A randomly generated 100BCell population, how many of them are Pareto optimal solutions.
My answer is few, almost 0, and occasionally one or two, and I obviously feel like I'm mistaken somewhere.
We say that Bcell A is better than Bcell B when and only if a is not inferior to B on each target and at least one target is better than B, which is the definition of Pareto optimal solution. Therefore, the non-inferior solution in a group must be superior to the other, such as 99 individuals, the probability is very small. Still is
Prove that there is only one Pareto optimal solution in 100 groups
: If there are at least two optimal solutions in the population, then set them to a, B, by definition (with an optimization target of two)
1. a.obj1>= b.obj1 a.obj2>= b.obj2 and at least one equals sign is not established
2. B.obj1>=a.obj1 B.obj2>=a.obj2 and at least one equals sign is not established
Obviously there is a.obj1 = b.obj1 A.obj2 = b.obj2 and the equal sign is at least one that is not tenable, apparently contradictory
So as I understand the Pareto optimal, the front end is only one, this is the problem I need to solve
However, according to the source code given by Professor Xxcui, of course, this code has errors, running is not expected, but the concept should be good
Judge non-inferior solution omit related code, only part of summary
for (int i = 0; I < Pop_size;i + +)
... {
Flag = 1;//is true in the original text is obviously wrong
for (int j=0;j<pop_size;j++)
... {
if (I!=J)
... {
if (R[i][0] > R[j][0] && r[i][1] > r[j][1])
... {
Flag = 0;
Break
}
}
}
} According to Professor's judgment method and I understand should also be similar, seemingly not wrong, it is not the number of Pareto front-end is very few
One decision, according to my proof, can only produce a single?
In the middle of the puzzle ....... .....
