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或代替和约束

In general, solvers takes constraints with an implicit AND:

constraint 1 AND constraint 2 AND constraint 3 are all satisfied.

However, sometimes you want an OR:

constraint 1 OR constraint 2 OR constraint 3 is satisfied.

These formulations are not logically equivalent, and there is generally no way to express OR constraints in terms of AND constraints.

Tip

Fortunately, nonlinear constraints are extremely flexible. You get OR constraints simply by setting the nonlinear constraint function to the minimum of the constraint functions.

The reason that you can set the minimum as the constraint is due to the nature ofNonlinear Constraints: you give them as a set of functions that must be negative at a feasible point. If your constraints are

F1(x) ≤ 0 ORF2(x) ≤ 0 ORF3(x) ≤ 0,

then set the nonlinear inequality constraint functionc(x) as:

c(x) = min(F1(x),F2(x),F3(x)).

c(x) is not smooth, which is a general requirement for constraint functions, due to the minimum. Nevertheless, the method often works.

Note

You cannot use the usual bounds and linear constraints in an OR constraint. Instead, convert your bounds and linear constraints to nonlinear constraint functions, as in this example.

For example, suppose your feasible region is the L-shaped region:xis in the rectangle –1 ≤x(1) ≤ 1, 0 ≤x(2) ≤ 1 ORxis in the rectangle 0 ≤x(1) ≤ 1, –1 ≤x(2) ≤ 1.

Code for creating the figure

To represent a rectangle as a nonlinear constraint, instead of as bound constraints, construct a function that is negative inside the rectanglea≤x(1) ≤b,c≤x(2) ≤d:

functioncout = rectconstr(x,a,b,c,d)%Negative when x is in the rectangle [a,b][c,d]%First check that a,b,c,d are in the correct orderif(b <= a) || (d <= c) error('Give a rectangle a < b, c < d')endcout = max([(x(1)-b),(x(2)-d),(a-x(1)),(c-x(2))]);

            

            
Following the prescription of using the minimum of nonlinear constraint functions, for the L-shaped region, the nonlinear constraint function is:

            

             

              

               
function[c,ceq] = rectconstrfcn(x) ceq = [];%no equality constraintF(1) = rectconstr(x,-1,1,0,1);%one rectangleF(2) = rectconstr(x,0,1,-1,1);%another rectanglec = min(F);%for OR constraints

              

             

            

            

             

            

            

             

              

             

            

            
Code for creating the figure

            
            
Suppose your objective function is

            

             

              

               
fun = @(x)exp(x(1)) * (4*x(1)^2 + 2*x(2)^2 + 4*x(1)*x(2) + 2*x(2) + 1);

              

             

            

            
Minimizefunover the L-shaped region:

            

             

            

            

             

              

               
opts = optimoptions(@fmincon,'Algorithm','interior-point','Display','off'); x0 = [-.5,.6];%an arbitrary guess[xsol,fval,eflag] = fmincon(fun,x0,[],[],[],[],[],[],@rectconstrfcn,opts)

              

             

            

            

             

              

               
xsol = 0.4998 -0.9996 fval = 2.4650e-07 eflag = 1

              

             

            

            
Clearly, the solutionxsolis inside the L-shaped region. The exit flag is1, indicating thatxsolis a local minimum.

            
See Also

            

            
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