Post on 27-Oct-2014
A Report in Quantitative Management
WAITING LINES AND QUEUING THEORY
MODELS
Introduction Waiting Line Costs Characteristics of a Queuing System Single-Channel Queuing Model with
Poisson Arrivals and Exponential Service Times
Multichannel Queuing Model with Poisson Arrivals and Exponential Service Times
CHAPTER OUTLINE
Queuing theoryQueuing theory is the study of waiting lineswaiting linesIt is one of the oldest and most widely used
quantitative analysis techniquesWaiting lines are an everyday occurrence
for most peopleQueues form in business process as wellThe three basic components of a queuing
process are arrivals, service facilities, and the actual waiting line
Analytical models of waiting lines can help managers evaluate the cost and effectiveness of service systems
INTRODUCTION
Most waiting line problems are focused on finding the ideal level of service a firm should provide
In most cases, this service level is something management can control
When an organization doesdoes have control, they often try to find the balance between two extremes
A large stafflarge staff and manymany service facilities generally results in high levels of service but have high costs
WAITING LINE COSTS
Having the minimumminimum number of service facilities keeps service costservice cost down but may result in dissatisfied customers
There is generally a trade-off between cost of providing service and cost of waiting time
Service facilities are evaluated on their total total expected costexpected cost which is the sum of service costsservice costs and waiting costswaiting costs
Organizations typically want to find the service level that minimizes the total expected cost
WAITING LINE COSTS
Queuing costs and service level
WAITING LINE COSTS
*Optimal Service Level
Co
st
Service Level
Cost of Providing Service
Total Expected Cost
Cost of Waiting Time
Three Rivers Shipping operates a docking facility on the Ohio River
An average of 5 ships arrive to unload their cargos each shift
Idle ships are expensiveMore staff can be hired to unload the ships, but
that is expensive as wellThree Rivers Shipping Company wants to
determine the optimal number of teams of stevedores to employ each shift to obtain the minimum total expected cost
THREE RIVERS SHIPPING COMPANY EXAMPLE
Three Rivers Shipping waiting line cost analysis
THREE RIVERS SHIPPING COMPANY EXAMPLE
NUMBER OF TEAMS OF STEVEDORES WORKING
1 2 3 4
(a) Average number of ships arriving per shift
5 5 5 5
(b) Average time each ship waits to be unloaded (hours)
7 4 3 2
(c) Total ship hours lost per shift (a x b)
35 20 15 10
(d) Estimated cost per hour of idle ship time
$1,000 $1,000 $1,000 $1,000
(e) Value of ship’s lost time or waiting cost (c x d)
$35,000 $20,000 $15,000 $10,000
(f) Stevedore team salary or service cost
$6,000 $12,000 $18,000 $24,000
(g) Total expected cost (e + f) $41,000 $32,000 $33,000 $34,000
Optimal costOptimal cost
There are three parts to a queuing system1. The arrivals or inputs to the system
(sometimes referred to as the calling calling populationpopulation)
2. The queue or waiting line itself3. The service facility
These components have their own characteristics that must be examined before mathematical models can be developed
CHARACTERISTICS OF A QUEUING SYSTEM
Arrival Characteristics have three major characteristics, sizesize, patternpattern, and behaviorbehaviorSize of the calling population
Can be either unlimited (essentially infiniteinfinite) or limited (finitefinite)
Pattern of arrivals Can arrive according to a known pattern or can arrive randomlyrandomly Random arrivals generally follow a Poisson distributionPoisson distribution
CHARACTERISTICS OF A QUEUING SYSTEM
The Poisson distribution is
CHARACTERISTICS OF A QUEUING SYSTEM
4,... 3, 2, 1, 0, for
XX
eXP
X
!)(
where
P(X) = probability of X arrivalsX = number of arrivals per unit of time = average arrival ratee = 2.7183
We find the values of e–
If = 2, we can find the values for X = 0, 1, and 2
CHARACTERISTICS OF A QUEUING SYSTEM
!)(
Xe
XPX
%.)(.
!)( 1413530
1113530
02
002
e
P
%.)(.
!)( 2727060
1213530
12
12
1212
ee
P
%.)(.
