© the mcgraw-hill companies, inc., 2004 1. 2 technical note 6 waiting line management
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©The McGraw-Hill Companies, Inc., 2004
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Technical Note 6
Waiting Line Management
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• Waiting Line Characteristics
• Suggestions for Managing Queues
• Examples (Models 1, 2, 3, and 4)
OBJECTIVES
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Components of the Queuing System
CustomerArrivals
Servers
Waiting Line
Servicing System
Exit
Queue or
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Customer Service Population Sources
Population Source
Finite Infinite
Example: Number of machines needing repair when a company only has three machines.
Example: Number of machines needing repair when a company only has three machines.
Example: The number of people who could wait in a line for gasoline.
Example: The number of people who could wait in a line for gasoline.
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Service Pattern
ServicePattern
Constant Variable
Example: Items coming down an automated assembly line.
Example: Items coming down an automated assembly line.
Example: People spending time shopping.
Example: People spending time shopping.
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The Queuing System
Queue Discipline
Length
Number of Lines &Line Structures
Service Time Distribution
Queuing System
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Examples of Line Structures
Single Channel
Multichannel
SinglePhase Multiphase
One-personbarber shop
Car wash
Hospitaladmissions
Bank tellers’windows
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Degree of Patience
No Way!
BALK
No Way!
RENEG
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Suggestions for Managing Queues
1. Determine an acceptable waiting time for your customers
2. Try to divert your customer’s attention when waiting
3. Inform your customers of what to expect
4. Keep employees not serving the customers out of sight
5. Segment customers
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Suggestions for Managing Queues (Continued)
6. Train your servers to be friendly
7. Encourage customers to come during the slack periods
8. Take a long-term perspective toward getting rid of the queues
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Waiting Line Models
Model LayoutSourcePopulation Service Pattern
1 Single channel Infinite Exponential
2 Single channel Infinite Constant
3 Multichannel Infinite Exponential
4 Single or Multi Finite Exponential
These four models share the following characteristics: Single phase Poisson arrival FCFS Unlimited queue length
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Notation: Infinite Queuing: Models 1-3
linein tingnumber wai Average
server single afor
rate sevice torate arrival totalof Ratio = =
arrivalsbetween timeAverage
timeservice Average
rate Service =
rate Arrival =
1
1
Lg
linein tingnumber wai Average
server single afor
rate sevice torate arrival totalof Ratio = =
arrivalsbetween timeAverage
timeservice Average
rate Service =
rate Arrival =
1
1
Lg
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14Infinite Queuing Models 1-3 (Continued)
linein waitingofy Probabilit
systemin units exactly ofy Probabilit
channels service identical ofNumber =
system in the units ofNumber
served) be to time(including
systemin time totalAverage
linein waiting timeAverage = g
served) being those(including
systemin number Average = s
Pw
nPn
S
n
Ws
W
L
linein waitingofy Probabilit
systemin units exactly ofy Probabilit
channels service identical ofNumber =
system in the units ofNumber
served) be to time(including
systemin time totalAverage
linein waiting timeAverage = g
served) being those(including
systemin number Average = s
Pw
nPn
S
n
Ws
W
L
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Example: Model 1Assume a drive-up window at a fast food restaurant.Customers arrive at the rate of 25 per hour.The employee can serve one customer every two minutes.Assume Poisson arrival and exponential service rates.
Determine:A) What is the average utilization of the employee?B) What is the average number of customers in line?C) What is the average number of customers in the system?D) What is the average waiting time in line?E) What is the average waiting time in the system?F) What is the probability that exactly two cars will be in the system?
Determine:A) What is the average utilization of the employee?B) What is the average number of customers in line?C) What is the average number of customers in the system?D) What is the average waiting time in line?E) What is the average waiting time in the system?F) What is the probability that exactly two cars will be in the system?
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= 25 cust / hr
= 1 customer
2 mins (1hr / 60 mins) = 30 cust / hr
= = 25 cust / hr
30 cust / hr = .8333
= 25 cust / hr
= 1 customer
2 mins (1hr / 60 mins) = 30 cust / hr
= = 25 cust / hr
30 cust / hr = .8333
Example: Model 1
A) What is the average utilization of the employee?
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Example: Model 1
B) What is the average number of customers in line?
4.167 = 25)-30(30
(25) =
) - ( =
22
Lg 4.167 = 25)-30(30
(25) =
) - ( =
22
Lg
C) What is the average number of customers in the system?
5 = 25)-(30
25 =
- =
Ls 5 = 25)-(30
25 =
- =
Ls
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Example: Model 1
D) What is the average waiting time in line?
mins 10 = hrs .1667 =
=
LgWg mins 10 = hrs .1667 =
=
LgWg
E) What is the average waiting time in the system?
mins 12 = hrs .2 = =Ls
Ws mins 12 = hrs .2 = =Ls
Ws
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Example: Model 1
F) What is the probability that exactly two cars will be in the system (one being served and the other waiting in line)?
p = (1-n
n
)( )p = (1-n
n
)( )
p = (1- = 2
225
30
25
30)( ) .1157p = (1- =
2
225
30
25
30)( ) .1157
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Example: Model 2
An automated pizza vending machine heats and dispenses a slice of pizza in 4 minutes.
Customers arrive at a rate of one every 6 minutes with the arrival rate exhibiting a Poisson distribution.
Determine:
A) The average number of customers in line.B) The average total waiting time in the system.
Determine:
A) The average number of customers in line.B) The average total waiting time in the system.
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Example: Model 2
A) The average number of customers in line.
