Queueing Models 4.
13
Average Number of Customers in the Queue:
k
Lq E ( N 1) (n 1) Pn
n 1
k k
nPn Pn
n 0 n 1
E( N ) 1 P0
Lq Ls
But in this model, we define the overall effective arrival rate,
denoted as
1 P0 1 P0
Hence (2) becomes
Lq Ls
Model
Formulae:
1) Server Utilization
n
2) Pn 1 (P 0 no customers in the system)
3) Ls
1
2
4) Lq
1
1
5) Ws
1
4.14 Probability and Queueing Theory
6) Wq
1
7) Probability that the waiting time of a customer in the
system exceeds t is P(ws t ) e ( )t .
Problem 4.1:
Customers arrive at a sale counter managed by a single person
according to a Poisson process with mean rate of 20 per hour.
The time required to serve a customer has an exponential
distribution with a mean of 100 seconds. Find the average
waiting time of a customer. (N/D 2013)
Solution:
-I (M/M
Arrival Rate:
20 1
20 per hr per min per min
60 3
Service Rate:
1 60 3
100 sec per sec per min per min
100 100 5
Server Utilization:
1/ 3 1 5 5
3/ 5 3 3 9
i) Average waiting time of a customer in the system:
1 1 1 15
Ws 3.75 mins
1 3 5 3 4 4
1
5 9 5 9
Queueing Models 4. 15
ii) Average waiting time of a customer in the queue:
5 5
9 9 25
Wq 2.08 mins
1 3 5 3 4 12
1
5 9 5 9
Problem 4.2:
Customer arrive at a one man barber shop according to a Poisson
process with a mean inter arrival time of 20 minutes. Customers
spend an average of 15 minutes in the barber chair. The service
time is exponentially distributed. If an hour is used as a unit of
time, then (M/J 2013)
(i) What is the probability that a customer need not wait for a
haircut?
(ii) What is the expected number of customer in the barber
shop and in the queue?
(iii) How much time can a customer expect to spend in the
barber shop?
(iv) Find the average time that a customer spend in the queue.
(v) Estimate the fraction of the day that the customer will be
idle?
(vi) What is the probability that there will be 6 or more
customers?
(vii) Estimate the percentage of customers who have to wait
Solution:
-
4.16 Probability and Queueing Theory
1
Arrival Rate: 20 min per min
20
1
Service Rate: 15 min per min
15
1 / 20 15 3
Server Utilization:
1 / 15 20 4
i) Probability that a customer need not wait:
P A customer need not wait P0 1
3 1
1
4 4
ii) Expected number of customers in the shop and queue:
3/ 4 3/ 4
Ls 3
1 1 3/ 4 1/ 4
2 2
3/ 4
Lq
1 1 3/ 4
9 / 16 9 4 9
2.25
1/ 4 16 1 4
iii) Expected waiting time of a customer in the shop:
1 1 1
Ws 60 mins
(1 ) 1 3 1 1
1
15 4 15 4
iv) Average waiting time of a customer in the queue:
3/ 4 3/ 4
Wq 45 mins
1 1 3 1 1
1
15 4 15 4
Queueing Models 4. 17
v) Fraction of the day that the server be idle:
3 1
Fraction of server be idle Po 1 1
4 4
vi) Probability that there will be 6 or more customers:
7
7 3
P( N 6) 0.1335
4
vii) Percentage of customers who have to wait prior to
3
Non-empty System 1 P0 1 1
4
3
Percentage Value 100 75%
4
Problem 4.3:
If people arrive to purchase cinema tickets at the average rate of
6 per minute, it takes an average of 7.5 seconds to purchase a
ticket. If a person arrives 2 min before the picture starts and it
takes exactly 1.5 min to reach the correct seat after purchasing
the ticket, (N/D 2010),(N/D 2014)
(i) Can he expect to be seated for the start of the picture?
(ii) What is the probability that he will be seated for the start of
the picture?
(iii) How early must he arrive in order to be 99% sure of being
seated for the start of the picture?
