mgs 3100 business analysis final exam review

1

MGS 3100 Business AnalysisFinal Exam Review

Julie Liggett De [email protected]/~mgs3100

Chapter 1Introduction to Modeling

Julie Liggett De [email protected]/~mgs3100

Applied to the first two stages of decision making

THE MODELING PROCESS

AnalysisResults

terp

reta

tion

Abst

ract

ion

Model

Real World

Symbolic World Managerial

Judgment

IntuitionManagementSituation

In

Decisions

AWorld

Take a “Black Box” View of a Model

Decisions(Controllable)

ParametersModel

PerformanceMeasure(s)

ConsequenceParameters(Uncontrollable)

Exogenous variables Endogenous variables

Consequence Variables

2

Take a “Black Box” View of a Model

Model ProfitUnit Cost, Filling

Unit Pie Processing CostFixed Cost

Pie Price

Unit Cost, Dough

Exogenous variables Endogenous variables

Fixed Cost

TYPES OF MODELS

Physical Model Analog Model Symbolic Model

Tangibility Tangible Intangible IntangibleTangibility Tangible Intangible Intangible

Comprehension Easy Harder Hardest

Duplicate & Share Difficult Easier Easiest

Modify & Manipulate

Difficult Easier Easiest

Scope of Use Lowest Wider Widestp

Examples Model AirplaneModel HouseModel City

Road MapSpeedometer

Pie Chart

Simulation ModelAlgebraic Model

Spreadsheet Model

Modeling Techniques Modeling

Techniques

Deterministic Models

ProbabilisticModels

Profit Model Time Series Forecasting Decision Analysis SimulationForecasting

BreakevenPricing for Max ProfitCrossover / Indifferencepoint

Naive / Exp SmoothingRegression (trend)Classical Decomposition(trend + seasonality)

Alternatives,States,PayoffsDecision CriteriaDecision TreesBayes Theorem

Random numbersDistributionsDiscrete VariablesContinuous Variables

Deductive Modeling

3

Deductive Modeling

Top down/ data poor

Focuses on variables

Assumes mathematical relationships & parameter& parameter values

Focuses on variables

Assumes mathematical relationships & parameter& parameter values

Depends on modeler’s knowledge &knowledge & judgments of mathematical relationships & data values& data values

4

Inferential Modeling

Inferential Modeling

Bottom up / data rich

Focuses on variables in existing data

ll ticollections

Analyzes data to determine relationships & torelationships & to estimate parameter values

Focuses on variables in existing data

ll ticollections

Analyzes data to determine relationships & torelationships & to estimate parameter values

5

Places premium on accurate, readily available d t & j d tdata & judgments about future applicability of data

Model building is an iterative process

DEDUCTIVE MODELING

D i i M d li D i i M d li

PROBABILISTICMODELS

DETERMINISTICMODELS

Low Certainty High Certainty

Data PoorDecision Modeling(‘What If?’ Projections, Decision

Analysis, Decision Trees)

Decision Modeling(‘What If?’ Projections, Profit

Model, Optimization)

Model BuildingProcess

INFERENTIAL MODELING

Data Rich

Data Analysis(Forecasting, Simulation,

Statistical Analysis,Parameter Estimation)

Data Analysis(Data Warehouses,

Parameter Evaluation

Chapter 2Spreadsheet ModelingModeling •Selling Price (SP) •Cost

Basic Profit Model

•Quantity (Q)Sales VolumeProduction Volume

Overhead CostSunk CostFixed Cost (FC)Variable Cost (VC)Total Cost (TC)

•Demand (D)•Revenue•Profit

•Contribution margin•Breakeven point•Crossover point.

6

Basic Profit ModelProfit = Revenue – Total CostProfit = SP*Q – (VC*Q + FC)P fit SP*Q VC*Q FCProfit = SP*Q – VC*Q – FCProfit = (SP – VC)*Q - FC

Contribution Margin (CM) = SP-VC, soProfit=CM*Q – FC, and Q=(FC+Profit)/CM

If a profit model has fixed costs of $150K, a sales price of $400, and variable costs of $250, at what quantity will profit be $300k?

Breakeven Point

Set Profit = 0 and solve for Q, finds breakeven quantity:q y

0 = CM*QBE – FCFC = CM*QBEFC / CM = QBE

If a profit model has fixed costs of $150,000, variable costs of $250, and a sales price of $400, what is its breakeven quantity?

Breakeven Point

Set Profit = 0 and solve for Q, finds breakeven quantity:q y

0 = CM*QBE – FCFC = CM*QBEFC / CM = QBE

If a profit model has a breakeven point of 400 units, a sales price of $300 and a variable cost of $75, what are its fixed costs?

Cross-over Point a/k/a Indifference PointProfit =CM *Q FC

Crossover Point

ProfitA=CMA*QA – FCAProfitB=CMB*QB - FCB

Total Cost of Process A = Total Cost of Process B

SetProfitA=ProfitBSolve for Q

CMA*QA – FCA = CMB*QB – FCBCMA*QA – CMB*QB = FCA – FCBQAtoB = (FCA – FCB) / (CMA - CMB)

7

Plan A Plan B Plan C

FC 150 000 450 000 2 850 000

Class ExerciseAnalyze the following plans

FC 150,000 450,000 2,850,000

VC 250 150 100

SP 400 400 400

• How many units must be sold to breakevenHow many units must be sold to breakeven under each?

