How to model exponential holding times? Please help.

In my code, after performing some action (e.g. changing var=0 to var=1), I need to trigger a time which is representative of how long the var should be 1. I want to model it as exponential holding time as below

exprnd(5)   
%outputs vary like 0.3414 or 1.9376 or 2.1120 or 8.8250...

Question: how can I know when to make the var=1 back to var=0 using this kind of output. The ouput is not a fixed integer count that I can keep decrementing. Any idea on how can I do this? Should I make use of system time?

Background: I have to make this var change multiple times. I.e. after trigerring for the above, I go back, get another input, perform the action (var=1) and then also do an exprnd(5) for that one. It is a dynamic system. Actions get removed (var reverted back) after a certain time and that time is modelled as exponential distribution.

Answer 1

[Walter Roberson]

You are modeling discrete events. At a particular time step, the event fires (or expires, depending on what you are modeling). Your events do not occur at times "between" the discrete time steps, and your events do not get "partly used up" each time step -- you are not randomly determining an integer "dwell time" and "counting down" at each time step.

In such situations, you should be using Poisson distribution.

Unless, that is, you are determining an "dweel time" at the beginning and counting down from that. If that is what you are doing, then consider taking ceil() of the exponential random value.

[Jaya]

If I use Poisson distribution, then I would just use something like poissrnd command, isn't so?. How would that give me any values pertaining to how long that particular action should take place? I mean, aren't arrival times and holding times different for my case? I am confused. Can you please provide any example?

[Walter Roberson]

The below truncates the distribution a bit, by killing everything off in the last generation.

pborn is the probability of an item being created. Additional births are tried for at the same probability until one fails.

pdie is the probability that an alive item will die.

This simulation has no built-in life-span. Any one "alive" entity is memoryless in this simulation -- it does not get less or more "vital" over a distribution of ages.

pdie = 0.09;
pborn = 0.1;
N = 2000;
state = zeros(0,1);        %birthday
num_alive_at = zeros(1,N);
num_died_at = zeros(1,N);
num_born_at = zeros(1,N);
death_age_count = zeros(1,1);
for K = 1 : N
   n_alive = size(state,1);
   num_alive_at(K) = n_alive;
   if n_alive > 0
      if K == N; death_idx = 1 : n_alive; end  %kill all at end
      death_idx = find(rand(n_alive,1) < pdie);
      num_funerals = length(death_idx);
      num_died_at(K) = num_funerals;
      age_at_death = K - state(death_idx,1);
      oldest_death = max(age_at_death);
      if oldest_death > length(death_age_count); death_age_count(oldest_death) = 0; end
      for J = 1 : num_funerals
          death_age_count(age_at_death(J)) = death_age_count(age_at_death(J)) + 1;
      end
      state(death_idx,:) = [];
   end
   num_birth = 0;
   while K ~= N && rand < pborn; num_birth = num_birth + 1; end  %no births at end
   num_born_at(K) = num_birth;
   if num_birth == 0; continue; end
   state(end+1:end+num_birth, 1) = K;
end

subplot(4,1,1)
total_lives = sum(death_age_count);
bar(death_age_count / total_lives * 100);
xlabel('age'); ylabel('% prob'); title('prob died at age');

subplot(4,1,2)
plot(num_alive_at)
xlabel('iteration'); ylabel('lives'); title('# alive at iteration');

subplot(4,1,3)
histogram(num_born_at, 0:max(num_born_at))
xlabel('# births'); ylabel('counts'); title('births per iteration');

subplot(4,1,4)
histogram(num_died_at, 0:max(num_died_at))
xlabel('# deaths'); ylabel('counts'); title('deaths per iteration');

[Jaya]

Thanks for the detailed answer. I had found some other way of solving my problem. Not exactly the one I asked here. So it is ok now.