ACCA PM Syllabus C. Decision-making Techniques - Formulating and Solving Multiple Scarce Resource Problems - Notes ^{3 / 5}

When there are two or more resources in short supply, linear programming is required to find the solution.

Linear programming is used to:

1. maximise contribution and/or
2. minimise costs

Linear programming - assumptions

1. a single quantifiable objective

2. each product always uses the same quantity of the scarce resource per unit.

3. the contribution (or cost) per unit is constant for each product, regardless of the level of activity. Therefore, the objective function is a straight line.

4. products are independent

5. the focus is short-term

6. all costs either vary with a single volume-related cost driver or they are fixed for the period under consideration.

Linear programming - limitations

1. It may be difficult to identify which resources are likely to be in short supply and what the amount of their availability will be.

2. Management may have objectives other than the maximisation of profit or contribution.

3. Linear relationships may not exist, apart from over very small ranges.

4. The linear programming model is essentially static and is therefore not suitable for analysing the effects of changes over time.

5. Some practical solutions derived from a linear programming model may be of limited use.

Illustration

A company manufactures two types of boxes, Box A and Box B.

Contribution / box A = $20 Contribution / box B = $45

Two materials, X and Y are used in the manufacturing of each box. 
Each material is in short supply.

Material X = 3,000 kg available         Material Y = 2,700 kg available

Each box A uses 20 kg of Material X and 15 kg of Material Y
Each box B uses 30 kg of Material X and 50 kg of Material Y

The objective of this company is to maximize profits.

• Variables

  Let A be the number of boxes of A produced and sold
  Let B be the number of boxes of B produced and sold

• Objective function

  Max C = 20A + 45B

• Constraints

  Mat X = 20A + 30B ≤ 3,000 
  Mat Y = 15A + 50B ≤ 2,700

  Non-negativity = A,B ≥ 0

• Material X

  20A + 30B = 3,000
  When A is 0, B = 100
  When B is 0, A = 150

• Material Y

  15A + 50B = 2700
  When A = 0, B = 54
  When B = 0, A = 180