Chapter 9
Queuing Models
1
9.1 Introduction
• Queuing is the study of waiting lines, or queues.
• The objective of queuing analysis is to design
systems that enable organizations to perform
optimally according to some criterion.
• Possible Criteria
– Maximum Profits.
– Desired Service Level.
2
9.1 Introduction
• Analyzing queuing systems requires a clear
understanding of the appropriate service
measurement.
• Possible service measurements
– Average time a customer spends in line.
– Average length of the waiting line.
– The probability that an arriving customer must wait
for service.
3
9.2 Elements of the Queuing Process
• A queuing system consists of three basic
components:
– Arrivals: Customers arrive according to some arrival
pattern.
– Waiting in a queue: Arriving customers may have to wait
in one or more queues for service.
– Service: Customers receive service and leave the system.
4
The Arrival Process
• There are two possible types of arrival
processes
– Deterministic arrival process.
– Random arrival process.
• The random process is more common in
businesses.
5
The Arrival Process
• Under three conditions the arrivals can be modeled as a
Poisson process
– Orderliness : one customer, at most, will arrive during any
time interval.
– Stationarity : for a given time frame, the probability of arrivals
within a certain time interval is the same for all time intervals of
equal length.
– Independence : the arrival of one customer has no influence
on the arrival of another.
6
The Poisson Arrival Process
ke- lt
(lt)
P(X = k) =
k!
Where
l = mean arrival rate per time unit.
t = the length of the interval.
e = 2.7182818 (the base of the natural logarithm).
k! = k (k -1) (k -2) (k -3) … (3) (2) (1).
7
HANK’s HARDWARE – Arrival Process
• Customers arrive at Hank’s Hardware according to a
Poisson distribution.
• Between 8:00 and 9:00 A.M. an average of 6
customers arrive at the store.
• What is the probability that k customers will arrive
between 8:00 and 8:30 in the morning (k = 0, 1, 2,…)?
8
HANK’s HARDWARE –
An illustration of the Poisson distribution.
• Input to the Poisson
distribution
l = 6 customers per hour.
t = 0.5 hour.
lt = (6)(0.5) = 3.
0 1 2 3 4 5 6 7 8
10k23
(lt)
e
P(X = 01k23 )=
k2!
1!
0!
3!!
lt
= 0.224042
0.149361
0.049787
0.224042
9
HANK’s HARDWARE –
Using Excel for the Poisson probabilities
• Solution
– We can use the POISSON function in Excel to
determine Poisson probabilities.
– Point probability: P(X = k) = ?
• Use Poisson(k, lt, FALSE)
• Example: P(X = 0; lt = 3) = POISSON(0, 1.5, FALSE)
– Cumulative probability: P(Xk) = ?
• Example: P(X3; lt = 3) = Poisson(3, 1.5, TRUE)
10
HANK’s HARDWARE –
Excel Poisson
11
The Waiting Line Characteristics
• Factors that influence the modeling of queues
– Line configuration
– Priority
– Jockeying
– Tandem Queues
– Balking
– Homogeneity
12
Line Configuration
• A single service queue.
• Multiple service queue with single waiting line.
• Multiple service queue with multiple waiting
lines.
• Tandem queue (multistage service system).
13
Jockeying and Balking
• Jockeying occurs when customers switch lines
once they perceived that another line is moving
faster.
• Balking occurs if customers avoid joining the line
when they perceive the line to be too long.
14
Priority Rules
• These rules select the next customer for service.
• There are several commonly used rules:
–
–
–
–
First come first served (FCFS).
Last come first served (LCFS).
Estimated service time.
Random selection of customers for service.
15
Tandem Queues
• These are multi-server systems.
• A customer needs to visit several service
stations (usually in a distinct order) to complete
the service process.
• Examples
– Patients in an emergency room.
– Passengers prepare for the next flight.
16
Homogeneity
• A homogeneous customer population is one in
which customers require essentially the same
type of service.
• A non-homogeneous customer population is one
in which customers can be categorized according
to:
– Different arrival patterns
– Different service treatments.
17
The Service Process
• In most business situations, service time varies
widely among customers.
• When service time varies, it is treated as a
random variable.
• The exponential probability distribution is used
sometimes to model customer service time.
18
The Exponential Service Time Distribution
f(t) = me-mt
m = the average number of customers
who can be served per time period.
Therefore, 1/m = the mean service time.
The probability that the service time X is less than some “t.”
P(X  t) = 1 - e-mt
19
Schematic illustration of the exponential
distribution
The probability that service is completed
within t time units
P(X  t) = 1 - e-mt
X=t
20
HANK’s HARDWARE – Service time
• Hank’s estimates the average service time to be
1/m = 4 minutes per customer.