• Find the Indifference Points between A & B; B & C; A & C

Profit

Influence DiagramInfluence Diagram

Revenue Total Cost

ProcessingCost

IngredientCost

RequiredIngredientQuantities

Pies Demanded

Pie PriceUnit Pie

Processing Cost Fixed CostUnit Cost

FillingUnit Cost

Dough

Quantities

Capstone Computers assembles and sells personal computers. Sales volume depends on

Exercise: Capstone Computer

personal computers. Sales volume depends on the sales price. At $1,000/each, 5,000 computers will be sold; every increase (or decrease) of $100, sales will decrease (or increase, respectively) by 1,000 units.

• Compute the y intercept• Compute the y-intercept.• Compute the slope.• What is the linear equation that describes the

relationship between demand and price?

Rules of Good Model Design and Layoutand Layout

8

Present clearly labeled input variables together.

Clearly label model results.

Include units of measure where appropriate.

Store input variables in separate cells & refer to addresses in formulas.

9

Format the spreadsheet to simplify interpretation.

Separate physical results from financial or economic results.

CHAPTER 9: Simulation

Simulation allows you to quickly and inexpensively acquire knowledge concerning a problem that is usually

Introduction

acquire knowledge concerning a problem that is usually gained through experience (which is often costly and time consuming).

An experimental device (simulator) that “acts like” (simulates) the system of interest.

Goal: To create an environment where information about alternative actions canbe obtained in a quick, cost-effective manner through experimentation.

10

Simulation is frequently used because:

When should simulation be used?

1. Analytical models can be difficult or impossible toobtain due to complicating factors.

2. Analytical models typically predict only average or“steady-state” (long-run) behavior.

3. Simulation models can be developed on a PC or workstation, with a minimum of computing and mathematical skill.

• Simulation Models

Terminology to Know

• Probability Distributions • Cumulative Probabilities• Random Variables/Numbers (RN)−discrete vs. continuous

• Expected Value

Calculating Expected Values

To calculate expected profit (or mean profit), multiply profits by respective probabilities for each item and then sum to get the expectedeach item and then sum to get the expected profit.

Calculating Expected Values

What is the expected profit for the following table?g

11

Calculating Expected ValuesCalculating Expected Values

Question 4 of Sample Exam 11. Construct a Cumulative Probability

Simulation

Distribution2. Generate Random Numbers3. Project a Random Observation of

the Variable

1. Construct a Cumulative Probability Distribution

Cases Prob. Cumulative

3 .156 .156

4 .287 .443

5 .362 .805

6 .195 1.000

2. Generate Random Numbers

Week Random Number

1 .587

2 .266

3 7023 .702

4 .307

12

3. Project Random Observation of Variable

Cases Prob. Cum Range

3 .156 .156 .000-.155

Week Random Number

Cases

3 .156 .156 .000 .155

4 .287 .443 .156-.442

5 .362 .805 .443-.804

6 .195 1.000 .805-1.000

1 .587 52 .266 43 .702 54 .307 4

Simulation & Random NumbersQuestion 5 of Sample Exam1

Simulation & Random NumbersSimulation & Random NumbersQuestion 6 of Sample Exam1

STOP HERE

13

Chapter 13Forecasting

a) Curve Fitting

1.1.Causal Forecasting ModelsCausal Forecasting Models

2.2.TimeTime--Series Forecasting Models:Series Forecasting Models:

Quantitative Forecasting ModelsQuantitative Forecasting Models

a) Curve Fitting

b) Moving Averages (Naive)i. Simple n-Period Moving Averageii. Weighted n-Period Moving Average

c) Exponential Smoothing) p gi. Basic modelii. Holt’s Model (exponential smoothing

with trend)

d) Seasonality

Causal Forecasting Models Requirements

14

I d d t d d d tIndependent and dependent variables must share a relationship

W t k th l f thWe must know the values of the independent variables when we make the forecast

: Independent variable

Important Variables:

X

: Value of actual dependent variable

: Average of dependent variable values (Y bar).Y

Y

: Forecast of dependent variable (Y hat).Y

1. Curve Fitting

2. Moving Averages a Nai e

TimeTime--Series Forecasting ModelsSeries Forecasting Models

a. Naiveb. Simple n-Period Moving Averagec. Weighted n-Period Moving Average

3. Exponential Smoothinga. Basic modela. Basic modelb. Holt’s Model (exponential smoothing with

trend)

4. Seasonality

15

MONTH

ACTUAL SALES ($000s)

THREE-MONTH SIMPLE MOVING

AVERAGE FORECAST

FOUR-MONTH SIMPLE MOVING AVERAGE

FORECASTJan. 20Feb. 24Mar. 27Apr. 31 (20 + 24 + 27)/3 = 23.67p ( )May 37 (24 + 27 + 31)/3 = 27.33 (20 + 24 + 27 + 31)/4 = 25.50June 47 (27 + 31 + 37)/3 = 31.67 (24 + 27 + 31 + 37)/4 = 29.75July 53 (31 + 37 + 47)/3 = 38.33 (27 + 31 + 37 + 47)/4 = 35.50Aug. 62 (37 + 47 + 53)/3 = 45.67 (31 + 37 + 47 + 53)/4 = 42.00Sep. 54 (47 + 53 + 62)/3 = 54.00 (37 + 47 + 53 + 62)/4 = 49.75Oct. 36 (53 + 62 + 54)/3 = 56.33 (47 + 53 + 62 + 54)/4 = 54.00Nov. 32 (62 + 54 + 36)/3 = 50.67 (53 + 62 + 54 + 36)/4 = 51.25Dec. 29 (54 + 36 + 32)/3 = 40.67 (62 + 54 + 36 + 32)/4 = 46.00

