Linear Programming

We all come across many target-based situations in day-to-day life can you think of any? Say a student has to complete a project in 15 days or a salesperson has to achieve a sales target within a month while another person has to buy an electronic gadget within a budget of ₹500 rupees, go through this situation and try to find out the main objective to achieve by each person individually. Let’s say what is the objective of the student in this case, yes she wants to achieve the maximum score in this project, can you tell me the objective of the salesperson in this case? Yes, he would aim to achieve the maximum possible sales in a month. What do you think would be the objective of a person buying a gadget, she would try to minimize the cost as much as possible, and she buys a gadget that Falls within a budget we can see that is each of these cases above the objective of each of the situation was to maximize the benefits or minimize the cost such types of problem are college Optimisation problems. 

In mathematics an Optimisation problem may involve finding maximum profit, minimum cost, or maybe minimum use of resources there can be many more examples in our day-to-day life that need to be solved using the Optimisation techniques, the problems can be as simple as stated above but could get complicated depending on the situation, we have already discussed the objective of the three given situation now we can look at the important factors we will identify the limiting factor in each case. what is that mean? well in each case there is a scarcity of some resources like is the first case, the time limit to complete the project is limited time to be allotted for completing the project is limited to 15 days only likewise in case two the time in the limiting factor the person has to sell the maximum possible product in a period of one month what can you say about the third situation what the limiting factor in this case the person has to buy the gadget within a predetermined budget that means amount to spend your money is the limiting factor in this case this limiting factor that is the scarcity of resources acts as constraints in finding the best solutions of the given problems, But how are these Optimisation problems solving in mathematics. They’re all different problems solving techniques to solve such problems the main technique we will discuss is linear programming. How it is used to solve Optimisation problems we’ll discuss above.

What is Linear Programming? 

linear programming is a technique that helps us to find the optimum solution for a given problem, an optimum solution is a solution that is the best possible outcome of a given particular problem. In simple terms, it is the method to find out how to do something in the best possible way with given limited resources you need to do the optimum utilization of resources to achieve the best possible result in a particular objective. such as least cost, highest margin, or least time on those resources have alternate uses The situation which requires a search for the best values of the variables subject to certain constraints is amendable programming analysis. These situations cannot be handled by the usual tools of Calculus or marginal analysis. The calculus technique can only handle exactly equal constraints while this limitation does not exist in the case of linear programming problems. A linear programming problem has two basic parts:

• First Part: It is the objective function that describes the primary purpose of the formation to maximize some return or to minimize some.
• Second Part: It is a constant set, It is the system of equalities or inequalities which describe the condition or constraints of the restriction under which Optimisation is to be accomplished.

Types of Linear Programming Problems

Basically, there are many different linear programming problems but we will deal with three major linear programming problems in this article.

1. Manufacturing Problems: Manufacturing problems are a problem that deals with the number of units that should be produced or sold in order to maximize profits when each product requires fixed manpower, machine hours, and raw materials.
2. Diet Problems: It is used to calculate the number of different kinds of constituents to be included in the diet in order to get the minimizing of cost and subject to the availability of food and their prices.
3. Transportation Problems: It is used to determine the transportation schedule to find the cheapest way of transporting a product from plants /factories situated at different locations to different markets.

Terms Related to Linear Programming Problems

In order to solve linear programming problems you need to be clear your concept about the basic terminologies used in solving the first linear programming problems are given below:

• Decision Variable: Variables that compete with each other to share limited resources such as product services etc. They are interrelated and have a linear relationship which is capable of deciding what is the best optimum solution called decision variable.
• Objective Function: The problem must have been a clear and well-defined objective that can be stated quantitatively such as maximization of profit or minimization of cost etc all examples fall under the category of the objective function.
• Constraints: These are the restrictions imposed on the resources available such as a restricted number of machines, labor material, etc.
• Redundant Constraint: Some constraints that are visibly present but do not hinder the process of the problem under study is called a redundant constraint.
• Feasible Solution: These are the set of all possible solutions in the form of variables that satisfy the constants.
• Optimum Solution: This is the best solution among all of the possible one that supports the objective of the problem in the best manner.

Sample Question of LPP

Question 1: A company manufactures and sells two types of products and the cost of production of each unit a and b is rupees 200 and 150 respectively each unit of product yields a profit of 20 rupees and each unit of product b yields a profit of 15 rupees on selling. The company estimates the monthly demand of A and B to b at a maximum of the harvested unit in all the production budget for the month is set at rupees 50000. How many units should the company manufacture in order to earn maximum profit from its monthly sales from a and b?

Solution:

Let x = number of units of type A

y = Number of units of type B

Maximize Z = 40x + 50y

Subject to the constraints

3x + y ≤ 9

x + 2y ≤ 8

and x, y ≥ 0

Consider the equation,

3x + y = 9

x = 3

y = 0

and x + 2y = 8

x = 8

y = 0

Now, we can determine the maximum value of Z by evaluating the value of Z at the four points (vertices) is shown below

Vertices	Z = 40x + 50y
(0, 0)	Z = 40 × 0 + 50 × 0 = Rs. 0
(3, 0)	Z = 40 × 3 + 50 × 0 = Rs. 120
(0, 4)	Z = 40 × 0 + 50 × 4 = Rs. 200
(2, 3)	Z = 40 × 2 + 50 × 3 = Rs. 230

Maximum profit, Z = Rs. 230

∴ The number of units of type A is 2 and the number of units of type B is 3.

Question 2: Maximize Z = 3x + 4y.

Subject to constraints , x + y ≤ 450, 2x + y ≤ 600 and x, y ≤ 0.

Solution: 

We have from the given

Constraints (1)

X + Y = 450

Putting x = 0, ⇒ 0 + y = 450 ⇒ y = 450

Putting y = 0, ⇒ x + 0 = 450 ⇒ x = 450

From, Constraints (2)

2x + y = 600

Putting x = 0, ⇒ 0 + y = 600 ⇒ y = 600

Putting  y = 0, ⇒ 2x + 0 = 600 ⇒ x = 300

Now, we have the points co-ordinate Z = 3x + 4y

Vertices	Z = 3x + 4y
(0, 0)	Z = 3 × 0 + 4 × 0 = 0
(300, 0)	Z = 3 × 300+ 4 × 0 = 900
(150, 300)	Z = 3 × 150 + 4 × 300 = 1650
(0, 450)	Z = 3 × 0 + 4 × 450 = 1800

Therefore, the optimal solution maximum Z = 1800 at co-ordinate x = 0 and y = 450. The graph is given below.