# Equation Method . Linear Programming

Topic Contents | |||

1. Introduction | 2. Steps | 3. Example |

## Introduction

Equation method can be used to solve linear programming problems with a greater accuracy than the graphical method. Unlike graphical method in which the coordinates of the optimum point are found by measuring directly from the graph, the equation method determines the precise coordinates by mathematically solving the two equations on which the optimum point lies.

*We explained the 7 Steps approach to solving linear programming problems using the graphical method. The first 6 steps of the graphical method are to be applied in the exact same manner in the equation method as well. This article will therefore only explain Step 7 onwards as the previous steps have already been covered. To review the first 6 steps please review the article on graphical method.*

Following steps describe process of solving linear programming problems using the equation method.

**Step 1 - 6**

Same as graphical method

**Step 7 Identify the 2 equations**

You need to identify the 2 equations on which the optimum point lies. These 2 equations will be solved in the subsequent steps to find the co-ordinates of the optimum point.

Example

The two equations on which the optimum point lies are:

1S = 0.2J

10J + 25S = 1000

**Step 8 Re-arrange equations**

Both equations must be written in the same format in order to solve them mathematically.

Example

1S = 0.2J could be re-written as -0.2J + 1S = 0 so that it is arranged in the same manner as the equation 10J + 25S = 1000.

**Step 9 Find the first co-ordinate by the process of elimination**

In order to find the value of the first co-ordinate (variable) of the optimum point, you first need to cancel out the other variable.

This can be done by multiplying one of the equations by a number that causes the positive value of a variable in one equation to equal the negative value of the same variable in the second equation.

You then proceed by calculating the sum of the two equations. The value of the remaining variable can then be found using simple algebra.

Example

-0.2J + 1S = 0 10J + 25S = 1000 | ||

50[-0.2J + 1S = 0] 10J + 25S = 1000 | Multiplying the first equation by 50 would make the value of J in the first equation to equal the negative value of J in the second equation. | |

-10J + 50S = 0 10J + 25S = 1000 | ||

0J + 75S = 1000 | This is a simple sum of the two equations. As you can see, J has been cancelled out. You could have canceled out S instead of J by changing the multiplication factor to -25. | |

0J + 75S = 1000 | Now we can use simple algebra to find the value of S. | |

75S = 1000 | ||

S = 1000 ÷ 75 | ||

S = 13.33 |

**Step 10 Find the second co-ordinate by the process of substitution**

To find the value of the second co-ordinate of the optimum point, you need to substitute the value of the variable obtained in Step 9 in any one of the two equations.

**Example**

10J + 25S = 1000

10J + 25(13.33) = 1000 Substituting 13.33 found in Step 9 in place of S.

10J + 333.25 = 1000

10J = 1000 - 333.25

10J = 666.75

J = 666.75 ÷ 10

**J = 66.67**

**The co-ordinates of the optimum point are 13.33S and 66.67J.**

**Tip**

You may be wondering why we need to draw a graph if we can just find the co-ordinates of all points of intersection between the constraint equations mathematically as above and determine the optimum point by trial and error (e.g. which of the points result in highest contribution or lowest cost).

The reason you have to draw a graph even in equation method is to avoid a solution that is not feasible in the first place. Drawing a graph will ensure that the optimum solution is always feasible.