Engineering Optimization: Theory and Practice, Fourth Edition

13.2 Genetic Algorithms 695

String of length 20

1 0 0 1 0 0 0 0 1 1 0 0 0 0 1 0 0 1 0 0

x 1 x 2 x 3 x 4

In general, if a binary number is given bybqbq− 1 · · ·b 2 b 1 b 0 , wherebk= 0 or 1,
k= 0 , 1 , 2 ,... , q, then its equivalent decimal numbery(integer) is given by

y=

∑q

k= 0

2 kbk (13.1)

This indicates that a continuous design variablexcan only be represented by a set
of discrete values if binary representation is used. If a variablex(whose bounds are
given byx(l)andx(u)) is represented by a string ofqbinary numbers, as shown in
Eq. (13.1), its decimal value can be computed as

x=x(l)+

x(u)−x(l)

2 q− 1

∑q

k= 0

2 kbk (13.2)

Thus if a continuous variable is to be represented with high accuracy, we need to use a
large value ofqin its binary representation. In fact, the number of binary digits needed
(q)to represent a continuous variable in steps (accuracy) ofxcan be computed from
the relation

2 q≥

x(u)−x(l)

x

+ 1 (13.3)

For example, if a continuous variablexwith bounds 1 and 5 is to be represented with
an accuracy of 0.01, we need to use a binary representation withqdigits where

2 q≥

5 − 1

0. 01

+ 1 =401 or q= 9 (13.4)

Equation (13.2) shows why GAs are naturally suited for solving discrete optimization
problems.

Example 13.1 Steel plates are available in thicknesses (in inches) of
1
32 ,

1

16 ,

3

32 ,

1

8 ,

5

32 ,

3

16 ,

7

32 ,

1

4 ,

9

32 ,

5

16 ,

11

32 ,

3

8 ,

13

32 ,

7

16 ,

15

32 ,

1

2

from a manufacturer. If the thickness of the steel plate, to beused in the construction
of a pressure vessel, is considered as a discrete design variable, determine the size of
the binary string to be used to select a thickness from the available values.

SOLUTION The lower and upper bounds on the steel plate (design variable,x) are
given by 321 and^12 in., respectively, and the resolution or difference between any two
adjacent thicknesses is 321 in. Equation (13.3) gives

2 q≥

x(u)−x(l)

x

+ 1 =

1

2 in.−

1

32 in.

1

32 in.

+ 1 = 15

from which the size of the binary string to be used can be obtained asq=4.