Evolution and Optimum Seeking

The Two Membered Evolution Strategy 107
better survives, but far more accurately that an individual with more favorable properties
produces on average more descendants than one less well adapted to the environment.
Neither does the present work go more deeply into the connections between cause and
e ect in the transmission of inherited information, of which somuch has been revealed
by molecular biology. Mutation is used in the widest biological sense as a synonym for
all types of alteration of the substance of inheritance. In his book Evolutionsstrategie,
Rechenberg (1973) examines in more detail the analogy between natural evolution and
technical optimization. He compares in particular the biological with the technical parameter
space, and interprets mutations as steps in the nucleotide space.
Expressed in mathematical language, the rules are as follows:
Step 0: (Initialization)
There should be storage allocated in a (digital) computer for two points of
an n-dimensional Euclidean space. Each point ischaracterized by a position
vector consisting of a set of n components.
Step 1: (Variation)
Starting from point E (g) , with position vector x (g)
E , in iteration g, a second
point N (g) , with position vector x (g)
N
, is generated, the components x(g)
Ni of
which di er only slightly from the x (g)
Ei. The di erences come about by the
addition of (pseudo) random numbers z (g)
i ,whicharemutually independent.
Step 2: (Filtering)
The two points or vectors are associated with di erent values of an objective
function F (x). Only one of them serves as a starting point for the new
variation in the next iteration g + 1: namely the one with the better (for
minimization, smaller) value of the objective function.
Taking account of constraints in the form of a barrier penalty function, this algorithm
can be formalized as follows:
Step 0: (Initialization)
De ne x (0)
E = fx (0)
Ei i = 1(1)ng T , such that Gj(x (0)
E ) 0 for all j =1(1)m.
Set g =0.
Step 1: (Mutation)
Construct x (g)
N
x (g)
Ni = x (g)
Ei + z (g)
i
= x(g)
E + z(g) with components
for all i = 1(1)n.
Step 2: (Selection)
Decide
x (g+1)
( (g)
(g)
x N if F (x(g)
N ) F (x(g)
E )andGj(x N ) 0 for all j = 1(1)m
E =
x (g)
E otherwise:
Increase g g + 1 and go to step 1 as long as the termination criterion does
not hold.