• Service time follows an exponential distribution.
• What is the probability that it will take less than 3
minutes to serve the next customer?
21
Using Excel for the Exponential Probabilities
• We can use the EXPDIST function in Excel to
determine exponential probabilities.
• Probability density: f(t) = ?
– Use EXPONDIST(t, m, FALSE)
• Cumulative probability: P(Xk) = ?
– Use EXPONDIST(t, m, TRUE)
22
HANK’s HARDWARE –
Using Excel for the Exponential Probabilities
• The mean number of customers served per
minute is ¼ = ¼(60) = 15 customers per hour.
• P(X < .05 hours) = 1 – e-(15)(.05) = ?
3 minutes = .05 hours
• From Excel we have:
– EXPONDIST(.05,15,TRUE) = .5276
23
HANK’s HARDWARE –
Using Excel for the Exponential Probabilities
=EXPONDIST(B4,B3,TRUE)
f(t)
Exponential Distribution for Mu = 15
16.000
14.000
12.000
10.000
8.000
6.000
4.000
2.000
0.000
0.000
0.075
0.150
0.225
0.300
0.375
t
=EXPONDIST(A10,$B$3,FALSE)
Drag to B11:B26
24
The Exponential Distribution Characteristics
• The memoryless property.
– No additional information about the time left for the completion of a
service, is gained by recording the time elapsed since the service
started.
– For Hank’s, the probability of completing a service within the next 3
minutes is (0.52763) independent of how long the customer has been
served already.
• The Exponential and the Poisson distributions are related to
one another.
– If customer arrivals follow a Poisson distribution with mean rate l,
their interarrival times are exponentially distributed with mean time
25
1/l.
9.3 Performance Measures of
Queuing System
• Performance can be measured by focusing on:
– Customers in queue.
– Customers in the system.
• Performance is measured for a system in steady
state.
26
9.3 Performance Measures of
Queuing System
• The transient period
occurs at the initial
time of operation.
• Initial transient
behavior is not
indicative of long run
performance.
n
Roughly, this
is a transient
period…
Time
27
9.3 Performance Measures of
Queuing System
• The steady state
period follows the
transient period.
• Meaningful long run
performance
measures can be
calculated for the
system when in
steady state.
n
Roughly, this
is a transient
period…
This is a
steady state
period………..
Time
28
9.3 Performance Measures of
Queuing System
In order to achieve steady state, the
effective arrival rate must be less than
the sum of the effective service rates .
k servers
l< m
For one server
l< m1 +m2+…+mk
For k servers
with service rates mi
l< km
Each with
service rate of m
29
Example :
• Suppose
– customers arrrive at the rate of l=20 per hour
– there are 3 servers, each serving an average of 4
customers per hour
• You will
– serve an average of 12 customers per hour
– add 8 customers each hour to your waiting line
30
Steady State Performance Measures
P0 = Probability that there are no customers in the system.
Pn = Probability that there are “n” customers in the system.
L = Average number of customers in the system.
Lq = Average number of customers in the queue.
W = Average time a customer spends in the system.
Wq = Average time a customer spends in the queue.
Pw = Probability that an arriving customer must wait
for service.
r = Utilization rate for each server
(the percentage of time that each server is busy).
31
Little’s Formulas
• Little’s Formulas represent important relationships
between L, Lq, W, and Wq.
• These formulas apply to systems that meet the
following conditions:
– Single queue systems,
– Customers arrive at a finite arrival rate l, and
– The system operates under a steady state condition.
L=lW
Lq = l Wq
For the case of an infinite population
L = Lq + l/m
32
Classification of Queues
• Queuing system can be classified by:
–
–
–
–
–
Arrival process.
Service process.
Number of servers.
System size (infinite/finite waiting line).
Population size.
Example:
M / M / 6 / 10 / 20
• Notation
– M (Markovian) = Poisson arrivals or exponential service time.
– D (Deterministic) = Constant arrival rate or service time.
– G (General) = General probability for arrivals or service time.
33
9.4 M/M/1 Queuing System Assumptions
– Poisson arrival process.
– Exponential service time distribution.
– A single server.
– Potentially infinite queue.
– An infinite population.
34
M / M /1 Queue - Performance Measures
P0 = 1 – (l/m)
Pn = [1 – (l/m)](l/m)n
L = l /(m – l)
Lq = l2 /[m(m – l)]
W = 1 /(m – l)
Wq = l /[m(m – l)]
Pw = l / m
r =l/m
The probability that
a customer waits in
the system more than
“t” is P(X>t) = e-(m - l)t
35
MARY’s SHOES
• Customers arrive at Mary’s Shoes every 12
minutes on the average, according to a Poisson
process.
• Service time is exponentially distributed with an
average of 8 minutes per customer.
