Binary variables allow us to formulate logical constraints. For example, if y
is 0 then sum(i, x(i))
should also be 0. The |Big M method allows us to formulate this with with linear constraints (here we assume the x
variables are non-negative):
yBin.. sum(i, x(i)) =l= bigM*y;
How big should the scalar bigM be? Inexperienced user just use scalar bigM /1e9/;
and cause a lot of numerical trouble in the solver. Moreover, solver work with integer tolerances, e.g. |epInt in GAMS/CPLEX.
In general the answer should be to choose bigM as small as possible. For example, you might limit x
in the following way:
cap.. sum(i, x(i)) =l= maxCap;
Here, we can combine these two equations in one:
yBin.. sum(i, x(i)) =l= maxCap*y;
Which both sets bigM to the smallest possible value and reduces the number of constraints. It might not be always that simple to determine a small bigM but in almost all cases bigM can be calculated from the input data rather than setting it to a data independent insane large value.
There are situations where it is not possible to find a finite bigM. In these rare cases one can use another trick to accomplish the original task, i.e. formulating the constraint if y
is 0 then sum(i, x(i))
should also be 0. We can do this with a SOS1 constraint where the SOS set contains only two elements: the binary and a slack variable:
positive variable slack; yBin.. sum(i, x(i)) =e= slack;
and SOS1 set containing (slack, 1-y). So if y=0
then slack has to be 0 (because 1-y
is non-zero) and SOS1 only allows one member to be non-zero. The GAMS way of formulating SOS constraints does not make it easy for this example, here is how you can do it anyway:
positive variable slack; binary variable y; set s / one, two /; SOS1 variable s1(s); yBin.. sum(i, x(i)) =e= slack; defs11.. s1('one') =e= slack; defs12.. s1('two') =e= 1-y;