送交者: wasguru 于 2006-4-08, 03:51:18:
回答: to Steve about your probability distriubtion problem 由 wasguru 于 2006-4-08, 03:34:45:
I'm not in the field of queuing. But here's some of my thoughts:
Assume at time T_i the (random) number of people waiting for a train is X_i. Further assume that the probability P(X_i = x) = P_i(x). Let's denote the Poisson distribution mass function as PP(x). Now consider the probability distribution of X_i+1, the number of people waiting for a train at time T_i+1:
Obviously, if X_i <= C, we have
P(X_i+1 = x) = PP(x).
Otherwise, we have X_i+1 = X_i - C + dx, where dx is the number of people arrived after the previous train had left. Therefore, under the condition X_i > C, we have
P(X_i+1 = x | X_i) = P(dx = x + C - X_i) = PP(x + C - X_i)
Combine the above two cases, we have
P(X_i+1 = x) = PP(x) * P(X_i <= C) + sum_{y=C+1}^{x+C} PP(x + C - X_i) * P(X_i = y).
If a stationary solution exists, the above equation is a discret "integral equation". I don't know how to rigorously solve the above equation. Intuition tells me that C > mu will be the necessary condition to have a stationary solution, where mu is the expectation of the Poisson distribution. I did some numerical calculation, and it seems that C > mu is also a sufficient condition.
Notice that when mu is sufficiently large, Poisson distibution can be well approximated as Normal(mu, mu). That means when C >= mu + 2*sqrt(mu), the first cases mentined above will be the dominant case, and the limit distribution will be very close to the original Poisson distribution. Further increasing capacity is a waste of resources. I guess setting C = mu + sqrt(mu) will be a good compromise.