COB 291
Introduction to Management Science
Michael E. Busing
CIS/OM Program, James Madison University
What is Management Science?
 field
of study that uses computers, statistics and
mathematics to solve business problems.
 sometimes referred to as operations research or
decision science.
 formerly, the field was available to only those with
advanced knowledge of mathematics and computer
programming languages. PC’s and spreadsheets have
made the tools available to a much larger audience.
Air New Zealand, “Optimized Crew Scheduling at Air
New Zealand”
The airline crew scheduling problem consists of the pairings
problem involving the generation of minimum-cost pairings
(sequences of duty periods) to cover all scheduled flights, and the
rostering problem involving the assignment of pairings to
individual crewmembers. Over the past fifteen years, eight
application-specific optimization-based computer systems have
been developed in collaboration with the University of Auckland to
solve all aspects of the pairings and rostering processes for Air
New Zealand's National and International operations. These
systems have produced large savings, while also providing crew
rosters that better respect crew preferences.
Federal Aviation Administration, “Ground Delay
Program Enhancements (GDPE) under
Collaborative Decision Making (CDM)”
When airport arrival capacity is reduced, the demand placed by
arriving aircraft may be too great. In these cases a ground delay
program (GDP) is used to delay flights before departure at their
origin airport, keeping traffic at an acceptable level for the
impacted arrival airport. However, GDPs sometimes lacked
current data and a common situational awareness. Working with
the FAA and airline community, Metron, Inc. and Volpe National
Transportation Systems Center improved the process by utilizing
real-time data exchange between all users, new algorithms to
assign flight arrival slots, and new software in place at FAA
facilities and airlines.
Fingerhut Companies, Inc., “Mail Stream
Optimization”
Fingerhut mails up to 120 catalogs per year to each of its 6 million
customers. With this dense mail plan and independent mailing
decisions, many customers were receiving redundant and
unproductive catalogs. To find and eliminate these unproductive
catalogs, optimization models were developed to select the
optimal sequence of catalogs, called a "mail stream", for each
customer. With mail streams, Fingerhut is able to make mailing
decisions at the customer level as well as increase profits. Today,
this application runs weekly using current data to find the most
profitable mail stream for each of its 6 million customers.
Ford, “Rightsizing and Management of Prototype
Vehicle Testing at Ford Motor Company”
Prototype vehicles are used to verify new designs and represent a
major annual investment at Ford Motor Company. Engineering
managers studying in a Wayne State University master's degree
program adapted a classroom set covering example and launched
the development of the Prototype Optimization Model (POM). The
POM is used for both operational and strategic planning. The
modeling approach was lean and rapid and was designed to
maintain the role of the experienced manager as the ultimate
decision-maker.
IBM, “Matching Assets with Demand in Supply
Chain Management at IBM Microelectronics”
The IBM Microelectronics Division is a leading-edge producer of
semiconductor and packaged solutions supplying a wide range of
customers inside and outside IBM. A critical component of
customer responsiveness is matching assets with demand to
correctly assess anticipated supplies linked with demand and
provide manufacturing guidelines. A suite of tools was developed
to handle matching in a division-wide "best can do", division wide
ATP, and daily individual manufacturing location MRPs. The key
modeling advance is the dynamic interweaving of linear
programming, traditional MRP explosion and implosion-based
heuristics and the ability to harness deep computing to solve large
linear programming problems.
For More Information
http://www.lionhrtpub.com/ORMS.shtml
http://www.informs.org/
http://dsi.gsu.edu/
Application to InfoSec?
Linear
Programming
Queuing Theory/Simulation
