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© 2007 Pearson Education
Waiting Lines
Supplement C
© 2007 Pearson Education
Waiting Lines
• Waiting line: One or more “customers” waiting for service.
• Customer population: An input that generates potential customers.
• Service facility: A person (or crew), a machine (or group of machines), or both, necessary to perform the service for the customer.
• Priority rule: A rule that selects the next customer to be served by the service facility.
• Service system: The number of lines and the arrangement of the facilities.
© 2007 Pearson Education
Waiting Line ModelsBasic Elements
Service systemService system
Customer Customer populationpopulation
Waiting line
Priority rule
Service facilities
Served Served customerscustomers
© 2007 Pearson Education© 2007 Pearson Education
Waiting Line Arrangements
Single lineSingle line
Service facilitiesService facilities
Multiple linesMultiple lines
Service facilitiesService facilities
© 2007 Pearson Education
Service Facility Arrangements
Channel: One or more facilities required to perform a given service.
Phase: A single step in providing a service.
Priority rule: The policy that determines which customer to serve next.
© 2007 Pearson Education
Service Facility Arrangements
Single channel, single phaseSingle channel, single phase
Service facility
© 2007 Pearson Education
Single channel, multiple phaseSingle channel, multiple phase
Service Facility Arrangements
Service facility 1
Service facility 2
© 2007 Pearson Education
Multiple channel, single phaseMultiple channel, single phase
Service Facility Arrangements
Service facility 1
Service facility 2
© 2007 Pearson Education
Multiple channel, multiple phaseMultiple channel, multiple phase
Service Facility Arrangements
Service facility 3
Service facility 4
Service facility 1
Service facility 2
© 2007 Pearson Education
Service facility 3
Service facility 4
Service facility 1
Service facility 2
Routing for : 1–2–4Routing for : 1–2–4Routing for : 2–4–3Routing for : 2–4–3Routing for : 3–2–1–4Routing for : 3–2–1–4
Service Facility Arrangements
Mixed ArrangementMixed Arrangement
© 2007 Pearson Education
Priority Rule
The priority rule determines which customer to serve next.
Most service systems use the first-come, first-serve (FCFS) rule. Other priority rules include:
Earliest promised due date (EDD)
Customer with the shortest expected processing time (SPT)
Preemptive discipline: A rule that allows a customer of higher priority to interrupt the service or another customer.
© 2007 Pearson Education
Probability DistributionsArrival Times
Example C.1Example C.1 Arrival rate = Arrival rate = 22/hour/hour
Customer ArrivalsCustomer Arrivals are usually random and can be are usually random and can be described by a Poisson distribution.described by a Poisson distribution.
Probability that Probability that nn customers customers will arrive…will arrive…
PPnn = e = e--TT((TT))nn
nn!!
PP44 = e = e-2(1)-2(1)[[22(1)](1)]44
44!! PP44 = e = e-2-2 = 0.090 = 0.09016162424
Interarrival times: The time between customer arrivals.
Probability that Probability that 44 customers customers will arrive…will arrive…
Mean = T Variance = T
© 2007 Pearson Education
The exponential distribution describes the probability that the service time will be no more than T time periods.
Probability DistributionsService time
If the customer service rate is three per hour, what is the probability If the customer service rate is three per hour, what is the probability that a customer requires less than 10 minutes of service?that a customer requires less than 10 minutes of service?
PP((t ≤Tt ≤T)) = = 1 – 1 – ee--TTμ = average number of customers completing service per periodt = service time of the customerT = target service time
PP((t ≤ t ≤ 0.167 hr)0.167 hr) = = 1 – 1 – ee-3(0.167) -3(0.167) = 1 – 0.61 = 0.39= 1 – 0.61 = 0.39
Example C.2 Example C.2
Mean = 1/ Variance = (1/ )2
© 2007 Pearson Education
Operating Characteristics
Line Length: Number of customers in line. Number of Customers in System: Includes
customers in line and being serviced. Waiting Time in Line: Waiting for service to begin. Total Time in System: Elapsed time between
entering the line and exiting the system. Service Facility Utilization: Reflects the percentage
of time servers are busy.
