I think I misinterpreted the question. Now I think you mean when I increase the number of hidden nodes from 4 to 5 why do I start the H=5 from scratch rather than just add a new node to either all or just the better H=4 solutions?
ANSWER: It is easier and faster.
That was good enough for me.
If you want to make a convincing TIME AND ERROR comparison of the two techniques, please use MATLAB datasets and alert me when you post it.