M670_7(Simulation)

# M670_7(Simulation) - Simulation k MGMT 670 Business...

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1 Simulation k MGMT 670: Business Analytics Krannert School of Management Purdue University

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2 Simulation Models § Computer representation of a problem that involves random variables § Forecasts the consequences of various management decisions before such decisions must be made. § Used in various management settings where things vary with uncertainty (e.g., interest rates, market share, yield, business process re-engineering and design).
3 Example: New Product Introduction § Market uncertainty Demand is estimated by two scenarios: low (60,000 units/yr) or high (100,000 units/yr) Unit price depends on competition, and is expected to be either \$10 or \$8 under low or high volume scenario, respectively. § Production cost uncertainty Variable cost is normally distributed with mean of \$7.50/unit and standard deviation of \$1/unit. Fixed cost is \$30,000/yr, and this cost is not expected to change much .

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4 Management’s Questions 1. What is the expected profit? 2. What are the best and worst scenarios? 3. What is the most likely outcome (or range) for the annual profit? 4. How likely the company will lose money from the new product introduction?
5 Simulation Approach B C D E F G H I 2 Data Simulation 3 Market Scenario 4 Low High High or Low 0 5 Probability 0.5 0.5 Market Scenario Low 6 Units 60000 100000 Sales Volume 60000 7 Price \$10 \$8 Price \$10 8 Unit Cost \$7.50 9 Unit Cost Fixed Cost \$30,000 10 Mean Std. Dev 11 \$7.50 \$1.00 Profit \$120,000.00 12 13 Fixed Cost 14 \$30,000 15 16 FORMULAS EXPLANATION 17 G4. =RiskDiscrete({0,1},C5:D5) Generate 0 or 1 with 0.5 probability each 18 G5. =IF(G4=0,C4,D4) If above is 0 then low demand else high 19 G6. =IF(G5=C4,C6,D6) If demand = low then 60000 else 100000 20 G7. =IF(G5=C4,C7,D7) If demand = low then \$10 else \$8 21 G8. =RiskNormal(C11,D11) Generate unit cost from Normal(7.5,1) 22 G9. =C14 Fixed Cost 23 G11. =(G7-G8)*G6-G9 (Price-Cost)*Sales-Fixed Cost

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6 Results from 50000 simulations Distribution for Profit/G11 Values in 10^ -6 Values in Thousands 0.000 0.500 1.000 1.500 2.000 2.500 3.000 3.500 4.000 4.500 5.000 208.3011 -400 -300 -200 -100 0 100 200 300 400 500 5% 90% 5% -107.9119 208.3011 Mean=69995.91 Name Profit Cell G11 Minimum -361725.5 Mean69995.91 Maximum 438276.9 Std Dev 96270.48 Variance 9.268005E+09 Skewness -0.5223645 Kurtosis 3.170453 Mode35744.46 Left X -107911.9 Left P 5% Right X 208301.1 Right P 95% Diff. X 316213 Diff. P 90% 5th Perc. -107911.9 95th Perc. 208301.1 #Errors 0 Filter Min Filter Max #Filtered 0
7 Expected Value and Standard Deviation of Profit 2 ( ) ( ) ( ) ( ) 700000 7.5 80000 30000 Var( ) Var( ) Var( ) 2 Cov( , ) Cov( , ) ( ) ( ) ( ) = ( ) ( ) ( ) ( ) 70000 = E PV CV F E PV E C E V F PV CV PV CV PV CV PV CV E PVCV E PV E CV E C E PV E PV E V - - = - × - = - × - = - = + - × = - × × - 2 2 2 2 2 2 2 11 9 2 9 ( ) Cov( , ) Var( ) ( ) ( ) Var( ) ( ) ( ) Var(V) Combining, Var( ) Var( ) Var( ) ( ) ( ) Var( ) ( ) Cov( , ) Var( ) 3.25 10 1 6.8 10 7.5 5.2 10 7.5 4.1 10 E C PV V CV E C V E CV C E V E C PV CV PV C E V E C V E C PV V PV CV × = - = × + × - = + × + × + × - = × + × × + × × + × × 10 = 96436.51 uses independence of C with P and V Here, P: Price, V: Volume, C: per Unit Cost

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8 Answers to Management’s Questions § Shape of the probability distribution of the annual profit  Constructing a histogram, descriptive statistics § Probability that the company will lose money  Finding the proportion of negative profits  22.27% from our simulation Comparison Expected annual profit (\$) Standard deviation of the annual profit (\$) Simulation results 69995.91 96270.48 Analytical solution 70,000 96436.51
9

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