Managerial Analysis & Decision Making Slides - Shinawatra MBA - Semester 2-2010 CC

11An International University with an Emphasis on Research

MF1002 Managerial Analysis & Decision Making

Introduction to Managerial Decision Modeling Introduction to Managerial Decision Modeling (Chapter 1 of (Chapter 1 of BalakrishnanBalakrishnan et al.et al., 2007), 2007)

Assoc. Prof. Dr. Chuvej ChansaAssoc. Prof. Dr. Chuvej Chansa--ngavejngavejProgram Director - Ph.D. in Management Science

Shinawatra University (SIU International)


Session InstructorAssociate Prof. Dr. Chuvej Chansa-ngavej

Program Director, PhD in Management Science

Graduate Building, SIU

Room 317, 3rd Floor, BBD-Viphavadi Building

Tel. 02-650-6035; 081-912-1535

Email: [email protected]

PhD (Management Science in Capital Investment) Ohio State University, USA

M.Eng. (Management Science in Marketing and Operations) University of New South Wales, Australia

B.Eng. (1st Class Honors) in Industrial Engineering University of New South Wales, Australia


Learning ObjectivesKnow the historical development and origin of managerial decision modeling

Recognize how managerial decision modeling can be applied

Able to explain significant development directions of managerial decision modeling


What is Managerial Decision Modeling?

A scientific approach to managerial decision making

• The development of a (mathematical) model of a real-world problem scenario

• The model provides insight into the solution of the managerial problem

• Also known as Quantitative Analysis, Management Science, Operations Research


Actual Applications in Business

Frequently used in such organizations as:

• American Airlines

• IBM

• Merrill Lynch

• AT&T


Used in All Kinds of Enterprises

• Business and Industry

• Government

• Health Care

• Education

• Agriculture

• Military

• Etc.


Origin of Managerial Decision ModelingEve of World War II in British military operations

Recommending the optimum location of radar signal masts

Optimum size of merchant ship convoys to avoid enemy detection

Optimum detonation of depth charges to destroy u-boat submarines

Decision Modeling played a significant role in helping allied forces won the war


Development of Managerial Decision Modeling

After World War II, applications spread to business and industry, especially in USA

Courses were quickly established in prestigious universities (Massachusetts Institute of Technology, Case Western Reserve University, Ohio State University)


Areas of Modern-day ApplicationsDesigning optimum factory layout

Minimum cost construction of telecommunication network

Road traffic management and ‘one-way’street allocation

Determining optimal school bus routes

Minimum-time design of computer-chip layout

Efficient customer response tactics


Areas of Modern-day Applications

Flow management of raw materials and products in a supply chain


An Example of Modern-day Applications

Blending of raw materials in oil refineriesX2

X1

250

200

150

100

50

00 50 100 150 200 250

objective function 350X1 + 300X2 = 35000


optimal solution

X2

X1

250

200

150

100

50

00 50 100 150 200 250



optimal solution


Example of Managerial Decision Modeling in Queueing ProblemsWaiting line situations are a part of everyday life

People often react in an adverse manner to excessive waits in queues Lining up for movie tickets in 1920


Characteristics of Queueing Problems“Customers” arrive at some types of systems according to some type of probability distribution

Arrival rate (number of customers per time period)

“Servers” provide the serviceService rate (number of customers served per time period)

Customer departs the system


Fields Invented by MS

Marketing Science

Decision AnalysisDelay Launch

1 Year

Launch as PlannedIncrease R&D $

Market Reaction to Delay

Feature Delivery

Mild

Moderate

Extreme

Mild

Moderate

Extreme

Fully Featured

De-featured

Decision to Launch as Planned or Delay

0.3

0.7

0.4

0.2

0.4

0.2

0.4

0.4

$3.6 M

($2.0 M)

$0.3 M

EV = $1.2

EV = $6.2 M$17.0 M


Fields Invented by MS

Search Theory

Financial Engineering

Transportation Science


Types of Decision ModelsDeterministic Models

Assume that all the input data value are known with certainty

Probabilistic ModelsAssume that some input data values are not known with

certainty. Hence probability is used to represent the uncertainty

Examples of probabilistic modeling techniques: queueing theory, decision analysis, simulation


Quantitative vs. Qualitative DataThe modeling process begins with data

Quantitative DataNumerical factors such as costs and revenues

Qualitative DataFactors that effect the environment which are difficult to quantify into numerical measures, and subjective opinion must be used instead


Spreadsheets in Decision MakingComputers are used to create and solve models

Spreadsheets are a convenient alternative to specialized software

Microsoft Excel has extensive modeling capability via the use of “add-ins” (Every student must buy a new copy of the textbook to get these essential add-ins)

Premium Solver, Tree Plan, Crystal Ball, ExcelModules


Steps in Managerial Decision Modeling

1. FormulationTranslating a problem scenario from words to a mathematical model

2. SolutionSolving the model to obtain the optimal solution

3. Interpretation and Sensitivity AnalysisAnalyzing results and implementing a solution


Steps in Managerial Decision Modeling


The Apprentice QuestionSuppose you were one of the contestants, how would you use managerial decision modeling to help the team in the decision making. Answer this based on what you have learned about managerial decision modeling.

