unit-v advanced queueing models · 1. an automatic car wash facility operates with only one bay....
TRANSCRIPT
Unit-V
ADVANCED QUEUEING MODELS
NON-MARKOVIAN QUEUES AND QUEUE NETWORK (M | G | 1): | |GD - Pollaczek-
Khintchine (P-K) Formula:
The characteristics of the model are
M ------ > Number of arrivals in time “t” follows Poisson Process
G ------ > General output distribution
I ------ > Single server
------ > Infinite Capacity
GD ------ > General discipline
Pollaczek-Khintchine (P-K) Formula:
Let N and 'N be the number of customers in the system at time t and t + T, when two
consecutive customers have just left the system after getting service.
Let T ------ > The continuous random variable denoting service time
f(t) ------ > The Pdf of T
E(T) ------ > The Mean of T
Var(T) ------ > The variance of T
Let “M” be the number of customers arrival in the system during the service time T.
, if N = 0'
N- 1 +M, if N >0
MN
Where „M‟ is the discrete random variable taking values 0, 1, 2……
' 1 ----- (1)N N M
' ( 1) N N M
1, If N = 0Where =
0, If N > 0
' 1 ----- (2)E N E N E M E
When the system is steady state the probability of number of customers in the system will be
constant.
2 2' ' ----- (3)E N E N and E N E N
From (2) E 1 ----- (4)E M
Squaring both sides of equation 1, we get
22 2 2 N' = N 1 2 1 2 1 2 ------ (5)M N M M N
2 2 2Now If 0, 0, If 1, 1Since
N 0 1 If N = 0and
= 0 If N 0N
N 0
Substituting in equation 5
22 2 N' = N 1 2 1 2 1 0 M N M M
2 2 22 1 N N' 2 1 2 2N M M M M
2 2 2 = N N' 2 1 (2 1)M M M
2 2 22 1 N N' 2 1 2 1 By equa (3)E N M E E E M E M E M
22E 1 2 1 2 1N E M E M E M E M
2 2 1 2 1 E
2 1
E M E M E E MN
E M
2 2 1 1 2 1 = by equation 4
2 1
E M E M E M E M
E M
2 22 2 1 2 1 =
2 1
E M E M E M E M E M
E M
2 22 E ------ (6)
2 1
E M E M E MN
E M
Since the number of arrivals M in time T follows a Poission process with parameter
| and |E M T T Var M T T
2 |E M T T T
E | |Now M E E M T E E y x E y
= E T
E EM T
2 2 E |M E E M T
2 2 = E T T
2 2 = E T E T
2 2 = ( ) ( )Var T E T E T
2 2 2 E ( ) ( ) M Var T E T E T
Substituting in equation 6 we get,
2 2 2 2 2
s
( ) ( ) - 2 ( ) ( ) L = E
2 1
Var T E T E T E T E TN
E T
2 2( ) ( ) 2 ( ) 1 =
2 1
Var T E T E T E T
E T
2 2( ) ( ) =
2 1
Var T E TE T
E T
2 2
s
( ) ( ) L
2 1
Var T E TE T
E T
Which is called Pollaczek-Khinchine formula FormulaP K
Pollaczek- Khinchine (or) (M | G | 1): | |GD Model:
(i) The average number of customers in the system: sL
2
2 2
2 Where = V(T)
2 1
sL
(ii) The average Number of customers in the queue: qL
q sL L
(iii) The average waiting time of a customer spend in the system : sW
ss
LW
(iv) The average waiting time of a customer spend in the queue : qW
q
q
LW
Problems:
1. An Automatic car wash facility operates with only one bay. Cars arriving according to a Poisson
distribution with a mean of 4 cars/hr. and may wait in the facility‟s parking lot if the bay is busy.
