送交者: 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 | X_i) = P(dx = x + C - Xi) = PP(x + C - Xi)
Combine the above two cases, we have
P(X_i+1 = x) = PP(x) * P(X_i mu, where mu is the expectation of the Poisson distribution, a stationary solution exists.
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, or even this capacity, is a waste of resources. I guess setting C = mu + sqrt(mu) will be a good compromise.