Video: The Apprentice UK, Series 6, Episode 3, October 2010


Example Spreadsheet Model: Tax Computation

Self employed couple must estimate and

pay quarterly income tax (joint return)

Income amount is uncertain

5% of income to retirement account, up to $4000 max

Personal exemption = 2 x $3200 = $6400

Standard deduction = $10,000

No other deductions

File 1-1.xls


Tax Brackets

Percent of Taxable Income Taxable Income

up to $14,600 10%$14,601 to $59,400 15%$59,401 to $119,950 25%


Example Spreadsheet Model: Break-Even Analysis

Profit = Revenue – Costs

Revenue = (Selling price) x (Num. units)

Costs = (Fixed cost) +

(Cost per unit) x (Num. units)

File 1-2.xls


The Break Even Point (BEP) is the number of units where:

Profit = 0, so

Revenue = Costs

BEP = Fixed cost

(Selling price) – (Cost per unit)


Possible Problems inDeveloping Decision Models

Defining the Problem – four typical roadblocks:

• Conflicting viewpoints

• Impact on other departments

• Beginning assumptions

• Solution outdated



Developing a Model• A manager’s perception of a problem does

not always match the textbook approach

• Managers do not use the results of a model they do not understand

Acquiring Input Data• Problem in using accounting data

• Available data must often be distilled and manipulated



Developing a Solution• Hard-to-understand mathematics

• Most managers would like to have a range of options instead of only one answer

Testing the Solution• Assumptions should be reviewed

Analyzing the Results

Implementation• Management support and user involvement are important



Session 2 Session 2 -- Linear Programming Models:Linear Programming Models:

Graphical and Computer MethodsGraphical and Computer Methods(Chapter 2 of (Chapter 2 of BalakrishnanBalakrishnan et al.et al., 2007), 2007)




Steps in Developing a Linear Programming (LP) Model

1) Formulation

2) Solution

3) Interpretation and Sensitivity Analysis


Properties of LP Models1) Seek to minimize or maximize

2) Include “constraints” or limitations

3) There must be alternatives available

4) All equations are linear


Example LP Model Formulation:The Product Mix Problem

Decision: How much to make of > 2 products?

Objective: Maximize profit

Constraints: Limited resources


Example: Flair Furniture Co.Two products: Chairs and Tables

Decision: How many of each to make this month?



Flair Furniture Co. Data

$5$7Profit Contribution

10001 hr2 hrsPainting

24004 hrs3 hrsCarpentry

Hours Available

Chairs(per chair)

Tables(per table)

Other Limitations:• Make no more than 450 chairs• Make at least 100 tables


Decision Variables:

T = Num. of tables to make

C = Num. of chairs to make

Objective Function: Maximize Profit

Maximize $7 T + $5 C


Constraints:

Have 2400 hours of carpentry time available

3 T + 4 C < 2400 (hours)

Have 1000 hours of painting time available

2 T + 1 C < 1000 (hours)


More Constraints:

Make no more than 450 chairs

C < 450 (number of chairs)

Make at least 100 tables

T > 100 (number of tables)

Nonnegativity:Cannot make a negative number of chairs or tables

T > 0

C > 0


Model Summary

Max 7T + 5C (profit)Subject to the constraints:

3T + 4C < 2400 (carpentry hours)

2T + 1C < 1000 (painting hours)

C < 450 (max # chairs)

T > 100 (min # tables)

T, C > 0 (nonnegativity)


Graphical SolutionGraphing an LP model helps provide insight into LP models and their solutions.

While this can only be done in two dimensions, the same properties apply to all LP models and solutions.


CarpentryConstraint Line

3T + 4C = 2400

Intercepts

(T = 0, C = 600)

(T = 800, C = 0)

0 800 T

C

600

0

Feasible< 2400 hrs

Infeasible> 2400 hrs

3T + 4C = 2400


PaintingConstraint Line

2T + 1C = 1000

Intercepts

(T = 0, C = 1000)

(T = 500, C = 0)

0 500 800 T

C1000

600

0

2T + 1C = 1000


0 100 500 800 T

C1000

600

450

0

Max Chair Line

C = 450

Min Table Line

T = 100

Feasible

Region

1515An International University with an Emphasis on Research0 100 200 300 400 500 T

C

500

400

300

200

100

0

Objective Function Line

7T + 5C = Profit

7T + 5C = $2,1007T + 5C = $4,040

Optimal Point(T = 320, C = 360)7T + 5C = $2,800

1616An International University with an Emphasis on Research0 100 200 300 400 500 T

C

500

400

300

200

100

0

Additional Constraint

Need at least 75 more chairs than tables

C > T + 75

Or

C – T > 75

T = 320C = 360

No longer feasible

New optimal pointT = 300, C = 375


LP CharacteristicsFeasible Region: The set of points that satisfies all constraints

Corner Point Property: An optimal solution must lie at one or more corner points

Optimal Solution: The corner point with the best objective function value is optimal


Special Situation in LP1. Redundant Constraints - do not affect

the feasible region

Example: x < 10

x < 12

The second constraint is redundant because it is less restrictive.