Project management
Forecasting
Decision Analysis/Decision Trees
A Visual Model of the
Problem-Solving Process
Identify
Problem
Formulate and
Implement
Model
Analyze
Model
unsatisfactory results
Test
Results
Implement
Solution
Linear Programming
 Problems
characterized by
 limited
resources.
 decisions about how best to utilize the limited resources
available to an individual or a business.
 maximization or minimization of profits or costs.
 Mathematical
Programming (MP) is a field of
management science that finds the optimal, or most
efficient, way of using limited resources to achieve the
objectives of an individual or a business. MP is
sometimes referred to as optimization.
Optimization Example
Seuss’s Sandwich Shop sells two types of sandwiches: green
eggs and ham (GEH) and ham and cheese (HC). A green eggs
and ham sandwich consists of 2 slices bread, 1 green egg, and 1
slice ham. It takes an employee 3 minutes to make one of these
sandwiches. A ham and cheese sandwich consists of 2 slices
bread, 2 slices ham and 1 slice cheese. It takes 2 minutes to
make a ham and cheese sandwich. The sandwich shop presently
has available 400 slices of bread, 80 slices cheese, 120 green
eggs and 200 slices of ham. The shop also has one employee
scheduled for 7 hours to make all of the sandwiches. If a green
egg and ham sandwich sells for $5 and a ham and cheese
sandwich sells for $4, how many of each type should be prepared
to maximize revenue? (Assume that demand is great enough to
ensure that all sandwiches made will be sold).
220
N
u
m
b
e
r
200
o
f
180
160
140
120
100
H
a
m
a
n
d
C
h
e
e
s
e
80
(H
C)
60
t
o
40
M
a
k
e
20
0
20
40
60
80
100
120
140
160
Number of Green Eggs and Ham Sandwiches (GEH) to Make
MAX 5 GEH + 4 HC
Subject to
2 GEH + 2 HC < 400
HC < 80
GEH
< 120
GEH + 2 HC < 200
3 GEH + 2 HC < 420
GEH, HC > 0
(BREAD)
(CHEESE)
(GREEN EGGS)
(HAM)
(TIME)
180
200
220
Post-Optimality Analysis
 Range
of Optimality – tells range that a decision
variable’s coefficient can take on in the objective function
without affecting current optimal solution (note that the
objective function value WILL change).
 Shadow/dual price – tells how receiving additional units
of a resource affects the objective function value (for <
constraints). Also tells how requiring more of something
(for > constraints) affects the objective function value.
Note that the changes in right hand side values are only
good for a relevant range.
220
N
u
m
b
e
r
200
o
f
180
160
140
120
100
H
a
m
a
n
d
C
h
e
e
s
e
80
(H
C)
60
t
o
40
M
a
k
e
20
0
20
40
60
80
100
120
140
160
Number of Green Eggs and Ham Sandwiches (GEH) to Make
MAX 5 GEH + HC (REVISED)
Subject to
2 GEH + 2 HC < 400
(BREAD)
HC < 80
(CHEESE)
GEH
< 120
(GREEN EGGS)
GEH + 2 HC < 200
(HAM)
3 GEH + 2 HC < 420
(TIME)
GEH, HC > 0
180
200
220
220
N
u
m
b
e
r
200
o
f
180
160
140
120
100
H
a
m
a
n
d
C
h
e
e
s
e
80
(H
C)
60
t
o
40
M
a
k
e
20
0
20
40
60
80
100
120
140
160
180
200
Number of Green Eggs and Ham Sandwiches (GEH) to Make
MAX 5 GEH + 4 HC
Subject to
2 GEH + 2 HC < 400
HC < 80
GEH
< 120
GEH + 2 HC < 220
3 GEH + 2 HC < 420
GEH, HC > 0
(BREAD)
(CHEESE)
(GREEN EGGS)
 (HAM REVISED)  
(TIME)
220
Spreadsheet Solution to Linear
Programming Problems
Green Eggs and Ham Production Problem (Zeuss' Sandwich Shop)
Decision Variables
No. to produce:
Green Eggs and Ham
0
Ham and Cheese
0
Objective
Unit Profits:
Max
5
4
Total Profit
=SUMPRODUCT(B8:C8,B5:C5)
Constraints:
Bread (slices)
Cheese (slices)
Eggs (units)
Ham (slices)
Time (minutes)
2
0
1
1
3
2
1
0
2
2
Used
=SUMPRODUCT(B12:C12,$B$5:$C$5)
=SUMPRODUCT(B13:C13,$B$5:$C$5)
=SUMPRODUCT(B14:C14,$B$5:$C$5)
=SUMPRODUCT(B15:C15,$B$5:$C$5)
=SUMPRODUCT(B16:C16,$B$5:$C$5)
Available
400
80
120
200
420
Other Applications in LP
1.
(Operations planning) Diagnostic Corporation assembles two types of electronic
calculators. The DC1 Calculator provides basic math functions, while the DC2 also
provides trigonometric computations needed by engineers. Due to a winter blizzard,
incoming shipments of components have been delayed. Ron Beckman, Manager of the
plant, has assembled his production staff to plan an appropriate response. Bob Driscoll, in
charge of supplies, reports that only three items are in short supply and likely to run out
