Heizer mod d

Operations
Management
Module D –
Waiting-Line Models

PowerPoint presentation to accompany
Heizer/Render
Principles of Operations Management, 7e
Operations Management, 9e
© 2008 Prentice Hall, Inc. D–1

Outline
 Characteristics of a Waiting-Line
System
 Arrival Characteristics
 Waiting-Line Characteristics
 Service Characteristics
 Measuring a Queue’s Performance
 Queuing Costs


Outline – Continued
 The Variety of Queuing Models
 Model A(M/M/1): Single-Channel
Queuing Model with Poisson Arrivals
and Exponential Service Times
 Model B(M/M/S): Multiple-Channel
Queuing Model
 Model C(M/D/1): Constant-Service-Time
Model
 Model D: Limited-Population Model


Outline – Continued

 Other Queuing Approaches


Learning Objectives
When you complete this module you
should be able to:
1. Describe the characteristics of
arrivals, waiting lines, and service
systems
2. Apply the single-channel queuing
model equations
3. Conduct a cost analysis for a
waiting line


Learning Objectives
When you complete this module you
should be able to:
4. Apply the multiple-channel
queuing model formulas
5. Apply the constant-service-time
model equations
6. Perform a limited-population
model analysis


Waiting Lines
 Often called queuing theory
 Waiting lines are common situations
 Useful in both
manufacturing
and service
areas


Common Queuing
Situations
Situation Arrivals in Queue Service Process
Supermarket Grocery shoppers Checkout clerks at cash
register
Highway toll booth Automobiles Collection of tolls at booth
Doctor’s office Patients Treatment by doctors and
nurses
Computer system Programs to be run Computer processes jobs
Telephone company Callers Switching equipment to
forward calls
Bank Customer Transactions handled by teller
Machine Broken machines Repair people fix machines
maintenance
Harbor Ships and barges Dock workers load and unload

Table D.1

Characteristics of Waiting-
Line Systems
1. Arrivals or inputs to the system
 Population size, behavior, statistical
distribution
2. Queue discipline, or the waiting line
itself
 Limited or unlimited in length, discipline
of people or items in it
3. The service facility
 Design, statistical distribution of service
times

Arrival Characteristics
1. Size of the population
 Unlimited (infinite) or limited (finite)
2. Pattern of arrivals
 Scheduled or random, often a Poisson
distribution
3. Behavior of arrivals
 Wait in the queue and do not switch
lines
 No balking or reneging
© 2008 Prentice Hall, Inc. D – 10

Parts of a Waiting Line
Population of Arrivals Queue Service Exit the system
dirty cars from the (waiting line) facility
general
population …
Dave’s
Car Wash

Enter Exit

Arrivals to the system In the system Exit the system

Arrival Characteristics Waiting Line Service Characteristics
 Size of the population Characteristics  Service design
 Behavior of arrivals  Limited vs.  Statistical distribution
 Statistical distribution unlimited of service
of arrivals  Queue discipline
Figure D.1

Poisson Distribution

e-λ λ x
P (x ) = for x = 0, 1, 2, 3, 4, …
x!

where P(x) = probability of x
arrivals
x = number of arrivals per
unit of time
λ = average arrival rate
e = 2.7183 (which is the base
of the natural logarithms)


Poisson Distribution
e-λ λ x
Probability = P(x) =
x!