Special Situation in LP2. Infeasibility – when no feasible solution

exists (there is no feasible region)

Example: x < 10

x > 15


Special Situation in LP3. Alternate Optimal Solutions – when

there is more than one optimal solution

Max 2T + 2CSubject to:

T + C < 10T < 5

C < 6T, C > 0

0 5 10 T

C

10

6

0

2T + 2C = 20All points onRed segment are optimal


Special Situation in LP4. Unbounded Solutions – when nothing

prevents the solution from becoming infinitely large

Max 2T + 2CSubject to:

2T + 3C > 6T, C > 0

0 1 2 3 T

C

2

1

0

Directio

n

of solution


Using Excel’s Solver for LP

Recall the Flair Furniture Example:Max 7T + 5C (profit)

Subject to the constraints:

3T + 4C < 2400 (carpentry hrs)

2T + 1C < 1000 (painting hrs)

C < 450 (max # chairs)

T > 100 (min # tables)

T, C > 0 (nonnegativity)Go to file 2-1.xls



Session 4 Session 4 -- Linear Programming Modeling Linear Programming Modeling

ApplicationsApplications

(Chapter 3 of (Chapter 3 of BalakrishnanBalakrishnan et al.et al., 2007), 2007)




Linear Programming (LP) Can Be Used for Many Managerial Decisions:

Product mix

Make-buy

Media selection

Marketing research

Portfolio selection

Shipping & transportation

Multiperiod scheduling


For a particular application we begin with

the problem scenario and data, then:

1) Define the decision variables

2) Formulate the LP model using the decision variables

• Write the objective function equation

• Write each of the constraint equations

3) Implement the model in Excel

4) Solve with Excel’s Solver


Product Mix Problem: Fifth Avenue Industries

Produce 4 types of men's ties

Use 3 materials (limited resources)

Decision: How many of each type of ties to make per month?



1,250$9Cotton

2,000$6Polyester

1,000$20Silk

Yards availableper monthCost per yardMaterial

Resource Data

Labor cost is $0.75 per tie


Product Data

8,50016,00014,0007,000Monthly Maximum

Type of Tie

0.100.100.080.125Total material(yards per tie)

6,00013,00010,0006,000Monthly Minimum

$4.81$4.31$3.55$6.70Selling Price(per tie)

Blend 2Blend 1PolyesterSilk


Material Requirements(yards per tie)

Type of Tie

0.070.0500Cotton

0.030.050.080Polyester

0000.125Silk

Blend 2(30/70)

Blend 1(50/50)PolyesterSilk

Material

0.100.100.080.125Total yards


Decision VariablesS = number of silk ties to make per month

P = number of polyester ties to make per month

B1 = number of poly-cotton blend 1 ties to make per month

B2 = number of poly-cotton blend 2 ties to make per month


Profit Per Tie Calculation

Profit per tie =

(Selling price) – (material cost) –(labor cost)

Silk Tie

Profit = $6.70 – (0.125 yds)($20/yd) - $0.75

= $3.45 per tie


Objective Function (in $ of profit)

Max 3.45S + 2.32P + 2.81B1 + 3.25B2


Material Limitations (in yards)

0.125S < 1,000 (silk)

0.08P + 0.05B1 + 0.03B2 < 2,000 (poly)

0.05B1 + 0.07B2 < 1,250 (cotton)


Min and Max Number of Ties to Make

6,000 < S < 7,000

10,000 < P < 14,000

13,000 < B1 < 16,000

6,000 < B2 < 8,500

Finally nonnegativity S, P, B1, B2 > 0

Go to file 3-1.xls


Media Selection Problem:Win Big Gambling Club

Promote gambling trips to the Bahamas

Budget: $8,000 per week for advertising

Use 4 types of advertising

Decision: How many ads of each type?

Objective: Maximize audience reached


Data

Advertising Options

2025512Max AdsPer week

$380$290$925$800Cost(per ad)

2,8002,4008,5005,000AudienceReached(per ad)

Radio(afternoon)

Radio(prime time)NewspaperTV Spot


Other RestrictionsHave at least 5 radio spots per week

Spend no more than $1800 on radio

Decision Variables

T = number of TV spots per week

N = number of newspaper ads per week

P = number of prime time radio spots per week

A = number of afternoon radio spots per week


Objective Function (in number of audience reached)

Max 5000T + 8500N + 2400P + 2800A


Budget is $8000

800T + 925N + 290P + 380A < 8000

At Least 5 Radio Spots per Week

P + A > 5


No More Than $1800 per Week for Radio

290P + 380A < 1800

Max Number of Ads per Week

T < 12 P < 25

N < 5 A < 20

Finally nonnegativity T, N, P, A > 0

Go to file 3-3.xls


Portfolio Selection:International City Trust

Has $5 million to invest among 6 investments

Decision: How much to invest in each of 6 investment options?