before new shipments arrive. The items in short supply are diodes (16,000 available),
digital displays (10,000 available), and resistors (18,000 on hand). The quantities of each
of these components that are required by each calculator are shown below. Mr. Beckman
states that he would like to plan production to maximize the profits that can be realized
using the available supply of parts.
Number of Parts Required
Profit Per
Model
Diodes Displays Resistors
Calculator
DC1
6
2
6
$10
DC2
4
3
1
$12
Other Applications in LP
2.
(Sales promotion)
Riverside Auto wants to conduct an advertising campaign
where each person who comes to the lot to look at a car receives $1. To advertise this
campaign, Riverside can buy time on two local TV stations. The advertising agency has
provided Riverside with the following data:
Cost per
Number of
Number of
Maximum
Station
Ad
Serious Buyers
Freeloaders
Spots
Ch. 47
$90
80
480
20
Ch. 59
$100
100
360
20
Riverside wants to minimize costs, while giving away a maximum of $15,000, and ensuring
that the ads attract at least 2400 serious buyers. Formulate Riverside’s problem as a linear
program.
Other Applications in LP
3.
(Staff Scheduling)
Palm General Hospital is concerned with the staffing of
its emergency department. A recent analysis indicated that a typical day may be divided
into six periods with the following requirements for nurses:
Time Period
7AM – 11AM
11AM – 3PM
3PM – 7PM
7PM – 11PM
11PM – 3AM
3AM – 7AM
Nurses Needed
8
6
12
6
4
2
Other Applications in LP
4.
(Agriculture) A farming organization operates 3 farms of comparable productivity. The output of
each farm is limited both by the usable acreage and by the amount of water available for irrigation.
Data for the upcoming season are as follows:
Farm
1
2
3
Usable Acreage
400
600
300
Water Available (Acre ft.)
1500
2000
900
The organization is considering 3 crops for planting, which differ in their expected profit/acre and in
their consumption of water. Furthermore, the total acreage that can be devoted to each of the crops
is limited by the amount of appropriate harvesting equipment available:
Max
Crop Acreage
A
700
B
800
C
300
Water Consumption
Expected Profit
in Acre ft./Acre
Per Acre
5
$400
4
$300
3
$100
In order to maintain a uniform workload among farms, the percentage of usable acreage planted
must be the same at each farm. However, any combination of crops may be grown at any of the
farms. Hw much of each crop should be planted at each farm to maximize expected profit?
Other Applications in LP
5. (Blending)
Suppose that an oil refinery wishes to blend 4 petroleum constituents into 3 grades of
gasoline A, B, and C. The availability and costs of the 4 constituents are as follows:
Constituent
W
X
Y
Z
Max Availability (bbls/day)
3,000
2,000
4,000
1,000
Cost per Barrel
$3
$6
$4
$5
To maintain the required quality for each grade of gasoline it is necessary to maintain the following
maximum and/or minimum percentages of the constituents in each blend. Determine the mix of the 4
constituents that will maximize profit.
Grade
A
B
C
Specification
no more than 30% of W
at least 40% of X
no more than 50% of Y
no more than 50% of W
at least 10% of X
no more than 70% of W
Selling Price per Barrel
$5.50
$4.50
$3.50
Queuing Theory/Simulation
 Probabilistic
Models – must make decision today, but
don’t know for sure what will happen.
 Americans are reported to spend 37 billion hours a year
waiting in line!
 E-mail often waits in a queue (i.e., line) on the Internet at
intermediate computing centers before sent to final
destination.
 Subassemblies often wait in a line at machining centers
to have the next operations performed.
 Queuing theory represents the body of knowledge
dealing with waiting lines.
Queuing Theory
Conceived in
the early 1900s when Danish
telephone engineer, A.K. Erlang, began studying
the congestion and waiting times occurring in the
completion of telephone calls.
Since that time, a number of quantitative models
have been developed to help business people
understand waiting lines and make better
decisions about how to manage them.
Queuing Theory
Any time
there is a finite capacity for service, you
have a queuing system.

