fmincon nonlinear constrain problem set-up
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One of my initial conditions is leaving me scratching my head. Since R & C are defined by summing functions (R will have w components and C will have d components), it may be simpliest to represent R & C as functions. Since there is a function f embedded in R, I run in to problems. I can see how to establish the A,b,Aeq and beq inputs, but I have no clue how to represent the nonlinear constraint of f and pass it in to R. Any help would be much appreciated.
P=R-C; % Objective Function
-P <= 0;
R= sum(U*f(sum(sqrt(((Xb+X)-Xw)^2)));
C= sum(Md*(X-Xb)+B);
f=@d h*exp(1/d);
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Matt J
2016 年 8 月 18 日
編集済み: Matt J
2016 年 8 月 18 日
Something like the following, perhaps?
Pfun=@(X) objectiveFull(X, Xb,Xw, U Md,B,h);
nonlcon=@(X) nonlconFull(X,Pfun);
Xopt = fmincon(Pfun,X0, A,b,Aeq,beq,[],[],nonlcon)
function P=objectiveFull(X,Xb,Xw, U Md,B,h)
f=@(d) h*exp(1/d);
R= sum(U*f( norm(X-(Xw-Sb) ) ) );
C= sum(Md*(X-Xb)+B);
P=R-C;
function [c,ceq]=nonlconFull(X,Pfun)
ceq=[];
c=-Pfun(X);
3 件のコメント
Matt J
2016 年 8 月 19 日
編集済み: Matt J
2016 年 8 月 19 日
Glad it worked out, but the purpose of my suggestion to use nested functions was to spare you the long argument lists. Below, the nested functions Pfun and nonlcon both share access to the workspace of newOptiCode, so there is no longer any need to pass extra arguments to them (like tempWalkerPos, etc...).
function [Profit,Xopt]=newOptiCode(tempBeePos,tempWalkerPos,...
UnitRewards,BeeAdj,DistRewardFunc)
M=BeeAdj(:,1);
B=BeeAdj(:,2);
Xopt = fmincon(@(x) -Pfun(x),tempBeePos, [],[],[],[],...
[],[],@nonlcon,options);
Profit=Pfun(Xopt);
function P=Pfun(x)
R=[];
for w=1:size(tempWalkerPos,1)
f=DistRewardFunc{w};
BWdist=sum(sqrt(((tempBeePos+x)-tempWalkerPos(w,:)).^2));
R(w,1)=UnitRewards(w)*f(BWdist);
end
R=sum(R);
% R= sum(U*f( norm(X-(Xw-Xb) ) ) );
% C= sum(Md*(X-Xb)+B);
C=[];
for d=1:size(tempBeePos,2)
if x(d) > 0;
C(d,1)=M(d)*(x(d)-tempBeePos(d))+B(d);
else
C(d,1)=B(d);
end
end
C=sum(C);
P=R-C;
end
function [c,ceq]=nonlcon(x)
ceq=[];
c=-Pfun(x);
% changed the sign on c=-Pfun(x)
end
end
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