In my problem there are 7 number of peoples and each one have different salary.Based on a condition they are divide as lenders,borrowers and common.I have to find out the optimal expenditure of each people.Based on this solution they have to find the cost.Based on this updated cost,the expenditure of each people have to be updated.This is a iterative compution, it stops only when the error converges nearly to zero
k=1;
a1=3;b1=8;
a(k)=a1;b(k)=b1;
G=[50 100 150 175 200 250 300];
Gbar=mean(G);
H=[20 15 10 5 1 0.5 0.1];
itmax=5;
x=[];
y=[];
z=[];
for k=1:itmax
for i=1:length(G)
if G(i)>Gbar
x=[x i];
elseif G(i)<Gbar
y=[y i];
else
z=[z i];
end
end
A=(H(x)/a(k))-1;
A=(H(y)/b(k))-1;
A=G(z);
if s>r
a(k+1)=sqrt((sum(a1.*H(:,x)))/(p1));
if (a1>=a(:,k+1)) && (a(:,k+1)>=b1)
a(k+1)=a1+(s/beta1);
else
end
this problem end when following condition satisfied
if(a(k+1)-a(k)<=0.01)&&(b(k+1)-b(k)<=0.01)&&(A(k+1)-A(k)<=0.01)
end
end
But I didn't get the correct output.This doesn't undergo iteration.
I have to get the updated cost price of each people on each iteration.
Also updated value of expenditure of each people.My error is I get the second iterative value of expenditure of first man is the first expenditure value of second man.How to correct this code?

1 comentario

Jan
Jan el 31 de Mzo. de 2022
"updated value of expenditure of each people"
"value of expenditure of first man is the first expenditure value of second man"
Prefer to use the names of the variables in the description. What is "people", "man", "expenditure"?

Iniciar sesión para comentar.

 Respuesta aceptada

Jan
Jan el 31 de Mzo. de 2022

1 voto

I did not understand, what you are asking for (see my comment). but this piece of the code looks suspicious:
A=(H(x)/a(k))-1; % lenders expenditure
A=(H(y)/b(k))-1; % borrowers expenditure
A=G(z); % common expenditure
You are overwriting the value of A twice. If this value matters, overwriting should be a problem. If it does not matter, why do you calculate it?

Más respuestas (0)

Etiquetas

Preguntada:

el 31 de Mzo. de 2022

Editada:

el 14 de Jun. de 2022

Community Treasure Hunt

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

Start Hunting!

Translated by