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`ga`

can solve problems when certain variables
are integer-valued. Give `IntCon`

, a vector of the *x* components
that are integers:

[x,fval,exitflag] = ga(fitnessfcn,nvars,A,b,[],[],... lb,ub,nonlcon,IntCon,options)

`IntCon`

is a vector of positive integers that
contains the *x* components that are integer-valued.
For example, if you want to restrict `x(2)`

and `x(10)`

to
be integers, set `IntCon`

to `[2,10]`

.

Restrictions exist on the types of problems that `ga`

can
solve with integer variables. In particular, `ga`

does
not accept any equality constraints when there are integer variables.
For details, see Characteristics of the Integer ga Solver.

`ga`

solves integer problems best when you
provide lower and upper bounds for every *x* component.

This example shows how to find the minimum of Rastrigin's function restricted so the first component of *x* is an integer. The components of *x* are further restricted to be in the region .

**Set up the bounds for your problem**

lb = [5*pi,-20*pi]; ub = [20*pi,-4*pi];

**Set a plot function so you can view the progress of ga**

opts = optimoptions('ga','PlotFcn',@gaplotbestf);

**Call the ga solver where x(1) has integer values**

rng(1,'twister') % for reproducibility IntCon = 1; [x,fval,exitflag] = ga(@rastriginsfcn,2,[],[],[],[],... lb,ub,[],IntCon,opts)

Optimization terminated: average change in the penalty fitness value less than options.FunctionTolerance and constraint violation is less than options.ConstraintTolerance. x = 16.0000 -12.9325 fval = 424.1355 exitflag = 1

ga converges quickly to the solution.

There are some restrictions on the types of problems that `ga`

can
solve when you include integer constraints:

No linear equality constraints. You must have

`Aeq = []`

and`beq = []`

. For a possible workaround, see No Equality Constraints.No nonlinear equality constraints. Any nonlinear constraint function must return

`[]`

for the nonlinear equality constraint. For a possible workaround, see Example: Integer Programming with a Nonlinear Equality Constraint.Only

`doubleVector`

population type.No custom creation function (

`CreationFcn`

option), crossover function (`CrossoverFcn`

option), mutation function (`MutationFcn`

option), or initial scores (`InitialScoreMatrix`

option). If you supply any of these,`ga`

overrides their settings.`ga`

uses only the binary tournament selection function (`SelectionFcn`

option), and overrides any other setting.No hybrid function.

`ga`

overrides any setting of the`HybridFcn`

option.`ga`

ignores the`ParetoFraction`

,`DistanceMeasureFcn`

,`InitialPenalty`

, and`PenaltyFactor`

options.

The listed restrictions are mainly natural, not arbitrary. For example:

There are no hybrid functions that support integer constraints. So

`ga`

does not use hybrid functions when there are integer constraints.To obtain integer variables,

`ga`

uses special creation, crossover, and mutation functions.

You cannot use equality constraints and integer constraints in the same problem. You can try to work around this restriction by including two inequality constraints for each linear equality constraint. For example, to try to include the constraint

3*x*_{1} –
2*x*_{2} = 5,

create two inequality constraints:

3*x*_{1} –
2*x*_{2} ≤ 5

3*x*_{1} –
2*x*_{2} ≥ 5.

To write these constraints in the form `A x`

≤ `b`

, multiply
the second inequality by `-1`

:

–3*x*_{1} +
2*x*_{2} ≤ –5.

You can try to include the equality constraint using `A`

= `[3,-2;-3,2]`

and `b`

= `[5;-5]`

.

Be aware that this procedure can fail; `ga`

has
difficulty with simultaneous integer and equality constraints.

**Example: Integer Programming with a Nonlinear Equality Constraint. **This example attempts to locate the minimum of the Ackley function
in five dimensions with these constraints:

`x(1)`

,`x(3)`

, and`x(5)`

are integers.`norm(x) = 4`

.

The Ackley function, described briefly in Resuming ga From the Final Population, is difficult to minimize. Adding integer and equality constraints increases the difficulty.

