Unrecognized variable, want to create a list of variable

1 次查看(过去 30 天)
Hello,
I'm trying to run an optimization model but I always get the error "Unrecognized function or variable 'c1_std'.
I want to call d = (c1_std, c2_std, c3_std) to run my optimization model and find the best parameters. How should I define d to call d(1)=c1_std in my different functions?
I'm not an expert in Matlab so please excuse me if my code is quite rude.
Thank you in advance.
Paul
muL = 2000;
sigL = 200;
R1 = 1-9.92*10^-5;
R2 = 1-1.2696*10^-4;
R3 = 1-3.87*10^-6;
d = [c1_std,c2_std,c3_std];
Sr1_min = sqrt(((((1.5-1)*muL)/norminv(R1))^2)-(sigL)^2);
Sr1_max = sqrt(((((2.5-1)*muL)/norminv(R1))^2)-(sigL)^2);
Sr2_min = sqrt(((((1.5-1)*muL)/norminv(R2))^2)-(sigL)^2);
Sr2_max = sqrt(((((2.5-1)*muL)/norminv(R2))^2)-(sigL)^2);
Sr3_min = sqrt(((((1.5-1)*muL)/norminv(R3))^2)-(sigL)^2);
Sr3_max = sqrt(((((2.5-1)*muL)/norminv(R3))^2)-(sigL)^2);
sqrt((((1.5-1)*muL)/norminv(R1)^2)-(sigL)^2)
norminv(R1)
lb = [Sr1_min,Sr2_min,Sr3_min];
ub = [Sr1_max,Sr2_max,Sr3_max];
A = [];
B = [];
Aeq = [];
Beq = [];
d0 = (lb+ub)/2;
fun = @(d) parameterfun(d,muL,sigL);
const = @(d) nonlcon(d,muL,sigL);
[d,fval] = fmincon(fun,d0,A,B,Aeq,Beq,lb,ub,const);
function Rs = parameterfun(d,muL,sigL)
muL = 2000;
sigL = 200;
R1 = 1-9.92*10^-5;
R2 = 1-1.2696*10^-4;
R3 = 1-3.87*10^-6;
d = [c1_std,c2_std,c3_std];
mu_Sr1 = muL+norminv(R1)*sqrt((sigL)^2+(d(1))^2);
mu_Sr2 = muL+norminv(R2)*sqrt((sigL)^2+(d(2))^2);
mu_Sr3 = muL+norminv(R3)*sqrt((sigL)^2+(d(3))^2);
Y1_mean = muL-mu_Sr1;
Y2_mean = muL-mu_Sr2;
Y3_mean = muL-mu_Sr3;
Y1_std = sqrt((d(1))^2+(sigL)^2);
Y2_std = sqrt((d(2))^2+(sigL)^2);
Y3_std = sqrt((d(3))^2+(sigL)^2);
Y_mean = [Y1_mean Y2_mean Y3_mean];
Y_std = [(Y1_std^2) (sigL)^2 (sigL)^2; (sigL)^2 (Y2_std)^2 (sigL)^2; (sigL)^2 (sigL)^2 (Y3_std)^2];
y = zeros(size(d));
Rs = mvncdf(y, Y_mean, Y_std);
end
function [c,ceq] = nonlcon(d,muL,sigL)
muL = 2000;
sigL = 200;
R1 = 1-9.92*10^-5;
R2 = 1-1.2696*10^-4;
R3 = 1-3.87*10^-6;
d = [c1_std,c2_std,c3_std];
c(1) = ns_min(1) - ((muL+norminv(R1)*sqrt((d(1)^2)+(L_std^2)))/muL);
c(2) = ns_min(2) - ((muL+norminv(R2)*sqrt((d(2)^2)+(L_std^2)))/muL);
c(3) = ns_min(3) - ((muL+norminv(R3)*sqrt((d(3)^2)+(L_std^2)))/muL);
c(4) = ((muL+norminv(R1)*sqrt((d(1)^2)+(L_std^2)))/muL) - ns_max(1);
c(5) = ((muL+norminv(R2)*sqrt((d(2)^2)+(L_std^2)))/muL) - ns_max(2);
c(6) = ((muL+norminv(R3)*sqrt((d(3)^2)+(L_std^2)))/muL) - ns_max(3);
c(7) = c_min(1) - (c1_std/((muL+norminv(R1)*sqrt((d(1)^2)+(L_std^2)))));
c(8) = c_min(2) - (c2_std/((muL+norminv(R2)*sqrt((d(2)^2)+(L_std^2)))));
c(9) = c_min(3) - (c3_std/((muL+norminv(R3)*sqrt((d(3)^2)+(L_std^2)))));
c(10) = (c1_std/((muL+norminv(R1)*sqrt((d(1)^2)+(L_std^2))))) - c_max(1);
c(11) = (c2_std/((muL+norminv(R2)*sqrt((d(2)^2)+(L_std^2))))) - c_max(2);
c(12) = (c3_std/((muL+norminv(R3)*sqrt((d(3)^2)+(L_std^2))))) - c_max(3);
format long
c
ceq = [];
end
  2 个评论
Jan
Jan 2022-11-21
In the line:
d = [c1_std,c2_std,c3_std];
neither Matlab nor the readers in the forum can guess, what c1_std, c2_std, c3_std is. What are these variables or functions? Where are they defined?
Paul AGAMENNONE
Paul AGAMENNONE 2022-11-21
编辑:Paul AGAMENNONE 2022-11-21
This is what the optimization model is supposed to search when it optimizes Rs.
Thus I don't know how to call them in my example

