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Re: What is error(19) or error(15): out of bound(1)...how to resolve?
From: |
Satyaballi |
Subject: |
Re: What is error(19) or error(15): out of bound(1)...how to resolve? |
Date: |
Sun, 28 Oct 2018 14:05:02 -0500 (CDT) |
Hi
thank you for your reply. But in my code, I've defined 'x' to be 4x1
matrix. Where is the problem?
please help me resolve this.
My code is:
*clear all;
format short e;
% global variable declaration for all constants
global P=6000 L=14 E=30e6 Delmax=0.25 G=12e6 Taumax=13.6e3 Sigmax=30e3;
%defining initial values, limiting values, max no.of iterations and
tolerance
x0=[0.5;10;0.5;1];
lb=[0.125; 0.1; 0.01; 0.01];
ub=[inf; inf; 10; 2];
maxiter=1000;
tol=1e-8;
%defining objective function
function obj=f(x)
obj=1.10471*x(1)^2*x(2)+0.04811*x(3)*x(4)*(14+x(2));
endfunction
%definging gradient of objective function
function gradf=df(x)
gradf=[2.20942*x(1)*x(2);1.10471*x(1)^2+0.04811*x(3)*x(4);0.04811*x(4)*(14+x(2));
0.04811*x(3)*(14+x(2))];
endfunction
%defining Hessian of objective function
function Hessobj=ddf(x)
Hessobj=[2.20942*x(2) 2.20942*x(1) 0 0;2.20942*x(1) 0 0.04811*x(4)
0.04811*x(3);
0 0.04811*x(4) 0 0.04811*(14+x(2));0 0.04811*x(3)
0.04811*(14+x(2)) 0];
endfunction
%defining constraints
function ineqcons=g(x)
global P L E Delmax G Taumax Sigmax;
%display(Taumax);
Tau1=P/(sqrt(2)*x(1)*x(2));
M=P*(L+x(2)/2);
R=sqrt(x(2)^2/4+((x(1)+x(3))/2)^2);
J=2*sqrt(2)*x(1)*x(2)*(x(2)^2/12+((x(1)+x(3))/2)^2);
Tau2=M*R/J;
Tau=sqrt(Tau1^2+Tau1*M*x(2)/J+Tau2^2);
Sigma=6*P*L/(x(4)*x(3)^2);
Delta=4*P*L^3/(E*x(3)^3*x(4));
Pc=(4.103*E/L^2)*(x(3)*x(4)^3/6)*(1-(x(3)/2*L)*sqrt(E/(4*G)));
ineqcons=[-(Tau-Taumax); -(Sigma-Sigmax); -(x(1)-x(4)); -(Delta-Delmax);
-(0.10471*x(1)^2+0.04811*x(3)*x(4)(14+x(2))-5); -(P-Pc)];
endfunction
%Supplying Objective function including Gradient and Hessian
Objfun={@(x)f(x),@(x)df(x),@(x)ddf(x)};
[x,obj,info,iter,nf,lambda] = ...*
sqp(x0,Objfun,[],@(x)g(x),lb,ub, maxiter, tol)
--
Sent from: http://octave.1599824.n4.nabble.com/Octave-General-f1599825.html