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;