Download - MGS 3100 Business Analysis Final Exam Review
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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
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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
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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
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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
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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.
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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)
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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
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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.
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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.
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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
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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
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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
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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
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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
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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
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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.
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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
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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
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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
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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
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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
Decisions under uncertainty
Multiple states of nature
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Decisions under uncertainty
Multiple states of nature
Don’t know what state of nature will occur
Decisions under uncertainty
LaplaceMaximinMaximaxMinimax regretMinimax regret
Maximin Maximin
extremely yconservative or pessimistic approach to making decisions
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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
maximum possibleexpected returnwithout sample
i f ti
THE EXPECTED VALUE OF SAMPLE INFORMATION
information information
Q8 & Q9 of sample exam3
EVPI = -expected return
with perfect information
maximum possibleexpected returnwithout sample
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:
Length size weightLength, size, weight, height, time, velocity
34
Variable measures
Measurable product characteristic
Length size weightLength, size, weight, height, time, velocity
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
NUMBER OF PROPORTION
p-Chart Example ~Western Jeans Company20 samples of 100 pairs of jeans
NUMBER OF PROPORTIONSAMPLE DEFECTIVES DEFECTIVE
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
NUMBER OF PROPORTION
p-Chart Example ~Western Jeans Company20 samples of 100 pairs of jeans
NUMBER OF PROPORTIONSAMPLE DEFECTIVES DEFECTIVE
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