Waiting Line SystemsQueuing Models

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BUAD 306 Operations Management William V.
Gehrlein University of Delaware
Waiting Line Systems/Queuing Models
Part 1 Introduction through Evaluating Data to
Determine if the Assumptions that we Need to make
to use Basic Queuing Models are Valid
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University of Delaware
WL-1 Components of a Waiting Line System
Waiting Line (Queue)
Population
Service Facility
Channel 1
Channel 2
Arrivals
Exit
Channel M -1
Channel M
Arrival Pattern
Queue Features
Selection Discipline
Service Facility
Service Time
Population
- Finite - Infinite

• Random -Scheduled

- Max Line Length
- FCFS
- Channels
- Random
- Phases
- Constant
- Priority
- Number of Lines
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University of Delaware
WL-2 Distribution of Random Arrivals
Expected Product
Number of Arrivals (n) During a 1 Minute Interval
Frequency from 100 Different Intervals
Probability
0
6
.06
.00
1
13
.13
.13
2
22
.22
.44
3
24
.24
.72
4
18
.18
.72
5
8
.08
.40
6
5
.05
.30
7
3
.03
.21
8
1
.01
.08
1.00
E(n) 3.00/min.
General Shape of a Poisson Distribution
Parameter E(n)
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University of Delaware
WL-3 Poisson Distribution Assumption for Random
Arrivals
With a Poisson Distribution, the Probability that
We Observe n Arrivals for a Given Time Period is
given by
is the Expected Arrival Rate with
The Given Time Period is determined by the
Units of E(n).
Using the data from our example,
Poisson Probability
Observed Probability
5
University of Delaware
WL-4 Distribution of Service Times
P(t) is the probability it takes time t or longer
for service
Service Time Interval for 100 Customers (Minutes)
Interval Midpoint
Interval Probability
Expected Product
Start (t)
Finish
Frequency
P(t)
0
.50
40
.40
1.00
.100
.25
.50
1.00
.24
24
.60
.180
.75
1.00
1.50
12
.12
.36
.150
1.25
1.50
2.00
8
.08
.24
.140
1.75
2.00
2.50
7
.07
.16
.158
2.25
2.50
3.00
5
.05
.09
.138
2.75
3.00
3.50
3
.03
.04
.098
3.25
3.50
4.00
1
.01
.01
.038
3.75
E(t) 1.000 min.
General shape of a Negative Exponential
Distribution
One parameter is based on Expected Time for
Service E(t)
6
University of Delaware
WL-5 Negative Exponential Distribution
Assumption for Service Times
With the Assumption of a Negative Exponential
Distribution, the Probability that it Takes a
Given Time t or Longer to Service an Item is
Given by
is the expected service rate, with
For the sample of 100 customers, with
Negative Exponential Probability
Observed Probability
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BUAD 306 Operations Management William V.
Gehrlein University of Delaware
Waiting Line Systems/Queuing Models
Part 2 Queuing Models and Examples of Their Use
8
University of Delaware
WL-6 Properties of Some Specific Queuing Models
Source Arrival
Service Model Channels Population
Pattern Pattern 1 Single
Infinite Poisson Negative Exponential
2 Single Infinite Poisson
Constant 3 Multiple
Infinite Poisson Negative Exponential
4 Single/ Finite Poisson Negative
Exponential Multiple
All Models are Single Phase, FCFS Queue
Discipline, with Unlimited Queue Length
For Models 1, 2 and 3
M Number of Channels
Arrival Rate
Service Rate per Channel
Average Time in Line (Queue)
Average Time in System
Average Number in Line (Queue)
Average Number in System
Utilization
Probability there are n in the system
9
University of Delaware
WL-7 General Relationships for M Channels
For Models 1, 2 and 3
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University of Delaware
WL-8 Example 1
American Vending Inc. supplies vended food to a
large university. Because students kick the
machines at every opportunity out of anger and
frustration, management has a constant repair
problem. Machines break down at an average rate
of three machines per hour, and the number of
breakdowns per hour follows a Poisson
distribution. Downtime costs the company
25/hour per machine for all of the time that the
equipment is not operating, including the time
during which it is being repaired. Maintenance
workers who repair the machines are paid
12/hour. One worker can service machines at a
rate of five per hour, with service times
following a negative exponential distribution.
Two workers, working as a single team, can
service seven machines per hour and three
workers can service eight machines per hour, with
all service times following a negative
exponential distribution. Based on expected
total cost per hour, what crew size should be
used to minimize the expected cost for this
waiting line system.
11
University of Delaware
WL-9 Model 1
One Channel, Infinite Source Population, Poisson
Arrivals, Negative Exponential Service Times
From General Relationships for
12
University of Delaware
WL-10 Example 1 - Solution
Total Cost/Hr. Labor Cost/Hr. Downtime
Cost/Hr.
Total Cost/Hr.
12/Hr.
25/Hr.
Workers

