waiting line analysis (operation managment)

[Khb_taylors]

Joe, a bartender at one of most popular pubs in St. John’s, can serve drinks at the rate of one every 50 seconds. During a hot evening recently, the bar was particularly busy and every 55 seconds someone was at the bar asking for a drink.
a) Assuming that everyone in the bar drank at the same rate and that Joe served people on a first-come, first-served basis, how long would you expect to have to wait for a drink?
b) How many people would you expect to be waiting for drinks?
c) What is the probability that three or more people are waiting for drinks?
d) What is the probability you could get your drink immediately when asking?

Here what i got:
lamda = 55 seconds
mean = 50 seconds (not sure)
should use model two because: one server, constant service time...

 

[tkhunny]

Well, just walk through it.

1) You'll need Traffic intensity: \(\displaystyle \rho = \frac{\lambda}{\mu}\)

2) This gives immediately the complete probability distribution of the queue. \(\displaystyle p(n) = \rho^{n}\cdot (1-\rho)\)

3) Out pops, \(\displaystyle p(no\;one\;in\;the\;queue) = p(0) = 1-\rho\)

4) Easily then, \(\displaystyle p(3\;or\;more\;in\;the\;queue) = 1- [p(0) + p(1) = p(2)]\)

5) How many can you expect to find in line: \(\displaystyle Mean\;queue\;length = E[queue\;pdf\;variable] = \frac{\rho}{1-\rho}\)

6) It seems logical that one could expect to wait (Mean Service Time)*(Expected Number of Folks in Line)

Are we getting anywhere?

 

[Denis]

should use model two because: one server, constant service time...

Whadda heck is a "model two"? Another Newfie joke?