Three- and Four- Month Simple Moving Averages

MONTH

ACTUAL SALES ($000s)

THREE-MONTH SIMPLE MOVING

AVERAGE FORECAST

FOUR-MONTH SIMPLE MOVING AVERAGE

FORECASTJan. 20Feb. 24Mar. 27Apr. 31 (20 + 24 + 27)/3 = 23.67p ( )May 37 (24 + 27 + 31)/3 = 27.33 (20 + 24 + 27 + 31)/4 = 25.50June 47 (27 + 31 + 37)/3 = 31.67 (24 + 27 + 31 + 37)/4 = 29.75July 53 (31 + 37 + 47)/3 = 38.33 (27 + 31 + 37 + 47)/4 = 35.50Aug. 62 (37 + 47 + 53)/3 = 45.67 (31 + 37 + 47 + 53)/4 = 42.00Sep. 54 (47 + 53 + 62)/3 = 54.00 (37 + 47 + 53 + 62)/4 = 49.75Oct. 36 (53 + 62 + 54)/3 = 56.33 (47 + 53 + 62 + 54)/4 = 54.00Nov. 32 (62 + 54 + 36)/3 = 50.67 (53 + 62 + 54 + 36)/4 = 51.25Dec. 29 (54 + 36 + 32)/3 = 40.67 (62 + 54 + 36 + 32)/4 = 46.00

412131415

16ˆ yyyyy

+++=

Question 6 of Sample Exam2:Using a Naïve forecast, what is the forecast for Q1, 2000?

A B C D E F

1 Year Quarter Enroll-ment Forecast Error Abs Error

2 1997 1 313 313 3 2 285 313 ? 4 3 312 292 20 20 5 4 339 307 32 32 6 1998 1 359 331 28 28 7 2 320 ? 8 3 356 328 28 28 9 4 385 349 36 ? 10 1999 1 396 376 20 20 11 2 367 ? 12 3 397 373 24 24 13 4 423 391 32 32 14 2000 1 415 15 Bias = ? 16 Alpha = 0.75 MAD =

Recent data is more important than old data

425160ˆ yyyy ααα ++=

16

Constraints:

Weights are positive numbersWeights are positive numbersAssign smaller weights to older dataWeights sum to 1

alpha2 = 0.167 Month Actual Sales (000) 3month WMA Fcst Absolute Erroralpha1 = 0.333 January 20alpha0 = 0.500 February 24SUM OF WTS= 1.00 March 27

April 31 24.83 6.17May 37 28.50 8.50June 47 33.33 13.67July 53 41.00 12.00August 62 48.33 13.67September 54 56.50 2.50October 36 56.50 20.50November 32 46.34 14.34December 29 37.01 8.01

Sum = 99.35MAD = 11.04

Use Solver to find the optimal weights

Exponential SmoothingForecast for t + 1 Observed in t Forecast for t

tt1t y)1(yy α−+α=+

Where is a user-specified constant: α 10 ≤≤ α

Questions 3-5 of Sample Exam2:Forecast for 1998 Q2 (cell D7)?Error for 1997 Q2 (cell E3)?Absolute error for 1998 Q4 (cell F9)?

A B C D E F

1 Year Quarter Enroll-ment Forecast Error Abs Error

2 1997 1 313 313 3 2 285 313 ? 4 3 312 292 20 20 5 4 339 307 32 32 6 1998 1 359 331 28 28 7 2 320 ? 8 3 356 328 28 28 9 4 385 349 36 ? 10 1999 1 396 376 20 20 11 2 367 ? 12 3 397 373 24 24 13 4 423 391 32 32 14 2000 1 415 15 Bias = ? 16 Alpha = 0.75 MAD =

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Measures of Comparison

forecastsofnumber

salesforecastsalesactualMAD forecastsall

−=

∑forecastsofnumber

forecastsofnumbersalesactual

salesforecastsalesactual

MAPE forecastsall∑ ∗

−

=%100

forecastsofnumber

salesforecastsalesactualMSE

n

t∑

=

−= 1

2)(

A B C D E F G1 Alpha 0.92

3 Time Demand

Exponential Smoothing Forecast Error Error2 Abs Error MAPE

Errors, Absolute Errors, & Errors SquaredErrors, Absolute Errors, & Errors Squared

3 Time Demand Forecast Error Error2 Abs Error MAPE4 1 10 8.005 2 14 9.80 4.20 17.64 4.20 30.006 3 19 13.58 5.42 29.38 5.42 28.537 4 26 18.46 7.54 56.88 7.54 29.018 5 31 25.25 5.75 33.11 5.75 18.569 6 35 30.42 4.58 20.93 4.58 13.07

10 7 39 34.54 4.46 19.87 4.46 11.4311 8 44 38.55 5.45 29.66 5.45 12.3812 9 51 43.46 7.54 56.92 7.54 14.7912 9 51 43.46 7.54 56.92 7.54 14.7913 10 55 50.25 4.75 22.60 4.75 8.6414 11 61 54.52 6.48 41.93 6.48 10.6215 12 54 60.35 -6.35 40.35 6.35 11.7616 13 54.641718 SUM = 49.82 369.28 62.52 188.7919 MAD = 5.6820 MAPE = 17.1621 MSE = 33.57

1. Look at original data to see seasonal pattern. Examine the data & hypothesize an m-period seasonal pattern.

2. Deseasonalize the Data

18

3. Forecast using deseasonalized data 4. Seasonalize the forecast to account for the seasonal pattern

Gillett Coal Mine

Coal Receipts Over a Nine-Year Period

2,000

2,500

3,000

0 To

ns)

