Optimisation, constraints and resource allocation

Translate the decision into quantities with units

Let x be the number of type X kits and y the number of type Y kits in a planning period. In our original classroom model, X uses 2 hours in area A and 1 in B; Y uses 1 in A and 2 in B. Each area has 8 available hours. Constraints are 2x+y≤8 and x+2y≤8, with nonnegative x and y. Both sides of each resource inequality are hours.

Suppose the declared objective is to maximise 3x+4y priority points. These invented points encode a chosen preference, not revenue or a universal social value. The feasible region is the set satisfying every constraint. A preferred but infeasible combination is not an eligible solution. Demand limits, compatibility, quality and human requirements would need additional representation if they govern the actual decision.

A continuous solution is a useful bound

First allow fractional quantities as a mathematical relaxation. The feasible polygon has vertices(0, 0),(4, 0),(8/3, 8/3),(0, 4). Their scores are 0, 12, 56/3 and 16. A linear objective on this bounded polygon attains a maximum at a vertex, so the intersection gives about 18.67 points in the continuous model. Both resource constraints are active there.

If kits must be whole, 8/3 of each is not an executable plan. Enumerate integer x from 0 through 4. The largest feasible y values are 4, 3, 3, 2, 0, giving scores 16, 15, 18, 17, 12. Since the coefficient of y is positive, a smaller y for each fixed x cannot improve its score. The best whole-unit pair is therefore x=2,y=3, score 18.

Optimal means best under this model

The integer plan uses 2×2+3=7 hours in A and 2+2×3=8 in B. One hour remains in A, but B is full. Unused capacity in one resource can coexist with a model optimum. Forcing all resources to appear fully occupied is a different objective and can require infeasible or unnecessary work.

Changing the objective, available hours or requirements can change the best plan. A priority score can conceal fairness or service obligations if they are omitted; one solution should not make them disappear. Check the formulation, input uncertainty and implementation conditions independently. A solver can return a precise optimum to a poorly posed question just as a hand calculation can.

Test a tempting high-output proposal

At x=3,y=3, the score would be 21, but each area would require 9 hours. It violates both 8-hour limits, so the larger score cannot make it acceptable. At x=2,y=3, score 18 is feasible with resource totals 7 and 8.

Rounding the continuous solution to x=3,y=3 creates the infeasible proposal. Rounding is not a general integer-optimisation method. Enumeration here is small enough to verify every possible x; a larger model needs an appropriate method and checks rather than simply rounding each output.

Try the reasoning

1. What makes a plan feasible?

   Show the answer and reasoning

   It satisfies every stated constraint, including resource bounds, nonnegativity and any whole-unit requirement.

2. Why can one resource have an unused hour at the best integer plan?

   Show the answer and reasoning

   The other resource and whole-unit restrictions prevent adding a beneficial kit within all constraints.

3. Does optimality establish that the priority points are the right objective?

   Show the answer and reasoning

   No. Choosing and justifying the objective is a separate design and stakeholder decision.

What to carry forward

An optimum is conditional on the objective, constraints and variable definitions; verify that the resulting plan can be implemented.