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© 1998 by Prentice-Hall IncRussell/Taylor Oper Mgt 2/e
Chapter 16
Waiting Line Models for Service Improvement
Ch 16 - 2© 1998 by Prentice-Hall IncRussell/Taylor Oper Mgt 2/e
Elements Of Waiting Line Analysis
Queue a single waiting line
Waiting line system arrivals servers waiting line structures
Ch 16 - 3© 1998 by Prentice-Hall IncRussell/Taylor Oper Mgt 2/e
Calling population– source of customers– an infinite population assumes such a large
number of customers that it is always possible for one more customer to arrive to be served
– a finite population consists of a countable number of potential customers
Arrival rate, – frequency of customer arrivals to system – typically follows Poisson distribution
Ch 16 - 4© 1998 by Prentice-Hall IncRussell/Taylor Oper Mgt 2/e
Service time– often follows negative exponential distribution– average service rate =
Arrival rate must be less than service rate or system never clears out
(
Ch 16 - 5© 1998 by Prentice-Hall IncRussell/Taylor Oper Mgt 2/e
Components Of A Queuing System
Source of
customersArrivals Server Served
customersWaiting Line
or queue
Ch 16 - 6© 1998 by Prentice-Hall IncRussell/Taylor Oper Mgt 2/e
Queue Discipline And Length
Queue discipline– order in which customers are served– first come, first served is most common
Length can be infinite or finite– infinite is most common– finite is limited by some physical
structure
Ch 16 - 7© 1998 by Prentice-Hall IncRussell/Taylor Oper Mgt 2/e
Basic Waiting Line Structures
Channels are the number of parallel servers
Phases denote number of sequential servers the customer must go through
Ch 16 - 8© 1998 by Prentice-Hall IncRussell/Taylor Oper Mgt 2/e
Single-Channel Structures
ServersWaiting line
Single-channel, multiple phases
Waiting line Server
Single-channel, single-phase
Ch 16 - 9© 1998 by Prentice-Hall IncRussell/Taylor Oper Mgt 2/e
Multi-Channel Structures
Servers
ServersWaiting line
Multiple-channel, single phase
Multiple-channel, multiple-phase
Waiting line
Ch 16 - 10© 1998 by Prentice-Hall IncRussell/Taylor Oper Mgt 2/e
Operating Characteristics The mathematics of queuing theory does not
provide optimal or best solutions
Instead, operating characteristics are computed that describe system performance
Steady state provides the average value of performance characteristics that the system will reach after a long time
Ch 16 - 11© 1998 by Prentice-Hall IncRussell/Taylor Oper Mgt 2/e
Operating Characteristics
Notation Description
L Average number of customers in the system(waiting and being served)
Lq Average number of customers in the waiting line
W Average time a customer spends in the system (waiting and being served)
Wq Average time a customer spends waiting in line
Ch 16 - 12© 1998 by Prentice-Hall IncRussell/Taylor Oper Mgt 2/e
P0 Probability of no (zero) customers in the system
Pn Probability of n customers in the system
Utilization rate; the proportion of time the system is in use
Ch 16 - 13© 1998 by Prentice-Hall IncRussell/Taylor Oper Mgt 2/e
Cost Relationship In Waiting Line Analysis
Exp
ecte
d co
sts
Level of service
Total cost
Service cost
Waiting Costs
Ch 16 - 14© 1998 by Prentice-Hall IncRussell/Taylor Oper Mgt 2/e
Waiting Line Analysis And Quality
Traditional view - the level of service should coincide with the minimum point on the total cost curve
TQM view - absolute quality service is the most cost-effective in the long run
Ch 16 - 15© 1998 by Prentice-Hall IncRussell/Taylor Oper Mgt 2/e
Single-Channel, Single-Phase Models
All assume Poisson arrival rate Variations
– exponential service times– general (or unknown) distribution of service times– constant service times– exponential service times with finite queue length– exponential service times with finite calling
population
Ch 16 - 16© 1998 by Prentice-Hall IncRussell/Taylor Oper Mgt 2/e
Basic Single-Server Model Assumptions:
– Poisson arrival rate– exponential service times– first-come, first-served queue discipline– infinite queue length– infinite calling population
= mean arrival rate = mean service rate
Ch 16 - 17© 1998 by Prentice-Hall IncRussell/Taylor Oper Mgt 2/e
Formulas for Single-Server Model
P0 =
(1 - )
Pn =
nP0
=( )
(1 - )
Lq =
L =
Probability that no customersare in system
Probability of exactly n customers in system
Average number of customers in system
Average number of customers in queue
n
)(
Ch 16 - 18© 1998 by Prentice-Hall IncRussell/Taylor Oper Mgt 2/e
W = =L
Wq =
=
= = = P0
(1 - )
Average time customerspends in system
Average time customer spends in queue
Probability that serveris busy, utilization factor
Probability that server is idle & customer can be served