© 2007 Pearson Education
Single-ServerModel
The simplest waiting line model involves a single server and a single line of customers.
Assumptions: The customer population is infinite and patient. The customers arrive according to a Poisson distribution,
with a mean arrival rate of The service distribution is exponential with a mean
service rate of The mean service rate exceeds the mean arrival rate.
Customers are served on a first-come, first-served basis.
The length of the waiting line is unlimited.
© 2007 Pearson Education
= Average utilization of the system =
L = Average number of customers in the service system = –
Lq = Average number of customers in the waiting line = L
W = Average time spent in the system, including service =1
–
Wq = Average waiting time in line = W
n = Probability that n customers are in the system = (1 – )n
Single-ServerModel
© 2007 Pearson Education
Single-Channel, Single-Channel, Single-Phase SystemSingle-Phase System
Arrival rate (= 30/hour, Service rate ( = 35/hour
Average time in line = Average time in line = WWqq = = 0.8570.857(0.20) = 0.17 hour, or 10.28 minutes(0.20) = 0.17 hour, or 10.28 minutes
Average time in system = Average time in system = WW = = 0.20 hour, or 12 minutes = = 0.20 hour, or 12 minutes11
3535 – – 3030
Average number in line = Average number in line = LLqq = = 0.8570.857(6) = 5.14 customers(6) = 5.14 customers
Average number in system = Average number in system = LL = = 6 customers = = 6 customers3030
3535 – – 3030
Utilization = Utilization = = = = = = = 0.8570.857, or 85.7%, or 85.7%
30303535
Example C.3
© 2007 Pearson Education
Single-Channel, Single-Channel, Single-Phase SystemSingle-Phase System
Arrival rate (= 30/hour Service rate ( = 35/hour/hour
© 2007 Pearson Education
Application C.1
8.025
20
42025
20
L
2.348.0 LLq
2.02025
11
W
16.02.08.0 WWq
© 2007 Pearson Education
Example C.4
© 2007 Pearson Education
Example C.4
© 2007 Pearson Education
Example C.4
© 2007 Pearson Education
Application C.2
17.01
W
117.0
88.25
17.0
2017.01
20
© 2007 Pearson Education
Multiple-ChannelMultiple-Channel,, Single-Phase SystemSingle-Phase System
With the multiple-server model, customers form a single line and choose one of s servers when one is available.The service system has only one phase.
There are s identical servers.
The service distribution for each is exponential.
Mean service time is 1/The service rate (s exceeds the arrival rate ().
© 2007 Pearson Education
American Parcel Service is concerned about the amount of time the American Parcel Service is concerned about the amount of time the company’s trucks are idle, waiting to be unloaded. company’s trucks are idle, waiting to be unloaded.
The terminal operates with four unloading bays. Each bay requires The terminal operates with four unloading bays. Each bay requires a crew of two employees, and each crew costs $30/hr.a crew of two employees, and each crew costs $30/hr.
The estimated cost of an idle truck is $50/hr. Trucks arrive at an The estimated cost of an idle truck is $50/hr. Trucks arrive at an average rate of three per hour, according to a Poisson distribution.average rate of three per hour, according to a Poisson distribution.
Unloading a truck averages one hour with exponential service Unloading a truck averages one hour with exponential service times.times.