To include the nonlinear equality constraint, give a small tolerance `tol`

that
allows the norm of `x`

to be within `tol`

of `4`

.
Without a tolerance, the nonlinear equality constraint is never satisfied,
and the solver does not realize when it has a feasible solution.

Write the expression

`norm(x) = 4`

as two “less than zero” inequalities:`norm(x) - 4`

≤`0`

`-(norm(x) - 4)`

≤`0`

.Allow a small tolerance in the inequalities:

`norm(x) - 4 - tol`

≤`0`

`-(norm(x) - 4) - tol`

≤`0`

.Write a nonlinear inequality constraint function that implements these inequalities:

function [c, ceq] = eqCon(x) ceq = []; rad = 4; tol = 1e-3; confcnval = norm(x) - rad; c = [confcnval - tol;-confcnval - tol];

Set options:

`MaxStallGenerations = 50`

— Allow the solver to try for a while.`FunctionTolerance = 1e-10`

— Specify a stricter stopping criterion than usual.`MaxGenerations = 300`

— Allow more generations than default.`PlotFcn = @gaplotbestfun`

— Observe the optimization.

opts = optimoptions('ga','MaxStallGenerations',50,'FunctionTolerance',1e-10,... 'MaxGenerations',300,'PlotFcn',@gaplotbestfun);

Set lower and upper bounds to help the solver:

nVar = 5; lb = -5*ones(1,nVar); ub = 5*ones(1,nVar);

Solve the problem:

rng(1,'twister') % for reproducibility [x,fval,exitflag] = ga(@ackleyfcn,nVar,[],[],[],[], ... lb,ub,@eqCon,[1 3 5],opts); Optimization terminated: stall generations limit exceeded and constraint violation is less than options.ConstraintTolerance.

Examine the solution:

x,fval,exitflag,norm(x) x = -3.0000 0.8233 1.0000 -1.1503 2.0000 fval = 5.1119 exitflag = 3 ans = 4.0001

The odd

`x`

components are integers, as specified. The norm of`x`

is`4`

, to within the given relative tolerance of`1e-3`

.Despite the positive exit flag, the solution is not the global optimum. Run the problem again and examine the solution:

opts = optimoptions('ga',opts,'Display','off'); [x2,fval2,exitflag2] = ga(@ackleyfcn,nVar,[],[],[],[], ... lb,ub,@eqCon,[1 3 5],opts);

Examine the second solution:

x2,fval2,exitflag2,norm(x2) x2 = -1.0000 -0.9959 -2.0000 0.9960 3.0000 fval2 = 4.2359 exitflag2 = 1 ans = 3.9980

The second run gives a better solution (lower fitness function value). Again, the odd

`x`

components are integers, and the norm of`x2`

is`4`

, to within the given relative tolerance of`1e-3`

.

Be aware that this procedure can fail; `ga`

has
difficulty with simultaneous integer and equality constraints.

Integer programming with `ga`

involves several
modifications of the basic algorithm (see How the Genetic Algorithm Works). For integer programming:

Special creation, crossover, and mutation functions enforce variables to be integers. For details, see Deep et al. [2].

The genetic algorithm attempts to minimize a penalty function, not the fitness function. The penalty function includes a term for infeasibility. This penalty function is combined with binary tournament selection to select individuals for subsequent generations. The penalty function value of a member of a population is:

If the member is feasible, the penalty function is the fitness function.

If the member is infeasible, the penalty function is the maximum fitness function among feasible members of the population, plus a sum of the constraint violations of the (infeasible) point.

For details of the penalty function, see Deb [1].

`ga`

does not enforce linear constraints when there are integer constraints. Instead,`ga`

incorporates linear constraint violations into the penalty function.

[1] Deb, Kalyanmoy. *An efficient
constraint handling method for genetic algorithms.* Computer
Methods in Applied Mechanics and Engineering, 186(2–4), pp.
311–338, 2000.

[2] Deep, Kusum, Krishna Pratap Singh, M.L.
Kansal, and C. Mohan. *A real coded genetic algorithm for
solving integer and mixed integer optimization problems.* Applied
Mathematics and Computation, 212(2), pp. 505–518, 2009.

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