0.25 – 0.25 –

0.02 – 0.02 –
Probability

Probability
0.15 – 0.15 –

0.10 – 0.10 –

0.05 – 0.05 –

– –

0 1 2 3 4 5 6 7 8 9 x 0 1 2 3 4 5 6 7 8 9 10 11 x
Distribution for λ = 2 Distribution for λ = 4
Figure D.2


Waiting-Line Characteristics

 Limited or unlimited queue length
 Queue discipline - first-in, first-out
(FIFO) is most common
 Other priority rules may be used in
special circumstances


Service Characteristics
 Queuing system designs
 Single-channel system, multiple-
channel system
 Single-phase system, multiphase
system
 Service time distribution
 Constant service time
 Random service times, usually a
negative exponential distribution

Queuing System Designs
A family dentist’s office

Queue
Service Departures
Arrivals facility after service

Single-channel, single-phase system

A McDonald’s dual window drive-through

Queue
Phase 1 Phase 2 Departures
Arrivals service service
after service
facility facility

Single-channel, multiphase system
Figure D.3


Most bank and post office service windows

Service
facility
Channel 1
Queue
Service Departures
Arrivals facility
after service
Channel 2

Service
facility
Channel 3

Multi-channel, single-phase system
Figure D.3


Some college registrations

Phase 1 Phase 2
service service
Queue facility facility
Channel 1 Channel 1
Departures
Arrivals after service
Phase 1 Phase 2
service service
facility facility
Channel 2 Channel 2

Multi-channel, multiphase system
Figure D.3


Negative Exponential
Distribution
Probability that service time is greater than t = e-µt for t ≥ 1
-µt

µ = Average service rate
1.0 – e = 2.7183
Probability that service time ≥ 1

0.9 –
0.8 – Average service rate (µ) = 3 customers per hour
⇒ Average service time = 20 minutes per customer
0.7 –
0.6 –
0.5 –
0.4 –
Average service rate (µ) =
0.3 – 1 customer per hour
0.2 –
0.1 –
0.0 |– | | | | | | | | | | | |
0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00 2.25 2.50 2.75 3.00
Figure D.4 Time t (hours)

Measuring Queue
Performance
1. Average time that each customer or object
spends in the queue
2. Average queue length
3. Average time each customer spends in the
system
4. Average number of customers in the system
5. Probability that the service facility will be idle
6. Utilization factor for the system
7. Probability of a specific number of customers
in the system

Queuing Costs
Cost

Minimum
Total Total expected cost
cost
Cost of providing service

Cost of waiting time

Low level Optimal High level
of service service level of service

Figure D.5

Queuing Models
The four queuing models here all assume:

 Poisson distribution arrivals
 FIFO discipline
 A single-service phase


Queuing Models
Model Name Example
A Single-channel Information counter
system at department store
(M/M/1)

Number Number Arrival Service
of of Rate Time Population Queue
Channels Phases Pattern Pattern Size Discipline
Single Single Poisson Exponential Unlimited FIFO

Table D.2

Queuing Models
Model Name Example
B Multichannel Airline ticket
(M/M/S) counter

Multi- Single Poisson Exponential Unlimited FIFO
channel

Table D.2

Queuing Models
Model Name Example
C Constant- Automated car
service wash
(M/D/1)

Single Single Poisson Constant Unlimited FIFO

Table D.2

Queuing Models
Model Name Example
D Limited Shop with only a
population dozen machines
(finite population) that might break

Single Single Poisson Exponential Limited FIFO

Table D.2

Model A – Single-Channel
1. Arrivals are served on a FIFO basis and
every arrival waits to be served
regardless of the length of the queue
2. Arrivals are independent of preceding
arrivals but the average number of
arrivals does not change over time
3. Arrivals are described by a Poisson
probability distribution and come from
an infinite population


4. Service times vary from one customer
to the next and are independent of one
another, but their average rate is
known
5. Service times occur according to the
negative exponential distribution
6. The service rate is faster than the
arrival rate


λ = Mean number of arrivals per time
period
µ = Mean number of units served per
time period
Ls λ = Average number of units
µ–λ
(customers) in the system (waiting and being
served)
=
1
Ws λ =
µ– Average time a unit spends in the
system (waiting time plus service time)
Table D.3
=

Lq = Average number of units waiting
in the queue
λ2
=
µ(µ – λ )
Wq = Average time a unit spends
waiting in the queue
λ
µ(µ – = )
λ
p = Utilization factor for the system
λ
µ =
Table D.3