4 Unloading bays4 Unloading bays Crew costs $30/hourCrew costs $30/hour2 Employees/crew2 Employees/crew Idle truck costs $50/hourIdle truck costs $50/hourArrival rate = 3/hourArrival rate = 3/hour Service time = 1 hourService time = 1 hour
Multiple-ServerMultiple-Server Model ModelExample C.5Example C.5
© 2007 Pearson Education© 2007 Pearson Education
Multiple-Server Model4 Unloading bays Crew costs $30/hour2 Employees/crew Idle truck costs $50/hourArrival rate = 3/hour Service time = 1 hour
Utilization = =3
1(4)= 0.75
0 = [∑ + ( )]-1(3/1)n
n!(3/1)4
4!1
1 – 0.75= 0.0377
Average trucks in line = Lq =0()ss!(1 – )2
0.0377(3/1)4(0.75)4!(1 – 0.75)2= = 1.53 trucks
Average time in line = Wq =Lq
1.53
3= 0.51 hours=
Average time in system = W = Wq + 1
= 0.51 + 11
= 1.51 hours
Average trucks in system = L = W = 3(1.51) = 4.53 trucks
© 2007 Pearson Education© 2007 Pearson Education
Multiple-Server ModelMultiple-Server Model
Labor costs: $30(s) = $30(4) = $120.00Idle truck cost: $50(L) = $50(4.53) = 226.50
Total hourly cost = $346.50
4 Unloading bays Crew costs $30/hour2 Employees/crew Idle truck costs $50/hourArrival rate = 3/hour Service time = 1 hour
© 2007 Pearson Education
Application C.3
© 2007 Pearson Education
Application C.3
8.05.122
20
s
11.0
8.01
1
2
5.12
20
5.12
201
1
1
11
120
s
Ps
408.18.01!2
8.05.12
2011.0
1! 2
2
20
s
PL
s
q
0704.020
408.1
q
q
LW hrs. (or 4.224 minutes)
© 2007 Pearson Education
Little’s Law
Little’s Law relates the number of customers in a waiting-line system to the waiting time of customers.
L = WL is the average number of customers in the
system.
is the customer arrival rate.
W is the average time spent in system, including service.
© 2007 Pearson Education
In the finite-source model, the single-server model assumptions are changed so that the customer population is finite, with N potential customers.
If N is greater than 30 customers, then the single-server model with an infinite customer population is adequate.
Finite-Source ModelFinite-Source Model
© 2007 Pearson Education
Finite-Source Model
= Average utilization of the server = 1 – 0
Lq = Average number of customers in line = N – (1 – 0) +
L = Average number of customers in the system = N – (1 – 0)
Wq = Average waiting time in line = Lq [(N – L)]–1
W = Average time in the system = L[(N – L)]–1
0 = probability of zero customers [∑ ( )n
]–1N!
(N – n)!
N
n=0
© 2007 Pearson Education© 2007 Pearson Education
Number of robots = 10 Loss/machine hour = $30Service time = 10 hrs Maintenance cost = $10/hrTime between failures = 200 hrs
Example C.6
© 2007 Pearson Education
Finite-Source ModelFinite-Source ModelExample C.6Example C.6 Solution Solution
= 1 – 0.538 = 0.462= 1 – 0.538 = 0.462
LLqq = 0.30 robots = 0.30 robots LL = = 0.76 robot0.76 robot
WW = 16.43 hours = 16.43 hours
Number of robots = 10 Loss/machine hour = $30Service time = 10 hrs Maintenance cost = $10/hrTime between failures = 200 hrs
00 = 0.538 = 0.538 WWqq = 6.43 hours = 6.43 hours
Labor cost: ($10/hr)(8 hrs/day)(0.462 utilization) = $ 36.96Idle robot cost: (0.76 robot)($30/robot hr)(8 hrs/day) = 182.40
Total daily cost = $219.36
© 2007 Pearson Education
Decision Areas for Management
Using waiting-line analysis, management can improve the service system in one or more of the following areas.
Arrival RatesNumber of Service FacilitiesNumber of PhasesNumber of Servers Per FacilityServer EfficiencyPriority RuleLine Arrangement
© 2007 Pearson Education
Application C.4
44.0
08.00
!808.0
!7
!808.0
!8
!8
1
!
!
810
1
00
N
n
n
nN
NP
56.044.011 0 P
144.0102.0
25.081 0 PNL
© 2007 Pearson Education
Application C.5
667.030
20
111.0667.0333.01
222.0667.0333.01
333.0667.0333.01
222
111
000
P
P
P
© 2007 Pearson Education
Application C.5
333.12030
20667.0
LLq
min6min601.02030
11
hrhrW