step 3) In this step, determine the valid side of each constraint line.

Step 4) Our next task would be to identify the feasible region.

Step 5) Plot the objective function to determine the direction of improvement.

Step 6) Find the most suitable optimum point.

Step 7) Determine the optimal solution algebraically by calculating the coordinates of the optimum point. 

Step 8) The final step would be to determine the value of the objective function. 

These linear programming problems graphical methods will be helpful to solve any problem. 

Linear Programming Graphical Method Problems With Solutions

Example 1)  let’s consider a furniture manufacturer that produces wooden tables and chairs. The unit profit for tables is \[\$\]6, whereas for chairs is \[\$\]8, and the only two resources that the company uses to manufacture tables and chairs are the woods (board feet) and labor (hours). On an estimate, it takes 30 bf & 5 hours to complete a table, and 20 bf & 10 hours to complete a chair. The company has 300 bf of wood and 110 hours of labor available. The objective function of the company is to maximize profit and the decision variables are the resources that are the woods and the laborers. The constraint set can be the limitations on resource availability, which is 300 bf of wood and 110 hours of labor. Using LP problem graphical methods, the management can come to a decision on how to allocate the limited resources to maximize profits.

Information for the Wooden Tables and Chairs Linear Programming Problem













Unit Profit                      \[\$\]6                           \[\$\]8

Table 1 gives us the information for the linear programming problem. We can go step-by-step for solving the linear programming problems graphically.

Step 1) The aforementioned table can help us to formulate the problem. The bottom row will serve the objective function. The objective function of the company is to maximize unit profit. The woods and the laborers are the constraint set. The nonnegativity conditions are also stated.

Maximize Z = \[6x_{2} + 8x_{2}\]     (is the objective function)

Subject to:  \[30x_{1} + 20x_{2}\] ≤ 300      (300 bf available) 

                   \[5x_{1} + 10x_{2}\] ≤ 110        (110 hours available)

                   \[x_{1} + x_{2}\] ≥ 0                (non - negative conditions)

The two variables (wood and labor) in this problem, can be solved graphically. 

Step 2) this is the graph plotting step. With the x-axis as the number of tables and y-axis as the number of chairs, we can find the two constraint lines. This can be found if we find the x and y-intercepts for the two constraint equations. But before that, we have to rewrite the constraint inequalities as equalities. 

Wood                                                                              Labour 

\[30x_{1} + 20x_{2}\] = 300                                         \[50x_{1} + 10x_{2}\] = 110

Setting x₂ = 0 to solve x₁                                               Setting x₂ = 0 to solve x₁

30x₁ = 300                                               5x₁ = 110

x₁ = \[\frac{300}{30}\]                                                   x₁ = 110/5

= 10 tables                                                                     = 22 tables

(Wood used to make tables)                                       (labors used to make tables)


Setting x₁ = 0 to solve x₂                                             Setting x₁ = 0 to solve x₂ 

20x₂ = 300                                                                     10x₂ = 110

x₂ = \[\frac{300}{20}\]                                      x₂ = 110/10

= 15 chairs                                                                      = 11 chairs

(Wood used to make chairs)                                        (labors used to make chairs) 

Now, plot the wood constraint line \[(x_{1}\] = 10 and \[x_{2} = 15)\] and labor constraint line \[(x_{1} = 22\] and \[x_{2} = 11)\]

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Step 3) To check the valid side for both constraint lines use the origin (0,0).

30(0) + 20(0) < 300 is the valid side of the wood constraint line. In the same way  5(0) + 10(0) < 110 also is a valid side of the labor constraint line. now, draw the arrows indicating the valid side of each constraint line. 

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Step 4) Identify the feasible region which is the area on the valid side of both constraint lines. 

Step 5) Find \[x_{1}\] and \[x_{2}\] using Z = 48 and 72.

In the first case, the values will be \[x_{1}\] = 8 and \[x_{2}\] = 12

In the second case, the values will be \[x_{1}\] = 6 and \[x_{2}\] = 9

Plot the objective function lines when Z = 48 and Z = 72.

The two objective function lines move away from the origin (0,0), Z increases.

(image will be uploaded soon)

Step 6) Find the most attractive corner. 

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Step 7) Calculate the coordinates and find the values of x and y. 

Therefore, according to the company’s optimal solution four tables and nine chairs can be manufactured.

Step 8) finally, determine the value of the objective function for the optimal solution by plugging in the number of tables and chairs and solve for Z: 

Z = \[\$\]6(4) + \[\$\]8(9) = \[\$\]96 

Thus, by producing four tables and nine chairs we can achieve the maximum profit of \[\$\]96.

FAQ (Frequently Asked Questions)

Question 1) What is the Use of the Graphical Method?

Answer 1) A graphical method of linear programming is used for solving the problems by finding out the maximum or minimum point of the intersection between the objective function line and the feasible region on a graph.

Question 2) What are Graphical Methods?

Answer 2) Graph is a powerful tool for data evaluation as they provide us with quick and visual summaries of essential data characteristics. Examples of graphs can be a box plot or a histogram. Graphical methods are basically used for qualitative statistical evaluations.

Question 3) What are Linear Programming Used for?

Answer 3) Linear programming is used for analyzing the supplies in the manufacturing industries. It is used for shelf space optimization. Linear programming is also used for optimizing daily routes. It is used in machine learning.