)(!)( 2727060
2413530
124
22
2222
ee
P
Two examples of the Poisson distribution for arrival rates
CHARACTERISTICS OF A QUEUING SYSTEM
0.30 –
0.25 –
0.20 –
0.15 –
0.10 –
0.05 –
0.00 –
Pro
bab
ilit
y
= 2 Distribution
|0
|1
|2
|3
|4
|5
|6
|7
|8
|9
X
0.25 –
0.20 –
0.15 –
0.10 –
0.05 –
0.00 –P
rob
abil
ity
= 4 Distribution
|0
|1
|2
|3
|4
|5
|6
|7
|8
|9
X
Behavior of arrivalsMost queuing models assume customers are patient and will wait in the queue until they are served and do not switch lines
BalkingBalking refers to customers who refuse to join the queue
RenegingReneging customers enter the queue but become impatient and leave without receiving their service
That these behaviors exist is a strong argument for the use of queuing theory to managing waiting lines
CHARACTERISTICS OF A QUEUING SYSTEM
Waiting Line CharacteristicsWaiting lines can be either limitedlimited or unlimitedunlimitedQueue discipline refers to the rule by which customers in the line receive service
The most common rule is first-in, first-outfirst-in, first-out (FIFOFIFO)
Other rules are possible and may be based on other important characteristics
Other rules can be applied to select which customers enter which queue, but may apply FIFO once they are in the queue
CHARACTERISTICS OF A QUEUING SYSTEM
Service Facility Characteristics Basic queuing system configurations Service systems are classified in terms of the number of channels,
or servers, and the number of phases, or service stops A single-channel systemsingle-channel system with one server is quite common MultichannelMultichannel systemssystems exist when multiple servers are fed by one
common waiting line In a single-phase systemsingle-phase system the customer receives service form just
one server If a customer has to go through more than one server, it is a
multiphase systemmultiphase system
CHARACTERISTICS OF A QUEUING SYSTEM
Four basic queuing system configurations
CHARACTERISTICS OF A QUEUING SYSTEM
Single-Channel, Single-Phase System
ArrivalsArrivalsDepartures Departures after Serviceafter Service
Queue
Service Facility
Single-Channel, Multiphase System
ArrivalsArrivalsDepartures Departures after after ServiceService
Queue
Type 1 Service Facility
Type 2 Service Facility
Four basic queuing system configurations
CHARACTERISTICS OF A QUEUING SYSTEM
Multichannel, Single-Phase System
ArrivalsArrivals
Queue
Service Facility
1DeparturesDepartures
afterafterService Facility
2
ServiceServiceService Facility
3
Four basic queuing system configurations
CHARACTERISTICS OF A QUEUING SYSTEM
Multichannel, Multiphase System
ArrivalsArrivals
Queue
Departures Departures after after ServiceService
Type 2 Service Facility
1
Type 2 Service Facility
2
Type 1 Service Facility
2
Type 1 Service Facility
1
Service time distribution Service patterns can be either constant or
random Constant service times are often machine
controlled More often, service times are randomly
distributed according to a negative exponential negative exponential probability distributionprobability distribution
Models are based on the assumption of particular probability distributions
Analysts should take to ensure observations fit the assumed distributions when applying these models
CHARACTERISTICS OF A QUEUING SYSTEM
Two examples of exponential distribution for service times
CHARACTERISTICS OF A QUEUING SYSTEM
f (x) = e–x
for x ≥ 0 and > 0
= Average Number Served per Minute
Average Service Time of 1 Hour
Average Service Time of 20 Minutes
f (x)
Service Time (Minutes)
|0
|30
|60
|90
|120
|150
|180
–
–
–
–
–
–
–
–
D. G. Kendall developed a notation for queuing models that specifies the pattern of arrival, the service time distribution, and the number of channels
It is of the form
IDENTIFYING MODELS USING KENDALL NOTATION
Specific letters are used to represent probability distributions
M = Poisson distribution for number of occurrencesD = constant (deterministic) rateG = general distribution with known mean and
variance
Arrival distribution
Service time distribution
Number of service channels open
So a single channel model with Poisson arrivals and exponential service times would be represented by
M/M/1If a second channel is added we would have
M/M/2A three channel system with Poisson arrivals
and constant service time would beM/D/3
A four channel system with Poisson arrivals and normally distributed service times would be
M/G/4
IDENTIFYING MODELS USING KENDALL NOTATION
