LINEAR PROGRAMMING

Linear Programming



Linear Programming

Introduction

This paper will be presenting a report on the company Brass Ltd. That produces two products, the Masso and the Russo. We will be discussing the optimal production plan for August, and the profit earned by the company. By using the company's data provided we will be highlighting the optimal product mix and the level of the profit that can be earned by the company.

Discussion Analysis

Brass Ltd. produces two products, the Masso and the Russo. Budgeted data relating to these products on a unit basis for August are as follows:

Determining the best product mix that maximizes profits is one of the most fundamental decisions that a company should make. If a company does not have sufficient capacity to satisfy the demand for its products, the best action would be that it should use all of its existing resources and/or expand capacity through capital investment to produce products with the highest profit. The conventional approach to the product mix problem is given below:

Where:

Xi : the number of units of product i (Pi) that is produced in a specific period

N: the total number of different kinds of products that can be produced in the company

mi : material cost per unit of product i

dli: direct labor cost per unit of product i

MT: the capital available for material purchase

DL: the direct labor dollars available

li: the lower limit of the number of units of product i that the company must produce during a specific period

ui: the upper limit of the number of units of product i that can be produced during a specific period

Ri: the return (profit) per unit of product i

Steps that were required to calculate the optimal product mix for the company

Step 1: define the variables

Let w = number of units of Masso to produce and sell

Let E = number of units of Russo to produce and sell

Step 2: Define and formulate the objective function

With C contribution, we aim to maximize C = 40W + 50E

Step 3: Formulate the constraints

Subject to

w + 2E = 700 (machining capacity)

2.5 w + 2E = 1,000 (assembly capacity)

W = 400 (Maximum output of Masso units)

E = 400 (Maximum output of Russo units)

W, E > 0

Step 4: Graph

The constraints are plotted on the graph as follows:

Machining constraint: if w= 0, E= 350 and if E= 0, w 700

Assembly constraint: If w = 0, E = 500 and if E = 0, w = 400

Output of Masso units constraints: w = 400

Output of Russo units constraints: w = 400

Feasible region = OABC

Step 5: Solve for the optimum production plan

B is the optimum point in the graph, where 200 Masso units and 250 Russo units should be output. This can be read from the graph, or solved using simultaneous equations as follows:

2.5 W + 2E = 1,000 (1)

W + 2E = 700 (2)

Subtract equation (2) from equation (1): ...