.6667 = 10)-(2)(15)(15
(10) =
) - (2 =
22
Lg .6667 = 10)-(2)(15)(15
(10) =
) - (2 =
22
Lg
B) The average total waiting time in the system.
mins 4 = hrs .06667 = 10)-51)(15(2
10 =
) - (2 =
Wg mins 4 = hrs .06667 = 10)-51)(15(2
10 =
) - (2 =
Wg
mins 8 = hrs .1333 = 15/hr
1 + hrs .06667 =
1 + =
WgWs mins 8 = hrs .1333 = 15/hr
1 + hrs .06667 =
1 + =
WgWs
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Example: Model 3Recall the Model 1 example:Drive-up window at a fast food restaurant.Customers arrive at the rate of 25 per hour.The employee can serve one customer every two minutes.Assume Poisson arrival and exponential service rates.
If an identical window (and an identically trained server) were added, what would the effects be on the average number of cars in the system and the total time customers wait before being served?
If an identical window (and an identically trained server) were added, what would the effects be on the average number of cars in the system and the total time customers wait before being served?
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Example: Model 3Average number of cars in the system
ion)interpolatlinear -using-TN6.10(Exhibit
1760= .Lgion)interpolatlinear -using-TN6.10(Exhibit
1760= .Lg
1.009 = 30
25 + .176 = + =
LgLs 1.009 = 30
25 + .176 = + =
LgLs
Total time customers wait before being served
)( = mincustomers/ 25
customers .176 = = Wait! No
LgWg mins .007
)( =
mincustomers/ 25
customers .176 = = Wait! No
LgWg mins .007
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Notation: Finite Queuing: Model 4
channels service ofNumber
linein units ofnumber Average
)( system
queuingin thoseless source Population =
served being units ofnumber Average
linein wait tohaving
ofeffect theof measure a factor, Efficiency
linein must wait arrivalan y that Probabilit =
S
L
n-N
J
H
F
D
channels service ofNumber
linein units ofnumber Average
)( system
queuingin thoseless source Population =
served being units ofnumber Average
linein wait tohaving
ofeffect theof measure a factor, Efficiency
linein must wait arrivalan y that Probabilit =
S
L
n-N
J
H
F
D
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Finite Queuing: Model 4 (Continued)
required timeservice of proportionor factor, Service
linein time waitingAverage
tsrequiremen servicecustomer between timeAverage
service theperform to timeAverage =
system queuingin units exactly ofy Probabilit
source populationin units ofNumber
served) being one the(including
system queuingin units ofnumber Average =
X
W
U
T
nPn
N
n
required timeservice of proportionor factor, Service
linein time waitingAverage
tsrequiremen servicecustomer between timeAverage
service theperform to timeAverage =
system queuingin units exactly ofy Probabilit
source populationin units ofNumber
served) being one the(including
system queuingin units ofnumber Average =
X
W
U
T
nPn
N
n
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Example: Model 4
The copy center of an electronics firm has four copymachines that are all serviced by a single technician.
Every two hours, on average, the machines require adjustment. The technician spends an average of 10minutes per machine when adjustment is required.
Assuming Poisson arrivals and exponential service, how many machines are “down” (on average)?
The copy center of an electronics firm has four copymachines that are all serviced by a single technician.
Every two hours, on average, the machines require adjustment. The technician spends an average of 10minutes per machine when adjustment is required.
Assuming Poisson arrivals and exponential service, how many machines are “down” (on average)?
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Example: Model 4N, the number of machines in the population = 4M, the number of repair people = 1T, the time required to service a machine = 10 minutesU, the average time between service = 2 hours
X =T
T + U
10 min
10 min + 120 min= .077X =
T
T + U
10 min
10 min + 120 min= .077
From Table TN6.11, F = .980 (Interpolation)From Table TN6.11, F = .980 (Interpolation)
L, the number of machines waiting to be serviced = N(1-F) = 4(1-.980) = .08 machines
L, the number of machines waiting to be serviced = N(1-F) = 4(1-.980) = .08 machines
H, the number of machines being serviced = FNX = .980(4)(.077) = .302 machines
H, the number of machines being serviced = FNX = .980(4)(.077) = .302 machines
Number of machines down = L + H = .382 machinesNumber of machines down = L + H = .382 machines
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Queuing Approximation• This approximation is quick way to analyze a queuing situation. Now, both
interarrival time and service time distributions are allowed to be general.• In general, average performance measures (waiting time in queue, number
in queue, etc) can be very well approximated by mean and variance of the distribution (distribution shape not very important).
• This is very good news for managers: all you need is mean and standard deviation, to compute average waiting time
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Define:
Standard deviation of Xcoefficient of variation for r.v. X =
Mean of XVariance
squared coefficient of variation (scv) = mean
x
x x
C
C C
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Queue Approximation
2( 1) 2 2
1 2
Sa s
q
C CL
s qL L S
Compute S
2 2,a sC CInputs: S, , ,
(Alternatively: S, , , variances of interarrival and service time distributions)
as before, , and q sq s
L LW W
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Approximation Example• Consider a manufacturing process (for example making plastic parts) consisting of a single stage with
five machines. Processing times have a mean of 5.4 days and standard deviation of 4 days. The firm operates make-to-order. Management has collected date on customer orders, and verified that the time between orders has a mean of 1.2 days and variance of 0.72 days. What is the average time that an order waits before being worked on?
Using our “Waiting Line Approximation” spreadsheet we get:Lq = 3.154 Expected number of orders waiting to be completed.
Wq = 3.78 Expected number of days order waits.
Ρ = 0.9 Expected machine utilization.
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End of Technical Note 6