Find sL , qL , sW , qW . If the service time
(i) Is constant and equal to 10 minutes
(ii) Follows uniform distribution between 8 and 12 minutes
(iii) Follows Normal distribution with mean 12 minutes and S.D 3 minutes
(iv) Follows discrete distribution with values 4, 8 and 15 minutes with corresponding
probabilities 0.2, 0.6 and 0.2
Solution:
Given, One bay ---- > Single server
Cars ----- > Infinite capacity
The given problem is (M | G | 1): | |GD Model
Given, Arrival rate 4 /cars hr
(i) If the service time is constant an equal to 10 minutes
Given, Service time = 10 minutes = 1
60Per hour = 6 cars/hr10
2Since Service time is constant, Variance = 0
s( ) To find L :a
2
2 2
2 1
sL
2
2 2 44 0
64
462 1
6
22
2 2 23 =
23 3 32
6
4 Cars
3sL
q( ) To find L :b
q sL L
4 2
3 3qL
2
3qL car
( ) To find :sc W
ss
LW
4
3 4
qW
1 20 minutes
3sW hr
( ) To find :qd W
q
q
LW
2
3 4
qL
1 10 minutes
6qW hr
(ii) If the service time follows uniform distribution between 8 and 12 minutes
Given, Arrival rate 4 /cars hr
1 14 Per minute
60 15
Given a =8, b = 12
8 12
102 2
a bE T
1 1
Per minute10E T
2 22 ( ) (12 8) 4
( )12 12 3
b aV T
2 4
3
s( ) To find L :a
2
2 2
2 1
sL
2
21
1 4 15115 31
1015 1 1
10 152 11
10
21 4 2
2 225 3 3
232 1
3
4 4
2 675 9 13
23
2 304 3
3 675 2
1.3422 Cars sL
q( ) To find L :b
q sL L
1
215 1.3422 1.34221 3
10
qL
0.6755 cars qL
( ) To find :sc W
ss
LW
1.3422 1.3422 15
1
15
qL
20.14 minutessW
( ) To find :qd W
q
q
LW
0.6755
1
15
qW
10.13 minutesqW
(iii) If the service time follows Normal distribution with mean 12 minutes and S.D 3 minutes
Given, Arrival rate 1
Per minute15
1Service rate Per minute
12
S.D = 3
s( ) To find L :a
2
2 2
2 1
sL
2
22
1
1 1531151
1215 1 1
12 152 11
12
21 4
94 225 5
15
25
9 16
4 225 25 25
5
4 153 5
5 225 2
2.5 Cars sL
q( ) To find L :b
q sL L
1
415 2.5 2.5 1.71 5
12
qL cars
1.7 cars qL
( ) To find :sc W
ss
LW
2.5 2.5 15
1
15
sW
37.5 minutessW
( ) To find :qd W
q
q
LW
1.7
1
15
qW
25.5 minutesqW
(iv) If the service time follows a discrete distribution with values 4, 8 and 15 minutes with
corresponding probabilities 0.2, 0.6 and 0.2
Given: T = t: 4 8 15
P(t): 0.2 0.6 0.2
( )E T tP t
=4 0.2 +8 0.6 +15 0.2
8.6E T
2 2 ( )E T t P t
2 2 2 = (4) 0.2 +(8) 0.6 +(15) 0.2
2 86.6E T
1 1
8.6E T
2 22 2 86.6 8.6 12.64E T E T
, 1
Per mi nute15
Given Arrival rate
1 Service rate Per minute
8.6
2 S.D = 12.64
s( ) To find L :a
2
2 2
2 1
sL
2
21
1 1512.641151
8.615 1 1
8.6 152 11
8.6
2(0.00444)(12.64) 0.5733
0.57332 1 0.5733
0.38485 0.5733
0.8534
1.024 Cars sL
q( ) To find L :b
q sL L
1
15 1.024 1.024 0.57331
8.6
qL
0.4527 cars qL
( ) To find :sc W
ss
LW
1.024 1.024 15
1
15
sW
37.5 minutessW
( ) To find :qd W
q
q
LW
0.4507 0.4507 15
1
15
qW
6.7605 minutesqW
2. A one man barber shop takes exactly 25 minutes to complete one haircut. If customer arrive at the
barber shop in a Poisson fashion at an average rate of one every 40 minutes, how long on the
average a customer spends in the shop? Also find the average time a customer must wait for
service.
Solution:
Given: One man barber shop ----- > Single server
Customers ----- > Infinite Capacity
The given problem is (M | G | 1): | |GD Model
, 1
Per mi nute40
Given Arrival rate
1 Service rate Per minute
25
Since the service times is constant, Variance = 2 0
(i) To find the average number of customer spends in the shop : sW
We know that ss
LW
sTo find L :
2
2 2
2 1
sL
2
22
1
1 40014012540
1 1
25 402 11
25
25
05 8
38
28
25
5 5 25 30 2564 68 8 48 48
8
55
48sL
55
5548 401 48
40
ss
LW
45.8333sW
(ii) To find the average time of a customer must wait for service : qW
We know that q
q
LW
qTo find L :
q sL L
1
55 55 5 55 30 2540 148 48 8 48 48
25
qL
25
48qL
q
q
LW
25
2548 401 48
40
qW
20.8333qW
Queueing Networks
Networks:
A network of queues is a group of “k” nodes (or stations) where each node represents a service
facility with iC servers at node “i” (i = 1, 2, …..k). Customers may enter the system at one node, and
after completion of service at one node May move to another node for further service and may leave the
system from some other node. They may return to previously visited nodes skip some nodes and May stay
in the system forever.