Ch 16 - 19© 1998 by Prentice-Hall IncRussell/Taylor Oper Mgt 2/e
Single-Server Example
Given: = 24 per hour, = 30 customers per hour
P0 =
(1 - )
Lq =
L =
Probability that no customersare in system
Average number of customers in system
Average number of customers in queue
= 1 - (24/30) = 0.20
= 24/(30-24) = 4
= 242/30(30-24) = 3.2
Ch 16 - 20© 1998 by Prentice-Hall IncRussell/Taylor Oper Mgt 2/e
W =
Wq =
=
I =
Average time customerspends in system
Average time customer spends in queue
Probability that serveris busy, utilization factor
Probability that server is idle & customer can be served
= 1(30-24) = 0.167 hr = 10 min
= 1 - 0.80 = 0.20
= 24/30(30-24) = 0.133 hr = 8 min
= 24/30 = 0.80
Ch 16 - 21© 1998 by Prentice-Hall IncRussell/Taylor Oper Mgt 2/e
Waiting Line Cost Analysis
Management wants to test two alternatives to reduce customer waiting time:
1. Hire another employee to pack up purchases
2. Open another checkout counter
Ch 16 - 22© 1998 by Prentice-Hall IncRussell/Taylor Oper Mgt 2/e
Alternative 1 Extra employee costs $150 / week Each one-minute reduction in customer waiting time avoids $75
in lost sales Extra employee will increase service rate to 40 customers per
hour Recompute operating characteristics Wq = 0.038 hours = 2.25 minutes, originally was 8 minutes 8.00 - 2.25 = 5.75 minutes 5.75 x $75/minute/week = $431.25 per week New employee saves $431.25 - 150.00 = $281.25 / week
Ch 16 - 23© 1998 by Prentice-Hall IncRussell/Taylor Oper Mgt 2/e
Alternative II New counter costs $6000 plus $200 per week for checker Customers divide themselves between two checkout lines Arrival rate is reduced from = 24 to = 12 Service rate for each checker is = 30 Recompute operating characteristics Wq = 0.022 hours = 1.33 minutes, originally was 8 minutes 8.00 - 1.33 = 6.67 minutes 6.67 x $75/minute/week = $500.00/wk - 200.00 = $300/wk Counter is paid off in 6000/300 = 20 weeks Counter saves $300/wk; choose alternative II
Ch 16 - 24© 1998 by Prentice-Hall IncRussell/Taylor Oper Mgt 2/e
Constant Service Times
Constant service times occur with machinery and automated equipment
Constant service times are a special case of the single-server model with general or undefined service times
Ch 16 - 25© 1998 by Prentice-Hall IncRussell/Taylor Oper Mgt 2/e
Operating Characteristics For Constant Service Times
Wq =Lq
(1 - )P0 =
Lq =
L = Lq +
Probability that no customersare in system
Average number of customers in system
Average number of customers in queue
Average time customer spends in queue
Ch 16 - 26© 1998 by Prentice-Hall IncRussell/Taylor Oper Mgt 2/e
Probability that serveris busy, utilization factor
W = Wq +Average time customerspends in system
=
When service time isconstant and = 0, formula can be simplified
Lq =
=
=
=
Ch 16 - 27© 1998 by Prentice-Hall IncRussell/Taylor Oper Mgt 2/e
Constant Service Time Example
Automated car wash with service time = 4.5 min Cars arrive at rate = 10/hour (Poisson) = 60/4.5 = 13.3/hour
2Lq =
(10)2
2(13.3)(13.3-10)= = 1.14 cars waiting
Wq =Lq
=1.14/10 = .114 hour or 6.84 minutes
Ch 16 - 28© 1998 by Prentice-Hall IncRussell/Taylor Oper Mgt 2/e
Finite Queue Length A physical limit exists on length of queue M = maximum number in queue Service rate does not have to exceed arrival rate to
obtain steady-state conditions ()
P0 =
L =
M
Pn = (P0 )( )n
for n M
(M + 1() M + 1
1 - ( )M+1
Probability that no customers are in system
Probability of exactly n customers in system
Average number of customers in system
Ch 16 - 29© 1998 by Prentice-Hall IncRussell/Taylor Oper Mgt 2/e
Average number of customers in queue
Average time customerspends in system
Average time customer spends in queue
LW = (1 - PM)
Lq =(1- PM)
L
WWq =
Let PM = probability a customer will not join the system
Ch 16 - 30© 1998 by Prentice-Hall IncRussell/Taylor Oper Mgt 2/e
Finite Queue ExampleQuick Lube has waiting space for only 3 cars = 20, = 30, M = 4 cars (1 in service + 3 waiting)
Probability that no cars are in system
Probability of exactly n cars in system
Average number of cars in system
P0 =M
1 - 20/30
20/305= = 0.38
Pm = (P0 )( )n=M (= (0.38) )420
30= 0.076
L =
(M + 1() M + 1
1 - ( )M+1
= 20/30
1 -20/30(5(20/30) 5
1 - (20/30)5= 1.24
Ch 16 - 31© 1998 by Prentice-Hall IncRussell/Taylor Oper Mgt 2/e
Average number of cars in queue
Average time car spends in system
Average time carspends in queue
LW = (1 - PM)
Lq =(1- PM)
WWq =
L - 20(1-0.076)
30
= 1.24 - = 0.62
1.24
20 (1-0.076)= = 0.67 hours
= 4.03 min
30
0.067 -= = 0.033 hours
= 2.03 min
Ch 16 - 32© 1998 by Prentice-Hall IncRussell/Taylor Oper Mgt 2/e
Finite Calling Population Arrivals originate from a finite (countable) population N = population size
Lq =+
N -
Pn = P0
( )nN!