Objective: Maximize interest earned


Data

Risk ScoreInterest

RateInvestment

1.77%Trade credits

1.210%Corp. bonds

2.914%Construction loans

2.08%Mortgage securities

2.412%Platinum stocks

3.719%Gold stocks


Constraints

Invest up to $ 5 million

No more than 25% into any one investment

At least 30% into precious metals

At least 45% into trade credits and corporate bonds

Limit overall risk to no more than 2.0


Decision Variables

T = $ invested in trade credit

B = $ invested in corporate bonds

G = $ invested gold stocks

P = $ invested in platinum stocks

M = $ invested in mortgage securities

C = $ invested in construction loans


Objective Function (in $ of interest earned)

Max 0.07T + 0.10B + 0.19G + 0.12P

+ 0.08M + 0.14C


Invest Up To $5 Million

T + B + G + P + M + C < 5,000,000


No More than 25% into Any One Investment

T < 0.25 (T + B + G + P + M + C)

B < 0.25 (T + B + G + P + M + C)

G < 0.25 (T + B + G + P + M + C)

P < 0.25 (T + B + G + P + M + C)

M < 0.25 (T + B + G + P + M + C)

C < 0.25 (T + B + G + P + M + C)


At Least 30% into Precious Metals

G + P > 0.30 (T + B + G + P + M + C)

At Least 45% into Trade Credits and Corporate Bonds

T + B > 0.45 (T + B + G + P + M + C)


Limit Overall Risk to No More Than 2.0Use a weighted average to calculate portfolio risk

1.7T + 1.2B + 3.7G + 2.4P + 2.0M + 2.9C < 2.0

T + B + G + P + M + C

or

1.7T + 1.2B + 3.7G + 2.4P + 2.0M + 2.9C <

2.0 (T + B + G + P + M + C)

Finally, nonnegativity: T, B, G, P, M, C > 0

Go to file 3-5.xls


Labor Planning:Hong Kong Bank

Number of tellers needed varies by time of day

Decision: How many tellers should begin work at various times of the day?

Objective: Minimize personnel cost


172 - 3181 – 2

1411 – 121210 – 11109 – 10

104 – 5153 – 4

1612 – 1

Min Num. TellersTime Period

Total minimum daily requirement is 112 hours


Full Time TellersWork from 9 AM – 5 PM

Take a 1 hour lunch break, half at 11, the other half at noon

Cost $90 per day (salary & benefits)

Currently only 12 are available


Part Time Tellers• Work 4 consecutive hours (no lunch break)

• Can begin work at 9, 10, 11, noon, or 1

• Are paid $7 per hour ($28 per day)

• Part time teller hours cannot exceed 50% of the day’s minimum requirement (50% of 112 hours = 56 hours)


Decision Variables

F = num. of full time tellers (all work 9–5)

P1 = num. of part time tellers who work 9–1






Objective Function (in $ of personnel cost)

Min 90 F + 28 (P1 + P2 + P3 + P4 + P5)


Part Time Hours Cannot Exceed 56 Hours

4 (P1 + P2 + P3 + P4 + P5) < 56


Minimum Num. Tellers Needed By Hour

Time of Day

F + P1 > 10 (9-10)

F + P1 + P2 > 12 (10-11)

0.5 F + P1 + P2 + P3 > 14 (11-12)

0.5 F + P1 + P2 + P3+ P4 > 16 (12-1)

F + P2 + P3+ P4 + P5 > 18 (1-2)

F + P3+ P4 + P5 > 17 (2-3)

F + P4 + P5 > 15 (3-4)

F + P5 > 10 (4-5)


Only 12 Full Time Tellers Available

F < 12

Finally, nonnegativity: F, P1, P2, P3, P4, P5 > 0

Go to file 3-6.xls


Vehicle Loading:Goodman Shipping

How to load a truck subject to weight and volume limitations

Decision: How much of each of 6 items to load onto a truck?

Objective: Maximize the value shipped


Data

0.448

$4.15

3500

$14,525

4

0.144

$3.45

3000

$10,350

3 6521Item

0.0180.0480.0640.125Cu. ft. per lb

$2.75$3.25$3.20$3.10$ / lb3500400045005000Pounds

$9,625$13,000$14,400$15,500Value


Decision Variables

Wi = number of pounds of item i to load onto truck, (where i = 1,…,6)

Truck Capacity

15,000 pounds

1,300 cubic feet


Objective Function (in $ of load value)

Max 3.10W1 + 3.20W2 + 3.45W3 + 4.15W4 + 3.25W5 + 2.75W6


Weight Limit of 15,000 Pounds

W1 + W2 + W3 + W4 + W5 + W6 < 15,000


Volume Limit of 1300 Cubic Feet

0.125W1 + 0.064W2 + 0.144W3 +0.448W4 + 0.048W5 + 0.018W6 < 1300

Pounds of Each Item AvailableW1 < 5000 W4 < 3500W2 < 4500 W5 < 4000W3 < 3000 W6 < 3500

Finally, nonnegativity: Wi > 0, i=1,…,6

Go to file 3-7.xls


Blending Problem:Whole Food Nutrition Center

Making a natural cereal that satisfies minimum daily nutritional requirements

Decision: How much of each of 3 grains to include in the cereal?

