Application of Linear Programming Problem in an laboratory experiment

 Use of linear programming problem

In our day-to-day life, we often come across situations where we need to maximize the profit utilizing less amount of money or time. In statistics, there is a technique to handle such situations, linear programming problems, or simply LLP.

A linear programming problem (LPP) is a statistical technique to optimize the desired objective function for a given system of linear inequalities. There are several methods like graphical method, simplex table method, big M method, etc. to solve linear programming problems. 

In this article, let us understand how to solve a linear programming problem using the graphical method with the help of an example.

Example

An experiment involves placing the males and females of a laboratory animal species in two separate controlled environments. There is a limited time (in minutes) available in these environments, and the experimenter wants to maximize the number of animals subject to the constraints described.

	Males	Females	Time Available
Environment A	20	25	800
Environment B	20	15	600

The number of male animals and female animals that will maximize the 

Solution

Let x be the number of male animals and y be the number of female animals. The experimenter wants to maximize the number of animals. There is a total (x+y) number of animals, therefore we maximize the objective function z=x+y. In environment A, for 20 males and 25 females at most 800 minutes are available. In environment B, for 20 males and 15 females at most 600 minutes are available. Since the number of males and females can never be negative values, we need to add non-negativity conditions.

Step 1: Formulation of LPP

We can formulate this problem into a linear programming problem as follows.

Maximize z = x+y

Subject to constraints as follows.

20x + 25y ≤ 800

20x + 15y ≤ 600  

With x ≥ 0 and y ≥ 0 (non-negativity condition).

Step 2: Graphical representation of constraints

Now we plot the inequalities on the graph as follows.

To plot the inequalities, we consider the equations of straight lines, 20x+25y=800 and 20x+15y=600. Then we plot these lines on the graph.

Consider the equation, 20x+25y=800. Plugging x=0 we get y=32 this gives point (0, 32). Plugging y=0 we get x=40 this gives point (40, 0).

x	0	40
y	32	0

Consider the equation, 20x+15y=600. Plugging x=0 we get y=40 this gives point (0, 40). Plugging y=0 we get x=30 this gives point (30, 0).

x	0	30
y	40	0

 

Step 3: Choosing a feasible region

After plotting these lines, we are now ready to shade the feasible region. The feasible region is the region within which all the inequalities are satisfied.

In the given problem, 20x+25y ≤ 800; 20x+15y ≤ 600; x≥0 and y≥0 are the constraints. Therefore, the feasible region is the area below the lines 20x+25y=800 and 20x+15y=600 that lies in the first quadrant. Consider the following figure.

Solving the equations, 20x+25y=800 and 20x+15y=600 we get a corner point as (15, 20). Moreover, from the above figure, we observe that (0, 0), (0, 32), and (30, 0) are also the corner point of the feasible region.

Step 4: Finding the optimal solution

Now we need to find the optimum solution. To get the optimum solution we plug the (x, y) values in the objective function, z = x+y.

Corner Points	Objective function (z=x+y)	Optimum Value (z)
(0, 0)	0+0	0
(0, 32)	0+32	32
(15, 20)	15+20	35
(30, 0)	30+0	30

Hence, the maximum of z = 35 lies at (15, 20).

Conclusion

The experimenter should take 15 male animals and 20 female animals for the experiment to suit both the environment.