Hello Ioannis.
You may be correct. As I indicated earlier, very often the objective
function will drive D down to zero unless the constraints force it
otherwise. In the models I build that invariably seems to be the case.
If that doesn't apply to you or you are not 100% sure, you could add
further constraints. For example, you could have:
D <= A
D <= U+K
I think this would do what you need
regards
Norman Jessup
On 28/03/2015 9:56 am, john tass wrote:
Very good idea! Thank you very much.
I would like to add though, that because I want to get D = 0 in case A
= 0 or ((U <= 0 AND K <= 0), I have to add the two constraints : D <=
A , D <= E.
Am I correct?
Thanks again
2015-03-28 0:24 GMT+02:00 Norman Jessup <[email protected]
<mailto:[email protected]>>:
Hello Ioannis.
If I understand your problem correctly, you can achieve the result
you need using the "Big M" method, though I think you may need to
introduce some new integer variables
Let E in {0, 1} binary
M*E >= U
M*E> = K
Where M is some "large" positive constant, though in this
particular case it just needs to be greater than 6. This will
force E to be positive if either U or K is positive.
D >= A + E - 1
This will require D to be positive if both A and E are positive.
Note that for this to work the objective must prefer to drive D to
zero if possible, which typically turns out to be the case. If
not then you will need to add complementary constraints to drive D
to zero if the conditions are not met.
On 28/03/2015 3:00 am, [email protected]
<mailto:[email protected]> wrote:
Good evening to every one,
I have a problem in modelling the following situation:
Let U, K in {-6, .. , 6} integers
Let A in {0, 1} binary
Let D in {0, 1} binary
What I want to do is to model the condition:
D = 1, iff (U > 0 OR K > 0) AND A = 1
Otherwise, D should equal 0.
I can not figure out how to model this situation.
Can any one give me an answear or even a hint? It would be very welcome.
Thanks a lot
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