Channels and Stages:
Channel: How many servers available for initial
operation step?
Stage: How many servers must an individual entity
see before service is completed?
Queuing Theory
Channels and Stages
Single Channel (#
of servers available at each
stage) / Single Stage (how many servers the
entity must see before service is complete.
Input
Source
Service Facility
Queuing Theory
Channels and Stages
Multiple Channel (#
of servers available at each
stage) / Single Stage (how many servers the
entity must see before service is complete.
Input
Source
Service Facility
Note: Every entity joins the same line and waits for the
Next available server.
Queuing Theory
Channels and Stages
Question:
What is the difference between the
following?
Multiple
Channel/Single Stage
Input
Source
Multiple
Service facility
– Single Channel Single Stage Systems
Input Service facility
Source
Input Service facility
Source
Other Forms
Queuing Systems
Single
Channel/Multiple Stage
Input
Source
Multiple
Channel/Multiple Stage
Input
Source
Queuing Theory
 Managers
use queuing theory to answer:
 How
many servers should we have?
 How long should a customer wait, on average?
 Queue
Discipline:
 Infinite
Calling Population
 Infinite Queue Capacity
 No Balking or Reneging
 First Come First Served
 Requires that service rate is greater than or equal to arrival
rate.
 In multiple server case, all servers are of equal capability.
Queuing Theory
Processes
 Input
process can be either deterministic (D), general
(G), or it can follow a Poisson Process (M). Poisson
Process says that I know, on average, how many
customers arrive per unit of time. Average arrival RATE
is represented by l.
 Service process can be either deterministic, general or
follow a Poisson Process. Average service RATE is
represented by m.
 Number of servers is represented by K.
Queuing Theory
Kendall’s Notation
 There
are infinite numbers of possible queuing systems:
3 Possible Arrival Processes X 3 Possible Service Processes X
Infinite Possible Number of Servers
 To
make sense of this, we use standard notation –
Kendall’s Notation.
Input Process Service Process # of Servers Queue Capacity
M/M/1/
M/M/2/
M/M/K/
M/G/K/
G/M/K/
G/G/K/
Queuing Theory
M/M/1 Example
Rate l=45 customers per hour
 Average Service Rate m=60 customers per hour
Questions of interest:
Average time a customer spends waiting in line?
Average number of customers waiting in line?
Average time a customer spends in the system?
Average number of customers in the system?
Probability that the system is empty and idle?
 Average Arrival
Little’s Flow Equations
for M/M/1/
P0  1 -
l
m
i
l
Pi    P0
m
l
Lq 
m m  l 
Wq 
Lq
l
W  Wq 
L  lW
1
m
Little’s Flow Equations
for M/M/1/Example
l45,m60
P0  1 -
l