0

500

1,000

1,500

1-1

1-3

2-1

2-3

3-1

3-3

4-1

4-3

5-1

5-3

6-1

6-3

7-1

7-3

8-1

8-3

9-1

9-3

Tim e (Year and Quarter)

Coa

l (00

0

1. Look at original data to see seasonal pattern. Examine the data & hypothesize an m-period seasonal pattern.

19

Deseasonalized Data

1,500.0

2,000.0

2,500.0

3,000.000

0 To

ns)

-

500.0

1,000.0

1-1

1-2

1-3

1-4

2-1

2-2

2-3

2-4

3-1

3-2

3-3

3-4

4-1

4-2

4-3

4-4

5-1

5-2

5-3

5-4

6-1

6-2

6-3

6-4

7-1

7-2

7-3

7-4

8-1

8-2

8-3

8-4

9-1

9-2

9-3

9-4

Tim e (Year & Qtr)

Coa

l (0

2. Deseasonalized the Data

2.Deseasonalize the Data

a) Calculate a series of m-period moving averages, where m is the length of the seasonal patternseasonal pattern.

b) Center the moving average in the middle of the data from which it was calculated.

c) Divide the actual data at a given point in the series by the centered moving average corresponding to the same point.

d) Develop seasonal index e) Divide actual data by the seasonal index

Time Coal 4 Period Year-Qtr Receipts Moving Average

1-1 2,159 -----1-2 1,203 -----1-3 1,094 1,6131-4 1,996 1,5942-1 2,081 1,6262-2 1,332 1,721, ,2-3 1,476 1,8562-4 2,533 1,8983-1 2,249 1,9483-2 1,533 2,0633-3 1,935 2,0603-4 2,523 2,0504-1 2,208 2,066

(2,159+1,203+1,094+1,996)/4 = 1,613

a)Calculate a series of m-period moving averages, where m is the length of the seasonal pattern.

Time Coal 4 Period CenteredYear-Qtr Receipts Moving Average Moving Average

1-1 2,1591-2 1,2031-3 1,094 1,613 1,6031-4 1,996 1,594 1,6102-1 2,081 1,626 1,6742-2 1,332 1,721 1,7882-3 1,476 1,856 1,877

(1613 + 1594)/2 = 16032-4 2,533 1,898 1,923

3-1 2,249 1,948 2,0053-2 1,533 2,063 2,0613-3 1,935 2,060 2,0553-4 2,523 2,050 2,0584-1 2,208 2,066 2,0644-2 1,597 2,061 2,0874-3 1,917 2,112 2,1634-4 2,726 2,213 2,255

1603

b)Center the moving average in the middle of the data from which it was calculated.

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Time Coal 4 Period Centered Ratio of Coal Receipts toYear-Qtr Receipts Moving Average Moving Average Centered Moving Average

1-1 2,1591-2 1,2031-3 1,094 1,613 1,603 0.6821-4 1,996 1,594 1,610 1.2402-1 2,081 1,626 1,674 1.2442-2 1,332 1,721 1,788 0.7452-3 1,476 1,856 1,877 0.7872 4 2 533 1 898 1 923 1 317

1,094 / 1,603 = 0.6822-4 2,533 1,898 1,923 1.3173-1 2,249 1,948 2,005 1.1223-2 1,533 2,063 2,061 0.7443-3 1,935 2,060 2,055 0.9423-4 2,523 2,050 2,058 1.226

c) Divide the actual data at a given point in the series by the centered moving average corresponding to the same point.

Time Coal 4 Period Centered Ratio of Coal Receipts to SeasonalYear-Qtr Receipts Moving Average Moving Average Centered Moving Average Indices

1-1 2,159 1.1121-2 1,203 0.7861-3 1,094 1,613 1,603 0.682 0.8631-4 1,996 1,594 1,610 1.240 1.2382-1 2,081 1,626 1,674 1.244 1.1122-2 1,332 1,721 1,788 0.745 0.7862-3 1,476 1,856 1,877 0.787 0.8632-4 2,533 1,898 1,923 1.317 1.2383-1 2,249 1,948 2,005 1.122 1.1123-2 1,533 2,063 2,061 0.744 0.786

d)Develop seasonal index for each quarter• Group ratios by quarter

3-3 1,935 2,060 2,055 0.942 0.8633-4 2,523 2,050 2,058 1.226 1.238

p y q• Average all of the ratios to moving

averages quarter by quarter• Add Seasonal Indices data to table• Normalize the seasonal index

Time Coal 4 Period Centered Ratio of Coal Receipts to Seasonal DeseasonalizedYear-Qtr Receipts Moving Average Moving Average Centered Moving Average Indices Data

1-1 2,159 1.112 1,941.0 1-2 1,203 0.786 1,529.8 1-3 1,094 1,613 1,603 0.682 0.863 1,267.7 1-4 1,996 1,594 1,610 1.240 1.238 1,611.9 2-1 2,081 1,626 1,674 1.244 1.112 1,870.9 2-2 1,332 1,721 1,788 0.745 0.786 1,693.8 2-3 1,476 1,856 1,877 0.787 0.863 1,710.3 2-4 2,533 1,898 1,923 1.317 1.238 2,045.6 3-1 2,249 1,948 2,005 1.122 1.112 2,021.9 3-2 1,533 2,063 2,061 0.744 0.786 1,949.4 3-3 1,935 2,060 2,055 0.942 0.863 2,242.2 3-4 2,523 2,050 2,058 1.226 1.238 2,037.5

e)Divide actual data by the seasonal index

See Q1 & Q2 on Sample Exam2.