请先登录,再进行评论。

采纳的回答

Stephen23
Stephen23 2022-11-21
编辑:Stephen23 2022-11-21
The basic problem appears to be your attempt to re-define these parameters again inside every function, thus obliterating the input values. The simple solution is to define them only once and then pass them as input arguments:
The d value seems to be what you are trying to optimize, so you should not specify any values for that, because the d values are provided by FMINCON.
muL = 2000;
sigL = 200;
R1 = 1-9.92*10^-5;
R2 = 1-1.2696*10^-4;
R3 = 1-3.87*10^-6;
Sr1_min = sqrt(((((1.5-1)*muL)/norminv(R1))^2)-(sigL)^2);
Sr1_max = sqrt(((((2.5-1)*muL)/norminv(R1))^2)-(sigL)^2);
Sr2_min = sqrt(((((1.5-1)*muL)/norminv(R2))^2)-(sigL)^2);
Sr2_max = sqrt(((((2.5-1)*muL)/norminv(R2))^2)-(sigL)^2);
Sr3_min = sqrt(((((1.5-1)*muL)/norminv(R3))^2)-(sigL)^2);
Sr3_max = sqrt(((((2.5-1)*muL)/norminv(R3))^2)-(sigL)^2);
lb = [Sr1_min,Sr2_min,Sr3_min];
ub = [Sr1_max,Sr2_max,Sr3_max];
A = [];
B = [];
Aeq = [];
Beq = [];
d0 = (lb+ub)/2;
fun = @(d) parameterfun(d,muL,sigL,R1,R2,R3);
const = @(d) nonlcon(d,muL,sigL,R1,R2,R3);
[d,fval] = fmincon(fun,d0,A,B,Aeq,Beq,lb,ub,const)
Converged to an infeasible point. fmincon stopped because the size of the current step is less than the value of the step size tolerance but constraints are not satisfied to within the value of the constraint tolerance. Consider enabling the interior point method feasibility mode.
d = 1×3
179.9441 186.3380 103.6089
fval = 0.9998
That works without error. Only you can check if it makes sense.
function Rs = parameterfun(d,muL,sigL,R1,R2,R3)
%
mu_Sr1 = muL+norminv(R1)*sqrt((sigL)^2+(d(1))^2);
mu_Sr2 = muL+norminv(R2)*sqrt((sigL)^2+(d(2))^2);
mu_Sr3 = muL+norminv(R3)*sqrt((sigL)^2+(d(3))^2);
%
Y1_mean = muL-mu_Sr1;
Y2_mean = muL-mu_Sr2;
Y3_mean = muL-mu_Sr3;
%
Y1_std = sqrt((d(1))^2+(sigL)^2);
Y2_std = sqrt((d(2))^2+(sigL)^2);
Y3_std = sqrt((d(3))^2+(sigL)^2);
%
Y_mean = [Y1_mean Y2_mean Y3_mean];
Y_std = [(Y1_std^2) (sigL)^2 (sigL)^2; (sigL)^2 (Y2_std)^2 (sigL)^2; (sigL)^2 (sigL)^2 (Y3_std)^2];
%
y = zeros(size(d));
Rs = mvncdf(y, Y_mean, Y_std);
%
end
function [c,ceq] = nonlcon(d,muL,sigL,R1,R2,R3)
%
% You did not define these, so I will presume all have value 1:
L_std = 1;
ns_min = ones(1,3);
ns_max = ones(1,3);
c_min = ones(1,3);
c_max = ones(1,3);
c1_std = 1;
c2_std = 1;
c3_std = 1;
%
c(1) = ns_min(1) - ((muL+norminv(R1)*sqrt((d(1)^2)+(L_std^2)))/muL);
c(2) = ns_min(2) - ((muL+norminv(R2)*sqrt((d(2)^2)+(L_std^2)))/muL);
c(3) = ns_min(3) - ((muL+norminv(R3)*sqrt((d(3)^2)+(L_std^2)))/muL);
c(4) = ((muL+norminv(R1)*sqrt((d(1)^2)+(L_std^2)))/muL) - ns_max(1);
c(5) = ((muL+norminv(R2)*sqrt((d(2)^2)+(L_std^2)))/muL) - ns_max(2);
c(6) = ((muL+norminv(R3)*sqrt((d(3)^2)+(L_std^2)))/muL) - ns_max(3);
c(7) = c_min(1) - (c1_std/((muL+norminv(R1)*sqrt((d(1)^2)+(L_std^2)))));
c(8) = c_min(2) - (c2_std/((muL+norminv(R2)*sqrt((d(2)^2)+(L_std^2)))));
c(9) = c_min(3) - (c3_std/((muL+norminv(R3)*sqrt((d(3)^2)+(L_std^2)))));
c(10) = (c1_std/((muL+norminv(R1)*sqrt((d(1)^2)+(L_std^2))))) - c_max(1);
c(11) = (c2_std/((muL+norminv(R2)*sqrt((d(2)^2)+(L_std^2))))) - c_max(2);
c(12) = (c3_std/((muL+norminv(R3)*sqrt((d(3)^2)+(L_std^2))))) - c_max(3);
%
ceq = [];
%
end
  1 个评论
Paul AGAMENNONE
Paul AGAMENNONE 2022-11-21
编辑:Paul AGAMENNONE 2022-11-21
Thank you Stephen for your answer, I finally understand and the program works perfectly :)

请先登录,再进行评论。

更多回答(0 个)

类别

Help CenterFile Exchange 中查找有关 Introduction to Installation and Licensing 的更多信息

产品


版本

R2022a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by