m
i
l
Pi    P0
m
l
Lq 
m m  l 
Wq 
Lq
l
W  Wq 
L  lW
1
m
1-
45
 0.25
60
1
 45 
P1    0.25  0.1875
 60 
45

 2.25 customers
6060  45

2.25
 0.05 hours  3 minutes
45

0.05 

45(0.067)  3 customers
1
 0.067 hours  4 minutes
60
Little’s Flow Equations
for M/M/n/
P0 

l

 
1
k

m

 i 0 i!


i
1
   l k 

   

   m 


  k!1 - l  
   km  

 
  l i
  
  m  P0
 i!

Pi  
  l i
   P0
 m 
i-k

 k k!
if i  k
(or use table)
NOTE: These are
same as for M/M/1/
Wq 
Lq
l
W  Wq 
if i  k
L  lW





0 





k
l 
lm
m 
Lq  P
(k - 1)! (km  l )











2 





1
m
l/m
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
0.55
0.6
0.65
0.7
0.75
0.8
0.85
0.9
0.95
1
1.2
1.4
1.6
1.8
2
2.2
2.4
2.6
2.8
3
3.2
3.4
3.6
3.8
4
4.2
4.4
4.6
4.8
2 channels
0.8605
0.8182
0.7778
0.7391
0.7021
0.6667
0.6327
0.6000
0.5686
0.5385
0.5094
0.4815
0.4545
0.4286
0.4035
0.3793
0.3559
0.3333
0.2500
0.1765
0.1111
0.0526
3 channels
0.8607
0.8187
0.7788
0.7407
0.7046
0.6701
0.6373
0.6061
0.5763
0.5479
0.5209
0.4952
0.4706
0.4472
0.4248
0.4035
0.3831
0.3636
0.2941
0.2360
0.1872
0.1460
0.1111
0.0815
0.0562
0.0345
0.0160
4 channels
0.8607
0.8187
0.7788
0.7408
0.7047
0.6703
0.6376
0.6065
0.5769
0.5487
0.5219
0.4965
0.4722
0.4491
0.4271
0.4062
0.3863
0.3673
0.3002
0.2449
0.1993
0.1616
0.1304
0.1046
0.0831
0.0651
0.0521
0.0377
0.0273
0.0186
0.0113
0.0051
5 channels
0.8607
0.8187
0.7788
0.7408
0.7047
0.6703
0.6376
0.6065
0.5769
0.5488
0.5220
0.4966
0.4724
0.4493
0.4274
0.4065
0.3867
0.3678
0.3011
0.2463
0.2014
0.1646
0.1343
0.1094
0.0889
0.0721
0.0581
0.0466
0.0372
0.0293
0.0228
0.0174
0.0130
0.0093
0.0063
0.0038
0.0017
Values for P0 for Multiple Channel
Waiting Lines with Poisson Arrivals
and Exponential Service Times
P0
Queuing Theory
M/M/2 Example
Rate l=45 customers per hour
Average Service Rate m=60 customers per hour
Number of Channels, K, = 2
Average Arrival
P0 
1
1