Time Coal 4 Period Centered Ratio of Coal Receipts to Seasonal DeseasonalizedYear-Qtr Receipts Moving Average Moving Average Centered Moving Average Indices Data Forecast

1-1 2,159 1.108 1,948.1 1,948.1 1-2 1,203 0.784 1,535.4 1,948.1 1-3 1,094 1,613 1,603 0.682 0.860 1,272.3 1,678.5 1-4 1,996 1,594 1,610 1.240 1.234 1,617.8 1,413.1 2-1 2,081 1,626 1,674 1.244 1.108 1,877.8 1,546.8 2-2 1,332 1,721 1,788 0.745 0.784 1,700.0 1,763.0 2-3 1,476 1,856 1,877 0.787 0.860 1,716.6 1,721.9 2-4 2,533 1,898 1,923 1.317 1.234 2,053.1 1,718.4 3-1 2,249 1,948 2,005 1.122 1.108 2,029.3 1,937.1 3 2 1 533 2 063 2 061 0 744 0 784 1 956 5 1 997 4

3. Forecast method in deseasonalized terms• Review the graphed deseasonalized data to

3-2 1,533 2,063 2,061 0.744 0.784 1,956.5 1,997.4 3-3 1,935 2,060 2,055 0.942 0.860 2,250.4 1,970.7 3-4 2,523 2,050 2,058 1.226 1.234 2,045.0 2,153.4

g preveal pattern

• Use forecasting method that accounts for the pattern in the deseasonalized data

• Use Excel’s Solver to minimize the MSE

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Time Coal 4 Period Centered Ratio of Coal Receipts to Seasonal Deseasonalized SeasonalizeYear-Qtr Receipts Moving Average Moving Average Centered Moving Average Indices Data Forecast Forecast

1-1 2,159 ----- ----- ----- 1.108 1,948.1 1,948.1 2,159.000 1-2 1,203 ----- ----- ----- 0.784 1,535.4 1,948.1 1,526.409 1-3 1,094 1,613 1,603 0.682 0.860 1,272.3 1,678.5 1,443.212 1-4 1,996 1,594 1,610 1.240 1.234 1,617.8 1,413.1 1,743.439 2-1 2,081 1,626 1,674 1.244 1.108 1,877.8 1,546.8 1,714.276 2-2 1,332 1,721 1,788 0.745 0.784 1,700.0 1,763.0 1,381.390 2-3 1,476 1,856 1,877 0.787 0.860 1,716.6 1,721.9 1,480.540 2-4 2,533 1,898 1,923 1.317 1.234 2,053.1 1,718.4 2,120.128 3-1 2,249 1,948 2,005 1.122 1.108 2,029.3 1,937.1 2,146.723 3-2 1,533 2,063 2,061 0.744 0.784 1,956.5 1,997.4 1,564.974 3-3 1,935 2,060 2,055 0.942 0.860 2,250.4 1,970.7 1,694.495 3 4 2 523 2 050 2 058 1 226 1 234 2 045 0 2 153 4 2 656 854

4. Reseasonalize the forecast to account for the seasonal pattern• Multiply the deseasonalized forecast by the

3-4 2,523 2,050 2,058 1.226 1.234 2,045.0 2,153.4 2,656.854

seasonal index for the appropriate period. • Graph the actual Coal Receipts and

Seasonalized Forecast

Qualitative Forecasting

Models

Expert Judgment

Consensus Panel

22

Delphi Method

Coordinator requests forecasts

Coordinator receives Individual forecasts

Coordinator determines(a) Median response(b) Range of middle

50% of answers

Delphi Method

50% of answers

Coordinator requests explanations from any expert whose estimate

is not in the middle 50%

Coordinator sends to all experts(a) Median response

(b) Range of middle 50%(c) Explanations

Grassroots Forecasting Market Research

23

Chapter 8Decision AnalysisAnalysis

Terminology

States of natureStates of nature

Payoff / payoff table

Probability

The payoff tablepayoff table is a fundamental component in decision analysis models

State of Nature1 2 … mDecision

d1 r11 r12 … r1m

d2 r21 r22 r2d2 r21 r22 … r2m

dn rn1 rn2 … rnm

… … … … …

Three Classes of Decision Models

Decisions under:

certainty

uncertainty

risk

24

Decisions under certainty:

If I know for sure that it will be raining when I leave work this afternoon, should I take my

mbrella to ork toda ?umbrella to work today?

Decisions under certainty:

If I know for sure that it will be raining when I leave work this afternoon, should I take my

mbrella to ork toda ?

Rain

Take Umbrella 0

umbrella to work today?

Take Umbrella 0Do Not -7.00

Decisions under risk:Decisions under risk:

Multiple states of natureProbabilities used to capture likelihoodHistorical frequencies vs subjective estimates

We calculate Expected ReturnsExpected Returns

25

The expected return (ERi) associated with decision i is

ER Σm

The decision is based on the maximum expected return. In other

ERi = Σrijpj = ri1p1 + ri2p2 + … + rimpmj=1

maximum expected return. In other words, i* is the optimal decision where:

ERi* = maximum overall i of ERi

A B C D E F1 Selling Price 752 Purchase Cost 403 Goodwill Cost 50

The Newsvendor ModelThe Newsvendor ModelThe Newsvendor ModelThe Newsvendor Model

3 Goodwill Cost 5045 States of Nature6 Decision 0 1 2 3 Expected Return7 0 0 -50 -100 -150 -858 1 -40 35 -15 -65 -12.59 2 -80 -5 70 20 22.5

10 3 -120 -45 30 105 7.5111112 Probabilities 0.1 0.3 0.4 0.2

Calculate the Expected Return

Decisions under uncertainty


Multiple states of nature

26


Multiple states of nature

Don’t know what state of nature will occur


LaplaceMaximinMaximaxMinimax regretMinimax regret

Maximin Maximin

extremely yconservative or pessimistic approach to making decisions

27

Maximin

Evaluate minimumminimum possible return associated with each decision.

