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I am trying to work out an if else statement for the following problem, which should be mathematically linear programmed:

when both item 1 and item 2 are picked, both their costs are reduced with 20%. I have more items than only item 1 and 2, and there costs never change. if only one of item 1 or 2 are picked, there regular costs will be implemented.

I thought of the following, however I don't know how to rewrite it in a way it is mathematical and thus not contains any ifs and whens:

$C_i$: reduced cost of item $i$ for $i \in \{1,2\}$

$K_i$: regular cost of item $i$ for $i \in \{1,2,3,4\}$

$X_i$: whether or not item $i$ is picked for $i \in \{1,2,3,4\}$

the sum of $X_i C_i$ should be smaller than or equal to $1300$

$C_1 = 0.8K_1$ if $X_1+X_2=2$

$C_2= 0.8K_2$ if $X_1+X_2=2$

$C_i=K_i$ for $i\in \{3,4\}$

Is there anyone who can help me with this?

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    $\begingroup$ For some basic information about writing mathematics at this site see, e.g., here, here, here and here. $\endgroup$ Commented Jun 14, 2022 at 9:43

3 Answers 3

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Normally in these cases we use an auxiliar variable that takes values $\{0,1\}$. Thus we add the variable $\delta \in \{0,1\}$ that is $1$ if both items are bought and $0$ otherwise. Add the corresponding price to the objective function times $\delta$ and you should be done.

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I agree with @Gowexx that you should use a binary variable $Y_{1,2}$. To enforce this, try using these constraints:

$$Y_{1,2} \ge (X_1 + X_2)-(Y_1 + Y_2)$$ $$Y_{1,2} \ge X_1 - Y_1$$ $$Y_{1,2} \ge X_2 - Y_2$$ $$|Y_1-X_2| =0$$ $$|Y_2-X_1| =0$$ $$Y_1,Y_2,X_1,X_2 \in \{0,1\}$$ $$Y_{1,2} \in \{0,2\}$$

$Y_1$ and $Y_2$ are used to account for one of the $X$'s being $0$ and the other being $1$. If one of the $X$'s is not turned on, the corresponding $Y$ will turn on and turn the RHS of Constraint 1 to 0. If both are turned on, RHS should be 2.

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\begin{align} \sum_{i=1}^4 K_i X_i - 0.2(K_1+K_2)Y &\le 1300 \\ Y &\le X_1 \\ Y &\le X_2 \end{align}

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