Unable to perform assignment because the left and right sides have a different number of elements.

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Jonathan Louie
Jonathan Louie 2022-6-7
回答： Image Analyst 2022-6-7
This is my auto-regulated construct. Does anyone know why prod_ratea=g_ar*((theta.^n_hill)./((theta.^n_hill)+(x_ar.^n_hill))); does not change and update when x_ar increases or decreases? That's why I get the error message.
t_max= 1200; % in s
k=0.01; % in s^-1
step_counter=1;
t=0;
g_nr=2;
x_nr=5;
g_ar=4;
n_hill=8;
x_ar=5;
t_ar=0;
theta=(g_nr)/k;
% Perturbations
perturbation_magnitude=150; % in molecules
perturbation_time_400=400;% in s
perturbation_time_800=800;% in s
is_perturbation_400=0;
is_perturbation_800=0;
while t<t_max
prod_ratea=g_ar*((theta.^n_hill)./((theta.^n_hill)+(x_ar.^n_hill)));
deg_rate=k*x_ar;
wait_time=-log(rand)./((prod_ratea)+(deg_rate(step_counter)));
prob_prod=prod_ratea./((prod_ratea)+(deg_rate(step_counter))); % Propensity of production
t=t+wait_time; % Update current time
step_counter=step_counter+1; % Update the number of steps (reactions) associated with the experiment.
t_ar(step_counter)=t; % Add the current time to the time log
if rand < prob_prod % Defines whether production takes place based on Monte Carlo method.
x_ar(step_counter)=x_ar(step_counter-1)+1; % Implements production
else
x_ar(step_counter)=x_ar(step_counter-1)-1; % Implements degradation
end
if t > perturbation_time_400 && is_perturbation_400==0
x_ar(step_counter)=x_ar(step_counter)+perturbation_magnitude;
is_perturbation_400=1;
end
if t > perturbation_time_800 && is_perturbation_800==0
x_ar(step_counter)=x_ar(step_counter)-perturbation_magnitude;
is_perturbation_800=1;
end
if t>t_max
t_ar(end)=[];
x_ar(end)=[];
end
end

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Image Analyst
Image Analyst 2022-6-7

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