Maximin

Select decisionSelect decision yielding maxmaximum value of minminimum returnsreturns.

Maximin

ReviewReview homework problem

See Q3 of sample exam3

A B C D E F1 Selling Price 752 Purchase Cost 403 Goodwill Cost 5045 States of Nature6 Decision 0 1 2 3 Expected Return6 Decision 0 1 2 3 Expected Return7 0 0 -50 -100 -150 -858 1 -40 35 -15 -65 -12.59 2 -80 -5 70 20 22.5

10 3 -120 -45 30 105 7.51112 Probabilities 0.1 0.3 0.4 0.2

EVPI = -expected return

with perfectinformation

maximum possibleexpected returnwithout sample

information

28

A B C D E F1 Selling Price 752 Purchase Cost 403 Goodwill Cost 5045 States of Nature6 Decision 0 1 2 3 Expected Return6 Decision 0 1 2 3 Expected Return7 0 0 -50 -100 -150 -858 1 -40 35 -15 -65 -12.59 2 -80 -5 70 20 22.5

10 3 -120 -45 30 105 7.51112 Probabilities 0.1 0.3 0.4 0.2

C l l t th EVPICalculate the EVPI

Q6 of sample exam3

Prior probabilities: PPrior probabilities: Probabilities that are initial estimates, such as P(S) and P(W).

Sonorola has estimated the prior probabilities

DECISIONS UNDER UNCERTAINTY

Sonorola has estimated the prior probabilities as P(S) = 0.45 and P(W) = 0.55.

Joint probabilities Joint probabilities

Marginal probabilitiesMarginal probabilities

Posterior probabilitiesPosterior probabilities: Conditional probabilities, such as P(S|E).

Bayes’ Theorem is used to determine the posterior probabilities.

Calculating Posterior Probabilities:

1. Enter given Reliabilities (conditional probabilities).

2. Calculate Joint Probabilities by multiplying y p y gReliabilities by Prior Probabilities.

3. Compute Marginal Probabilities by summing the entries in each row.

4. Generate Posterior Probabilities by dividing each row entry of joint probability table by its row sum.

P(E|W)

P(D|W)

P(S) P(W)

P(E&S)

P(W|E)P(W|D)

29

Good Credit Risk Bad Credit RiskSteady Job 0.90 0.25

ReliabilitiesP(S|B)

P(S|G)

Question 2 of Sample Exam3

No Steady Job 0.10 0.75

Prior Probabilities 0.25 0.75

Joint & Marginal ProbabilitiesSteady Job 0.225 0.1875 0.4125No Steady Job 0.025 0.5625 0.5875

P(G) P(B)

P(N|B)P(S|G)

P(N|G)

Posterior ProbabilitiesSteady Job 0.55 0.45No Steady Job 0.04 0.96

Decision Trees

GGraphical device for analyzing decisions under risk

Useful when there is are sequences of decisions.

Bayes’ Theorem is used to incorporate new information (posterior probabilities) into the processprocess.

CREATING A DECISION TREECREATING A DECISION TREE

A square nodesquare node represents a point at which a decision must be made.a decision must be made.Each line (branch) leading from the square represents a possible decision.

A circular nodecircular node represents an event (a situation when the outcome is not

Each line (branch) leading from the circle represents a possible outcome.

certain).

To use the decision tree to find the optimal decision, we must append:

• Probabilities for each branch emanating from h i l d

•• Terminal valuesTerminal values (the return associated with each terminal position).

each circular node.

30

FOLDING BACK

To solve a decision tree, one works backward (i.e., from right to left) by folding backfolding back the tree.

1. Fold back the terminal branches by calculating an expected value for each terminal node.

Expected terminal value = 30(0.45) + (-8)(0.55) = 9.10

Question 10 of sample exam3

Conditional Probability: For two events A and B, the conditional probability [P(A|B)], is the probability of event A occurs given that event B

Incorporating New Information

probability of event A occurs given that event B will occur.

For example, P(E|S) is the conditional probability that marketing gives an encouraging report given given that the market is in fact going to be strongthat the market is in fact going to be strong.

Marketing has the following “track record” in predicting the market:

P(E|S) = 0.6P(E|S) = 0.6P(D|S) = 1 P(D|S) = 1 -- P(E|S) = 0.4P(E|S) = 0.4

P(D|W) = 0.7P(D|W) = 0.7P(E|W) = 1 P(E|W) = 1 -- P(D|W) = 0.3P(D|W) = 0.3

31

INCORORATING POSTERIOR PROBABILITIES IN THE DECISION TREE

IVSW

P(W|E)

30

-820

I

II V

VIE

AB

C

S

WS

W

S

P(W|E)

20

75

1530

III

VII

VIII

IX

D

AB

C

S

WS

WS

W

P(W|D)

P(W|D)

-820

75

15

EVSI = -maximum possible

expected returnwith samplei f ti


i f ti

THE EXPECTED VALUE OF SAMPLE INFORMATION

information information

Q8 & Q9 of sample exam3

EVPI = -expected return

with perfect information


i f ti

THE EXPECTED VALUE OF PERFECT INFORMATION

information

EVPI = 30(0.45) + 15(0.55) – 12.85 ≈ 8.90

Q6 of sample exam3The EVPI is an upper bound of how much one would be willing to pay for sample information.