Assumptions of the modelArrivals are served on a FIFO basisNo balking or renegingArrivals are independent of each other but rate is constant over time
Arrivals follow a Poisson distributionService times are variable and independent but the average is known
Service times follow a negative exponential distribution
Average service rate is greater than the average arrival rate
SINGLE-CHANNEL MODEL, POISSON ARRIVALS, EXPONENTIAL SERVICE TIMES
(M/M/1)
When these assumptions are met, we can develop a series of equations that define the queue’s operating characteristicsoperating characteristics
Queuing EquationsWe let
SINGLE-CHANNEL MODEL, POISSON ARRIVALS, EXPONENTIAL SERVICE TIMES
(M/M/1)
= mean number of arrivals per time period = mean number of people or items served
per time period
The arrival rate and the service rate must be for the same time period
1. The average number of customers or units in the system, L
SINGLE-CHANNEL MODEL, POISSON ARRIVALS, EXPONENTIAL SERVICE TIMES
(M/M/1)
L
2. The average time a customer spends in the system, W
3. The average number of customers in the queue, Lq
1W
)(
2
qL
4. The average time a customer spends waiting in the queue, Wq
SINGLE-CHANNEL MODEL, POISSON ARRIVALS, EXPONENTIAL SERVICE TIMES
(M/M/1)
)(
qW
5. The utilization factorutilization factor for the system, , the probability the service facility is being used
6. The percent idle time, P0, the probability no one is in the system
SINGLE-CHANNEL MODEL, POISSON ARRIVALS, EXPONENTIAL SERVICE TIMES
(M/M/1)
10P
7. The probability that the number of customers in the system is greater than k, Pn>k
1
k
knP
Arnold’s mechanic can install mufflers at a rate of 3 per hour
Customers arrive at a rate of 2 per hour = 2 cars arriving per hour = 3 cars serviced per hour
ARNOLD’S MUFFLER SHOP CASE
2311
W 1 hour that an average car spends in the system
12
232
L 2 cars in the system
on the average
ARNOLD’S MUFFLER SHOP CASE
hour 32
)(
qW 40 minutes average
waiting time per car
)()()( 134
233222
qL 1.33 cars waiting in line
on the average
67032
. percentage of time
mechanic is busy
33032
110 .
P probability that there are 0 cars in the system
Probability of more than k cars in the system
ARNOLD’S MUFFLER SHOP CASE
k Pn>k = (2/3)k+1
0 0.667 Note that this is equal to 1 – Note that this is equal to 1 – PP00 = 1 – 0.33 = 0.667 = 1 – 0.33 = 0.667
1 0.444
2 0.296
3 0.198 Implies that there is a 19.8% chance that more Implies that there is a 19.8% chance that more than 3 cars are in the systemthan 3 cars are in the system
4 0.132
5 0.088
6 0.058
7 0.039
Introducing costs into the modelArnold wants to do an economic analysis of the queuing system and determine the waiting cost and service cost
The total service cost is
ARNOLD’S MUFFLER SHOP CASE
Total service cost = (Number of channels)
x (Cost per channel)
Total service cost = mCs
wherem = number of channelsCs = service cost of each channel
Waiting cost when the cost is based on time in the system
ARNOLD’S MUFFLER SHOP CASE
Total waiting cost = (W)Cw
Total waiting cost = (Total time spent waiting by all
arrivals) x (Cost of waiting)
= (Number of arrivals) x (Average wait per arrival)Cw
If waiting time cost is based on time in the queue
Total waiting cost = (Wq)Cw
So the total cost of the queuing system when based on time in the system is
ARNOLD’S MUFFLER SHOP CASE
Total cost = Total service cost + Total waiting cost
Total cost = mCs + WCw
And when based on time in the queue
Total cost = mCs + WqCw
Arnold estimates the cost of customer waitingwaiting time in line is $10 per hour
ARNOLD’S MUFFLER SHOP CASE
Total daily waiting cost = (8 hours per day)WqCw
= (8)(2)(2/3)($10) = $106.67 Arnold has identified the mechanics wage $7 per
hour as the serviceservice costTotal daily service cost = (8 hours per day)mCs
= (8)(1)($7) = $56 So the total cost of the system is
Total daily cost of the queuing system = $106.67 + $56 = $162.67
Arnold is thinking about hiring a different mechanic who can install mufflers at a faster rate
The new operating characteristics would be = 2 cars arriving per hour = 4 cars serviced per hour
ARNOLD’S MUFFLER SHOP CASE
2411
W 1/2 hour that an average car spends in the system
22
242
L 1 car in the system
on the average
ARNOLD’S MUFFLER SHOP CASE
hour 41
)(
qW 15 minutes average
waiting time per car
)()()( 184
244222
qL 1/2 cars waiting in line
on the average
5042
. percentage of time
mechanic is busy
5042
110 .
