MiguelL
MiguelL

Reputation: 81

LP / MILP formulation with OR logic

I'm solving a LP / MILP problem using ILOG CPLEX.

int n = ...;
range time =1..n;

dvar float+ c[time] in 0..0.3;
dvar float+ d[time] in 0..0.3;
dvar float+ x[time];

int beta[time]=...;
float pc[time]=...;
float pd[time]=...;

//Expressions

dexpr float funtion = sum(t in time) (d[t]*pd[t]-c[t]*pc[t]); 

//Model

maximize function;

subject to {

    x[1] == 0.5;
    c[1] == 0;
    d[1] == 0;

    forall(t in time)
        const1:
            x[t] <= 1;          

    forall(t in time: t!=1)
        const2:
            (x[t] == x[t-1] + c[t] - d[t]);         

    forall(t in time: t!=1)
        const3:         
            ( d[t] <= 0) || (c[t] <= 0); 

As you can see I've forced c[t] and d[t] to never be bigger than 0 at the same time with "const3".

My question is, how would this constraint be represented in a LP/MILP mathematical formulation?

Is adding this new variable enough? :

y[t]≤c[t]+d[t]

y[t]≥c[t]

y[t]≥d[t]

0≤y[t]≤M (M is the maximum value of both c or d)

Upvotes: 0

Views: 590

Answers (1)

Daniel Junglas
Daniel Junglas

Reputation: 5940

As far as I can tell, the constraints you suggested would allow this setting:

c[t] = 0.1
d[t] = 0.1
y[t] = 0.2

Which has c and d different from 0 simultaneously.

I can see these options to formulate your condition without logical constraints:

1) Use an SOS constraint that contains just c[t] and d[t]. By definition of SOS only one of the two can be non-zero in any feasible solution.

2) Use a boolean variable y[t] and add constraints

c[t] <= M * y[t]
d[t] <= M * (1 - y[t])

3) Again, use boolean y[t] and then indicator constraints

(y[t] == 0) => (c[t] == 0);
(y[t] == 1) => (d[t] == 0);

4) You can just state c[t] * d[t] == 0 but that will make the model non-linear.

In any case, a solver will probably be able to reduce your original formulation to either 2 or 3. So reformulating the constraint may not make things faster but only more obscure.

Upvotes: 3

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