University Agenda Some Review from Last Class Data Envelopment Analysis Revenue Management Game Theory Concepts Chapter 5 Advanced Linear Programming Applications Data Envelopment Analysis ID: 757953
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Slide1
Slides by
John
Loucks
St. Edward’s
UniversitySlide2
Agenda
Some Review from Last ClassData Envelopment AnalysisRevenue ManagementGame Theory ConceptsSlide3
Chapter 5 Advanced Linear Programming Applications
Data Envelopment Analysis
Compares one unit to similar othersIe branch of a bank, franchise of a chainRevenue Management
Maximize revenue with a fixed inventoryPortfolio Models and Asset Allocation
Determine best portfolio compositionGame Theory
Competition with a zero sumSlide4
Data Envelopment Analysis
Data envelopment analysis (DEA): used to determine the relative operating efficiency of units with the same goals and objectives.DEA creates a
hypothetical composite optimal weighted average (W1,
W2,…) of existing units
.E – Efficiency IndexAllows comparison between composite and unit“what the outputs of the composite would be, given the units inputs”
If E
< 1, unit is less efficient than the composite unit If E = 1, there is no evidence that unit k is inefficient.Slide5
Data Envelopment Analysis
The DEA Model
MIN
E
s.t. OUTPUTS INPUTS
Sum of weights = 1 E, weights > 0 Slide6
Data Envelopment Analysis
The Langley County School District is trying to
determine the relative efficiency of its three high schools
. In particular, it wants to evaluate
Roosevelt High.
Outputs:
performances on SAT scores, the number of seniors finishing high school
the number of students who enter collegeInputs number of teachers teaching senior classes the prorated budget for senior instruction number of students in the senior class. Slide7
Data Envelopment Analysis
Input
Roosevelt1 Lincoln2
Washington3 Senior Faculty 37 25 23 Budget ($100,000's) 6.4 5.0 4.7 Senior Enrollments 850 700 600
Slide8
Data Envelopment Analysis
Output
Roosevelt1
Lincoln2 Washington3 Average SAT Score 800 830 900 High School Graduates 450 500 400 College Admissions 140 250 370Slide9
Data Envelopment Analysis
Define the Decision Variables
E = Fraction of Roosevelt's input resources required by the composite high school
w1
= Weight applied to Roosevelt's input/output resources by the composite high school w2
= Weight applied to Lincoln’s input/output resources by the composite high school
w3 = Weight applied to Washington's input/output resources by the composite high schoolSlide10
Data Envelopment Analysis
Define the Objective Function
Since our objective is to DETECT INEFFICIENCIES
, we want to minimize the fraction of Roosevelt High School's input resources required by the composite high school:
MIN ESlide11
Data Envelopment Analysis
Define the Constraints
Sum of the Weights is 1: (1) w1 +
w2 + w
3 = 1Output Constraints
General form for each output: output for composite >= output for Roosevelt
Output for composite = (Output for Roosevelt * weight for Roosevelt ) +(output for Lincoln * weight for Lincoln )
+ (output for Washington * weight for Washington ) + Slide12
Data Envelopment Analysis
Output Constraints: Since w1
= 1 is possible, each output of the composite school must be at least as great as that of Roosevelt: (2) 800w1 + 830w
2 + 900w3
> 800 (SAT Scores) (3) 450w1
+ 500w2 + 400
w3 > 450 (Graduates) (4) 140w1 + 250w2 + 370
w3 > 140 (College Admissions)Slide13
Data Envelopment Analysis
Input ConstraintsGeneral Form
Input for composite <= input for Roosevelt * EInput for composite =
(Input for
Roosevelt * Input for Roosevelt ) +(Input
for Lincoln * Input for Lincoln ) +
(Input for Washington * Input for Washington ) (5) 37w
1 + 25w2 + 23w3 < 37E (Faculty) (6) 6.4w1 + 5.0w2 + 4.7w3 < 6.4E (Budget) (7) 850w1
+ 700w2
+ 600w3 < 850E (Seniors) Nonnegativity : E, w1, w2
, w3 > 0Slide14
Data Envelopment Analysis
MIN EST (1) w1 +
w2 + w3 = 1
(2) 800w1
+ 830w2 + 900w
3 > 800
(SAT Scores) (3) 450w1 + 500w2 + 400w
3 > 450 (Graduates) (4) 140w1 + 250w2 + 370w3 > 140 (College Admissions) (5) 37w1 + 25w2 + 23w
3 < 37
E (Faculty) (6) 6.4w1 + 5.0w2 + 4.7w3 < 6.4E (Budget) (
7) 850w1 + 700w2 + 600w
3 < 850E (Seniors)
(8)
E
,
w
1
,
w
2
,
w
3
>
0Slide15
Data Envelopment Analysis
Computer
Solution
OBJECTIVE FUNCTION VALUE = 0.765
VARIABLE VALUE REDUCED COSTS E 0.765 0.000
W1 (R) 0.000 0.235 W2 (L) 0.500 0.000 W3 (W) 0.500 0.000*Composite is 50% Lincoln, 50% Washington*Roosevelt is no more than 76.5% efficient as composite Slide16
Data Envelopment Analysis
Computer Solution
(continued
)
CONSTRAINT
SLACK/SURPLUS DUAL VALUES 1 0.000
-0.235 2 (SAT) 65.000 0.000 3 (grads) 0.000 -0.001 4 (college) 170.000 0.000 5 (fac
) 4.294 0.000
6 (budget) 0.044 0.000 7 (seniors) 0.000 0.001Zero Slack – Roosevelt is 76.5% efficient in this area (ie grads)
Positive slack – Roosevelt is LESS THAN 76.5% efficient (ie
SAT) ie SAT scores are 65 points higher in the composite school
Slide17
Revenue Management
Another LP application is revenue management.