P0 = Probability of 0 units in the
system (that is, the service unit is idle)
λ
=µ 1–

Pn > k = Probability of more than k units in the
system, where n is the number of units in
the system
k+1
λ
µ =

Table D.3


Single-Channel Example
λ = 2 cars arriving/hour
2 µ = 3 cars serviced/hour
λ
µ–λ 3-2
Ls = = = 2 cars
in the system on average
1 1
µ–λ 3-2
Ws = = = 1
λ2 22 hour average waiting time in
µ(µ – λ ) 3(3 - 2) the system

Lq = = =
1.33 cars waiting in line

λ = 2 cars arriving/hour
µ = 3 cars serviced/hour
λ 2
Wq = =
µ(µ – λ ) 3(3 - 2)
= 2/3 hour = 40 minute
average waiting time

p = λ /µ = 2/3 = 66.6%
λ
P0 = 1 of time mechanic is busy
- = .33 probability
µ
there are 0 cars in the system


Probability of more than k Cars in the System
k Pn > k = (2/3)k + 1
0 .667 ← Note that this is equal to 1 -
P0 = 1 - .33
1 .444
2 .296
3 .198 ← Implies that there is a 19.8%
chance that more than 3 cars are in the
system
4 .132
5 .088
6
© 2008 Prentice Hall, Inc. .058 D – 34

Single-Channel Economics
Customer dissatisfaction
and lost goodwill = $10 per hour
Wq = 2/3 hour
Total arrivals = 16 per day
Mechanic’s salary = $56 per day
Total hours
customers spend 2 2
= (16) = 10 hours
waiting per day 3 3
2
Customer waiting-time cost = $10 10 = $106.67
3

Total expected costs = $106.67 + $56 = $162.67


Multi-Channel Model
M = number of channels
open
λ = average arrival rate
µ = average service rate at
1
P0 = each channel for M µ > λ
M–1 n M
1 λ 1 λ Mµ
∑
n! µ
+
M! µ Mµ - λ
n=0 n=0

M
λ µ(λ /µ) λ
Ls = P +
(M - 1)!(Mµ - λ )
2 0 µ
Table D.4


Multi-Channel Model
M
λ µ(λ /µ) 1 Ls
Ws = P + =
(M - 1)!(Mµ - λ )
2 0 µ λ

λ
Lq = Ls –
µ

1 Lq
Wq = W s – =
µ λ
Table D.4


Multi-Channel Example
λ = 2 µ = 3 M = 2

1 1
P0 = =
1 n 2 2
2(3)
∑ 1
n!
2
3
+
1
2!
2
3 2(3) - 2
n=0

(2)(3(2/3)2 1 2 3
Ls = + =
1! 2(3) - 2 2 2 3 4

2 1 .083
Ws =
3/4
=
3 Lq = 3 – = Wq = = .0415
2 8 4 3 12 2


Multi-Channel Example
Single Channel Two Channels
P0 .33 .5
Ls 2 cars .75 cars
Ws 60 minutes 22.5 minutes
Lq 1.33 cars .083 cars
Wq 40 minutes 2.5 minutes


Waiting Line Tables
Poisson Arrivals, Exponential Service Times
Number of Service Channels, M
ρ 1 2 3 4 5
.10 .0111
.25 .0833 .0039
.50 .5000 .0333 .0030
.75 2.2500 .1227 .0147
1.0 .3333 .0454 .0067
1.6 2.8444 .3128 .0604 .0121
2.0 .8888 .1739 .0398
2.6 4.9322 .6581 .1609
3.0 1.5282 .3541
4.0 2.2164

Table D.5

Waiting Line Table Example
Bank tellers and customers
λ = 18, µ = 20
Lq
Utilization factor ρ = λ /µ = .90 Wq =
λ
From Table D.5

