1. In a waiting line situation, arrivals occur at a rate of 2 per minute, and the service times average 18 seconds. Assume the Poisson and exponential distributions.
a. What is l? - Average number of arrivals expected during some time period
b. What is m? - Average number that the system can serve during some time period
c. Find probability of no units in the system.
1-λ/ µ = 1 - 2/18=0.89
d. Find average number of units in the system.
Ls = λ/( µ- λ) = 2/(18-2) = 0.125
e. Find average time in the waiting line.
Wq= λ/ µ( µ- λ) = 2/18(18-2) = 0.006
f. Find average time in the system.
Ws= 1/ (µ- λ)= 1/(18-2) = 0.0625
g. Find probability that there is one person waiting.
(λ/ µ) = 2/18 = 0.11
h. Find probability an arrival will have to wait.
2/18(18-2) = 0.006
2. The Sea View Resort uses a multiple-channel queue registration system. If the average service time is 8 minutes, there are three registration clerks, and guests arrive at the rate of one every 5 minutes, find
a. l and m.
l 60/5=12 m= 60/8 = 7.5 customers/hr
b. the probability all three clerks are idle.
(THE PROBABILITY OF NO UNITS IN THE SYSTEM) 1/[(12/7.5)^0/0!+(12/7.5)^1/1!+ (12/7.5)^3/3!(1/1-12/3(7.5)]
P0 = .1872
c. the probability a guest will have to wait.
(PROBABILITY THAT AN ARRIVING UNIT HAS TO WAIT)
Pw = .2738
d. the average time a customer is in line.
Wq= Lq/l = .0261
e. the average number of customers in line.
Lq = 0.3129
3. The post office uses a multiple channel queue, where customers wait in a single line for the first available window. If the average service time is 1 minute and the arrival rate is 7 customers every five minutes, find, when two service windows are open,
a. the probability both windows are idle.
Po = .1765
b. the probability a customer will have to wait.
Pw = .5765
c. the average time a customer is in line.
Wq = .1922
d. the average time a customer is in the post office.
W= .3922