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I'm trying to model the number of new planes an airline must buy. There are 3 types: small, medium and big. Let $x_i$ be the number of new planes of type $i$ that the airline buys (1=small, 2=medium and 3=big).

One of the restrictions of the problem is that the maintenance crew can only keep 40 small planes. We also know that a medium size plane equals $\frac{4}{3}$ of a small one and that a big one equals $\frac{5}{3}$ of a small plane.

I'm not sure how to write this restriction. I wrote

\begin{equation*} x_1+\frac{3}{4}x_2+\frac{3}{5}x_3 \leq 40 \end{equation*}

but I'm not sure if it is the correct response.

MarianaM
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1 Answers1

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We need to make sure that all planes don't take more than 40 space. We can achieve this by adding the constraint \begin{equation*} x_1+\frac{4}{3}x_2+\frac{5}{3}x_3 \leq 40. \end{equation*}

In total we would have the following set of constraints:

$$x_1+\frac{4}{3}x_2+\frac{5}{3}x_3 \leq 40$$ $$x_1 \ge 0$$ $$x_2 \ge 0$$ $$x_3 \ge 0$$ $$x_i \in \mathbb{Z}, \; i = 1,2,3$$

The following constraints make sure that no single plane exceeds the total space available. Even though these constaints are valid, they do not have to be included because they are already expressed via $x_1+\frac{4}{3}x_2+\frac{5}{3}x_3 \leq 40$ with the non-negativity constraints.

$$x_1 \le 40$$ $$x_2 \le 40 \cdot \frac{3}{4} = 30$$ $$ x_3 \le 40 \cdot \frac{3}{5} = 24$$

These constraints are valid but will only make sure that we don't buy too much of a single kind of plane. With only these constraints a solution like (40,30,24) would still be feasible but would clearly violate the space requirement. This is why we need $x_1+\frac{4}{3}x_2+\frac{5}{3}x_3 \leq 40$.

YukiJ
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