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 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 : 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, address@hidden 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```