Thread Subject:

NaN issue. Formulation of the Binary Integer Programming Problem

Hello All,

I'm interested in how the BIP problem can be formulated for the following case. Let's say there is Delaunay Triangulation model of the 2D space. I'd like to find a shortest channel of triangles from the initial position to the goal position. I need to codify the following type of inequality constraints, where xi is the i-th triangle (0 - not selected; 1 - selected) and sum(xij) is the sum of adjacent triangles:

x1 + sum(x1j) <= 3

The problem is that sometimes xij can be equal to NaN meaning that the ith triangle has only 2 adjacent triangles instead of 3. I'm trying to use 'isnan' function to check if x is equal to NaN:

A = [x(1,1) + x(~isnan(neighbors(handles.dt, 1)),1)]

But it does not work. Please give me some advice on how to formulate such kind of constraints.

Thanks a lot.

"Liana" wrote in message <img42n$kbr$1@fred.mathworks.com>...
> Hello All,
>
> I'm interested in how the BIP problem can be formulated for the following case. Let's say there is Delaunay Triangulation model of the 2D space. I'd like to find a shortest channel of triangles from the initial position to the goal position. I need to codify the following type of inequality constraints, where xi is the i-th triangle (0 - not selected; 1 - selected) and sum(xij) is the sum of adjacent triangles:
>
> x1 + sum(x1j) <= 3
>
> The problem is that sometimes xij can be equal to NaN meaning that the ith triangle has only 2 adjacent triangles instead of 3. I'm trying to use 'isnan' function to check if x is equal to NaN:
>
> A = [x(1,1) + x(~isnan(neighbors(handles.dt, 1)),1)]
>
> But it does not work. Please give me some advice on how to formulate such kind of constraints.
>
> Thanks a lot.

I see that ~isnan() returns 0 or 1, therefore x(~isnan(...)) does not work correctly. Let me give one example:
x =
   9 16 NaN
   12 10 11
Now I need to change this code (A = [x(1,1) + x(~isnan(neighbors(handles.dt, 1)),1)]) in such a way that NaN cell is simply skipped.