Q6 of sample exam3

Chapter 15Statistical ProcessProcess Control

32

Take periodic samples from process

Plot sample points on control chart

Determine if process is within limits

Variation

1 Common Causes1. Common CausesVariation inherent in a processEliminated through system improvements

Variation

2 Special Causes2. Special CausesVariation due to identifiable factorsModified through operator or management action

Attribute measures

ProductProduct characteristic evaluated with a discrete choice:Good / bad Yes / NoPass / Fail

33

Attribute measures

ProductProduct characteristic evaluated with a discrete choice:Good / bad Yes / NoPass / Fail

Attribute measures

ProductProduct characteristic evaluated with a discrete choice:

Good / bad Yes / NoPass / Fail

Variable measures

Measurable product characteristic:

Length size weightLength, size, weight, height, time, velocity

Variable measures

Measurable product characteristic:


34

Variable measures

Measurable product characteristic


Control ChartsControl Charts

Graphs that establish process control limits

Process Control Chart

Uppercontrol

li it

Out of control

limit

Processaverage

Lowercontrol

1 2 3 4 5 6 7 8 9 10Sample number

controllimit

Figure 15.1

35

To develop Control Charts:

Use in-control data

If non-random causes are present, find them and discard data

Correct control chart limits

Control ChartsControl Charts Measures Description p Chart Attributes Calculates percent defectives

in samplep

r Chart (range chart)

Variables Reflects the amount of dispersion in a sample

x bar Chart (mean chart)

Variables Indicates how sample results relate to the process average

Cp Process Measures the capability of a p (Process Capability

Ratio)

Capability p y

process to meet design specifications

Cpk (Process Capability

Index)

Process Capability

Indicates if the process mean has shifted away from design target

p-ChartUCL = p + zσp

LCL = p - zσppwhere

z = the number of standard deviations from the process average

p = the sample proportion defective; an estimate of the process average

σp = the standard deviation of the sample proportion,σp the standard deviation of the sample proportion, computed as:

σp = p(1 - p)

n

20 samples of 100 pairs of jeans

NUMBER OF PROPORTION

p-Chart Example ~Western Jeans Company

NUMBER OF PROPORTIONSAMPLE DEFECTIVES DEFECTIVE

1 6 .062 0 .003 4 .04: : :: : :

20 18 .18200

Example 15.1

36


p-Chart Example ~Western Jeans Company20 samples of 100 pairs of jeans


1 6 .062 0 .003 4 .04: : :

p =total defectives

total sample observations

: : :20 18 .18

200

Example 15.1

= 200 / 20(100)= 0.10


p-Chart Example ~Western Jeans Company20 samples of 100 pairs of jeans


1 6 .062 0 .003 4 .04: : :

p = 0.10

UCL = p + z = 0.10 + 3p(1 - p)

n0.10(1 - 0.10)

100

UCL = 0.190: : :

20 18 .18200

Example 15.1

LCL = 0.010

LCL = p - z = 0.10 - 3p(1 - p)

n0.10(1 - 0.10)

100

0 16

0.18

0.20

UCL = 0.190

p-Chart Example ~Western Jeans Company

0.08

0.10

0.12

0.14

0.16

Prop

ortio

n de

fect

ive

p = 0.10

0.02

0.04

0.06

P

Sample number2 4 6 8 10 12 14 16 18 20

LCL = 0.010

Range ( R ) Chart

UCL = D R LCL = D RUCL = D4R LCL = D3R

R = ∑Rk

where:where:

R = range of each samplek = number of samples

37

n A2 D3 D4

SAMPLE SIZE FACTOR FOR x-CHART FACTORS FOR R-CHART

2 1.88 0.00 3.273 1.02 0.00 2.574 0 73 0 00 2 28

Factors for R-Chart: D3 & D4

4 0.73 0.00 2.285 0.58 0.00 2.116 0.48 0.00 2.007 0.42 0.08 1.928 0.37 0.14 1.869 0.44 0.18 1.82

10 0.11 0.22 1.7811 0.99 0.26 1.7412 0.77 0.28 1.7213 0.55 0.31 1.6914 0.44 0.33 1.6715 0.22 0.35 1.6516 0.11 0.36 1.6417 0.00 0.38 1.6218 0.99 0.39 1.6119 0.99 0.40 1.6120 0.88 0.41 1.59

Table 15.1

R-Chart Example ~ Goliath Tool Company

R-Chart Example ~ Goliath Tool Company

OBSERVATIONS (SLIP-RING DIAMETER, CM)SAMPLE k 1 2 3 4 5 x RSAMPLE k 1 2 3 4 5 x R

1 5.02 5.01 4.94 4.99 4.96 4.98 0.082 5.01 5.03 5.07 4.95 4.96 5.00 0.123 4.99 5.00 4.93 4.92 4.99 4.97 0.084 5.03 4.91 5.01 4.98 4.89 4.96 0.145 4.95 4.92 5.03 5.05 5.01 4.99 0.136 4.97 5.06 5.06 4.96 5.03 5.01 0.107 5.05 5.01 5.10 4.96 4.99 5.02 0.14