P probability that there are 0 cars in the system
Probability of more than k cars in the system
ARNOLD’S MUFFLER SHOP CASE
k Pn>k = (2/4)k+1
0 0.500
1 0.250
2 0.125
3 0.062
4 0.031
5 0.016
6 0.008
7 0.004
The customer waitingwaiting cost is the same $10 per hour
ARNOLD’S MUFFLER SHOP CASE
Total daily waiting cost = (8 hours per day)WqCw
= (8)(2)(1/4)($10) = $40.00 The new mechanic is more expensive at $9 per
hourTotal daily service cost = (8 hours per day)mCs
= (8)(1)($9) = $72 So the total cost of the system is
Total daily cost of the queuing system = $40 + $72 = $112
The total time spent waiting for the 16 customers per day was formerly
(16 cars per day) x (2/3 hour per car) = 10.67 hours
It is now is now
(16 cars per day) x (1/4 hour per car) = 4 hours
The total system costs are less with the new mechanic resulting in a $50 per day savings
$162 – $112 = $50
ARNOLD’S MUFFLER SHOP CASE
Reducing waiting time is not the only way to reduce waiting cost
Reducing waiting cost (Cw) will also reduce total waiting cost
This might be less expensive to achieve than reducing either W or Wq
ENHANCING THE QUEUING ENVIRONMENT
Assumptions of the modelArrivals are served on a FIFO basisNo balking or renegingArrivals are independent of each other but rate is constant over time
Arrivals follow a Poisson distributionService times are variable and independent but the average is known
Service times follow a negative exponential distribution
Average service rate is greater than the average arrival rate
MULTICHANNEL MODEL, POISSON ARRIVALS, EXPONENTIAL SERVICE TIMES
(M/M/M)
Equations for the multichannel queuing modelWe let
MULTICHANNEL MODEL, POISSON ARRIVALS, EXPONENTIAL SERVICE TIMES
(M/M/M)
m = number of channels open = average arrival rate = average service rate at each channel
1. The probability that there are zero customers in the system
m
mm
mn
Pmmn
n
n for
11
11
0
0
!!
2. The average number of customer in the system
MULTICHANNEL MODEL, POISSON ARRIVALS, EXPONENTIAL SERVICE TIMES
(M/M/M)
021P
mmL
m
)()!()/(
3. The average time a unit spends in the waiting line or being served, in the system
L
Pmm
Wm
1
1 02)()!()/(
4. The average number of customers or units in line waiting for service
MULTICHANNEL MODEL, POISSON ARRIVALS, EXPONENTIAL SERVICE TIMES
(M/M/M)
LLq
5. The average number of customers or units in line waiting for service
6. The average number of customers or units in line waiting for service
q
q
LWW
1
m
Arnold wants to investigate opening a second garage bay
He would hire a second worker who works at the same rate as his first worker
The customer arrival rate remains the same
ARNOLD’S MUFFLER SHOP REVISITED
m
mm
mn
Pmmn
n
n for
11
11
0
0
!!
500 .P
probability of 0 cars in the system
Average number of cars in the system
ARNOLD’S MUFFLER SHOP REVISITED
7501 02 .
)()!()/(
Pmm
Lm
minutes 2122hours
83
L
W
Average time a car spends in the system
Average number of cars in the queue
ARNOLD’S MUFFLER SHOP REVISITED
Average time a car spends in the queue
0830121
32
43
.
LLq
minutes 212hour 0.0415
208301
.
q
q
LWW
Effect of service level on Arnold’s operating characteristics
ARNOLD’S MUFFLER SHOP REVISITED
LEVEL OF SERVICE
OPERATING CHARACTERISTIC
ONE MECHANIC = 3
TWO MECHANICS = 3 FOR BOTH
ONE FAST MECHANIC = 4
Probability that the system is empty (P0)
0.33 0.50 0.50
Average number of cars in the system (L) 2 cars 0.75 cars 1 car
Average time spent in the system (W) 60 minutes 22.5 minutes 30 minutes
Average number of cars in the queue (Lq)
1.33 cars 0.083 car 0.50 car
Average time spent in the queue (Wq)
40 minutes 2.5 minutes 15 minutes
Table 14.2
Adding the second service bay reduces the waiting time in line but will increase the service cost as a second mechanic needs to be hired
ARNOLD’S MUFFLER SHOP REVISITED
Total daily waiting cost = (8 hours per day)WqCw
= (8)(2)(0.0415)($10) = $6.64
Total daily service cost = (8 hours per day)mCs
= (8)(2)($7) = $112
So the total cost of the system is
Total system cost = $6.64 + $112 = $118.64
The fast mechanic is the cheapest option