Example:
A factory may have many queues linked together by the logical sequence of the production
process.
Types of queuing networks:
1. Open queueing network:
2. Closed queueing network
1. Open queueing network:
An open queueing network is defined as customers may arrive from outside the system at
any node and may depart (or leave)from the system from any node.
2. Closed queueing network:
An closed queueing network is defined as customers are not allowed to enter from
outside or to leave the system.
Series Queues: (Tandem queues)
Series queues are special types of open network in which there are a series of service facilities
which each customer should visit before leaving the system. The nodes form a series system with flows
always in a single direction from node to node. Customers enter from outside only at node “l” and depart
only from node “k”
Diagram:
Example:
1. An admission process in a college where the student has to visit a series of officials or clerks
2. A master health check-up programme in a hospital where a patient has to undergo a series of
tests.
3. An assembly-line in which units must pass through a series of work stations
Series queues with blocking:
Two station series Model:
A queueing system with two stations in series, one server at each station and no queue allowed to
form at either station. A customer entering for service has to go through station 1 and then station 2. NO
queues are in front of station 1 or station 2.
System
The following points have to be noted:
1. If the customer is being served at station 1 then arriving customers are turned away even if station
2 is empty.
2. If a customer is in station 2 and service is completed in station 1, the station 1 customer must wait
there until the service for station 2 customer is completed. In this case we say that the system is
blocked when this happens arrivals at station1 are turned away.
Steady state probabilities:
Each station may be either free or busy and station 1 may also be blocked. If 0, 1 and b denote the
free, busy and blocked states respectively and i, j respectively the states of station 1 and station 2
respectively, then the possible states of the system are given below.
i, j Description
0,0 System is empty
1,0 Customer in process at station 1 and station 2 is empty.
0,1 Customer 1 is empty and customer in process at station 2.
1,1 Customer in process at station 1 and station 2
b,1 Customer in process at 2 and customer finished at 1 and is waiting to become
available. i.e The system is blocked
Station 1 Station 2
Let the arrival of customers to the system at station 1 be Poisson with parameters and service
is exponential with parameters 1 2 and respectively.
The steady state equation for series queues with blockings:
00 2 01P P
1 10 00 2 11P P P
01 2 01 1 10 2 1bP P P P
01 2 11 1 11P P P
2 1 1 11bP P
Series Queues with blocking formula:
1.
2
00 2 2
2
2 4 3P
2.
1,0 0,02
2
2P P
3. 0,1 0,0P P
4.
2
1,1 0,022P P
5.
2
,1 0,022bP P
Jackson Net Works
Jackson Net Works or The characteristics of Jackson Net Works:
1. Arrivals from outside through node “i” follow a Poisson process with mean arrival rate “ ir ”
2. Services times at each channel at node “i” are independent and exponentially distributed with
Parameter .i
3. The probability that a customer who has completed service at node “i” will go to next to node “j”
is ijP . Where i = 1, 2, 3, …..k, j = 0, 1, 2, ……
Also ioP denotes the probability that a customer will leave the system from the node i.
Open Jackson Networks:
Open Jackson Networks are referred as 0ir for any i (or) 0 0ir for any i
Where ( 1)
0 ( )i
ir
elsewhere
And
1 ( 1,1 1)
1 ( , 0)
0 ( )
ij
j i i k
r i k j
elsewhere
These networks are called series, as the nodes can be considered as forming a series system with
flow always in a single direction from node to node
Therefore, the customer may enter the system from outside only at node 1 and May leave only from node
K.
We denote the total arrival rate of customers to server j by j , then (the traffic equations for open
Jackson Network is,
1
, 1, 2,..... )k
j j i ij
i
r P j K
Where ijP is the probability that a departure from server i joins the queue at server j, The Traffic equations
are also know as Flow balance equations.