R = max – min = 5.02 – 4.94 = 0.08

8 5.09 5.10 5.00 4.99 5.08 5.05 0.119 5.14 5.10 4.99 5.08 5.09 5.08 0.15

10 5.01 4.98 5.08 5.07 4.99 5.03 0.1050.09 1.15

Example 15.3

∑RkR = = = 0.115

1.1510

UCL = D4R = 2.11(0.115) = 0.243LCL = D3R = 0(0.115) = 0

R-Chart Example ~Goliath Tool Company

LCL D3R 0(0.115) 0

UCL = 0.243

nge

R = 0.115

0.28 –

0.24 –

0.20 –

0.16 –

LCL = 0

Ra

Sample number

|1

|2

|3

|4

|5

|6

|7

|8

|9

|10

0.12 –

0.08 –

0.04 –

0 –

Example 15.3

38

x-Chart Calculations

UCL = x + A2R LCL = x - A2R= =

x = x1 + x2 + ... xk

k=

where

x = the average of the sample meansR bar = the average range values

=

OBSERVATIONS (SLIP-RING DIAMETER, CM)SAMPLE k 1 2 3 4 5 x R

x = = = 5.01 cm= ∑xk

50.0910

x-Chart Example

SAMPLE k 1 2 3 4 5 x R1 5.02 5.01 4.94 4.99 4.96 4.98 0.082 5.01 5.03 5.07 4.95 4.96 5.00 0.123 4.99 5.00 4.93 4.92 4.99 4.97 0.084 5.03 4.91 5.01 4.98 4.89 4.96 0.145 4.95 4.92 5.03 5.05 5.01 4.99 0.136 4.97 5.06 5.06 4.96 5.03 5.01 0.107 5.05 5.01 5.10 4.96 4.99 5.02 0.14

UCL = x + A2R = 5.01 + (0.58)(0.115) = 5.08

LCL = x - A2R = 5.01 - (0.58)(0.115) = 4.94

=

=

k 10

8 5.09 5.10 5.00 4.99 5.08 5.05 0.119 5.14 5.10 4.99 5.08 5.09 5.08 0.15

10 5.01 4.98 5.08 5.07 4.99 5.03 0.1050.09 1.15

Example 15.3

Question 15 from Sample Exam3:Average Thickness: 005 inchAverage Range: 0015 inchSample Size: 3

x-Chart:Sample

SizeA2 D3 D4

2 1.88 0 3.27

3 1 02 0 2 573 1.02 0 2.57

4 0.73 0 2.28

5 0.58 0 2.11

6 0.48 0 2.00

Using x- and R-charts together

Each measures the process differently Both process average (x bar chart) and variability (R chart) must be in y ( )control

39

Sample Size Determination

Attribute control charts (p chart)• 50 to 100 parts in a sample

Sample Size Determination

Variable control charts (R- & x bar- charts)• 2 to 10 parts in a sample

Process Capability

• Control limits (the “Voice of the Process” or the “Voice of the Data”): based on natural variations (common causes) ( )

• Tolerance limits (the “Voice of the Customer”): customer requirements

• Process Capability: A measure of how “ bl ” th i t t t“capable” the process is to meet customer requirements; compares process limits to tolerance limits

Process Capability Measures

Process Capability Ratio (Cp )

t lCCpp ==

==

tolerance rangeprocess range

upper specification limit -lower specification limit

66σ

40

Process Capability Measures

Process Capability Index ( Cpk )

Cpk = minimum

x - lower specification limit3σ

=

upper specification limit - x3σ

=,

D iDesign Specifications

Process

Computing Cpk

Net weight specification = 9.0 oz ± 0.5 ozProcess mean = 8.80 oz

Munchies Snack Food Company

Process standard deviation = 0.12 oz

Cpk = minimum

x - lower specification limit3σ

=

upper specification limit - x=,

= minimum , = 0.83

3σ

8.80 - 8.503(0.12)

9.50 - 8.803(0.12)

Example 15.7

Computing Cpk

Net weight specification = 5 inches. ± 0.004 inchesProcess mean = 5.001 inches

Question 14 of Sample Exam3

Process mean 5.001 inchesProcess standard deviation = 0.001 inches

The Process Capability Index

Cpk < 1 Not Capablepk p

Cpk > 1 Capable at 3σ

Cpk > 1.33 Capable at 4σ

Cpk > 1.67 Capable at 5σ

Cpk > 2 Capable at 6σ

41

Six Sigma equals 3.4 defects per million opportunities

Six Sigma Improvement MethodsDMAIC vs. DMADV

Define

Measure

AnalyzeContinuous Improvement Reengineering

Design

Validate

Improve

Control

Chapter 11Implementation Prototype

ModelInstitutionalized

Model

Institutionalized

Effort: 1X Effort: 10X-100X

Modeling Application

InstitutionalizedModeling Application

Effort: 10X Effort: 100X – 1000X

42

The Separation of Players Curse

Modeler,Project

Manager,Decision

Curse of

Player SeparationClient

Modeler

DecisionMaker

Maker,Client

Project Manager

Single player Multiple players

The Curse of Scope Creep

Narrow Wide

Model(s) Single MultipleModel(s) Single Multiple

Objective(s) Single Multiple

Activity Focused Diffused

Players Few Many

Stakeholders Few ManyStakeholders Few Many

Effort Low High

Cost Low High

Development Risk Low High

The Curse of Scope Creep

Narrow Wide

Coordination & project Informal FormalCoordination & project management

Informal Formal

Project visibility Low High

Economies of scale:

- Information systems None Many

- Model & database maint. None Many

Model use & support Deteriorates Wide

Potential org. impact Low High

mgs 3100 business analysis final exam review

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