Theory of traffic flow/ poisson arrivals and mean delay

[catcherintherye]

Hello, From the following formula in the following theorem I am to deduce the mean delay of a customer arriving at a queue.

Theorem 1

Suppose that customers arrive at a single server queue according to a Poisson process mean rate q and that service times are exponentially distributed mean rate Q^-1.
Then provided that the traffic intensity x = q/Q < 1

the queue length distribution in equilibrium is

P_n = (1-x)x^n (x >=0)

where n is the number of customers in the system including any in service

Some notes on notation

Q = capacity of the queue i.e. the maximum mean rate of service

q = the mean arrival rate (customers/sec)

d= mean delay incurred by a customer in the queue which is equal to the difference between the time it takes to pass through the queue and the time it would normally take in the absense of other customers.

This is how I have so far approached the solution.

Suppose that delay is proportional to the number of customers in the system. Then d=kn for some proportionality constant n.

now i proceeded by considering a particular case namely settin g q=0.3 and Q=1

P(n=0) = 700/1000 (so 700 times out of 1000 there is no delay)

P(n=1) = 210/1000 (210 " " a delay of k seconds)

P(n=2) = 63/1000 (63 " " " 2k secs)

but then i got stuck, how do i find the mean delay for a customer arriving at this queue, i know i am to take an average of the above values, but not really sure how to proceed...??

 

[Valkarie]

Solution:The mean delay of a customer arriving at a queue can be calculated using the equation d = kn, where k is the proportionality constant and n is the number of customers in the system. Given the information provided in the theorem, we can calculate the mean delay as follows: First, we need to calculate the proportionality constant k. To do this, we need to solve for P_n, which is the probability that there are n customers in the system including any in service. From the theorem, we know that P_n = (1-x)x^n, where x = q/Q. Substituting the values for q and Q given in the problem (q = 0.3 and Q = 1), we get x = 0.3. Therefore, P_n = (1-0.3)0.3^n. Next, we need to calculate the mean delay for each value of n. To do this, we use the equation d = kn, where k is the proportionality constant. For n = 0, the mean delay is 0. For n = 1, the mean delay is k. For n = 2, the mean delay is 2k. Now, we need to calculate the mean delay by taking an average of the above values. To do this, we first calculate the probability of each value of n occurring. This can be done by substituting the value of P_n into the equation. For n = 0, the probability is 0.7. For n = 1, the probability is 0.21. For n = 2, the probability is 0.063. Now, we can calculate the mean delay by taking the weighted average of the three values of n. The formula for calculating the weighted average is: Mean Delay = (P(n=0) x 0) + (P(n=1) x k) + (P(n=2) x 2k) Substituting the values for P(n=0), P(n=1), and P(n=2) from above, we get: Mean Delay = (0.7 x 0) + (0.21 x