Case with 1 Worker
Total Cost/Hr. 1(12.00) 1.5(25.00)
49.50/Hr.
Case with 2 Workers
Total Cost/Hr. 2(12.00) .75(25.00)
42.75/Hr.
Case with 3 Workers
Total Cost/Hr. 3(12.00) .60(25.00)
51.00/Hr.
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University of Delaware
WL-11 Example 2
A cafeteria serving line for a large
facility has one self-serve coffee urn. People
arrive into the system at a rate of 3 per minute,
with the arrival rates following a Poisson
distribution. With customers serving themselves,
the service times have a negative exponential
distribution, with an average time of 15
seconds. a. How many customers would you
expect to see on average in the system? b.
What is the average time in system for each
customer? c. What percent of the time is the
urn being used? d. If the cafeteria installs
an automatic vendor with a constant service time
of 15 seconds, how does this change the answers
to questions a and b above?
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University of Delaware
WL-12 Example 2 - Solution
Model 1
a.)
b.)
From General Relationships for
From General Relationships for
c.)
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University of Delaware
WL-13 Example 2 - Solution
Model 2
One Channel, Infinite Source Population,
Poisson Arrival Pattern, Constant Service Time
d.)
From General Relationships for and
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University of Delaware
WL-14 Example 3
A bank manager is trying to determine the
least total cost staffing option for the three
drive-in teller stations at a branch of the bank.
Customers arrive at a rate of 15 per hour, with
a Poisson arrival pattern. Customers are
serviced on average in 2 minutes, with a negative
exponential distribution of service times. Total
costs for staffing are 15 per hour for each
station that is open. Customer goodwill costs
are 90 per hour for each customer, during the
entire time that they are in the system. Based
on expected total cost per hour, how many
stations should be staffed?
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University of Delaware
WL-15 Example 3 - Solution
Evaluate with One Teller
Model 1
Cost/Hr. 15.00(1)90.00
Cost/Hr. 15.00(1)90.00(1.0) 105/Hr.
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University of Delaware
WL-16 Example 3 - Solution
Evaluate with Two Tellers
Model 3
M 2
From General Relationships for
Total Cost 15.00(2) 90.00(.533) 78.00/Hr.
Evaluate with Three Tellers
M 3
Total Cost 15.00(3) 90.00(.503) 90.27/Hr.
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Multiple Channel Queuing Table
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University of Delaware
WL-17 Model 4 - Finite
Population
M Number of Channels
N Population Size
T Average Time to Perform the Service
U Average Time Working Before a Service
Request is Made
X Service Factor
F Efficiency Factor - Comes from a Table, for
given N, M and X
L Average Number of Units in Line
L N(1-F)
H Average Number of Units Being Serviced
H FNX
J Average Number of Units Running
JNF(1-X)
W Average Waiting Time in Line
D Probability that an Arrival Must Wait in
Line - From Table
21
University of Delaware
WL-18 Example 4
An engineering firm retains a technical
specialist to assist five design engineers. The
help that the specialist provides varies widely
in terms of time for providing the service. The
specialist can provide some answers from memory,
but others require a substantial amount of work
time. Service times follow a negative
exponential distribution, and the average amount
of time required is one hour. The engineers
require help from the specialist on an average of
once each day. Since the average time for
assistance is one hour, the engineers can be
expected to work for seven hours without
assistance. Requests for service follow a
Poisson distribution. In addition, engineers in
need of help do not interrupt if the specialist
is already involved with another problem. a. How
many engineers, on average, are waiting for the
technical specialist for help? b. What is
the average amount of time that an engineer has
to wait for help from a specialist? c.
What is the probability that an engineer will
have to wait in line for the
specialist?
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University of Delaware
WL-19 Example 4 - Solution
Model 4
M 1
N 5
T 1 Hr.
U 7 Hrs.
Step 1 Calculate the Service Factor
Step 2 Get F from Finite Queuing Table
For X .125, M 1, N5 we get F .920
a.) Average Number Waiting in Line L
L N(1-F) 5(1-.920) .40
b.) Average Time Waiting in Line W
c.) Probability That Arrival Must Wait in Line
D
From Finite Queuing Table, D .473
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