@Chibo have you looked at Section 7.7 Elimination of products of variables in the AIMMS optimization modelling guide ? It contains an example on how to formulate products of continuous variables and binary variables (in your case, x(i, j, a) and R(i, j, c, a)). 

Consider lb(i, j, a) and ub(i, j, a) to be the lower and upper bounds of variable x(i, j, a). So, 

lb(i, j, a) <= x(i, j, a) <= ub(i, j, a) forall (i, j, a)

The binary variable is R(i, j, c, a) and the product you need to linearize is x(i, j, a) * R(i, j, c, a1). 

This product is equal to 0 when R = 0 and x when R = 1.

Introduce a continuous variable z(i, j, c, a, a1) with bounds 0, ub(i, j, a)]. Use index domain condition |ord(a1) <= ord(a).

Then introduce the below constraints -

forall (i, j, c, a, a1) | ord(a1) <= ord(a)

z(i, j, c, a, a1) <= x(i, j, a)

z(i, j, c, a, a1) <= R(i, j, c, a1) * ub(i, j, a)

z(i, j, c, a, a1) >= R(i, j, c, a1) * lb(i, j, a)

z(i, j, c, a, a1) >= x(i, j, a) - ub(i, j, c) * (1 - R(i, j, c, a1))

When R(i, j, c, a) = 0, these four constraints become 

z <= x, z <= 0, z >= 0, z >= -ub. z <= 0 and z >= 0 make z = 0

When R(i, j, c, a ) = 1, they become

z <= x, z <= ub, z >= lb, z >= x. z <= x and z >= x make z = x

The expression in your constraint can now be

sum<(c,a1)|ord(a1) <= ord(a), (z(i, j, c, a, a1) / B(c))].

You can drop the a1 index from z and use R(i, j, c, a) instead of R(i, j, c, a1) in the constraints if you need to linearize x(i, j, a) * R(i, j, c, a) instead of x(i, j, a) * R(i, j, c, a1)

Dear Mohansx,

thank you very much for your kind reply. I will study your suggestions and try to implement them!

I will write again if I have any questions.

Thanks very much!