Question

Turley Limited produces two products, product X and Product Y. Machine and

labour is used in the production process. The company has a maximum of 1000 machine

hours and 1600 labour hour per week. Other information are as follows:

                                                                                                              Product X                            Product Y

Selling price/unit                                                            $150                                     $136

Variable cost/unit                                                          $110                                     $89

Machine hours required                                              2 hours                                 4hours

Labour hours required                                 4 hours                                 4 hours

Required:

1. Calculate the contribution margin (CM) for product X and product Y.

2. Identify the decision variables and determine the ‘objective function’ (sum of contribution margin) for the above.

3. Determine the constraints based on the limited resources.

4. Calculate the optimal number of product X and product Y in order to maximise the contribution margin. Graphs are not necessary.

5. What is the value of the ‘objective function’ at the optimal point?

Answer:

Answer 1

Requirement 1:

	Product X	Product Y
Selling price per unit	$       150	$       136
Variable cost per unit	$       110	$         89
Contribution margin per unit	$         40	$         47

Requirement 2:

Decision Variables : Number of Units of Product X = X

Number of Units of Product Y = Y

Objective Function :

Maximise Z = ($40 * X) + ($47 * Y)

Requirement 3:

Constraints are:

Machine Hours required : 2X + 4Y 1,000 machine hours

Labor Hours Required : 4X + 4Y 1,600 labor hours

Product constraints :

X   0

Y   0

Requirement 4:

Let us equate the machine and labor hours constraints

2X + 4Y = 1,000

4X + 4Y = 1,600

By Subtracting the equation, we get

2X = 600

X = 600 / 2

X = 300

Now put the value of X in any equation

(2 * 300) + 4Y = 1,000

4Y = 1,000 - 600

Y = 400 / 4

Y = 100

The optimum number of product X is 300 and product Y is 100 to maximise contribution margin

Requirement 5:

Value of Objective function:

Z = ($40 * 300) + (47 * 100)

= $12,000 + $4,700

= $16,700