I want to convert an IF statement for my optimization problem. I want to minimize the total price. I want 800 tones of salt and 3 suppliers offer me their prices.

Supplier $1$ offers me $100$ tones at $\\\$150$ a tone.

Supplier $2$ offers me $400$ tones at $\\\$170$ a tone, if I take $300$ tones (or more)the price drops to $\\\$150$.

Supplier $3$ offers me $1600$ tones at $\\\$170$ a tone.

Here is my model without the if statement :

$x_i$ is an integer variable representing the quantity bought to the supplier $i$, $p_i$ is the price of the supplier $i$, $c_i$ is the maximal capacity offered by the supplier $i$.

$Min( p_1\times x_1 + p_2\times x_2 + p_3\times x_3)$

$s.t. x_1 + x_2 + x_3 = 800$

$ x_1 \leq 100$

$ x_2 \leq 400$

$ x_3 \leq 1600$

Now I want to use the Big M method the convert my if statement but I don't know how to do it. I tried this method : Express a "complex" IF-Statement to Linear Programming to find the constraints but I don't know how to modify my prices. I tried this :

If $x_2 \geq 300$, Then $p_2 = 150$, Else $p_2 = 170$.


If $x_2 \geq 300$, Then $b = 1$, Else $b = 0$ and $p_2 = 150b + 170(1-b)$

In addition to this I code in python the the mip library, do you know a better library to do this?

Thanks a lot !


Let $x_i$ represents the quantity bought from the $i$-th supplier, and consider the following scenario: the price of the 2nd supplier is 170\$ per tonne but, if you buy at least 300 tonnes, then you get to choose a non-negative quantity $d \le x_2 - 300$ and you get a 20\$/tonne discount on $d$ tonnes, otherwise $d=0$ (i.e., you get no discount). Clearly this is equivalent to your original problem since, given $x_1, \dots, x_3$, it is always convenient to maximize $d$.

You can model this as follows:

$$ \min 150x_1 + 170 x_2 + 170x_3 - 20d \quad \mbox{s.t.}\\[6pt] \begin{align*} x_1 &\le 100 \\ x_2 &\le 400 \\ x_3 &\le 1600 \\ x_1 + x_2 + x_3 &= 800 \\[6pt] 100 y &> x_2 - 300 & (1)\\ 300 y &\le x_2 &(2)\\[6pt] d &\le x_2 - 300y &(3)\\ d &\le 100y &(4)\\[6pt] x_1, x_2, x_3,d &\ge 0 \\ y &\in \{0,1\} \end{align*} $$

Notice that $y$ is a binary variable that equals $1$ if and only if $x_2 \ge 300$. Indeed, if $x_2 \ge 300$ the constraint $(1)$ is satisfied only when $y=1$ (recall that $x_2 - 300$ is at most $100$), and the constraint $(2)$ simplifies to $300 \le x_2$, which is true. If $x < 300$ then the constraint $(2)$ forces $y=0$ and the constraint $(1)$ simplifies to $0 \ge x_2 - 300$, which is true.

If $y=0$, then $d$ must be $0$, and this is enforced by constraint $(4)$ which becomes $d \le 0$ (recall that $d$ is non-negative), while $(3)$ is always satisfied. If $y=1$, then constraint $(3)$ enforces $d \le x_2 - 300$, which also implies constraint $(4)$ since $d \le x_2 - 300 \le 100 = 100y$.

  • $\begingroup$ Thanks a lot, that's what I needed ! :) $\endgroup$ – xavier laurens Jan 12 at 17:55

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