Automobiles arrive at the drive-through window at a post office at the rate of 4 every 10 minutes. The average service t
Posted: Wed Apr 27, 2022 2:10 pm
Automobiles arrive at the drive-through window at a post
office at the rate of 4 every
10 minutes. The average service time is 2 minutes. The
Poisson distribution is
appropriate for the arrival rate and service times are
exponentially distributed.
a) What is the average time a car is in the
system?
Answer: 1/30-24= ⅙ * 60 min =10 min
b) What is the average number of cars in the
system?
Answer: 24/30-24= 24/6 = 4cars
c) What is the average time cars spend waiting to
receive service?
Answer: (24)2 / 30(20 - 24) = 3.2
minutes
d) What is the average number of cars in line behind the
customer receiving
Service?
Answer: 24/30(30-24) = 8 cars
e) What is the probability that there are no cars at the
window?
Answer: 1-0.8 = 0.2
f) What percentage of the time is the postal clerk
busy?
Answer: 0.4/0.5 = 80%
g) What is the probability that there are exactly two
cars in the system?
Answer: (0.4/0.5)^2 * 0.2 = 12.8%
For the post office in the previous problem, a second
drive-through window is being considered. A single line would be
formed and as a car reached the front of the line it would go to
the next available clerk. The clerk at the new window works at the
same rate as the current one.
a) What is the average time a car is in the system?
b) What is the average number of cars in the system? c)
What is the average time cars spend waiting to receive
service?
d) What is the average number of cars in line behind the
customer receiving service?
e) What is the probability that there are no cars in the
system?
f) What percentage of the time are the clerks busy?
g) What is the probability that there are exactly two cars in
the system?
office at the rate of 4 every
10 minutes. The average service time is 2 minutes. The
Poisson distribution is
appropriate for the arrival rate and service times are
exponentially distributed.
a) What is the average time a car is in the
system?
Answer: 1/30-24= ⅙ * 60 min =10 min
b) What is the average number of cars in the
system?
Answer: 24/30-24= 24/6 = 4cars
c) What is the average time cars spend waiting to
receive service?
Answer: (24)2 / 30(20 - 24) = 3.2
minutes
d) What is the average number of cars in line behind the
customer receiving
Service?
Answer: 24/30(30-24) = 8 cars
e) What is the probability that there are no cars at the
window?
Answer: 1-0.8 = 0.2
f) What percentage of the time is the postal clerk
busy?
Answer: 0.4/0.5 = 80%
g) What is the probability that there are exactly two
cars in the system?
Answer: (0.4/0.5)^2 * 0.2 = 12.8%
For the post office in the previous problem, a second
drive-through window is being considered. A single line would be
formed and as a car reached the front of the line it would go to
the next available clerk. The clerk at the new window works at the
same rate as the current one.
a) What is the average time a car is in the system?
b) What is the average number of cars in the system? c)
What is the average time cars spend waiting to receive
service?
d) What is the average number of cars in line behind the
customer receiving service?
e) What is the probability that there are no cars in the
system?
f) What percentage of the time are the clerks busy?
g) What is the probability that there are exactly two cars in
the system?