MATLAB: Array indices must be positive integers or logical values.

arrayslogical?

im sorry i dont know how this can happen, can someone help me

k=[0:20:1000]
for a=1:length(k);
   derivative_x(a)=2*kons_G*rho_model*t*(-(1/(z2((x(a)-x0)^2/z2^2+1)))+(1/(z1((x(a)-x0))^2/z1^2+1))); %sudah dalam mGal

   derivative_z(a)=((-2*kons_G*t*rho_model*((x(a)-x0)/(z1^2+(x(a)-x0)^2)))); %sudah dalam mGal
end

Best Answer

   derivative_x(a)=2*kons_G*rho_model*t*(-(1/(z2((x(a)-x0)^2/z2^2+1)))+(1/(z1((x(a)-x0))^2/z1^2+1))); %sudah dalam mGal

MATLAB has absolutely no implied multiplication. Not even in the symbolic mathematics languages.

Therefore your z2( and z1( expressions are interpreted as requests to index z2 and z1 at calculated locations that do not happen to come out as positive integers.

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MATLAB: Failure in initial user-supplied objective function evaluation. FSOLVE cannot continue

Hi,

this runs - not sure if it is really efficient, but it works:

x0=[1,1,1,1,1];
fsolve(@F,x0)
function m=F(x)
syms  y1 y2 y3 y4 z1 k11 k12 k13 k14 l11 
h=1/1000;
f1(y1,y2,y3,y4,z1)=(-(z1^3)/(y3^2))*(3*(y2-y1+y3^(-1)-z1/10)^2+(1/5)*(y2-y1+y3^(-1)-(z1)/10))-y4;
f2(y1,y2,y3,y4,z1)=1/10*z1-y4;
f3(y1,y2,y3,y4,z1)=(z1^3)*(3*(y2-y1+y3^(-1)-z1/10)^2+(1/5)*(y2-y1+y3^(-1)-(z1)/10));
f4(y1,y2,y3,y4,z1)=y1-y3^(-1);
g(y1,y2,y3,y4,z1)=(y1-y3^(-1))^2+y4^2-1/10*z1;
J1(y1,y2,y3,y4,z1)=jacobian([f1,f2,f3,f4],[y1,y2,y3,y4]);
J1=double(J1(2,2,1,0,10));
J2(y1,y2,y3,y4,z1)=jacobian([f1,f2,f3,f4],z1);
J2=double(J2(2,2,1,0,10));
J3(y1,y2,y3,y4,z1)=jacobian(g,[y1,y2,y3,y4]);
J3=double(J3(2,2,1,0,10));
J4(y1,y2,y3,y4,z1)=jacobian(g,z1);
J4=double(J4(2,2,1,0,10));
k11=x(1);
k12=x(2);
k13=x(3);
k14=x(4);
l11=x(5);
k1=[k11;k12;k13;k14];
m(1)=double(h*(f1(2,2,1,0,10)+0.44*(J1(1,:)*k1 +J2(1,:)*l11))-k11);
m(2)=double(h*(f2(2,2,1,0,10)+0.44*(J1(2,:)*k1 +J2(2,:)*l11))-k12);
m(3)=double(h*(f3(2,2,1,0,10)+0.44*(J1(3,:)*k1 +J2(3,:)*l11))-k13);
m(4)=double(h*(f4(2,2,1,0,10)+0.44*(J1(4,:)*k1 +J2(4,:)*l11))-k14);
m(5)=double(g(2,2,1,0,10)+0.44*(J3*k1 +J4(1,:)*l11));
end

Best regards

Stephan

MATLAB: Calculation in inter dependency matrix

You could use a 3D matrix to store the new values of A1, D1, P and Utility

 result=zeros(221,4,1000);
for (i=1:1000)
    A1= ( randn(221,1) * 0.1 ) + 0.45;
    X11= cell2mat(A)*A1;
    P= zeros(221,1);
    P(:)= X11(:)/sum(X11)
    D1= ( randn(221,1) * 0.1 ) + 0.45;
    Utility=0;
     for(j=1:221)
         Utility= (((1-P(j))*(D(j)/(D(j)+A1(j)))* V)-C +Utility;
     end
    result(:,1,i)=A1;
    result(:,2,i)=D1;
    result(:,3,i)=P;
    result(1,4,i)=Utility;
 end

To find the maximum Utility use:

 [Utility_max,index_max]=max(result(1,4,:));
 A1_max=result(:,1,index_max);
 D1_max=result(:,2,index_max);
 P_max=result(:,3,index_max);

where A1_max, D1_max, P_max are the values of A1, D1, P at maximum Utility

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MATLAB: Failure in initial user-supplied objective function evaluation. FSOLVE cannot continue

MATLAB: Calculation in inter dependency matrix

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