Revenue management
managing
the short-term demand for a fixed perishable inventory in order to
maximize revenue potential.
first used to determine how many airline seats to sell at an early-reservation discount fare and many to sell at a full fare.Slide18
Revenue Management
General FormMAX (revenue per unit * units allocated)STCAPACITY
DEMAND NONNEGATIVESlide19
Revenue Management
LeapFrog
Airways provides passenger service for
Indianapolis, Baltimore, Memphis, Austin, and Tampa.
LeapFrog
has two WB828 airplanes, one
based in Indianapolis and the other in Baltimore. Each morningthe Indianapolis based plane flies to Austin with a stopover in Memphis. The Baltimore based plane flies to
Tampa with a stopover in Memphis
. Both planes have a coach section with a 120-seat capacity. Slide20
LeapFrog
uses two fare classes: a discount fare D
class and a full fare F class. Leapfrog’s products, each
referred to as an origin
destination itinerary
fare (ODIF), are listed on the next slide with their fares and
forecasted demand. LeapFrog wants to determine how many seats it should allocate to each ODIF.
Revenue ManagementSlide21
IND
BAL
MEM
AUS
TAM
Each day a plane
Leaves both IND
And BAL for
AUS and TAM
Respectively.
Both flights lay over
In MEM
No return flights
(for simplicity)
Each plane holds 120
Leg 1
Leg 2
Leg 3
Leg 4Slide22
Orig
Dest
INDMEM
INDAUS
INDTAM
BAL
MEMBALAUSBALTAM
MEMAUSMEMTAM8 different origin-destination combinationsPlus two different fare classes: Discount and Full Fare8 Orig-Desination combinations * 2 fare classes = 16 combinations Slide23
ODIF
1234
5678
910
111213
1415
16OriginIndianapolisIndianapolis
IndianapolisIndianapolisIndianapolisIndianapolisBaltimoreBaltimoreBaltimoreBaltimoreBaltimoreBaltimoreMemphisMemphisMemphisMemphis
Destination
MemphisAustinTampaMemphisAustinTampaMemphisAustinTampaMemphisAustin
TampaAustinTampa AustinTampa
Fare
Class
D
D
D
F
F
F
D
D
D
F
F
F
D
D
F
F
ODIF
Code
IMD
IAD
ITD
IMF
IAF
ITF
BMD
BAD
BTD
BMF
BAF
BTF
MAD
MTD
MAF
MTF
Fare
175
275
285
395
425
475
185
315
290
385
525
490
190
180
310
295
Demand
44
25
40
15
10
8
26
50
42
12
16
9
58
48
1411
Revenue ManagementSlide24
Revenue Management
Define the Decision Variables
There are 16 variables, one for each ODIF:
IMD = number of seats allocated to Indianapolis-Memphis-
Discount class
IAD = number of seats allocated to Indianapolis-Austin- Discount class
ITD = number of seats allocated to Indianapolis-Tampa- Discount class
IMF = number of seats allocated to Indianapolis-Memphis- Full Fare class
IAF = number of seats allocated to Indianapolis-Austin-Full Fare classSlide25
Revenue Management
Define the Decision Variables (continued)
ITF = number of seats allocated to Indianapolis-Tampa-
Full Fare class
BMD = number of seats allocated to Baltimore-Memphis-
Discount class
BAD = number of seats allocated to Baltimore-Austin-
Discount class
BTD = number of seats allocated to Baltimore-Tampa-
Discount classBMF = number of seats allocated to Baltimore-Memphis- Full Fare class
BAF = number of seats allocated to Baltimore-Austin-
Full Fare classSlide26
Revenue Management
Define the Decision Variables (continued)
BTF = number of seats allocated to Baltimore-Tampa-
Full Fare class
MAD = number of seats allocated to Memphis-Austin-
Discount class
MTD = number of seats allocated to Memphis-Tampa-
Discount class
MAF = number of seats allocated to Memphis-Austin-
Full Fare classMTF = number of seats allocated to Memphis-Tampa- Full Fare classSlide27
Revenue Management
Define the Objective Function
Maximize total revenue:
Max (fare per seat for each ODIF)
x (number of seats allocated to the ODIF)