Open – Jackson Networks formulas:
(i) The average number of customer in the system sL :
1 2
1 1 2 2
( )sL or
1 2
sL
(ii) The average waiting time of a customer in the system sW :
ss
LW
(iii) P[n customers at server 1] = 1 1
1
n
(iv) P[M customers at server 2]=2 2
1
m
(v) P[n customers at server 1 and m customers at server 2] = ,n mP
Ie ,
1 1 2 2
1 1
n m
n mP
Closed Jackson Networks:
Closed Jackson networks are referred as 0ir for all i (ie. no customers may enter the system from
outside) and 0 0ir for all i (ie. no customers may leave the system)
The total arrival rate of customers to server j by j , then the traffic equations for closed Jackson
Network is
1
, 1, 2,....j j ij
i
P i j k
Where ijP is the probability that a departure
From server i joins the queue at server j The Traffic equations are also know as Flow balance
equations.
The steady state solution for Closed Jackson Network:
The steady state solution for Network is
31 2
1 2 3, , 1 2 3
nn n
n n n NP D
Where N = 1 2 3........ kn n n n
Where 1 2
1 2 3
1 2
........
......... k
k
nn n
N k
n n n n
D
And j
j
j
P
The Arrival Theorem:
In a Closed network system with N customers, the system as seen by arrivals to serves j is
distributed as the stationary distribution in the same network system where there are only N – 1
customers.
Problems:
1) In a big factory, there are a large number of operating machines and two sequential repair shops,
which do the service of the damaged machines exponentially with respective rates of 1/hour and
2/hour. If the cumulative failure rate of all the machines in the factory is 0.5/hour.
(i) Find the probability that both repair shops are idle.
(ii) Find the average number of machines in the service section of the factory
(iii) The average repair time of a machine.
Solution:
Given Arrival rate = = 0.5/ hour
1
2 Per hour
Service rate 1 = 1 per hour
Service rate 2 = 2 per hour
(i) To find P[Both the service stations are idle]: 0P
W.K.T ,
1 1 2 2
1 1
n m
n mP
0 0
0,0
1 1 2 2
1 1P
0 01 1 1 1
2 2 2 21 11 1 2 2
1 1 1 3
12 4 2 4
0,0
3
8P
(ii) To find the average number of Machines in services sL :
W.K.T 1 2
sL
1 1
2 21 1
1 22 2
1 112 2 1
1 3 3
2 2
4
3sL
(iii) To find the average repair time of a Machine sW :
W.K.T. 1 2
1 1sW
1 1
1 11 2
2 2
1 1
1 3
2 2
= 2
23
8
3sW
PROBLEMS
1. A car manufacturing plant uses one big crane for loading cars into a truck. Cars arrive for loading
by the Crane according to a Poisson distribution with mean of 5 cars per hour. Given that the
service time for all cars is constant and equal to 6 minutes. Determine , , , .q s s qL L W W
Solution:
= 5 per hour = 1
12 per minute
Service time „T‟ is a constant.
„T‟ follows a distribution with E(T) = 0 and Var(T) = 0.
By Pollaczek –Khinchine formula,
22 ( ) ( ). ( )
2 1 ( )s
Var T E TL E T
E T
2 2112
112
0 616
12 2 1 (6)
1144
12
361
12 2 1
1 1 3
2 4 4
3
4sL
By Little‟s Formula,
34
112
3 129
4 1
ss
LW
9sW Minutes
1
q sW W
= 1
( ) sW E T E T
= 9 – 6 = 3
3qW Minutes
ie., a customer has to spend 20 minutes in the system and 10 minutes in the queue.
.q qL W
1
312
1
0.254
0.25qL Customers
2. A car wash facility operates with only one bay cars arrive according to a Poisson distribution with
a mean of 4 cars per hour and may wait in the facility‟s parking lot if the bay is busy. The
parking lot is large enough to accommodate any number of cars. Find the average number of cars
waiting in the parking lot and the average time a car spends in the facility If
(a) The time for washing and cleaning a car follows uniform distribution between 8 and 12 minutes.
(b) The service time T is 10 minutes for a car.
Solution:
Mean = = 4/hour
4
60 Per minute
1
15 Per minute
(a)Let T be a continuous random variable.