• Management is interested in determining the
performance measures for this service system.
36
MARY’s SHOES - Solution
– Input
l = 1/12 customers per minute = 60/12 = 5 per hour.
m = 1/ 8 customers per minute = 60/ 8 = 7.5 per hour.
– Performance Calculations
m –l = 7.5 – 5 = 2.5 per hr.
P0 = 1 - (l/m) = 1 - (5/7.5) = 0.3333
P(X<10min) = 1 – e-2.5(10/60)
n
n
Pn = [1 - (l/m)](l/m) = (0.3333)(0.6667)
= .565
L = l/(m - l) = 2
Pw = l/m = 0.6667
2
Lq = l /[m(m - l)] = 1.3333
r = l/m = 0.6667
W = 1/(m - l) = 0.4 hours = 24 minutes
Wq = l/[m(m - l)] = 0.26667 hours = 16 minutes
37
• Why is L - Lq = 0.6667 and not 1 ?
– 1/3 of the time : no customers are present and thus
you serve 0 customer
– 2/3 of the time : one or more customers are present
and you serve 1 customer
– the weighted average number of customers being
served is :
0.3333 x (0) + 0.6667 x (1) = 0.6667
38
Relationship between System and Queue
Performance measures
Average time in system =
Average time in queue+ Average service time
W = Wq +1/ m
24 = 16 + 8 minutes
39
Relationship between System and Queue
Performance measures
Average number of customers in system =
Average number of customers in queue + Average number
of customers being served
L = Lq +l/ m
2 = 1.3333 + (5/7.5)
40
• Customer waiting time in the system follows an
exponential distribution with an average rate of
( m -l) = 7.5 - 5 = 2.5 per hour
• The probability that a customer will wait less than 10 or
20 minutes is :
P<10 min = 1 - e- 2.5 X (10/60) = 0.341
P<20 min = 1 - e- 2.5 X (20/60) = 0.565
41
• If µ would be 9 per hour, customer waiting time in the
system would follow an exponential distribution with an
new average rate of
( m -l) = 9 - 5 = 4 per hour
• The probability that a customer will wait less than 10 or
20 minutes is :
P<10 min = 1 - e- 4 X
P<20 min = 1 - e- 4 X
(10/60) =
0.487
(20/60) = 0.736
42
m
WINQSB Input Screen
l
43
Performance Measurements
Performance Measurements
Performance Measurements
Performance Measurements
Performance Measurements
44
45
12.5 M/M/k Queuing Systems
• Characteristics
– Customers arrive according to a Poisson process at a
mean rate l.
– Service times follow an exponential distribution.
– There are k servers, each of who works at a rate of m
customers (with km> l).
– Infinite population, and possibly infinite line.
46
M / M /k Queue - Performance Measures
P0 =
Pn =
Pn =
1
1l 
1  l   km 

 +

 


m
m




n= 0 n !
k!
 km  l 
n
k 1
 l 
 m
n!
n
 l 
 m
k !k
n k
k
P0
for n  k.
P0
for n > k.
n
47
M / M /k Queue - Performance Measures
k
 l  m
 m
1
W=
P +
2 0
m
(k  1) !(km l)
1  l   km 

 
Pw =
 P0
m


k!
 km  l 
k
l
r=
km
The performance measurements L, Lq, Wq,, can be obtained
from Little’s formulas.
48
LITTLE TOWN POST OFFICE
• Little Town post office is open on Saturdays between
9:00 a.m. and 1:00 p.m.
• Data
– On the average 100 customers per hour visit the office
during that period. Three clerks are on duty.
– Each service takes 1.5 minutes on the average.
– Poisson and Exponential distributions describe the
arrival and the service processes respectively.
49
LITTLE TOWN POST OFFICE
• The Postmaster needs to know the relevant
service measures in order to:
– Evaluate the current service level.
– Study the effects of reducing the staff by one clerk.
50
LITTLE TOWN POST OFFICE - Solution
• This is an M / M / 3 queuing system.
– Input
l = 100 customers per hour.
m = 40 customers per hour (60/1.5).
Does steady state exist (l < km )?
l = 100 < km = 3(40) = 120.
51
LITTLE TOWN POST OFFICE – solution
continued
• First P0 is found by
P0 =
1
n
3
1  100  1  100   3( 40) 


 + 
 

3!  40   3( 40)  100 
n= 0 n!  40 
1
=
= .045
6.25
1 + 2.5 +
+ 15.625
2
2
• P0 is used now to determine all the other
performance measures.
