Model Formulation Examples: Goal Programming

In the previous section, we provided the approach to formulate the goal programming model. Here we provide an example of model formulation.

Example: Model Formulation

The Japan Life Company produces two products- A and B. According to the past experience, production of either product A or product B requires an average of one hour in the plant. The plant has a normal production capacity of 300 hours a month. The marketing department of the firm reports that because of limited market, the maximum number of product A and product B that can be sold in a month are 140 and 200 respectively. The net profit from the sale of product A and product B are Rs. 600 and Rs. 200 respectively. The manager has set the following goals.

P₁: The first goal is to avoid any underutilization of normal production capacity.
P₂: He wants to sell maximum possible units of product A and B. Since the net profit from the sale of product A is thrice the amount from Product B, therefore, the manager has thrice as much desire to achieve sales for product A as for Product B.
P₃: He wants to minimize the overtime operation of the plant as much as possible.

Solution.

Production capacity

Let x₁ = number of units of product A
x₂ = number of units of product B

Since overtime operation of the plant is allowed to a certain extent, the constraint can be written as
x₁ + x₂ + d₁⁻  - d₁⁺ = 300

Where
d₁⁻ = underutilization (idle) of production capacity and
d₁+ = overtime operation of the normal production capacity.

Sales constraints

In this case, the maximum sales for product A and product B are set at 140 and 200 respectively. Hence, it is assumed that overachievement of sales beyond the maximum limits is impossible. Then, the sales (market) constraints can be expressed as

x₁ + d₂⁻ = 140
x₂ + d₃⁻ = 200

Where d₂⁻ and d₃⁻ are the underachievement of the sales goal for product A and B respectively.

Therefore, the complete goal programming model can be written as

Minimize z = P₁d₁⁻ + 3P₂d₂⁻ + P₂d₃⁻ + P₃d₁⁺

subject to

x₁ + x₂ + d₁⁻  - d₁⁺ = 300
x₁ + d₂⁻ = 140
x₂ + d₃⁻ = 200

and x₁, x₂, d₁⁻, d₂⁻, d₃⁻, d₁⁺ ≥ 0