Linear Program in Management Accounting
Linear Program is the optimization technique for a system of constraints and a linear objective function. It aims to find the value of the variable that maximizes the objective function. It is the mathematical technique to determine the optimal solution.
In management accounting, the linear program can be used to support management decisions when there are multiple resource limitations. For example, we have limited material and labor which can use to manufacture two product types. There are many combinations of these two products, so we need to use a linear program to calculate a certain level of both products, which can maximize our profit.
Liner program is the alternative technuiqe to analyze the optimal point for company beside of the limiting factor analysis.
Example of Linear Program
Company A produces product X and Y, which require the following input:
|Product||Direct Material||Direct Labor||Selling price|
|Product X||6 Kg||10 hours||250|
|Product Y||8 Kg||8 hours||200|
The company has a limitation of material and labor as follows:
- Material: 200,000 Kg with a cost of $ 5 per Kg
- Labor: 300,000 hours with a cost of $ 10 per hour
Calculate the amount of products X and Y, which the company needs to produce to maximize the profit.
- Assume the X and Y is the amount of product X and Y.
- Define the objective
We are maximizing the contribution, which we assume as the profit.
Contribution of product X & Y:
|Product||Selling price/unit (1)||Material cost/unit (2)||Labor /unit (3)||Contribution = (1)-(2)-(3)|
|Product X||250||(6 Kg * $ 5)||(10 h * $ 10)||$ 120|
|Product Y||200||(8 Kg * $ 5)||(8 h * $ 10)||$ 80|
Contribution = ($ 120 * X) + ($80 * Y)
C = 120X + 80Y
- Define Constraints
- 6X + 10Y £ 300,000 (limitation of direct material)
- 8X + 8Y £ 320,000 (limitation of direct labor)
X & Y must be greater or equal to zero, and we cannot produce a negative number.
- Solve the constraint
- Direct material
If X = 0, Y = 30,000
If Y = 0, X = 50,000
- Direct Labor
If X = 0, Y = 40,000
If Y = 0, X = 40,000
- Build the graph
The OABC is the feasible region.
The optimal point is X = 25,000 units and Y = 15,000 units that generate the contribution of $ 4,200,000 [($ 120 * 25,000) + ($ 80 * 15,000)]