Graphical Method Linear Programming

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 Programming is 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.

Methods used for Solving Linear Programming

Following two methods used for Graphical Method of Solving Linear Programming:

• Extreme point solution method 

• Iso-profit (cost) performance line method to find the perfect solution to the LP problem.

Important Points

• A set of variable values ​​xj (j = 1, 2,3..., N) that satisfies the issues of the LP problem is said to form the solution to that LP problem.

• Potential Solution Set of variable values ​​xj (j = 1, 2,3.., N) that satisfies all the barriers and non-negative conditions of the LP problem at the same time is said to form a possible solution to that LP problem.

• Unlikely solution A set of variable values ​​xj (j = 1, 2,3..., N) that do not satisfy all the barriers and non-negative conditions of the LP problem at one time is said to form an impossible solution for that. LP problem.

• With an m-group of simultaneous n (n> m) variables in the LP problem, the solution obtained by placing (n - m) zero-equal variables and solving the remaining m-variables of m is called the basic solution. of that LP problem. Variables (n - m) whose value did not appear in the basic solution are called non-core variables and the remaining variables of m are called the base variables.

• A possible basic solution A possible solution to the LP problem and a basic solution is called a possible basic solution. That is, all basic changes take non-negative values. 

The Basic Solution is two Types

• Degenerate A basic solution that is possible is called degenerate if the value of at least one basic element is zero. 

• Non-degradable The basic practical solution is called non-degenerate if the value of all the basic variables is zero and positive.

Ways to Solve the LP Problem Law

Point Method

To resolve the issue using the existing point method you need to follow these steps:

Step 1: Create statistical design from a given problem. If not provided.

Step 2: Now plan the graph using the given obstacles and find a region that you can use.

Step 3: Find the possible location links (vertices) that we found in step 2.

Step 4: Now check the objective function in each of the possible location areas. Assume that N and n represent the largest and smallest values ​​in these points.

Step 5: If the possible region is bound then N and n are the maximum and minimum values of the objective function. Or if the area is likely to have no boundaries then:

N is the highest value of the target function if the open half system is obtained with an ax + by> N without a common point in the possible area. If not, purpose work has no solution.

n a small amount of objective work if the open half system is found by the ax + by <n does not have a common point in the probable location. If not, purpose work has no solution. 

Iso-Cost Method

The term iso-cost or iso-profit method provides a combination of points that produce the same cost/profit as any other combination in the same line. This is done by arranging the lines corresponding to the slope of the equation.

Follow the following steps for the Iso-cost method solution:

Step 1: Create a statistical design from a given problem. If not provided.

Step 2: Now plan the graph using the given obstacles and find a region that you can use.

Step 3: Now find the possible location links we found in step 2.

Step 4: Find the correct Z (objective function) and draw a line for this objective activity.

Step 5: If the objective function is high type, draw a line corresponding to the objective performance line and this line is farther from the root and has only one common point in the possible area. Or if the target function is a small type then draw a line that corresponds to the target performance line and this line is very close to the source and has one common point in the possible position.

Step 6: Now find the links to the common point we find in step 5. Now, this point is used to find the right solution and the amount of objective work.

Information for the Wooden Tables and Chairs Linear Programming Problem

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 = 6x₂+8x₂ (is the objective function)

Subject to: 30x₁+20x₂≤ 300      (300 bf available) 

5x₁+10x₂ ≤ 110        (110 hours available)

x₁+x₂≥ 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

30x₁+20x₂= 300 

Setting x₂ = 0 to solve x₁         

30x₁ = 300   

x₁ =300/30 = 10 tables      (Wood used to make tables) 

Next: 

Setting x₁ = 0 to solve x₂  

20x₂ = 300  

x₂ =300/20  = 15 chairs (Wood used to make chairs)

Labor

50x₁+10x₂= 110

Setting x₂ = 0 to solve x₁

5x₁ = 110

x₁ = 110/5= 22 tables(labors used to make tables)

Setting x₁ = 0 to solve x₂ 

10x₂ = 110

x₂ = 110/10= 11 chairs (labors used to make chairs)

Now, plot the wood constraint line (x₁= 10 and x₂=15) and labor constraint line (x₁=22

and x₂=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 x₁ and x₂ using Z = 48 and 72.

In the first case, the values will be x₁ = 8 and x2= 12

In the second case, the values will be x₁= 6 and x₂= 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.

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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.