52
LITTLE TOWN POST OFFICE –
Spreadsheet Solution
M/M/k Queuing Model
INPUTS
Lambda =
Mu =
OUTPUTS
# Servers
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Value
100
40
INPUTS
Server Cost =
Goodwill Cost When Waiting =
Goodwill Cost While Being Served =
Value
L
Lq
W
Wq
Pw
Rho
Cost
6.011236
3.033095
2.630371
2.533889
2.50858
2.502053
2.50046
2.500096
2.500019
2.500003
2.500001
2.5
2.5
3.511236
0.533095
0.130371
0.033889
0.00858
0.002053
0.00046
9.59E-05
1.87E-05
3.4E-06
5.79E-07
9.28E-08
1.4E-08
0.060112
0.030331
0.026304
0.025339
0.025086
0.025021
0.025005
0.025001
0.025
0.025
0.025
0.025
0.025
0.035112
0.005331
0.001304
0.000339
8.58E-05
2.05E-05
4.6E-06
9.59E-07
1.87E-07
3.4E-08
5.79E-09
9.28E-10
1.4E-10
0.702247
0.319857
0.130371
0.047445
0.015443
0.004517
0.001195
0.000288
6.34E-05
1.29E-05
2.43E-06
4.27E-07
7.02E-08
0.833333
0.625
0.5
0.416667
0.357143
0.3125
0.277778
0.25
0.227273
0.208333
0.192308
0.178571
0.166667
0
0
0
0
0
0
0
0
0
0
0
0
0
Probabiility of n Customers
in the System where n =
0
1
0.044944
0.073695
0.0801
0.08162
0.08198
0.082063
0.082081
0.082084
0.082085
0.082085
0.082085
0.082085
0.082085
0.11236
0.184237
0.20025
0.204051
0.204951
0.205157
0.205201
0.20521
0.205212
0.205212
0.205212
0.205212
0.205212
2
0.140449
0.230297
0.250313
0.255063
0.256189
0.256446
0.256502
0.256513
0.256515
0.256516
0.256516
0.256516
53
0.256516
9.6 M/G/1 Queuing System
• Assumptions
– Customers arrive according to a Poisson process with a
mean rate l.
– Service time has a general distribution with mean rate m.
– One server.
– Infinite population, and possibly infinite line.
54
• Pollaczek - Khintchine Formula for L


l

(l) +  m
2
L=
2 1  l 
m

2
l
+
m
With  = standard deviation of the service time
Note: It is not necessary to know the particular service time distribution.
Only the mean and standard deviation of the distribution are needed.
55
TED’S TV REPAIR SHOP
• Ted’s repairs television sets and VCRs.
• Data
– It takes an average of 2.25 hours to repair a set.
– Standard deviation of the repair time is 45 minutes.
– Customers arrive at the shop once every 2.5 hours on the
average, according to a Poisson process.
– Ted works 9 hours a day, and has no help.
– He considers purchasing a new piece of equipment.
• New average repair time is expected to be 2 hours.
• New standard deviation is expected to be 40 minutes.
56
TED’S TV REPAIR SHOP
Ted wants to know the effects of using the new
equipment on –
1. The average number of sets waiting for repair;
2. The average time a customer has to wait
to get his repaired set.
57
TED’S TV REPAIR SHOP - Solution
• This is an M/G/1 system (service time is not
exponential (note that   1/m).
• Input
– The current system (without the new equipment)
l = 1/ 2.5 = 0.4 customers per hour.
m = 1/ 2.25 = 0.4444 costumers per hour.
 = 45/ 60 = 0.75 hours.
– The new system (with the new equipment)
m = 1/2 = 0.5 customers per hour.
 = 40/ 60 = 0.6667 hours.
58
9.7 M / M / k / F Queuing System
• Many times queuing systems have designs that
limit their line size.
• When the potential queue is large, an infinite
queue model gives accurate results, even
though the queue might be limited.
• When the potential queue is small, the limited
line must be accounted for in the model.
59
Characteristics of M/M/k/F Queuing System
• Poisson arrival process at mean rate l.
• k servers, each having an exponential service
time with mean rate m.
• Maximum number of customers that can be
present in the system at any one time is “F”.
• Customers are blocked (and never return) if
the system is full.
60
M/M/k/F Queuing System –
Effective Arrival Rate
•
•
•
A customer is blocked if the system is full.
The probability that the system is full is PF (100PF% of
the arriving customers do not enter the system).
The effective arrival rate = the rate of arrivals that
make it through into the system (le).
le = l(1 - PF)
61
RYAN ROOFING COMPANY
• Ryan gets most of its business from customers
who call and order service.
– When a telephone line is available but the secretary
is busy serving a customer, a new calling customer is
willing to wait until the secretary becomes available.
– When all the lines are busy, a new calling customer
gets a busy signal and calls a competitor.
62
RYAN ROOFING COMPANY
• Data
– Arrival process is Poisson, and service process is
Exponential.