Max 175IMD + 275IAD + 285ITD + 395IMF
+ 425IAF + 475ITF + 185BMD + 315BAD
+ 290BTD + 385BMF + 525BAF + 490BTF + 190MAD + 180MTD + 310MAF + 295MTFSlide28
Revenue Management
Define the Constraints
There are 4 capacity constraints, one for each flight leg:
Indianapolis-Memphis leg
(1)
IMD + IAD + ITD + IMF + IAF + ITF < 120
Baltimore-Memphis leg (2) BMD + BAD + BTD + BMF + BAF + BTF
< 120
Memphis-Austin leg
(3)
IAD + IAF + BAD + BAF + MAD + MAF
<
120
Memphis-Tampa leg
(4)
ITD + ITF + BTD + BTF + MTD + MTF
<
120Slide29
Revenue Management
Define the Constraints (continued)
Demand Constraints Limit the amount of seats for each ODIF
There
are 16 demand constraints, one for each ODIF:
(5) IMD < 44 (11) BMD < 26 (17) MAD <
58
(6) IAD < 25 (12) BAD <
50 (18) MTD < 48
(7) ITD
<
40 (13) BTD
<
42 (19) MAF
<
14
(8) IMF
<
15 (14) BMF
<
12 (20) MTF
<
11
(9) IAF
<
10 (15) BAF
<
16
(10) ITF
<
8 (16) BTF
<
9Slide30
Revenue Management
Max 175IMD + 275IAD + 285ITD + 395IMF + 425IAF + 475ITF + 185BMD + 315BAD
+ 290BTD + 385BMF + 525BAF + 490BTF + 190MAD + 180MTD + 310MAF + 295MTF
ST: IMD + IAD + ITD + IMF + IAF + ITF
< 120
BMD + BAD + BTD + BMF + BAF + BTF
< 120IAD + IAF + BAD + BAF + MAD + MAF < 120
ITD + ITF + BTD + BTF + MTD + MTF < 120IMD < 44, BMD < 26, MAD < 58, IAD < 25, BAD
< 50
MTD < 48, ITD < 40, BTD < 42, MAF < 14, IMF
< 15BMF <
12, MTF < 11, IAF
<
10, BAF
<
16, ITF
<
8
BTF
<
9
IMD, IAD, ITD, IMF, IAF, ITF, BMD, BAD, BTD, BMF, BAF, BTF, MAD, MTD, MAF, MTF
> 0Slide31
Revenue Management
Computer
Solution
Revenue Contribution is $96265
Slide32
Revenue Management
Computer Solution
(continued
)
IMD dual value is
90
IMF dual value is
310Slide33
Introduction to Game Theory
In
decision analysis
, a single decision maker seeks to select an optimal alternative.
In
game theory
, there are two or more decision makers, called players, who compete as adversaries against each other.
It is assumed that each player has the same information and will select the strategy that provides the best possible outcome from his point of view.Each player selects a strategy independently without knowing in advance the strategy of the other player(s). continueSlide34
Introduction to Game Theory
The combination of the competing strategies provides the
value of the game
to the players.
Examples of competing players are teams, armies, companies, political candidates, and contract bidders.Slide35
Two-person
means there are two competing players in the game.
Zero-sum
means the gain (or loss) for one player is equal to the corresponding loss (or gain) for the other player.
The gain and loss balance out so that there is a zero-sum for the game.
What one player wins, the other player loses.
Two-Person Zero-Sum GameSlide36
Competing for Vehicle Sales
Suppose that there are only two vehicle dealer-ships in a small city. Each dealership is considering
three strategies that are designed
to take
sales of
new
vehicles
from the other dealership over a four-month period. The strategies, assumed to be
the same
for both dealerships, are on the next slide.
Two-Person Zero-Sum Game ExampleSlide37
Strategy Choices
Strategy
1: Offer a
cash rebate
on a new vehicle. Strategy 2: Offer free optional
equipment on a
new vehicle.
Strategy
3: Offer a
0% loan
on
a new vehicle.
Two-Person Zero-Sum Game ExampleSlide38
2 2 1
Cash
Rebate
b
1
0%
Loan
b
3
FreeOptionsb2
Dealership B
Payoff Table: Number of Vehicle Sales
Gained Per Week by Dealership A
(or Lost Per Week by Dealership B)
-3 3 -1
3 -2 0
Cash Rebate
a
1
Free Options
a
2
0% Loan
a
3
Dealership A
Two-Person Zero-Sum Game ExampleSlide39
Step 1:
Identify the minimum payoff for each
row (for Player A).