Objective: Minimize cost of a 2 ounce serving of cereal


GrainCBA

6

9

25

21

$0.38

178Phosphorus per pound

32822Protein per pound

Minimum Daily

Requirement$0.47$0.33$ per pound

0.42505Magnesium per pound

21416Riboflavin per pound


Decision VariablesA = pounds of grain A to use

B = pounds of grain B to use

C = pounds of grain C to use

Note: grains will be blended to form a 2- ounce serving of cereal


Objective Function (in $ of cost)

Min 0.33A + 0.47B + 0.38C


Total Blend is 2 Ounces, or 0.125 Pounds

A + B + C = 0.125 (lbs)


Minimum Nutritional Requirements

22A + 28B + 21C > 3 (protein)

16A + 14B + 25C > 2 (riboflavin)

8A + 7B + 9C > 1 (phosphorus)

5A + 6C > 0.425 (magnesium)

Finally, nonnegativity: A, B, C > 0

Go to file 3-9.xls


Multiperiod Scheduling:Greenberg Motors

Need to schedule production of 2 electrical motors for each of the next 4 months

Decision: How many of each type of motor to make each month?

Objective: Minimize total production and inventory cost


Decision VariablesPAt = number of motor A to produce in

month t (t=1,…,4)PBt = number of motor B to produce in

month t (t=1,…,4)

IAt = inventory of motor A at end of month t (t=1,…,4)

IBt = inventory of motor B at end of month t (t=1,…,4)


Sales Demand Data

140010003 (March)12007002 (February)

140011004 (April)

10008001 (January)BA

MotorMonth


Production Data

0.91.3Labor hours

$6$10Production costBA

Motor

(values are per motor)

• Production costs will be 10% higher in months 3 and 4

• Monthly labor hours must be between2240 and 2560


Inventory Data

00Beginning inventory

(beginning of month 1)

300450Ending Inventory

(end of month 4)

$0.13$0.18Inventory cost

(per motor per month)

BAMotor

Max inventory is 3300 motors


Production and Inventory Balance(inventory at end of previous period)

+ (production the period)

- (sales this period)

= (inventory at end of this period)



Min 10PA1 + 10PA2 + 11PA3 + 11PA4

+ 6PB1 + 6 PB2 + 6.6PB3 + 6.6PB4

+ 0.18(IA1 + IA2 + IA3 + IA4)

+ 0.13(IB1 + IB2 + IB3 + IB4)


(see next slide)


Production & Inventory Balance

0 + PA1 – 800 = IA1 (month 1)

0 + PB1 – 1000 = IB1

IA1 + PA2 – 700 = IA2 (month 2)

IB1 + PB2 – 1200 = IB2

IA2 + PA3 – 1000 = IA3 (month 3)

IB2 + PB3 – 1400 = IB3

IA3 + PA4 – 1100 = IA4 (month 4)

IB3 + PB4 – 1400 = IB4


Ending Inventory

IA4 = 450

IB4 = 300

Maximum Inventory level

IA1 + IB1 < 3300 (month 1)

IA2 + IB2 < 3300 (month 2)

IA3 + IB3 < 3300 (month 3)

IA4 + IB4 < 3300 (month 4)


Range of Labor Hours2240 < 1.3PA1 + 0.9PB1 < 2560 (month 1)2240 < 1.3PA2 + 0.9PB2 < 2560 (month 2)2240 < 1.3PA3 + 0.9PB3 < 2560 (month 3)2240 < 1.3PA4 + 0.9PB4 < 2560 (month 4)

Finally, nonnegativity: PAi, PBi, IAi, IBi > 0

Go to file 3-11.xls



Session 6 Session 6 –– Integer, Goal, and Nonlinear Integer, Goal, and Nonlinear

Programming ModelsProgramming Models





Variations of BasicLinear Programming

Integer Programming

Goal Programming

Nonlinear Programming


Integer Programming (IP)

Where some or all decision variables are required to be whole numbers.

General Integer Variables (0,1,2,3,etc.)

Values that count how many

Binary Integer Variables (0 or 1)

Usually represent a Yes/No decision


General Integer Example:Harrison Electric Co.

Produce 2 products (lamps and ceiling fans) using 2 limited resources

Decision: How many of each product to make? (must be integers)



Decision Variables

L = number of lamps to make

F = number of ceiling fans to make

$700$600Profit Contribution

305 hr6 hrsAssembly Hours

123 hrs2 hrsWiring Hours

Hours Available

Fans(per fan)

Lamps(per lamp)


LP Model SummaryMax 600 L + 700 F ($ of profit)


2L + 3F < 12 (wiring hours)

6L + 5F < 30 (assembly hours)

L, F > 0


Graphical Solution


Properties of Integer SolutionsRounding off the LP solution might not yield the optimal IP solution

The IP objective function value is usually worse than the LP value

IP solutions are usually not at corner points


Using Solver for IP

IP models are formulated in Excel in the same way as LP models

The additional integer restriction is entered like an additional constraint

int - Means general integer variables

bin - Means binary variables

Go to file 6-1.xls


Binary Integer Example:Portfolio Selection

Choosing stocks to include in portfolio

Decision: Which of 7 stocks to include?