– Each phone call takes 3 minutes on the average.
– 10 customers per hour call the company on the
average.
– One appointment secretary takes phone calls from
3 telephone lines.
63
RYAN ROOFING COMPANY
• Management would like to design the following system:
– The fewest lines necessary.
– At most 2% of all callers get a busy signal.
• Management is interested in the following information:
– The percentage of time the secretary is busy.
– The average number of customers kept on hold.
– The average time a customer is kept on hold.
– The actual percentage of callers who encounter a busy signal.
64
RYAN ROOFING COMPANY - Solution
M/M/1/4
system
• This isM/M/1/5
an M/M/1/system
3 system
• Input
l = 10 per hour.
See spreadsheet next
= 20 per
(1/3 per
P0 = m0.516,
P1 =hour
0.258,
P2 =minute).
0.129, P3 = 0.065, P4 = 0.032
• Excel spreadsheet gives:
P0 = 0.508, PP1 ==0.254,
PP2 == 0.133,
0.127,PP3= =0.06
0.063, P4 = 0.032
0.533,
1
3.2% of 0the customers
get the3 busy signal
P5 = 0.016
Still
above
the
goal
of
2%
6.7% of the
customers
get
a
busy
signal.
1.6% of the customers get the busy signal
The goal
of 2%
has been achieved.
This is above
the goal
of 2%.
65
RYAN ROOFING COMPANY Spreadsheet Solution
66
12.8 M / M / 1 / / m Queuing Systems
• In this system the number of potential customers is
finite and relatively small.
• As a result, the number of customers already in the
system affects the rate of arrivals of the remaining
customers.
• Characteristics
– A single server.
– Exponential service time, Poisson arrival process.
– A population size of a (finite) m customers.
67
PACESETTER HOMES
• Pacesetter Homes runs four different development projects.
• Data
– At each site running a project is interrupted once every 20 working
days on the average.
– The V.P. for construction handles each stoppage.
• How long on the average a site is non-operational?
– If it takes 2 days on the average to restart a project’s progress (the
V.P. is using the current car).
– If it takes 1.875 days on the average to restart a project’s progress
(the V.P. is using a new car)
68
PACESETTER HOMES – Solution
• This is an M/M/1//4 system, where:
– The four sites are the four customers.
– The V.P. for construction is the server.
• Input
l = 0.05
m = 0.5
(1/2 days, using the current car)
(1/20)
m = 0.533
(1/1.875 days, using a new car).
69
PACESETTER HOMES – Solution
continued
Summary of results
Performance
Measures
Average number of customers in the system
Average number of customers in the queue
Average number of days a customer is in the system
Average number of days a customer is in the queue
The probability that an arriving customer will wait
Oveall system effective utilization-factor
The probability that all servers are idle
L
Lq
W
Wq
Pw
r
Po
Current
New
Car
Car
0.467
0.113
2.641
0.641
0.353
0.353
0.647
0.435
0.100
2.437
0.562
0.334
0.334
0.666
70
PACESETTER HOMES –
Spreadsheet Solution
M/M/1//m Queuing Model
INPUTS
Lambda =
Mu =
m=
Value
0.05
0.53333
4
OUTPUTS
# Servers
L
Lq
W
Wq
Pw
Rho
1
0.43451 0.10025 2.43734 0.56233 0.33427 0.33427
Probabiility of n Customers
in the System where n =
0
1
2
3
4
0.66573 0.24965 0.07021 0.01317 0.00123
71
9.9 Economic Analysis of
Queuing Systems
• The performance measures previously developed
are used next to determine a minimal cost queuing
system.
• The procedure requires estimated costs such as:
– Hourly cost per server .
– Customer goodwill cost while waiting in line.
– Customer goodwill cost while being served.
72
WILSON FOODS
TALKING TURKEY HOT LINE
• Wilson Foods has an 800 number to answer
customers’ questions.
• If all the customer representatives are busy when a
new customer calls, he/she is asked to stay on the
line.
• A customer stays on the line if the waiting time is
not longer than 3 minutes.
73
WILSON FOODS
TALKING TURKEY HOT LINE
• Data
–
–
–
–
–
On the average 225 calls per hour are received.
An average phone call takes 1.5 minutes.
How many customer
A customer will stay on the line waiting at most 3 minutes.
service
representatives
A customer service representative is paid $16 per hour.
be used
Wilson paysshould
the telephone
company $0.18 per minute when the
minimize
customer is to
on hold
or whenthe
being served.
– Customer goodwill
is $20
minute while on hold.
hourlycost
cost
of per
operation?
– Customer goodwill cost while in service is $0.05.