P
0 
 0.4545
i
k
0
1
2
 l   l


  45   45     45 
      


       




k
1
 m   m


  60   60     60 



 


 0!
  
1
!

45
i
!


l
 i 0
  k!1 
  2!1  
 

2(60)  

   km  





 

NOTE: l/m=45/60=0.75 – this value is on lookup table for k=2 0.4545
Lq  P
Wq 





0 










l
m





k
lm
(k - 1)!(km  l )
Lq
l
W  Wq 






2





Queuing Theory
M/M/2 Example












2

45
45(60) 

60
  0.1227 customers
 0.4545
2
(2 - 1)!(2(60)  45) 











0.1227
 0.0027 hours or 0.16 minutes
45
1
 0.0027 
m



1
 0.0194 hours or 1.16 minutes
60
L  lW  (45)(0.019 4)  0.8730 customers
Cost/Service Tradeoffs
To Recap Service Statistics:
Service Statistic
Probability No Entities in System
Average Number of Entities Waiting in Queue
Average Time Spent in Queue
Average Time Spent in System
Average Number of Entities in System
k=1, l45,m60
0.25
2.25 customers
3.00 minutes
4.00 minutes
3 customers
k=2, l45,m60
0.4545
0.1227 customers
0.16 minutes
1.16 minutes
0.8730 customers
So, is it worth the extra cost to add an additional
server? Suppose servers earn $15 per hour, but
customer cost of waiting (loss of goodwill, etc.) has been
estimated to be $25 per hour. 
Cost/Service Tradeoffs (cont’d)
 Customer
waiting cost can either be associated with
queue time or total time in system. We’ll go with “total
time in system.”
 Therefore, our per-hour total (system) cost is
represented by
TC=(Cs X K)+ (CW X L)
where Cs = hourly cost for server and
CW = hourly waiting cost
TCk=1 =(15 X 1) + (25 X 3) = $90
TCk=2 =(15 X 2) + (25 X 0.8730) = $51.83
Simulation
 Assumptions
of Queuing - Revisited
 Infinite
Calling Population
 Infinite Queue Capacity
 No Balking or Reneging
 First Come First Served
 Requires that service rate is greater than or equal to arrival
rate.
 In multiple server case, all servers are of equal capability.
 Note
that the above assumptions are fairly restrictive!
Simulation offers us flexibility that Queuing Theory does
not.
Simulation
 Simulation
is a quick way to model long periods of time.
 Simulation requires that we generate a stream of
numbers that are random and that have no relationship
to each other. The RANDOM NUMBER GENERATOR
IS KEY.
 In MS Excel, I can use
=rand() to generate a random
number greater than 0.00 but less than 1.00.
 Can humans generate random numbers?
Simulation Example
Salesperson
no purchase
0.75
no answer
0.30
House
answer
0.70
M
0.20
F
0.80
Purchase
0.25
purchase
0.15
no purchase
0.85
0.10
1
2
3
4
1
2
3
0.40
0.30
0.20
0.60
0.30
0.10
Simulation Example (cont’d)
We now need to construct probability distributions associated
with each event in our simulation example
Answer? Probability RNUM
Y
0.7
0.00-0.69
N
0.3
0.70-0.99
Gender Probability RNUM
M
0.2
0.00-0.19
F
0.8
0.20-0.99
Female Sale? Probability RNUM Male Sale? Probability RNUM
Y
0.15
0.00-0.14
Y
0.25
0.00-0.24
N
0.85
0.15-0.99
N
0.75
0.25-0.99
Number (F)? Probability RNUM Number (M)?
1
0.60
0.00-0.59
1
2
0.30
0.60-0.89
2
3
0.10
0.90-0.99
3
4
Probability RNUM
0.10
0.00-0.09
0.40
0.10-0.49
0.30
0.50-0.79
0.20
0.80-0.99
Simulation Example (cont’d)
Forecasting
A forecast is
an estimate of future demand
Forecasts contain error
Forecasts can be created by subjective means by
estimates from informal sources
OR forecasts can be determined mathematically
by using historical data
OR forecasts can be based on both subjective
and mathematical techniques.
Forecast Ranges
Long
Range (i.e., greater than one year)
production
capacity
automation needs
Quantitative Methods - L.S.
Regression Example
Perfect Lawns, Inc., intends to use sales of lawn
fertilizer to predict lawn mower sales. The store
manager feels that there is probably a six-week
lag between fertilizer sales and mower sales.
The pertinent data are shown below. =>
Quantitative Methods - L.S.
Regression Example
Period
1
2
3
4
5
6
7
8
9
10
11
12
Fertilizer Sales
(Tons)
1.7
1.4
1.9
2.1
2.3
1.7
1.6
2
1.4
2.2
1.5
1.7
Number of Mowers Sold
(Six-Week Lag)
11
9
11
13
14
10
9
13
9
16
10
10
A) Use the least squares method to obtain a linear regression line for the data.
Quantitative Methods - L.S.
Regression Example
Period
1
2
3
4
5
6
7
8
9
10
11
12
Fertilizer Sales
(Tons) (X)
1.7
1.4
1.9
2.1
2.3
1.7
1.6
2
1.4
2.2
1.5
1.7
Number of Mowers Sold (X) (Y)
(Six-Week Lag) (Y)
11
18.7
9
12.6
11
20.9
13
27.3
14
32.2
10
17.0
9
14.4
13
26.0
9
12.6
16
35.2
10
15.0
10
17.0
X2
Y2
2.89
1.96
3.61
4.41
5.29
2.89
2.56
4.00
1.96
4.84
2.25
2.89
121
81
121
169
196
100
81
169
81
256
100
100
Quantitative Methods - L.S.
Regression Example
Period
1
2
3
4
5
6
7
8
9
10
11
12
SUM
Fertilizer Sales
(Tons) (X)
1.7
1.4
1.9
2.1
2.3
1.7
1.6
2
1.4
2.2
1.5
1.7
21.5
Number of Mowers Sold (X) (Y)
(Six-Week Lag) (Y)
11
18.7
9
12.6
11
20.9
13
27.3
14
32.2
10
17.0
9
14.4
13
26.0
9
12.6
16
35.2
10
15.0
10
17.0
135
248.9
X2
Y2
2.89
1.96
3.61
4.41
5.29
2.89
2.56
4.00
1.96
4.84
2.25
2.89
121
81
121
169
196
100
81
169
81
256
100
100
39.55
1575
Quantitative Methods - L.S.
Regression Example
b
n ( X )(Y )   X  Y
n X 
2
 X 
2
Quantitative Methods - L.S.
Regression Example
(12)(248.9)  (215
. )(135)
b
2
(12)(39.55)  (215
.)
84.3

 6.826
12.35
Quantitative Methods - L.S.
Regression Example
Y b X 

a

n
n
135 (6.826)(215
.)


