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Modelling a linear programme

Going from a word problem to a linear programme: decision variables, objective function, constraints, sign conditions, feasible solution, and what makes a relation linear or not. Three typical problems: production, blending, transport.

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  • Linear programme: what is it?

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    A problem that maximises or minimises an expression of degree 1 in unknowns subject to equalities or inequalities of degree 1.

    Everything is of degree 1: no product of two unknowns, no square, no inverse. That regularity is what lets us solve it with certainty.

  • A table earns €30 and a chair €20. With x tables and y chairs, which objective function is to be maximised?

    • 30x + 20y
    • 20x + 30y
    • 30x × 20y
    • 50(x + y)
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    30x + 20y

    Each product contributes its unit profit times its quantity, and we add. Swapping the coefficients or multiplying them no longer describes the workshop's profit.

  • To build a model, first set the …, then the …, then the ….

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    To build a model, first set the decision variables, then the objective function, then the constraints.

    This order avoids mixing roles: neither the objective nor a constraint can be written before we know what x and y stand for.

  • Decision variable: what is it?

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    An unknown of the model whose value stands for a choice the decision maker controls, for example a quantity to make.

    They come first, with their unit: “x = number of tables”. A figure given in the statement, such as a price or a capacity, is not a decision variable.

  • A table takes 2 h of carpentry and a chair 1 h; the workshop has 100 h. Which constraint expresses this limit?

    • x + 2y ≤ 100
    • 2x + y ≤ 100
    • 2x + y ≥ 100
    • 2x + y ≤ 3
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    2x + y ≤ 100

    Hours used are 2 per table plus 1 per chair, and they must not exceed the stock: hence ≤ 100. x + 2y swaps the coefficients; 3 is the time to make one of each, not the stock of hours.

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