Application of Linear Programming: 3 Examples | Project Management

This article throws light upon the top three examples on the application of linear programming.

Example # 1. Production Allocation Problem:

A firm produces three products. These products are processed on three different machines. The time required to manufacture one unit of each of the three products and the daily capacity of the three machines are given in the table below.

It is required to determine the daily no. of units to be manufactured for each product. The profit per unit for product 1, 2 and 3 is Rs. 4, Rs. 3 & Rs. 6 respectively. It is assumed that all the amounts produced are consumed in the market.

Formulation of Linear Programming Model:

Step 1:

From the study of the situation find the key-decisions to be made. This connection, looking for variables helps considerably. In the given situation key decision is to decide the extent of products 1, 2 and 3, as the extents are permitted to vary.

Step 2:

Assume symbol for variable qualities noticed in step 1. Let the extents of product. 1, 2, and 3 manufactured daily be, x₁, x₂ and x₃ respectively.

Step 3:

Express the feasible alternatives mathematically in terms of variables. Feasible alternatives are those which are physically, economically and financially possible. In the given situation feasible alternatives are sets of values of x₁ x₂ and x₃.

Where x₁, x₂, x₃ > 0 …(1)

Since negative production has no meaning and is not feasible.

Step 4:

Mention the objective quantitatively and express it as a linear function of variables. In the present situation: objective is to maximize the profit.

i.e., which maximize Z = 4x₁ + 3x₂ + 6x₃ … (2)

Step 5:

Put into words the influencing factors or constraints. These occur generally because of constraints on availability or requirements. Express these constraints also as linear equalities / inequalities in terms of variable.

Here, constraints are on the capacities and can be mathematically expressed as

2x₁ + 3X₂ + 2X₃ ≤ 440

4x₁ + 01X₂ + 3x₃ ≤ 470 …(3)

2x₁ + 5x₃ + 0x₃ ≤ 430

Example # 2. Production Planning Problem:

A factory manufactures a product each unit of which consists of 5 units of part A and 4 units of part B. The two parts A & B require different raw materials of which 120 units & 240 units respectively are available. Three parts can be manufactured by three different methods. Raw material requirements per production run and the number of units for each part produced are given below.

Determines the number of production runs for each method so as to maximize the total no. of complete units of the final product.

Formulation of Linear Programming Model:

Step 1:

The key decision to be made is to determine the number of production runs for each method.

Step 2:

Let x₁, x₂, x₃ represents the number of production runs for method 1, 2 and 3 respectively

Step 3:

Feasible alternative are the sets of values of x₁, x₂, and x₃ where x_v x₂, x₃ ≥ 0 …(1)

Since negative no. of production runs has no meaning and is not feasible.

Step 4:

The objective is to maximize the total no. of units of the final product. Now the total no. of units of part A produced by different methods is (6x₁ + 5x₂ + 7x₃) and for part B is (4x₁ + 8x₂ + 3x₃). Since each unit of the final product requires 5 units of part A and 4 units of part B, it is evident that the maximum no of units of the final product cannot exceed the smaller value of;

Step 5:

Constraints are on the availability of raw material they are for raw material 1,7x₁ + 4x₂ + 2x₃ ≤ 120 …(3)

& raw material 2, 5x₁ + 7x₂ + 9x₃ ≤ 240

.^.. The LPP of this problem

Example # 3. Product Mix Problem:

A chemical company produces two products x and y, each unit of product x requires 3 hours on operation 1 & 4 hours on operation II. While each unit of product;’ requires 4 hours on operation 1 and 5 hours on operation II. Total available time for operation 1 and II is 20 hours and 26 hours respectively. The production of each unit of product, y also result in two units of a by-product z at no extra cost.

Product x sells at a profit of Rs. 10/ unit, Whiles sells at a profit of 20/unit. By product z brings a unit profit of Rs. 6 if sold; in case of it cannot be sold the destruction cost is Rs. 4 unit. Forecasts indicate that not more than 5 units of z can be sold. Determine the quantities of x and y to be produced keeping z in mind, so that the profit earned is maximum.

Formulation of L. P. Model:

Step 1:

The key decision to be made is to determine the no. of units of products x, y and z to be produced.

Step 2:

Let the no. of units of products x, y, z produced by x₁, x₂, x₃ where

x₃ = no. of units of z produced

= no. of units of z sold + ——- z destroyed

= x₃ + x₄ (say)

Step 3:

Feasible alternatives are sets of values of x₁, x_?, x₃ & x₄, where x₁, x₂, x₃, x₄ ≥ 0

Step 4:

Objective is to maximize the profit, objective function (profit function) for products x and y is a linear because the profits are constant irrespective of the no. of units produced.

Thus the objective function is maximize z = 10x₁ + 20x₂ + 6x₃ – 4x₄

Step 5:

Constraints are on the time available on operation I: 3x₁ + 4x₂ ≤ 20

——II: 4x₁ + 5x₂ ≤ 26

On the number of units of product z sold: x₃ ≤ 5 produced

2x₂ = x₃ + x₄

or -2x₂ + x₃ + x₄ = 0

Example 4: [Diet Problem]:

A person wants to decide the constituents of a diet which will fulfill his daily requirement of proteins, fats and carbohydrates at the minimum cost. The choice is to be made from four different types of foods. The yield per unit of those foods are given below.

Formulate linear programming model for the problem.

Formulation of L. P. Model

Step 1:

Key decision is to determine the number of units of food type 1, 2, 3, & 4 to be used.

Step 2:

Let three units be x₁, x₂, x₃ & x₄ respectively

Step 3:

Feasible alternatives are sets of value of x_j

Where x_j > 0, J = 1,2,3,4 …(1)

Step 4:

Objective is to minimize the cost i.e., minimize z = (54 x₁ +49 x₂ + 89 x₃ + 75 x₄) …(2)

Step 5:

Constraints are on the fulfillment of the duty requirements of the various constituents.

i. e., for proteins 5x, + 6x₂ + 9x_} + 3x₄ ≥ 900

For fats x₁ + 4x₂ + 4x₃ + 5x₄ ≥ 300 … (3)

For carbohydrates 3x₁+ 2x₂ + 6x₃ + 2x₄ ≥ 800

Thus the L.P. Model is to determine the no. of units of x₁, x₂, x₃ & x₄ that minimize eq. (2) Subject to constraints eq. (3) and non-negatively equations (1).

Example 5:

A ship was there cargo loads – forward after and centre, the capacity limits are:

The following cargoes are offered, the sheep owner accept all or any part of each commodity:

In order to preserve the trim of the ship, the weight in each load must be proportional to the capacity in tonnes. The cargo is to be distributed so as to maximize the profit. Formulate the problem as LP model.

Solution:

Consider the decision variable as:

X_iA, x_iB and x_ic – Weight (in kg.) of commodities A, B, & C to be accommodated in the direction i.e., (1, 2, 3 – forward, centre and after) respectively.