74
WILSON FOODS – Solution
• The total hourly cost model
Total hourly wages
Average hourly goodwill
cost for customers on hold
TC(K) = Cwk + CtL + gwLq + gs(L - Lq)
Total average
hourly Telephone charge
Average hourly goodwill
cost for customers in service
TC(K) = Cwk + (Ct + gs)Lq + (Ct + gs)(L – Lq)
75
WILSON FOODS – Solution continued
• Input
Cw= $16
Ct = $10.80 per hour
[0.18(60)]
gw= $12 per hour
[0.20(60)]
gs = $3 per hour
[0.05(60)]
– The Total Average Hourly Cost =
TC(K) = 16K + (10.8+3)L + (12 - 3)Lq = 16K + 13.8L + 9Lq
76
WILSON FOODS – Solution continued
• Assuming a Poisson arrival process and an Exponential
service time, we have an M/M/K system.
l = 225 calls per hour.
m = 40 per hour (60/1.5).
– The minimal possible value for K is 6 to ensure that
steady state exists (l<Km).
– Excel MMk worksheet was used to generate results for
L, Lq, and Wq.
77
WILSON FOODS – Solution continued
• Summary of results of the runs for k=6,7,8,9,10
K
6
7
8
9
10
L
18.1255
7.6437
6.2777
5.8661
5.7166
Lq
12.5
2.0187
0.6527
0.2411
0.916
Wq
0.05556
0.00897
0.0029
0.00107
0.00041
TC(K)
458.64
235.65
220.51
227.12
239.70
Conclusion: employ 8 customer service representatives.
78
WILSON FOODS – Spreadsheet Solution
M/M/k Queuing Model
INPUTS
Lambda =
Mu =
OUTPUTS
# Servers
1
2
3
4
5
6
7
8
9
10
11
12
13
Value
225
40
INPUTS
Server Cost =
Goodwill Cost When Waiting =
Goodwill Cost While Being Served =
Value
16
22.8
13.8
L
Lq
W
Wq
Pw
Rho
Cost
18.1255
7.643727
6.277703
5.866105
5.716569
5.659381
5.637531
5.629393
12.5005
2.018727
0.652703
0.241105
0.091569
0.034381
0.012531
0.004393
0.080558
0.033972
0.027901
0.026072
0.025407
0.025153
0.025056
0.02502
0.055558
0.008972
0.002901
0.001072
0.000407
0.000153
5.57E-05
1.95E-05
0.833367
0.493467
0.275586
0.144663
0.07122
0.032853
0.014202
0.00576
0.9375
0.803571
0.703125
0.625
0.5625
0.511364
0.46875
0.432692
458.6364
235.652
220.5066
227.1222
239.7128
254.4089
269.9107
285.7252
Probabiility of n Customers
in the System where n =
0
1
0.001184
0.002742
0.003291
0.003492
0.003565
0.003592
0.003602
0.003605
0.006659
0.015423
0.018514
0.019641
0.020056
0.020206
0.02026
0.020278
2
0.01873
0.043376
0.05207
0.055241
0.056407
0.056831
0.056981
0.057032
79
HARGROVE HOSPITAL MATERNITY
WARD
• Hargrove Hospital is experiencing cutbacks, and is
trying to reorganize operations to reduce operating
costs.
• There is a trade off between
– the costs of operating more birthing stations and
– the costs of rescheduling surgeries in the surgery room when
women give birth there, if all the birthing stations are
occupied.
• The hospital wants to determine the optimal number of
birthing stations that will minimize operating costs.
80
HARGROVE HOSPITAL MATERNITY
WARD
• Data
–
–
–
–
Cutting one birthing station saves $25,000 per year.
Building a birthing station costs $30,000.
Maternity in the surgery room costs $400 per hour.
Six women on the average need a birthing station a
day. The arrival process is Poisson.
– Every birthing process occupies a birthing station for
two hours on the average.
81
HARGROVE HOSPITAL
• Solution
• Analysis of the current situation
– Currently there are two birthing stations
– The current problem can be modeled as a M/G/2/2 queuing
system.
– Using the MGkk Excel worksheet with l = 6 and m = 12/day we
have:
•
•
•
•
•
r = .23077
W = .0833 days
Pw = .076923
L = .46154
P0 = .6154
7.7% of the arriving women are sent to
the surgery room to give birth.
82
HARGROVE HOSPITAL
• Solution – continued
• The birthing stations problem can be modeled as a
M/G/k/k queuing model.
• Input
– l = 6 women per day;
m = 12 women per day (24/2);
– k = the number of birthing stations used
• The total cost for the hospital is
TC(k) = Cost of using the surgery room for maternity
+ Additional cost of operating k stations
83
HARGROVE HOSPITAL
• Solution – continued
• Average daily cost of using the surgery room for maternity:
Pk(l)(average time in the surgery room)(hourly cost)
• Average additional daily cost of operating k stations
25,000/365 = $68.49 per day.
