In the Imperialist Competitive Algorithm, the normalization formula is given as C(n)=c(n)-max(c(i)) where C(n) is the normalized cost for nth imperialist, c(n) is the cost for nth imperialist. According to this formula, we start with N imperialists and one of them has the maximum cost value. This means that one of the imperialist has a normalized cost of 0.

The normalized costs are used for the calculation of the probability of the number of the colonies that would be assigned to the imperialist. The formula is p(n)=C(n)/sum(C(i)) This makes that the imperialist with the maximum cost have 0 colonies at all; making the imperialist being eliminated at the first iteration.

Likely, this logic is applied at each iteration according to Atashpaz-Gargari and Lucas* (2007). Hence, at each iteration a normalized cost is calculated and probabilities are recalculated. Using this formula means that each step we need to eliminate at least an imperialist (definitely one at this stage, others maybe at the imperialistic competition stage). As a result, according to these formulae, the algorithm should stop at an iteration value that is less than the number of imperialist. This means if we have 4 imperialists at the beginning of iterations, we may have 2-3 iterations. I know the algorithm can last longer. Where am I misinterpreting the algorithm? How many imperalists are accustomary?

*Atashpaz-Gargari, E. Lucas, C. (2007) Imperialist Competitive Algorithm: An Algorithm for Optimization Inspired by Imperialistic Competition, IEEE Congress on Evolutionary Computation.