linear program formulation

[progli]

Hi everyone,

Could you help me to model the following constraint in a LP

z=1 if x!=y
z=0 if x=y

knowing that x,y and z are binary decision variables (0/1)

we can also express it as:

z=x+y case1
z=x-y case2

Thank you in advance.

 

[Ishuda]

Hi everyone,

Could you help me to model the following constraint in a LP

z=1 if x!=y
z=0 if x=y

knowing that x,y and z are binary decision variables (0/1)

we can also express it as:

z=x+y case1
z=x-y case2

Thank you in advance.

I'm not understanding the constraints. If x and y are binary decision variables, then x! (assumed x factorial) is always 1. Therefore there case of (x,y)=(1,0), y is neither equal to x nor is it equal to x!, so what is z? In the case of (1,1), y is equal to both x and x! which means z is both 0 and 1. So, I would think something would have to be changed.

 

[progli]

Thank you Ishuda for your reply.

I mean X<>y (x different from y) not factorial.

If you have any idea please let me know.

 

[Ishuda]

Thank you Ishuda for your reply.

I mean X<>y (x different from y) not factorial.

If you have any idea please let me know.

Yes, makes more sense. I may should have seen that.

Seems to me you answered the question when you said either z=x+y [modular arithmetic I assume] or z=x-y. So I'm still not sure about the problem. What is statement of the complete problem as given?

 

[progli]

Actually, the LP has to minimise some function over a set of solutions by comparing them. Each solution is a vector of 0/1 bits. to be sure that it will generate and compare different solutions, I need to add a constraint in the form:

and

z_i=1 if x_i<>y_i

z_i=0 if x_i=y_i