• Average daily total cost
TC(k) = Pk(l)(24/m)(Hourly cost) + 68.49k
= Pk(6)(24/12)(400) + 68.49k
= 4800Pk + 68.49k
84
HARGROVE HOSPITAL - Solution
• From repeated runs of the MGkk worksheet to
determine Pk we got the following results:
Current
Optimal
k
Pk
$4800Pk
Additional
Cost of
stations
1
2
3
4
.333333
.076923
.012658
.001580
$1,600
369.23
60.76
7.58
$ – 68.49
0
82.19
163.38
Total net
average
Daily cost
$1,531.51
364.23
142.95
171.96
85
HARGROVE HOSPITAL –
Spreadsheet Solution
M/G/k/k Queuing Model
INPUTS
Lambda =
Mu =
k=
OUTPUTS
# Servers
L
2
0.461538
Value
6
12
2
Lq
0
W
0.083333
Wq
0
Pw
Rho
0.076923 0.230769
Probabiility of n Customers
in the System where n =
0
1
2
0.615385 0.307692 0.076923
3
86
9.10 Tandem Queuing Systems
• In a Tandem Queuing System a customer must visit
several different servers before service is completed.
Meats
Beverage
• Examples
– All-You-Can-Eat restaurant
87
9.10 Tandem Queuing Systems
• In a Tandem Queuing System a customer must visit
several different servers before service is completed.
Meats
Beverage
• Examples
– All-You-Can-Eat restaurant
88
9.10 Tandem Queuing Systems
• In a Tandem Queuing System a customer must visit
several different servers before service is completed.
Meats
Beverage
• Examples
– All-You-Can-Eat restaurant
– A drive-in restaurant, where first you place your order, then
pay and receive it in the next window.
– A multiple stage assembly line.
89
9.10 Tandem Queuing Systems
• For cases in which customers arrive according to
a Poisson process and service time in each
station is exponential, ….
Total Average Time in the System =
Sum of all average times at the individual stations
90
BIG BOYS SOUND, INC.
• Big Boys sells audio merchandise.
• The sale process is as follows:
– A customer places an order with a sales person.
– The customer goes to the cashier station to pay for
the order.
– After paying, the customer is sent to the pickup desk
to obtain the good.
91
BIG BOYS SOUND, INC.
• Data for a regular Saturday
– Personnel.
• 8 sales persons are on the job.
• 3 cashiers.
• 2 workers in the merchandise pickup area.
– Average service times.
• Average time a sales person waits on a customer is 10 minutes.
• Average time required for the payment process is 3 minutes.
• Average time in the pickup area is 2 minutes.
– Distributions.
• Exponential service time at all the service stations.
• Poisson arrival with a rate of 40 customers an hour.
92
BIG BOYS SOUND, INC.
Only 75% of the arriving customers make a purchase!
What is the average amount of time, a
customer who makes a purchase spends
in the store?
93
BIG BOYS SOUND, INC. – Solution
• This is a Three Station Tandem Queuing System
(.75)(40)=30
Sales Clerks
M/M/8
Pickup desk
M/M/2
Cashiers
M/M/3
W3 = 2.67 minutes
W1 = 14 minutes
W2 = 3.47 minutes
Total = 20.14 minutes.
94
Appendix : Assembly Line Balancing
• An Assembly Line can be thought of as a tandem
queue, because a product visits several workstations
in a given sequence.
• In a balanced assembly line the time spent in each of
the different workstations is about the same.
• The objective is to maximize production throughput
by allocating tasks to workstations
95
McMURRAY MACHINE COMPANY
• McMurray manufactures lawn mowers and snow
blowers.
• The assembly operation of a certain mower consists of
four workstations.
• The longest time spent at a workstation is 4 minutes.
Thus, the maximum number of mowers that can be
produced is 15 per hour.
• Management would like to increase productivity by better
96
balancing the assembly line.
• Data
Station
1
2
3
4
Average
Operations
Time (min)
Stamp mower body and handle; attach control bar to handle
2
Mount engine unit to body; attach backstop; blade; wheel; ….
4
Attach mower handle, control bar, cables, grass bag colar
3
Connect control bar; insert spark plug; quality inspection; packaging
3
• The entire operation takes 12 minutes
• Workstation 2 is a bottleneck.
97
SOLUTION
• Various options to balance the assembly line
– Try to schedule operations to take a total of three
minutes in each station .
– Assign workers to workstations as needed to
balance the station outputs.
– Assign multiple workstations to perform the operations.