We know that, if T is a continuous R.V. following as uniform distribution is (a,b), then
E(T) = 1
( )2
a b
and
2
( )12
b aVar T
Then, E(T) = mean of the uniform distribution
1 1
( ) (8 12)2 2
a b
1
(20) 102
2
( )12
b aVar T
212 8
12
24 16 4
12 12 3
By Pollaczek –Khinchine formula,
22 ( ) ( ). ( )
2 1 ( )s
Var T E TL E T
E T
2 21 415 3
115
10110
15 2 1 (10)
1 4225 3
23
1002
3 2 1
3041225 3
13
2
3 2
2 1 304 3
. .3 225 3 2
2 152 150 152 302
3 225 225 225
= 1.342 Cars
1sL Car
By Little‟s Formula,
q sL L
1
15
110
11.342 0.675
( )Cars
E T
1qL Car
(b) The Service time T is 10 minutes for a car.
Service time = 10 minutes.
4 |
6 |
hr
hr
4 2
6 3
2 2 2
2(1 )sL
22
16(0)2 3
232 1
3
[ Service time is constant variable2 0 ]
49
23
2
3
2 4 3 2 2
3 9 2 3 3
4
3sL Cars
4 2 2
.3 3 3
q sL L Car
43 1
4 3
ss
LW hr
= 20 Minutes
23 1
10 .4 6
q
q
LW hr Minutes
3. In an Ophthalmic Clinic, there are sections- one section for assessing the Power approximately
and the other for final assessment and prescription of glasses. Patients arrive at the clinic in a
Poisson fashion at the rate of 3 per hour. The assistant in the first section takes nearly 15 minutes
per patient for approximate assessment of power and the doctor in the second section takes nearly
6 minutes per patient for final Prescription. If the service times in the two sections are
approximately exponential, find the probability that there are 3 Patients in the first section and 2
Patients in the second section. Find also the average number of Patients in the clinic and average
waiting time of a Patient in the clinic. Assume that enough space is available for the Patients to
wait in front of both sections.
Solution:
The situation in this clinic is comparable with a 2-stage series queue with single server at each station.
i.e., both section comes under (M|M|1): ( | FCFS) model.
Given: Mean Arrival rate = 3/hour
Service time in the Doctor section = 1
1
= 15 minutes
= 1
.4
hr
Service rate in the Doctor section 1 = 4/hr.
Service time in the chief Doctor section = 2
1
= 6 minutes.
= 1
.10
hr
Service rate in the Doctor section 2 = 10 / hr
(1)We know that
s in
s inmn
m customer the I stationandP P
ncustomer the II station
1 1 2 2
1 1
m n
Hence
3 2
32
3 3 3 31 1
4 4 10 10P
= 27 1 9 7
* * *64 4 100 10
32
1701
256000P
(2) Average number of patients in the clinic sL
1 2
1 2
LS LS
Station Station
1 2
3 3
4 3 10 3
3 243
7 7
3.43 3.
(3) Average waiting time of a patient in the clinic .sW
1 2
1 2
WS WS
Station Station
1 2
1 1
1 1
4 3 10 3
1 8
17 7
hours
480
7hours
= 68.6 minutes (approximately).
4. In a big factory, there are a larger number of operating machines and two sequential repair shops.
Which do the service of the damaged machines exponentially with respective rates of 1 per hour
and 2 per hour. If the cumulative failure rate of all the machines in the factory is 0.5 per hour,
find
(1) The Probability that both repair shops are idle.
(2) The average number of machines in the service section of the factory.
(3) The average repair time of a machine.
Solution:
Here = 0.5 hr
= 1
/ .2
hr
1 = 1 / hr.
2 = 2 / hr.
(1) Probability that both repair shops are idle 00P
0 0
00
1 1 2 2
1 1P
0 01 1 1 1
1 12 2 4 4
1 3 3
1* *1*2 4 8
(2) Average number of machines in the service station
12
12 2
1 2*
2 2 3
1
3
(3) Average repair time of a machine.
1 2
1 1
1 1
1 11 2
2 2
312 2
1 1
22
3
8
3
22
3
2.66
5. A TVS company in MADURAI containing a repair section shared by a large number of machines
has 2 sequential stations with respective service rates of 3 per hour and 4 per hour. The
cumulative failure rate of all the machines is 1 per hour. Assuming that the system behavior can
be approximated by a 2 stage tandem queue, find
(i) The probability that both the service stations are idle.
(ii) The average repair time including the waiting time and
(iii) the bottle neck of the repair facility.
Solution:
The Situation under series queue as all stations has to be visited (ie., no station can be omitted)
The current situation comes under the sequence queue model. Since any number of machines can
be repaired, each station comes under the model (M|M|1)( | FCFS)
Cumulative failure rate = 1/hr
Service rate of station I 1 = 3/hr
Service rate of station II 2 = 4/hr.