 0.98
12
12
Quantitative Methods - L.S.
Regression Example
Ye  0.98  6.826( X )
Predict lawn mower sales for the first week in August, given fertilier
sales six weeks earlier of two tons.
Ye  0.98  6.826(2)  12.67 lawn mowers
Time Series Analysis
A time
series is a set of numbers where the order
or sequence of the numbers is important, e.g.,
historical demand
Analysis of the time series identifies patterns
Once the patterns are identified, they can be
used to develop a forecast
Time Series Models
Simple
moving average
Weighted moving average
Exponential smoothing (exponentially weighted
moving average)
Exponential
smoothing with random fluctuations
Exponential smoothing with random and trend
Exponential smoothing with random and seasonal
component
Time Series Models
Simple Moving Average
Sample Data (3-period moving average)
t
Dt
Quarter Actual Demand
1
100
2
110
3
110
4
?
Ft
Dt-Ft
| Dt-Ft |
Forecast
Error
Error
(100+110+110)/3=106.67
Time Series Models
Simple Moving Average
Sample Data (3-period moving average)
t
Dt
Quarter Actual Demand
Error
1
100
2
110
3
110
4
80
Ft
Dt-Ft
Forecast
Error
(100+110+110)/3=106.67
80-106.67=-26.67
| Dt-Ft |
26.67
Time Series Models
Simple Moving Average
Sample Data (3-period moving average)
t
Dt
Quarter Actual Demand
1
100
2
110
3
110
4
80
5
?
Ft
Dt-Ft
| Dt-Ft |
Forecast
Error
Error
(100+110+110)/3=106.67
(110+110+80)/3 = 100.00
80-106.67=-26.67
26.67
Time Series Models
Simple Moving Average
Sample Data (3-period moving average)
t
Dt
Quarter Actual Demand
1
100
2
110
3
110
4
80
5
100
Ft
Dt-Ft
| Dt-Ft |
Forecast
Error
Error
(100+110+110)/3=106.67
(110+110+80)/3 = 100.00
80-106.67=-26.67
0
26.67
0
Time Series Models
Exponential smoothing (exponentially weighted moving average)
St  Dt  (1   ) St 1
Ft 1  St
Time Series Models
Exponential smoothing (exponentially weighted moving average)
Where
t=time period
St=smoothed average at end of period t
Dt=actual demand in period t
=smoothing constant (0<<1)
Ft=forecast for period t
Time Series Models
Exponential smoothing (exponentially weighted moving average)
Sample Data (alpha = 0.2)
t
Dt
Quarter Actual Demand
0
St
Smoothed Average
100
Ft
Forecast
Dt-Ft
Error
Time Series Models
Exponential smoothing (exponentially weighted moving average)
Sample Data (alpha=0.2)
t
Dt
Quarter Actual Demand
0
1
?
St
Smoothed Average
100
Ft
Forecast
100
Dt-Ft
Error
Time Series Models
Exponential smoothing (exponentially weighted moving average)
Sample Data (alpha=0.2)
t
Dt
Quarter Actual Demand
0
1
100
St
Smoothed Average
100
Ft
Forecast
100
Dt-Ft
Error
Time Series Models
Exponential smoothing (exponentially weighted moving average)
Sample Data (alpha=0.2)
t
Dt
Quarter Actual Demand
0
1
100
St
Smoothed Average
100
Ft
Dt-Ft
Forecast
Error
100
100-100=0
Time Series Models
Exponential smoothing (exponentially weighted moving average)
Sample Data (alpha=0.2)
t
Dt
Quarter Actual Demand
0
1
100
St
Ft
Smoothed Average Forecast
100
.2(100)+.8(100)=100 100
Dt-Ft
Error
100-100=0
Time Series Models
Exponential smoothing (exponentially weighted moving average)
Sample Data (alpha=0.2)
t
Dt
Quarter Actual Demand
0
1
100
2
?
St
Ft
Smoothed Average Forecast
100
.2(100)+.8(100)=100 100
100
Dt-Ft
Error
100-100=0
Time Series Models
Exponential smoothing (exponentially weighted moving average)
Sample Data (alpha=0.2)
t
Dt
Quarter Actual Demand
0
1
100
2
110
St
Ft
Smoothed Average Forecast
100
.2(100)+.8(100)=100 100
100
Dt-Ft
Error
100-100=0
110-100=10
Time Series Models
Exponential smoothing (exponentially weighted moving average)
Sample Data (alpha=0.2)
t
Dt
Quarter Actual Demand
0
1
100
2
110
St
Ft
Smoothed Average Forecast
100
.2(100)+.8(100)=100 100
.2(110)+.8(100)=102 100
Dt-Ft
Error
100-100=0
110-100=10
Time Series Models
Exponential smoothing (exponentially weighted moving average)
Sample Data (alpha=0.2)
t
Dt
Quarter Actual Demand
0
1
100
2
110
3
?
St
Ft
Smoothed Average Forecast
100
.2(100)+.8(100)=100 100
.2(110)+.8(100)=102 100
102
Dt-Ft
Error
100-100=0
110-100=10