98
Work
Station 1
Work
Station 2
Work
Station 3
Work
Station 4











Work flow





Workers

Product
99
– Use Integer and Dynamic Programming optimization
techniques, to minimize the total amount of idle time at all
workstations.
– Use an heuristic techniques such as “The Ranked Position
Weight Technique,” to find the smallest number of
workstations needed to meet a prespecified cycle time.
• (Although a heuristic solution does not guarantee optimality, this
heuristic was found optimal in large number of applications.)
100
– The Ranked Position Weight technique
1. For each task, find the total job time for all tasks for which that task
is a predecessor.
2. Rank the jobs in descending order of these total times.
3. Consider workstation 1 the current workstation.
4. Assign the lowest ranked task to the current station provided the
following conditions are met
– The task is not already assigned.
– The time at the current workstation (with the new task added) will not exceed
the desired cycle time.
5. If the second condition in step 4 is not met, designate a new
station as a current station, and assign the task there.
6. Repeat step 4 until all tasks have been assigned to workstations.
101
McMURRAY - continued
• Demand for the mower increased, and as a
result the needed cycle time drops to 3 minutes
in the assembly line.
• McMurray would like to balance the line using
the smallest number of workstations.
102
• Data
Tasks Required to Manufacture a Lawn Mower
Tasks
A--Stamp out mower body
B--Stamp out mower handle
C--Bolt engine to mower body
D--Attach back stop to mower body
E--Attach left wheels and handle latch
F--Attach right wheels and handle latch
G--Attach grass bag collar
H--Attach blade
I--Attach throttle cable to engine
J--Attach control bar cable to engine
K--Insert spark club
L--Attach control bar to handle
M--Attach handle to mower
N--Attach throttle cable to handle
0--Attach control bar cable to control bar
P--Test mower
Q--Pack mower
Must Follow Job
None
None
A
A
C
C
E
C
F, G
I
H
B
L
M
None
D, J, K, O
P
Job Time
40
50
55
30
65
65
30
25
35
50
20
30
20
50
45
50103
60
SOLUTION
Tasks
Time
A--Stamp… 40
B--Stamp… 50
C--Bolt …
55
D--Attach … 30
E--Attach… 65
F--Attach… 65
G--Attach… 30
H--Attach… 25
I--Attach…
35
J--Attach… 50
K--Insert… 20
L--Attach… 30
M--Attach… 20
N--Attach… 50
0--Attach… 45
P--Test…
50
Q--Pack…
60
Remaining tasks
C,D,E,F,G,H,I,J,K,P,Q
L,M,N,O,P,Q
E,F,G,H,I,J,KP,Q
P,Q
G,I,J,P,Q
I,J,P,Q
I,J,P,Q
K,P,Q
J,P,Q
P,Q
P,Q
M,N,P,O,Q
N,O,P,Q
O,P,Q
P,Q
Q
None
Total Time
525
305
455
140
290
260
225
155
195
160
130
255
225
205
155
110
60
Rank
1
3
2
14
4
5
6.5
12.5
10
11
15
6.5
8
9
12.5
16
17
104
• Step 1 and 2 Tasks sorted by ranks
Tasks
Time
A--Stamp… 40
B--Stamp… 50
C--Bolt …
55
D--Attach … 30
E--Attach… 65
F--Attach… 65
G--Attach… 30
H--Attach… 25
I--Attach…
35
J--Attach… 50
K--Insert… 20
L--Attach… 30
M--Attach… 20
N--Attach… 50
0--Attach… 45
P--Test…
50
Q--Pack…
60
Remaining tasks
C,D,E,F,G,H,I,J,K,P,Q
L,M,N,O,P,Q
E,F,G,H,I,J,KP,Q
P,Q
G,I,J,P,Q
I,J,P,Q
I,J,P,Q
K,P,Q
J,P,Q
P,Q
P,Q
M,N,P,O,Q
N,O,P,Q
O,P,Q
P,Q
Q
None
Total Time
525
305
455
140
290
260
225
155
195
160
130
255
225
205
155
110
60
Rank
1
3
2
14
4
5
6.5
12.5
10
11
15
6.5
8
9
12.5
16
17
105
• Step 3 and 4
Workstation
1
1
1
1
2
Rank
1
2
3
4
4
And so on….
Task
Time
Total time
A
40
40
95
ActualC cycle time 55
= 170.
B be reduced
50 to 160, 145
This can
E
65
210
by moving
“K” from
E
65 station 4 65
to station 5.
Idle Time
140
85
35
Negative
115
Workstation Assignment for Lawn Mower Production
Workstation
1
2
3
4
5
Tasks
A,B,C
E,F,G
I,L,M,N
D,H,J,K
P,Q
Total Time
145
160
135
170
110
Idle Time
35
20
45
10
70
106