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Posted: **Wed Jun 02, 2021 3:19 pm**

Dear all ,

I want to have a linear function ,

y = f(x,y1,y2,y3)

The above function should give the following answers :

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y = y1 , if x = x1

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y = y2 , if x = x2

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y = y2 , if x = x3

The variable, x is an integer variable, which will only take values, x1, x2 and x3.

I can do it when x is binary (0,1) as follows:

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y = x*y1 + (1-x)*y2

this gives,

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y = y1 , if x = 1
y = y2 , if x = 0

But I want to do it for x , taking three / four values.

Please help me in this.

Thanks

Posted: **Wed Jun 02, 2021 5:44 pm**

First of all, you can model your integer variable as binary variables.

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set
i/1*3/
x = sum(i, i*xb(i));
sum(i, xb(i)) =e= 1;

The idea is that since x can take values 1,2,3, you can declare 3 binary variables xb('1'),xb('2'), and xb('3').

If x is 1, xb('1') is 1 and the rest are zero and so on.

You can then use xb(i) to impose conditions that you want (you know how to do that).

- Atharv

Posted: **Wed Jun 02, 2021 6:01 pm**

Thank you for your answer. Yes. I did the same.

It seems that linearization is more of an art and less of a science.

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