Objective: Maximize expected annual return (in $1000’s)


Stock DataOil Investment Opportunities (Table 6.2)


Decision Variables

Use the first letter of each stock’s name

Example for Trans-Texas Oil:

T = 1 if Trans-Texas Oil is included

T = 0 if not included


Restrictions

Invest up to $3 million

Include at least 2 Texas companies

Include no more than 1 foreign company

Include exactly 1 California company

If British Petro is included, then

Trans-Texas Oil must also be included


Objective Function (in $1000’s return)

Max 50T + 80B + 90D + 120H + 110L + 40S + 75C


Invest up to $3 Million

480T + 540B + 680D + 1000H

+ 700L + 510S + 900C < 3000


Include At Least 2 Texas Companies

T + H + L > 2

Include No More than 1 Foreign Company

B + D < 1

Include Exactly 1 California Company

S + C = 1


If British Petro is included (B=1), then

Trans-Texas Oil must also be included (T=1)

oknot okB=1

okokB=0

T=1T=0

B < Tallows the 3 acceptable combinations and prevents the unacceptable one

Go to file 6-3.xls

Combinationsof B and T


Mixed Integer Models:Fixed Charge Problem

Involves both fixed and variable costs

Use a binary variable to determine if a fixed cost is incurred or not

Either linear or general integer variables deal with variable cost


Fixed Charge Example:Hardgrave Machine Co.

Has 3 plants and 4 warehouses and is considering 2 locations for a 4th plant

Decisions:Which location to choose for 4th plant?How much to ship from each plant to each warehouse?

Objective: Minimize total production and shipping cost


Supply and Demand Data

9,000Los Angeles

$5214,000Pittsburgh15,000New York

35,00046,000Total

$506,000Kansas City12,000Houston

$4815,000Cincinnati10,000Detroit

Production Cost

(per unit)

Monthly

SupplyPlant

Monthly

DemandWarehouse

Note: New plant must supply 11,000 units per month


$325,000$49Birmingham

$400,000$53Seattle

Fixed Cost(per month)

Production Cost(per unit)

Possible Locations for New Plant


Shipping Cost DataHardgrave Machine’s (Table 6.6)


Decision VariablesBinary Variables

Ys = 1 if Seattle is chosen

= 0 if not

YB = 1 if Birmingham is chosen

= 0 if not

Regular Variables

Xij = number of units shipped from plant i to warehouse j



Min 73XCD + 103XCH + 88XCN + 108XCL + 85XKD + 80XKH + 100XKN + 90XKL + 88XPD + 97XPH + 78XPN + 118XPL + 113XSD + 91XSH + 118XSN + 80XSL + 84XBD + 79XBH + 90XBN + 99XBL +

400,000YS + 325,000YB


(see next slide)


Supply Constraints

-(XCD + XCH + XCN + XCL) = -15,000 (Cincinnati)

-(XKD + XKH + XKN + XKL) = - 6,000 (Kansas City)

-(XPD + XPH + XPN + XPL) = -15,000 (Pittsburgh)

Possible Locations for New Plant

-(XSD + XSH + XSN + XSL) = -11,000YS (Seattle)

-(XBD + XBH + XBN + XBL) = -11,000YB (B’ham)


Demand Constraints

XCD + XKD + XPD +XSD + XBD = 10,000 (Detroit)

XCH + XKH + XPH +XSH + XBH = 12,000 (Houston)

XCN + XKN + XPN +XSN + XBN = 15,000 (New York)

XCL + XKL + XPL +XSL + XBL = 9,000 (L. A.)

Choose 1 New Plant Location

YS + YB =1

Go to File 6-5.xls


Goal Programming Models

Permit multiple objectives

Try to “satisfy” goals rather than optimize

Objective is to minimize underachievement of goals


Goal Programming Example:Wilson Doors Co.

Makes 3 types of doors from 3 limited resources

Decision: How many of each of 3 types of doors to make?

Objective: Minimize total underachievement of goals


DataWilson Doors (Table 6.7)


Goals1. Total sales at least $180,000

2. Exterior door sales at least $70,000

3. Interior door sales at lest $60,000

4. Commercial door sales at least $35,000


Regular Decision Variables

E = number of exterior doors made

I = number of interior doors made

C = number of commercial doors made

Deviation Variables

di+ = amount by which goal i is overachieved

di- = amount by which goal i is underachieved


Goal Constraints

Goal 1: Total sales at least $180,000

70E + 110I + 110C + dT- - dT

+ = 180,000

Goal 2: Exterior door sales at least $70,000

70E + dE- - dE

+ = 70,000

Note: Each highlighted deviation variable measures goal underachievement


Goal 3: Interior door sales at least $60,000

110 I + dI- - dI

+ = 60,000

Goal 4: Commercial door sales at least $35,000

110C + dC- - dC

+ = 35,000


Objective FunctionMinimize total goal underachievement

Min dT- + dE

- + dI- + dC

-


The 4 goal constraints

The “regular” constraints (3 limited resources)

nonnegativity


Weighted Goals

When goals have different priorities, weights can be used

Suppose that Goal 1 is 5 times more important than each of the others

Objective Function

Min 5dT- + dE

- + dI- + dC

-


Properties of Weighted GoalsSolution may differ depending on the weights used

Appropriate only if goals are measured in the same units


Ranked GoalsLower ranked goals are considered only if all higher ranked goals are achieved