Number of Number
service windows M in queue Time in queue
1 window 1 8.1 .45 hrs, 27 minutes
2 windows 2 .2285 .0127 hrs, ¾ minute
3 windows 3 .03 .0017 hrs, 6 seconds
4 windows 4 .0041 .0003 hrs, 1 second


Constant-Service Model
Average length Lq = λ2
of queue 2µ(µ – λ )

Average waiting time λ
in queue Wq =
2µ(µ – λ )

Average number of λ
Ls = Lq +
customers in system µ

Average time 1
in the system Ws = W q +
µ
Table D.6


Constant-Service Example
Trucks currently wait 15 minutes on average
Truck and driver cost $60 per hour
Automated compactor service rate (µ) = 12 trucks per hour
Arrival rate (λ ) = 8 per hour
Compactor costs $3 per truck

Current waiting cost per trip = (1/4 hr)($60) = $15 /trip
8 1
Wq = = hour
2(12)(12 – 8) 12

Waiting cost/trip
with compactor = (1/12 hr wait)($60/hr cost) = $ 5 /trip
Savings with = $15 (current) – $5(new) = $10
new equipment
/trip
Cost of new equipment amortized = $ 3 /trip
Net savings = $ 7 /trip

Limited-Population Model
T
Service factor: X =
T+U
Average number running: J = NF(1 - X)
Average number waiting: L = N(1 - F)
Average number being serviced: H = FNX
T(1 - F)
Average waiting time: W = XF
Number of population: N = J + L + H

Table D.7


Limited-Population Model
D = Probability that a unit N = Number of potential
T
Service factor:inX =
will have to wait customers
queue T+U
F = Average number running: J = NF(1 - X) time
Efficiency factor T = Average service
H = Averagenumber of waiting: =LAverage time between
Average number units U = N(1 - F)
being served unit service
Average number being serviced: H = FNX
requirements
T(1 - F)
J = Average number of units W = Average time a unit
Average waiting time: W = waits in line
not in queue or in
service bay
XF
L = Number number of units X = =Service + H
Average of population: N J + L factor
waiting for service
M = Number of service
channels


Finite Queuing Table
X M D F
.012 1 .048 .999
.025 1 .100 .997
.050 1 .198 .989
.060 2 .020 .999
1 .237 .983
.070 2 .027 .999
1 .275 .977
.080 2 .035 .998
1 .313 .969
.090 2 .044 .998
1 .350 .960
.100 2 .054 .997
Table D.8
1 .386 .950

Limited-Population Example
Each of 5 printers requires repair after 20 hours (U) of use
One technician can service a printer in 2 hours (T)
Printer downtime costs $120/hour
Technician costs $25/hour
2
Service factor: X = = .091 (close to .090)
2 + 20
For M = 1, D = .350 and F = .960
For M = 2, D = .044 and F = .998
Average number of printers working:
For M = 1, J = (5)(.960)(1 - .091) = 4.36
For M = 2, J = (5)(.998)(1 - .091) = 4.54

Limited-Population Example
Average Average
Each of 5 printers require Cost/Hrafter 20 Cost/Hr(for of use
Number repair for hours U)
Number of Printers Downtime Technicians
One technician can service a printer in 2 hours (T) Total
Technicians Down (N - J) (N - J)$120 ($25/hr) Cost/Hr
Printer downtime costs $120/hour
Technician costs $25/hour $76.80
1 .64 $25.00 $101.80

2 .46 2 $55.20 $50.00 $105.20
Service factor: X = = .091 (close to .090)
2 + 20
For M = 1, D = .350 and F = .960
For M = 2, D = .044 and F = .998
Average number of printers working:
For M = 1, J = (5)(.960)(1 - .091) = 4.36
For M = 2, J = (5)(.998)(1 - .091) = 4.54

Other Queuing Approaches
 The single-phase models cover many
queuing situations
 Variations of the four single-phase
systems are possible
 Multiphase models
exist for more
complex situations


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