(i) s in
s inmn
m customer the I stationandP P
ncustomer the II station
1 1 2 2
1 1
m n
0 0
00
1 1 1 11 1
3 3 4 4P
2 3 1
*3 4 2
(ii) Average repair time (including waiting time)
1 2
1 1
1 1
3 1 4 1
1 1 5
2 3 6
5
6 Hours.
= 50 minutes.
(iii) Since 1
1
3
2
1
4
ie 1 2
we got the service station I is the bottleneck of the repair facility.
6. A repair facility shared by a large number of machines has 2 sequential stations with respective
service rates of 2 per hour and 3 per hour. The cumulative failure rate of all the machines is 1 per
hour. Assuming that the system behavior may be approximated by the 2 stage series queue find.
(i) The average repair time including the waiting time
(ii) The probability that both the service stations are idle.
(iii) The bottleneck of the repair facility.
Solution:
The Situation comes under series queue as all stations has to be visited (ie.no station can be
omitted)
The current situation comes under the sequence queue model. Since any number of machines can
be repaired, each station comes under the model (M|M|1)( |FCFS)
Cumulative failure rate = 1
Service rate of station I 1 = 2
Service rate of station II 2 = 3.
(i) Average number of machines in service at both the stations 1 2
1 1 3
2 1 3 1 2
Average repair time including the waiting time 1 2
1 1
1 1 3
2 1 3 1 2
hrs
= 90 minutes.
(ii) P(both service stations are idle)
0 0
00
1 1 2 2
1 1P
0 01 1 1 1
1 12 2 3 3
1 2 1
*2 3 3
(iii) 1 2
1 1,
2 3Since
ie 1 2
we got the service station I is the bottleneck of the repair facility.
7. In a network of 3 service stations 1,2,3 customers arrive 1,2,3 from outside, in accordance with
Poisson process having rates 5,10,15 respectively. The service times at the 3 stations are
exponential with respective rates 10, 50, 100. A customer completing service at station 1 is
equally like to.
1. Go to station 2.
2. Go to station 3.
3. Leave the system
A customer departing from service at station 2 always goes to station 3. A departure from service at
station 3 is equally like to go to station 2 or leave the system.
1. What is the average number of customers in the system consisting of all the three stations?
2. What is the average time a customer spends in the system?
Solution:
Since the customers entering the shop need not necessarily go to all sections, the situation in this
problem comes under Jackson‟s open queue system.
In view of Jackson‟s flow balance equations,
We have
3
1
1,2,3 (1)j j i ij
i
r P j
We have
1 2 35, 10, 15 (2)r r r
And
11 21 31 22 33
12 13 23 32
0, 0, 0, 0, 0
---- (3)1 1 1, , 1
3 3 2
P P P P P
P P P P
For j=1, (1) becomes
3
1 1 1
1
i i
i
r P
1 1 1 11 2 21 3 31r P P P
1 5 0 0 0 [Using (2) & (3) ]
1 5
For j=2, (1) becomes
3
2 2 2
1
i i
i
r P
2 2 1 12 2 22 3 32r P P P
2 1 3
1 110
3 2 [Using 2 & 3]
2 3 1
35 15
3 2
2 36 3 70 ------- (4)
For j = 3 in 1, we get
3 3 1 13 2 23 3 33r P P P
3 1 2
115
3
3 2 1
505
3
2 33 3 50 -------- (5)
(4) + (5)
23 120
2 40
Substituting 2 40 in (4) we get
3240 70 3
3
170
3
Resultant arrival rate at 1S , 1 5 hr
Resultant arrival rate at 2S , 2 40 hr
Resultant arrival rate at 3S , 3
170
3hr
Resultant Service rate at 1S , 1 10 hr
Resultant Service rate at 2S , 2 50 hr
Resultant Service rate at 3S , 3 100 hr
(i) Average number of customers in the system
1 2 3( 1) ( 2) ( 3)S S S SL L station L station L station
= 31 2
1 1 2 2 3 3
1705 40 3
17010 5 50 40100
3
= 17
1 43
= 6.3 (Approximately).
(ii) Average waiting time of a customer in the system
1 2 3,S
S
LW r r r
= 5+10+15 = 30
6.3
30SW hr = 6.3 * 2 minutes
= 12.6 minutes.