Suppose they added a 5th goal

Goal 5: Steel usage as close to 9000 lb as possible

4E + 3I + 7C + dS- = 9000 (lbs steel)

(no dS+ is needed because we cannot

exceed 9000 pounds)


Rank R1: Goal 1

Rank R2: Goal 5

Rank R3: Goals 2, 3, and 4

A series of LP models must be solved

1) Solve for the R1 goal while ignoring the other goals

Objective Function: Min dT-

Go to file 6-7.xls


2) If the R1 goal can be achieved (dT- = 0),

then this is added as a constraint and we attempt to satisfy the R2 goal (Goal 5)

Objective Function: Min dS-

3) If the R2 goal can be achieved (dS- = 0),

then this is added as a constraint and we solve for the R3 goals (Goals 2, 3, and 4)

Objective Function: Min dE- + dI

- + dC-


Nonlinear Programming ModelsLinear models (LP, IP, and GP) have linear objective function and constraints

If a model has one or more nonlinear equations (objective or constraint) then the model is nonlinear

Example nonlinear terms: X2, 1/X, XY


Characteristics of Nonlinear Programming (NLP) Models

Difficult to solve

Optimal solutions are not necessarily at corner points

There are both local and global optimal solutions

Solution may depend on starting point

Starting point is usually arbitrary


Nonlinear Programming Example:Pickens Memorial Hospital

Patient demand exceeds hospital’s capacity

Decision: How many of each of 3 types of patients to admit per week?



Decision Variables

M = number of Medical patients to admit

S = number of Surgical patients to admit

P = number of Pediatric patients to admit

Profit FunctionProfit per patient increases as the number of

patients increases (i.e. nonlinear profit function)


ConstraintsHospital capacity: 200 total patients

X-ray capacity: 560 x-rays per week

Marketing budget: $1000 per week

Lab capacity: 140 hours per week


Objective Function (in $ of profit)

Max 45M + 2M2 + 70S + 3S2 + 2MS +

60P + 3P2


M + S + P < 200 (patient cap.)

M + 3S + P < 560 (x-ray cap.)

3M + 5S + 3.5P < 1000 (marketing $)

(0.2+0.001M)x(3M+3S+3P) < 140 (lab hrs)

M, S, P > 0


Using Solver for NLP Models

Solver uses the Generalized Reduced Gradient (GRG) method

GRG uses the path of steepest ascent (or descent)

Moves from one feasible solution to another until the objective function value stops improving (converges)

Go to file 6-8.xls



Session 12 Decision AnalysisSession 12 Decision Analysis





Decision Analysis

For evaluating and choosing among alternatives

Considers all the possible alternatives and possible outcomes


Five Steps in Decision Making

1. Clearly define the problem

2. List all possible alternatives

3. Identify all possible outcomes for each alternative

4. Identify the payoff for each alternative & outcome combination

5. Use a decision modeling technique to choose an alternative


Thompson Lumber Co. Example

1. Decision: Whether or not to make and sell storage sheds

2. Alternatives:

• Build a large plant

• Build a small plant

• Do nothing

3. Outcomes: Demand for sheds will be high, moderate, or low


4. Payoffs

5. Apply a decision modeling method

Outcomes (Demand)

Alternatives

000No plant-20,00050,00090,000Small plant

-120,000100,000200,000Large plantLowModerateHigh


Decision Making Under Risk

Where probabilities of outcomes are available

Expected Monetary Value (EMV) uses the probabilities to calculate the average payoff for each alternative

EMV (for alternative i) =

∑(probability of outcome) x (payoff of outcome)


Outcomes (Demand)

Alternatives



0.20.50.3Probability of outcome

48,000

0

86,000EMV

Chose the large plant

Expected Monetary Value (EMV) Method


Perfect Information

Perfect Information would tell us with certainty which outcome is going to occur

Having perfect information before making a decision would allow choosing the best payoff for the outcome


Expected Value WithPerfect Information (EVwPI)The expected payoff of having perfect

information before making a decision

EVwPI = ∑ (probability of outcome)

x ( best payoff of outcome)


Expected Value ofPerfect Information (EVPI)

The amount by which perfect information would increase our expected payoff

Provides an upper bound on what to pay for additional information

EVPI = EVwPI – EMVEVwPI = Expected value with perfect information

EMV = the best EMV without perfect information


Demand

Alternatives



Payoffs in blue would be chosen based on perfect information (knowing demand level)

0.20.50.3Probability

EVwPI = $110,000


Expected Value of Perfect Information

EVPI = EVwPI – EMV

= $110,000 - $86,000 = $24,000

The “perfect information” increases the expected value by $24,000

Would it be worth $30,000 to obtain this perfect information for demand?


