# Linear Programming

Topic Contents | |||

1. Introduction | 2. Use | 3. Methods | |

4. Objective Function | 5. Constraints | 6. Examples |

## Introduction

In managerial accounting, linear programming refers to the application of various mathematical techniques to determine an optimum solution.

A common example of the use of linear programming is to find the optimum mix of products or services that shall lead to maximum profits (i.e. objective function) while taking into consideration any shortage of resources (i.e. constraints).

## Use

As we learned in the tutorial on single limiting factor analysis, financial problems involving only one limiting factor can be solved simply by ranking the alternatives according to their contribution per unit of limiting factor.

Unfortunately, such approach cannot be applied to problems involving **multiple** limiting factors. This is where the use of linear programming techniques such as the graphical method become necessary in order to find the best solution.

Linear programming can be used to solve financial problems involving multiple limiting factors and multiple alternatives. However, where the number of alternatives ( e.g. types of products) is greater than 2, only a specific method of linear programming (known as the simplex method) can be used to determine the optimum solution.

In the following sections, we will learn how to apply linear programming to problems involving only 2 alternatives. Simplex Method shall be covered later in a separate article.

## Methods

There are 2 methods of solving multiple limiting factor problems involving 2 alternatives:

a) Graphical Method

b) Equation Method

Whichever method is used, you will first need to define the objective function and constraints as explained below.

## Objective Function

Objective Function is an equation that defines what you want to achieve by solving the financial problem.

Objective of solving a linear programming problem could for example be to maximize contribution by the production and sale of optimum quantities of products. Objective function simply presents such objectives in the form of mathematical equations.

Example 1

Bob is a farmer.

Bob can grow wheat and barley on his land in the coming season.

Contribution per metric ton of Wheat and Barley is $200 and $100 respectively.

Although wheat seems more profitable to grow, Bob is confused which crop (or a combination of the two crops) to grow in order to maximize his income because growing barley requires less land and fertilizers compared to wheat.

Define Bob's objective function.

As Bob's objective is to maximize total income, his object function will be:

200W + 100B = C

Where:

W = Quantity of wheat (in metric tons) to be grown

B = Quantity of barley (in metric tons) to be grown

C = Maximum Contribution

## Constraints

Constraints are any limitations that prevent an organization from maximizing its profits.

Constraints have to be 'programmed' into a linear programming problem in the form of mathematical expressions so that the optimum solution is within feasible limits.

Some examples of constraints are as follows:

**Limiting factor constraints**

These are mathematical expressions of the scarce resources (e.g. land, labor, machine hours, etc.) that prevent a business from maximizing its sales.**Demand constraints**

These constraints quantify the maximum demand of products or services.**Minimum Supply constraints**

These constraints quantify the minimum supply of products or services (e.g. due to contractual commitments).**Non-negativity constraints**

If the objective of a business is to minimize cost, solving the linear programming problem without defining non-negativity constraints (e.g. the number of units of product X and Y should not be lower than zero) would suggest an optimum solution of producing negative infinite units of both products.

Non-negativity constraints simply ensure that the optimum solution always returns positive.

Example 2

Bob has 4000 acres of land.

Growing 1 Metric Ton each of wheat and barley requires 1 and 0.8 acres of land respectively.

Bob only uses organic fertilizers for his crops.

Bob estimates that he could procure a maximum of 10 Metric Tons of organic fertilizers for the next crop season.

Growing 1 Metric Ton each of wheat and barley requires 0.0035 and 0.002 Metric Tons of fertilizers respectively.

Define the constraints.

Constraints for the problem include land, fertilizer and non-negativity constraints.

As in Example 1, we shall use the following variables for our constraints:

W = Quantity of wheat (in metric tons) to be grown

B = Quantity of barley (in metric tons) to be grown

**Land**

A maximum of 4000 acres of land is available. Therefore, only such quantities of wheat and barley can be grown that their combined coverage of land does not exceed 4000 acres.

So the land constraint will be: 1W + 0.8B ≤ 4000

**Fertilizers**

A maximum of 10 MT of fertilizers is available. Hence, only such quantities of wheat and barley can be grown that their combined requirement of fertilizers does not exceed 10 MT.

Fertilizer constraint will therefore be: 0.0035W + 0.002B ≤ 10

**Non-Negativity Constraints**

Only positive quantities of wheat and barley can be produced.

Therefore:

W ≥ 0

and

B ≥ 0