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Chapter 4

Example 2, continued

Objective function: to maximise contribution C = 4x + 8y

Subject to:

 Department A time, 8x + 10y ≤ 11,000

 Department B time, 4x + 10y ≤ 9,000

 Department C time, 12x + 6y ≤ 12,000

 Non-negativity constraint: 0 ≤ x, y

 y ≤ 600.

The optimal solution (x = 625, y = 600) was found where the department A
and Product B constraints crossed.

We can therefore say:

 Department A time is a critical constraint with no slack.

 Product Y constraint is a critical constraint with no slack.

 For Department B time, actual amount used = 4x + 10y = 4 × 625 +10 ×
600 = 8,500. The maximum amount available was 9,000 giving slack of
500 hours.

 For Department C time, actual amount used = 12x + 6y = 12 × 625 + 6 ×
600 = 11,100. The maximum amount available was 12,000 giving slack
of 900 hours.

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