Decision TreesCan be used instead of a table to show alternatives, outcomes, and payofffs

Consists of nodes and arcs

Shows the order of decisions and outcomes


Decision Tree for Thompson Lumber


Folding Back a Decision TreeFor identifying the best decision in the tree

Work from right to left

Calculate the expected payoff at each outcome node

Choose the best alternative at each decision node (based on expected payoff)


Thompson Lumber Tree with EMV’s


Using TreePlan With ExcelAn add-in for Excel to create and solve decision trees

Load the file Treeplan.xla into Excel

(from the CD-ROM)


Decision Trees for Multistage Decision-Making Problems

Multistage problems involve a sequence of several decisions and outcomes

It is possible for a decision to be immediately followed by another decision

Decision trees are best for showing the sequential arrangement


Expanded ThompsonLumber Example

Suppose they will first decide whether to pay $4000 to conduct a market survey

Survey results will be imperfect

Then they will decide whether to build a large plant, small plant, or no plant

Then they will find out what the outcome and payoff are


Thompson LumberOptimal Strategy

1. Conduct the survey

2. If the survey results are positive, then build the large plant (EMV = $141,840)

If the survey results are negative, then build the small plant (EMV = $16,540)


Expected Value of Sample Information (EVSI)

The Thompson Lumber survey provides sample information (not perfect information)

What is the value of this sample information?

EVSI = (EMV with free sample information)

- (EMV w/o any information)


EVSI for Thompson LumberIf sample information had been free

EMV (with free SI) = 87,961 + 4000 = $91,961

EVSI = 91,961 – 86,000 = $5,961


EVSI vs. EVPI

How close does the sample information come to perfect information?

Efficiency of sample information = EVSIEVPI

Thompson Lumber: 5961 / 24,000 = 0.248


Estimating ProbabilityUsing Bayesian Analysis

Allows probability values to be revised based on new information (from a survey or test market)

Prior probabilities are the probability values before new information

Revised probabilities are obtained by combining the prior probabilities with the new information


Known Prior Probabilities

P(HD) = 0.30

P(MD) = 0.50

P(LD) = 0.30

How do we find the revised probabilities where the survey result is given?

For example: P(HD|PS) = ?


It is necessary to understand the Conditional probability formula:

P(A|B) = P(A and B)P(B)

P(A|B) is the probability of event A occurring, given that event B has occurred

When P(A|B) ≠ P(A), this means the probability of event A has been revised based on the fact that event B has occurred


The marketing research firm provided the following probabilities based on its track record of survey accuracy:

P(PS|HD) = 0.967 P(NS|HD) = 0.033

P(PS|MD) = 0.533 P(NS|MD) = 0.467

P(PS|LD) = 0.067 P(NS|LD) = 0.933

Here the demand is “given,” but we need to reverse the events so the survey result is “given”


Finding probability of the demand outcome given the survey result:

P(HD|PS) = P(HD and PS) = P(PS|HD) x P(HD)

P(PS) P(PS)

Known probability values are in blue, so need to find P(PS)

P(PS|HD) x P(HD) 0.967 x 0.30

+ P(PS|MD) x P(MD) + 0.533 x 0.50

+ P(PS|LD) x P(LD) + 0.067 x 0.20


Now we can calculate P(HD|PS):

P(HD|PS) = P(PS|HD) x P(HD) = 0.967 x 0.30

P(PS) 0.57

= 0.509

The other five conditional probabilities are found in the same manner

Notice that the probability of HD increased from 0.30 to 0.509 given the positive survey result


Utility Theory

An alternative to EMV

People view risk and money differently, so EMV is not always the best criterion

Utility theory incorporates a person’s attitude toward risk

A utility function converts a person’s attitude toward money and risk into a number between 0 and 1


Jane’s Utility Assessment

Jane is asked: What is the minimum amount that would cause you to choose alternative 2?


Suppose Jane says $15,000

Jane would rather have the certainty of getting $15,000 rather the possibility of getting $50,000

Utility calculation:

U($15,000) = U($0) x 0.5 + U($50,000) x 0.5

Where, U($0) = U(worst payoff) = 0

U($50,000) = U(best payoff) = 1

U($15,000) = 0 x 0.5 + 1 x 0.5 = 0.5 (for Jane)


The same gamble is presented to Jane multiple times with various values for the two payoffs

Each time Jane chooses her minimum certainty equivalent and her utility value is calculated

A utility curve plots these values


Jane’s Utility Curve


Different people will have different curves

Jane’s curve is typical of a risk avoider

Risk premium is the EMV a person is willing to willing to give up to avoid the risk

Risk premium = (EMV of gamble)

– (Certainty equivalent)

Jane’s risk premium = $25,000 - $15,000

= $10,000


Types of Decision MakersRisk Premium

Risk avoiders: > 0

Risk neutral people: = 0

Risk seekers: < 0


Utility Curves for Different Risk Preferences


Utility as aDecision Making Criterion

Construct the decision tree as usual with the same alternative, outcomes, and probabilities

Utility values replace monetary values

Fold back as usual calculating expected utility values


Decision Tree Example for Mark


Utility Curve for Mark the Risk Seeker


Mark’s Decision Tree With Utility Values

Managerial Analysis & Decision Making Slides - Shinawatra MBA - Semester 2-2010 CC

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