Info

Questa domanda è chiusa. Riaprila per modificarla o per rispondere.

how can I improve the logic of this code

1 visualizzazione (ultimi 30 giorni)
summyia qamar
summyia qamar il 8 Gen 2018
Chiuso: MATLAB Answer Bot il 20 Ago 2021
Im trying to build a relation between supply demand and price using 3 equations. the code I've generated is given below
clear all;
clc;
T=100;
P=zeros(1,T);
D=zeros(1,T);
R=zeros(1,T);
P(1)=10; %Initial Price
D(1)=10; %Initial Demend
R(1)=5; % Initial Resource available
K=100; %Maximum demand
M=10; %maximum resource
N=50; %Max Price K/N=2; N/K=0.5; N/M=5; M/K=0.1
for t=2:T
D(t)=D(t-1)+((1-(P(t-1)/N))*(K/N));
P(t)=P(t-1)-((1-(D(t-1)/K))*(N/K))+((1-(R(t-1)/M))*(N/M));
R(t)=R(t-1)+((1-(D(t-1)/K)*(M/K)));
end
Xvals=1:T;
plot(Xvals,D,'b',Xvals,P,'r',Xvals,R,'g')
legend('Demand','Price','Resource')
when I run the code, the values get below zero. I want that whatever the results are but the values remain positive since neither of price nor demand nor resources can be negative. can anybody help me in improving these equations?

Risposte (2)

Walter Roberson
Walter Roberson il 8 Gen 2018
You could do things like
D(t) = max(0, D(t-1)+((1-(P(t-1)/N))*(K/N)) );
This will substitute 0 if the value would have been negative.

Image Analyst
Image Analyst il 8 Gen 2018
To clip values to 0, use this code after the loop but before you call plot.
% If arrays are below zero, then clip arrays to 0.
D = max(D, 0);
P = max(P, 0);
R = max(R, 0);

Tag

Community Treasure Hunt

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

Start Hunting!

Translated by