(N - n)!where n = 1, 2, ..., N
(1- P0)
Probability that no customers are in system
Probability of exactly n customers in system
Average number of customers in queue
P0 =
nN!
(N - n)!N
n = 0
Ch 16 - 33© 1998 by Prentice-Hall IncRussell/Taylor Oper Mgt 2/e
W = Wq +
Lq +L = (1- P0)
Wq = (N - L) Lq
Average time customerspends in system
Average time customer spends in queue
Average number of customers in system
Ch 16 - 34© 1998 by Prentice-Hall IncRussell/Taylor Oper Mgt 2/e
Finite Calling Pop’n Example
20 machines which operate an average of 200 hrs before breaking down = 1/200 hr = 0.005/hr
Mean repair time = 3.6 hrs = 1/3.6 hr = 0.2778/hr
Probability that no machines are in system
P0 =1
nN!
(N - n)!N
n = 0
=1
(0.005/0.2778)n20!
(20 - n)!20
n = 0
= 0.652
Ch 16 - 35© 1998 by Prentice-Hall IncRussell/Taylor Oper Mgt 2/e
Lq =
0.005 + 0.2778 0.005
= 20 (1- 0.652)
Average number of machines in queue
W = Wq +1
L = Lq + (1-P0) = 0.169 + (1-0.62) = 0.520
Wq =(N - L)
Lq
Average time machinespends in system
Average time machinespends in queue
Average number of machines in system
= 0.169
+
N (1- P0)
(20 - 0.520) 0.005
0.169= = 1.74
1.74 +1
0.278= = 5.33 hrs
Ch 16 - 36© 1998 by Prentice-Hall IncRussell/Taylor Oper Mgt 2/e
Multiple-Channel, Single-Phase Models
Two or more independent servers serve a single waiting line
Poisson arrivals, exponential service, infinite calling population
s>
P0 =1
n = s - 1
n = 0] +
( )n1
n!1s!
( )s s
s - ( )
Ch 16 - 37© 1998 by Prentice-Hall IncRussell/Taylor Oper Mgt 2/e
L =
Pn = P0,( )n1
s! sn-sfor n > s
Pn = P0,( )n1
n!
Pw = P0
( )s1
s!
ss - ( )
( )
()s
(s - 1 ! (s - P0 +
Probability of exactly n customers in system
Average number of customers in system
Probability an arrivingcustomer must wait
for n <= s
Ch 16 - 38© 1998 by Prentice-Hall IncRussell/Taylor Oper Mgt 2/e
/s
Lq =
L
=
Wq =
1W
=
Lq
W =L
Average number of customers in queue
Average time customerspends in system
Average time customer spends in queue
Utilization factor
Ch 16 - 39© 1998 by Prentice-Hall IncRussell/Taylor Oper Mgt 2/e
Multiple-Server Example
Customer service area = 10 customers/area = 4 customers/hour per service reps = (3)(4) = 12
P0 =1
n = s - 1
n = 0] +
( )n1
n!1s!
( )s s
s - ( )=
1
]1( )1
1!13!
( )3 3(4)
3(4)-10( )( )10!
( )1
2! 20
+ + +
= 0.045
Ch 16 - 40© 1998 by Prentice-Hall IncRussell/Taylor Oper Mgt 2/e
L =( )
()s
(s - 1 ! (s - P0 +Average number of
customers in system
(10)(4) (10/4) 3
=(3-1)! [3(4)-10] 2
(0.045) + (10/4) = 6
Average time customerspends in system W =
L
= 6/10 = 0.60 hr = 36 min
Ch 16 - 41© 1998 by Prentice-Hall IncRussell/Taylor Oper Mgt 2/e
Average number of customers in queue
Average time customer spends in queue
Pw = P0
( )s1 s
s - ( )Probability an arrivingcustomer must wait
Wq = = 3.5/10 = 0.35 hrs = 21 min
Lq
Lq =
L = 6 - 10/4 = 3.5
s!
= (0.45) = 0.703104( )31 3(4)
3(4)-10( ) 3!
Ch 16 - 42© 1998 by Prentice-Hall IncRussell/Taylor Oper Mgt 2/e
Improving Service
Add a 4th server to improve service Recompute operating characteristics
Po = 0.073 prob of no customers
L = 3.0 customersW = 0.30 hour, 18 min in serviceLq = 0.5 customers waiting
Wq = 0.05 hours, 3 min waiting, versus 21 earlier
Pw = 0.31 prob that customer must wait