Solution:
-I (M/M/1) :
Arrival Rate: 6 per min
4.18 Probability and Queueing Theory
Service Rate:
1 60
7.5 sec per sec per min 8 per min
7.5 7.5
6 3
Server Utilization:
8 4
i)
1 1 1 1
Ws 0.5 mins
1 3 1 2
8 1 8
4 4
Expected time required to purchase the ticket and to reach
the seat Ws 1.5 0.5 1.5 2 mins
Hence he can just be seated for the start of the picture.
ii) Probability that he will be seated for the start of the
picture:
( )t
P W t e
P[he will be seated for the start of the picture]
P[total time 2 mins]
P W 0.5
1 PW 0.5
(8 6) 0.5
1 e
1
1 e 0.632
iii)
PW t 0.99
Queueing Models 4. 19
1 PW t 0.99
P W t 0.01
( )t
e 0.01
( )t log0.01
2t log0.01 8 and 6
2t 4.6
t 2.3
P ticket purchasing time 2.3 0.99
P total time to get the ticket
and to go to the seat (2.3 1.5) 0.99
The person must arrive at least 3.8 minutes early, so as to
be 99% sure of seeing the start of the picture.
Problem 4.4:
A T.V. repairman finds that the time spent on his job has an
exponential distribution with mean 30 minutes. If he repair sets
in the order in which they came in and if the arrival of sets is
approximately Poisson with an average rate of 10 per 8 hour day.
Find (M/J 2012),(N/D 2013),(N/D 2015)
(1)
(2) how many jobs are ahead of average set just brought?
Solution:
-
Arrival Rate:
10 10
10 per 8 hrs per hr per min
8 8 60
4.20 Probability and Queueing Theory
1
per min
48
1
Service Rate: 30 mins per min
30
1 / 48 30 5
Server Utilization:
1 / 30 48 8
(1) The R Expected Idle Time each day:
5 3
Idle Time P0 1 1
8 8
(2) Number of jobs are ahead:
2 2
5/8 25 / 64 25 8 25
Lq
1 1 5/8 3/8 64 3 24
Problem 4.5:
Customers arrive at the express checkout lane in a supermarket
in a Poisson process with a rate of 15 per hour. The time to
check out a customer is an exponential random variable with
mean of 2 minutes. Find the average number of customers
present. What is the expected waiting time for a customer in the
system? (M/J 2014)
Solution:
-
Arrival Rate:
15 1
15 per hr per min per min
60 4
1
Service Rate: 2 mins per min
2
Queueing Models 4. 21
1/ 4 2 1
Server Utilization:
1/ 2 4 2
i) Average number of customers present:
1/ 2 1/ 2
Ls 1
1 1 1/ 2 1/ 2
2
1/ 4 1/ 4 2 1
Lq
1 1 1/ 2 1/ 2 4 2
ii) Waiting time of a customer in the system:
1 1 1
Ws 4 mins
1 1 1 1 1
1
2 2 2 2
Problem 4.6:
Arrivals at a telephone booth are considered to be Poisson with
an average time of 12 mins. between one arrival and the next.
The length of a phone call is assumed to be distributed
exponentially with mean 4 mins.
a) Find the average number of persons waiting in the system.
b) What is the probability that a person arriving at the booth
will have to wait in the queue?
c) What is the probability that it will take hime more than 10
min. altogether to wait for the phone and complete his call?
d) Estimate the fraction of the day when the phone will be in
use.
e) The telephone department will install a second booth, when
convinced that an arrival has to wait on the average for at
least 3 min for phone. By how much the flow of arrivals
should increase in order to justify a second booth?
4.22 Probability and Queueing Theory
Solution:
-I (M/M/1) :
1
Arrival Rate: 12 mins per min
12
1
Service Rate: 4 mins per min
4
1 / 12 4 1
Server Utilization:
1/ 4 12 3
a) Average number of persons waiting in the system:
1/ 3 1/ 3 1
Ls
1 1 1/ 3 2/3 2
b) Probability that a person arriving at the booth will have
to wait in the queue:
P person wait and getting service 1 P0
1
1 1
3
c)
t
P Ws t e
1 1 3 1
(10) (10) 5
4 12 12
P Ws 10 e e e3 0.1889
d) Fraction of the day when the phone will be in use:
Fraction of the day phone will be in use 1 P0
1
1 1
3