Step 2:
For Player A, select the strategy that provides
the maximum of the row minimums (called
the
maximin).Two-Person Zero-Sum GameSlide40
Identifying Maximin and Best Strategy
Row
Minimum
1
-3
-2
2 2 1
Cash
Rebate
b
1
0%
Loan
b
3
Free
Options
b
2
Dealership B
-3 3 -1
3 -2 0
Cash Rebate
a
1
Free Options
a
2
0% Loan
a
3
Dealership A
Best Strategy
For Player A
Maximin
Payoff
Two-Person Zero-Sum Game ExampleSlide41
Step 3:
Identify the maximum payoff for each column
(for Player B).
Step 4:
For Player B, select the strategy that provides
the minimum of the column maximums
(called the
minimax).Two-Person Zero-Sum GameSlide42
Identifying Minimax and Best Strategy
2 2 1
Cash
Rebate
b
1
0%
Loan
b
3
FreeOptions
b2
Dealership B
-3 3 -1
3 -2 0
Cash Rebate
a
1
Free Options
a
2
0% Loan
a
3
Dealership A
Column Maximum
3 3 1
Best Strategy
For Player B
Minimax
Payoff
Two-Person Zero-Sum Game ExampleSlide43
Pure Strategy
Whenever an optimal
pure strategy
exists:
the maximum of the row minimums equals the minimum of the column maximums (Player A’s
maximin
equals Player B’s minimax
)the game is said to have a saddle point (the intersection of the optimal strategies)the value of the saddle point is the value of the game
neither player can improve his/her outcome by changing strategies even if he/she learns in advance the opponent’s strategySlide44
Row
Minimum
1
-3
-2
Cash
Rebate
b
1
0%
Loan
b
3
Free
Options
b
2
Dealership B
-3 3 -1
3 -2 0
Cash Rebate
a
1
Free Options
a
2
0% Loan
a
3
Dealership A
Column Maximum
3 3 1
Pure Strategy Example
Saddle Point and Value of the Game
2 2 1
Saddle
Point
Value of the
game is 1Slide45
Pure Strategy Example
Pure Strategy Summary
Player A should choose Strategy
a
1
(offer a cash rebate).
Player A can expect a
gain of at least 1 vehicle sale per week.Player B should choose Strategy b3 (offer a 0% loan).
Player B can expect a
loss of no more than 1 vehicle sale per week.Slide46
Mixed Strategy
If the maximin value for Player A does not equal the minimax value for Player B, then a pure strategy is not optimal for the game.
In this case, a
mixed strategy
is best.
With a mixed strategy, each player employs more than one strategy.
Each player should use one strategy some of the time and other strategies the rest of the time.
The optimal solution is the relative frequencies with which each player should use his possible strategies.Slide47
Mixed Strategy Example
b
1
b
2
Player B
11 5
a
1
a
2
Player A
4 8
Consider the following two-person zero-sum game. The maximin does not equal the minimax. There is not an optimal pure strategy.
Column
Maximum
11 8
Row
Minimum
4
5
Maximin
MinimaxSlide48
Mixed Strategy Example
p
= the probability Player A selects strategy
a
1
(1
-
p) = the probability Player A selects strategy a2
If Player B selects
b1:EV = 4
p + 11(1 –
p)
If Player B selects
b
2
:
EV = 8
p
+ 5(1 –
p
)Slide49
Mixed Strategy Example
4
p
+ 11(1 –
p
) = 8p
+ 5(1 – p)
To solve for the optimal probabilities for Player Awe set the two expected values equal and solve forthe value of p.
4
p + 11 – 11p = 8p + 5 – 5p
11 – 7
p = 5 + 3
p
-10
p
= -6
p
= .6
Player A should select:
Strategy
a
1
with a .6 probability and
Strategy
a
2
with a .4 probability.
Hence,
(1
-
p
) = .4Slide50
Mixed Strategy Example
q
= the probability Player B selects strategy
b
1
(1
-
q) = the probability Player B selects strategy b2
If Player A selects
a1:EV = 4
q + 8(1 –
q)
If Player A selects
a
2
:
EV = 11
q
+ 5(1 –
q
)Slide51
Mixed Strategy Example
Value of the Game
For Player A:
EV = 4
p
+ 11(1 –
p
) = 4(.6) + 11(.4) = 6.8
For Player B:
EV = 4
q
+ 8(1 –
q
) = 4(.3) + 8(.7) = 6.8
Expected gain
per game
for Player A
Expected loss
per game
for Player B