logical expression in objective function

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mohammad alquraan 2019-6-22

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评论： mohammad alquraan 2019-6-25

采纳的回答： Matt J

can i use logical expression inside the objective function of an optimization problem?

...

prob.Objective = (x(1)+x(2)+x(3)>= 0.0001)+(x(6)+x(7)+x(8)>= 0.0001);

....

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Answer 1

Matt J 2019-6-23

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编辑：Matt J 2019-6-23

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The way you would do this is to introduce additional binary variables,

b1=optimvar('boolean1','Type','integer','LowerBound',0,'UpperBound',1);
b2=optimvar('boolean2','Type','integer','LowerBound',0,'UpperBound',1);

and linear constraints,

prob.Constraint.Boolean1_Upper = b1>=(x(1)+x(2)+x(3)-0.0001)/U1;
prob.Constraint.Boolean1_Lower = b1<=(x(1)+x(2)+x(3))/0.0001;
prob.Constraint.Boolean2_Upper = b2>=(x(6)+x(7)+x(8)-0.0001)/U2;
prob.Constraint.Boolean2_Lower = b2<=(x(6)+x(7)+x(8))/0.0001;

where U1 and U2 are constant bounds large enough that the right hand sides can never be greater than 1. For example, if you have positive upper bounds on the x(i), the U can be the sum of those bounds.

These contraints combined with the binariness of b1 ensure that b1=(x(1)+x(2)+x(3)>=0.0001), and analogously for b2.

And now your objective would just be,

prob.Objective=b1+b2;

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Matt J 2019-6-23

编辑：Matt J 2019-6-23

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mohammad's reply moved here

Hi matt thanks for your assistance.

i tried to run the code using the commands you explained above, the code is as the following: (note: this is a part of my code)

.
.
.
prob = optimproblem('ObjectiveSense','max');
x = optimvar('x',10,1,'LowerBound',0,'UpperBound',1);
b1=optimvar('boolean1','Type','integer','LowerBound',0,'UpperBound',1);
b2=optimvar('boolean1','Type','integer','LowerBound',0,'UpperBound',1);
prob.Constraints.Boolean1_Upper = b1>=(x(1)+x(2)+x(3)-0.0001)/10;
prob.Constraints.Boolean1_Lower = b1<=(x(1)+x(2)+x(3)-0.0001);
prob.Constraints.Boolean2_Upper = b2>=(x(6)+x(7)+x(8)-0.0001)/10;
prob.Constraints.Boolean2_Lower = b2<=(x(6)+x(7)+x(8)-0.0001);
prob.Objective=b1+b2;
cons1 = A1*x <= RD ;
cons2 = A2*x <= Lx;
cons3 = A3*x <= ones(M,1);
cons5 = A5*x == beq;
prob.Constraints.cons1 = cons1;
prob.Constraints.cons2 = cons2;
prob.Constraints.cons3 = cons3;
prob.Constraints.cons5 = cons5;
.
.
.

but i got the following error

OptimizationVariables appearing in the same problem, constraint, or expression must have distinct "Name" properties.
Make a new variable with a different "Name" property, or retrieve the original variable using the Variables property.
 

And Walter's reply moved here:

Change

b2=optimvar('boolean1','Type','integer','LowerBound',0,'UpperBound',1);

to

b2=optimvar('boolean2','Type','integer','LowerBound',0,'UpperBound',1);

mohammad alquraan 2019-6-25

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i changed the constraints like the following:

x = optimvar('x',N*(M+2),'Type','integer','LowerBound',0,'UpperBound',1);
b1=optimvar('b1','Type','integer','LowerBound',0,'UpperBound',1);
b2=optimvar('b2','Type','integer','LowerBound',0,'UpperBound',1);
prob.Constraints.b1_Upper = b1>=(x(1)+x(2)+x(3))/M;
prob.Constraints.b1_Lower = b1<=(x(1)+x(2)+x(3));
prob.Constraints.b2_Upper = b2>=(x(6)+x(7)+x(8))/M;
prob.Constraints.b2_Lower = b2<=(x(6)+x(7)+x(8));
prob.Objective=b1+b2;

this code gave me correct result.

Thanks so much Mr. Matt J for your assistance.

mohammad alquraan 2019-6-25

the above code is a simple example.

how i can write general code were:

x=x(1),x(2),x(3),....,x(M)

and b=b1,b2,b3,....,bN

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Answer 2

Stephan 2019-6-22

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Hi,

use the inequality constraints A and b as input arguments for the solver.

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Walter Roberson 2019-6-23

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The code you posted is already an example of using logical expressions inside objective functions.

The difficulty is that only intlinprog and ga and gamultiobj are certain to handle discontinuities. patternsearch() might handle discontinuities; I would have to review how simannealbnd works to confirm whether it handles discontinuities or not.

fmincon and fminsearch and lsqnonlin do not handle discontinuities.

This does not mean that you definitely cannot use logical expressions for those functions, but you would have to be careful to retain continuity of the function, and continuity of the first derivative; you can violate continuity of the second and further derivatives.

You should probably be recoding

(x(1)+x(2)+x(3)>= 0.0001)+(x(6)+x(7)+x(8)>= 0.0001)

as a pair of linear constraints,

 x(1)+x(2)+x(3) <= 0.0001*(1-eps)
 x(6)+x(7)+x(8) <= 0.0001*(1-eps)
 

The 1-eps has to do with converting the >= to < form: if 0.0001 could be exactly represented in binary floating point, then a value of 0.0001 exactly would generate a logical value of true, which is numeric 1, which would greater than the logical value of false, which is numeric 0, so and for optimization you want minima, so you want x(1)+x(2)+x(3) == 0.0001 to be outside the constraint. And 0.0001 as a literal is the value 0.000100000000000000004792173602385929598312941379845142364501953125 in binary, so permitting equality would get you values that were larger than 0.0001 which you do not want. Permitting equality to 0.0001*(1-eps) is fine as that is 0.000099999999999999977687119290248318748126621358096599578857421875

mohammad alquraan 2019-6-23

编辑：mohammad alquraan 2019-6-23

thanks alot Mr. Walter Roberson for your assistance again.

let me explain my problem in more details.

my optimization problem try to determine maximum number of users that can be served at a given time. (every logical expression will return 0 or 1, then it will sum them to determine number of served users).

my objective function is as the following:

....

prob.Objective = (x(1)+x(2)+x(3)>0)+(x(6)+x(7)+x(8)>0)+.......;

.....

this code is a simplified part of my optimization code. But when i try to run the whole code i got the following error:

"Objective must be a scalar OptimizationExpression or a struct containing a scalar OptimizationExpression."

could you help on this.

thanx in advance.

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