An Introduction to Genetic Algorithms - Boente

(4.1)
The goal is to find n = n* that minimizes L(N n, n). This can be done by taking the derivative of L(N n, n)
with respect to n, setting it to zero, and solving for n:
(4.2)
To solve this equation, we need to express q in terms of n so that we can find dq/ dn. Recall that qis the
probability that A1(N, n) = A1. Suppose A1(N, n) indeed is A1; then A1 was given ntrials. Let S n 1be the sum of
the payoffs of the ntrials given to A1, and let s Nn 2 be the sum of the payoffs of the N n trials given to A 2.
Then
(4.3)
that is, the probability that the observed average payoff of A2 is higher than that of A1. Equivalently,
(4.4)
Chapter 4: Theoretical Foundations of Genetic Algorithms
(For subtle reasons, this is actually only an approximation; see Holland 1975, chapter 5.) Since S n 1 and S Nn 2
are random variables, their difference is also a random variable with a well−defined distribution. Pr((S n 1/n
S Nn 2/(N n)) < 0) is simply the area under the part of the distribution that is less than zero. The problem now
is to compute this area—a tricky task. Holland originally approximated it by using the central limit theorem to
assume a normal distribution. Dan Frantz (as described in chapter 10 of the second edition of Holland 1975)
corrected the original approximation using the theory of large deviations rather than the central limit theorem.
Here the mathematics get complicated (as is often the case for easy−to−state problems such as the
Two−Armed Bandit problem). According to Frantz, the optimal allocation of trials n * to the observed second
best of the two random variables corresponding to the Two−Armed Bandit problem is approximated by
where c1,c2, and c3 are positive constants defined by Frantz. (Here In denotes the natural logarithm.) The
details of this solution are of less concern to us than its form. This can be seen by rearranging the terms and
performing some algebra to get an expression for N n *, the optimal allocation of trials to the observed
better arm:
As n* increases, e n */ 2c 1dominates everything else, so we can further approximate (letting c = 1/2 c1):
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