Time Series Models
Exponential smoothing (exponentially weighted moving average)
Sample Data (alpha=0.2)
t
Dt
Quarter Actual Demand
0
1
100
2
110
3
110
Make forecasts for periods 4-12.
St
Ft
Smoothed Average Forecast
100
.2(100)+.8(100)=100 100
.2(110)+.8(100)=102 100
102
Dt-Ft
Error
100-100=0
110-100=10
110-102=8
Time Series Models
Forecast Error
2 error measures:
 (D  F )
Bias
n
tells direction (i.e., over or under forecast)
t
t
 D F
t
n
Mean Absolute Deviation
tells magnitude of forecast error
t
Characteristics of Good
Forecasts
Stability
Responsiveness
Data
Storage Requirements
BESM Example Cont’d
BESM - Expanded
The Basic Exponential Smoothing Model
(BESM)
is nothing more than a cumulative weighted
average of all past demand (and the initial
smoothed average).
Proof:
Decision Analysis/Decision
Trees
 Decision Analysis
(also called Bayesian Statistics) allows
the decision maker to make a choice today (based on
probability of future events) even though the future is
unclear. The problems are generally multi-stage.
 We can use Decision Trees to structure the above
problems. All decision analysis includes:
 Decision Alternatives
(must pick one and stick with it)
 State(s) of Nature (after the decision is made, something
happens – something which the decision maker does not
control).
 Outcomes or payoffs.
Decision Analysis Example
A famous news anchor, who is thinking seriously about running for the US Senate,
estimates the probability of being elected as 60%. While campaign expenditures are not
a problem, the anchor feels that the gain or loss of prestige needs to be evaluated. The
anchor estimates that, if prestige can be quantified, a win would be worth 100 “prestige
points” and a loss would be represented by negative 50 “prestige points.”
There is no shortage of political advisors who would love to have this anchor for a client.
One in particular offers to study the electorate and give an opinion of whether or not the
race could be won. If this analyst returns a positive opinion, the anchor’s chance of
winning rises to 92%. On the other hand, a negative opinion decreases the anchor’s
winning chance to 29%. There is a 49% probability that the analyst will return a positive
opinion.
If the analyst charges $15,000 for his opinion, what would the dollar amount of a “prestige
point” have to be in order for the anchor to consider hiring the analyst.
Draw Tree
Structure
win
Label the Tree
run
lose
“+”
don’t run
win
hire
run
“-”
lose
win
run
lose
don’t hire
don’t run
don’t run
Add Probabilities
win
.92
run
“+”
.49
lose
.08
don’t run
hire
run
“-”
.51
run
don’t hire
don’t run
win
.60
lose
.40
don’t run
win
.29
P(win|neg)
lose
.71
Add Values
run
“+”
.49
win
.92
100
lose
.08
-50
don’t run
hire
run
“-”
.51
run
don’t hire
don’t run
win
.60
lose
.40
don’t run
win
.29
lose
.71
0
100
-50
0
100
-50
0
Fold Back
run
88
“+” 88
.49 don’t run
hire
win
.92
100
lose
.08
-50
43.12
win
.29
run
-6.5
“-”
.51
run
0
100
0
40
don’t hire
win
.60
lose
.40
don’t run
lose
.71
-50
0
100
-50
40
don’t run
0
Trim Undesirable
Decisions Away
run
88
“+” 88
.49 don’t run
hire
win
.92
100
lose
.08
-50
43.12
win
.29
run
-6.5
“-”
.51
run
0
100
0
40
don’t hire
win
.60
lose
.40
don’t run
lose
.71
-50
0
100
-50
40
don’t run
0
Decision Analysis Example
 EVSI
= 43.12 – 40 = 3.12
What is analyst charging per point?
$15,000/3.12 = $4,807.69
 Perfect
Information: The analyst that is never wrong:
If the analyst knows that you will win, you will and he’ll tell you
to RUN
winrun100
If the analyst knows you will lose, you will and he’ll tell you to
NOT RUN
losenot run0
Perfect Analyst: 0.6(100)+0.4(0)=60 NOTE: This is EVWPI
EVPI = 60-40=20
NOTE: My analyst was only worth 3.12 points
Decision Analysis Example
 What
is the EFFICIENCY of my analyst?
EVSI / EVPI = 3.12 / 20 = 0.156
What if I thought 1 “prestige point” was worth $100?
How much would I then be worth to pay ANY analyst?
$2000  the PERFECT analyst can only get me
20 points above and beyond what I can do on my own.