Presentation Transcript

1. Review Review We will spend up to 30 minutes reviewing Exam 1 Know how your answers were graded. Know how to correct your mistakes. Your final exam is cumulative, and may contain similar questions.
2. Readings Readings Chapter 12 Simulation
3. Overview Overview
4. Overview
5. The Law of Large Numbers The Law of Large Numbers
6. The Law of Large Numbers Overview The Law of Large Numbers proves the average of many independent random variables converges to its expected value with certainty.
7. The Law of Large Numbers The Law of Large Numbers proves the average of many independent and identically-distributed random variables, with mean m and finite variance is a random variable with almost all of its probability weight on values near m. The more variables, the greater the concentration of probability weight near m. For example, the average (mean) height for all males in the U.S. is 5’10”. If you measured 2 randomly-selected males in the U.S., their average height may be a lot different from 5’10”. But if you randomly measured 2,000 males in the U.S., their average height will most likely be almost equal to 5’10”.
8. The Law of Large Numbers The expected value (mean) of the sum of any two random variables equals the sum of the expected values of each individual variable: E(X+Y) = E(X)+E(Y). The expected value (mean) of the product of two independent random variables equals the product of the expected values of each individual variable: E(XY) = E(X)E(Y).
9. Simulation Simulation
10. Simulation Overview Simulation models equations of random variables that are too complex to be solved with analytical formulas. (Like queuing models where arrivals are do not have a Poisson distribution.) One disadvantage of simulation is that there is no guarantee that you preformed enough trials so your conclusions are accurate.
11. Simulation Simulation is a common management science technique. It is typically used to model random processes that are too complex to be solved with analytical formulas. (Like queuing models where arrivals are do not have a Poisson distribution.) One disadvantage of simulation is that there is no guarantee that you preformed enough trials so your conclusions are accurate.
12. Simulation One begins a simulation by developing a mathematical statement of the problem. The model should be realistic yet solvable within the speed and storage constraints of the computer system being used. Input values for the model as well as probability estimates for the random variables must then be determined.
13. Simulation The simplest random variable used in simulation has a finite distribution is the simplest random input into a simulation. For example, a change in a stock price is random but it can take on only one of a finite number of possible values.
14. Simulation Another random variable used in simulation has the continuous uniform distribution The continuous uniform distribution has all intervals of the same length in the distribution's support are equally probable. The support is defined by the two parameters, a and b, which are its minimum and maximum values. The standard uniform distribution has support [0,1]. The probability of each interval subset of [0,1] equals the length of that interval, Prob[a,b] = b-a. The graph of the probability densityof any uniform distribution is flat:
15. Simulation The most common random variable used in simulation has the normal distribution. That is a continuous probability distribution that is convenient to manipulate when used in an equation: A linear function of a normal random variable is another normal random variable.
16. Simulation The normal distribution is a family of distributions of the same general form, differing in their location and scale parameters: the mean ("average")m and standard deviation ("variability") s. The probability density function is The standard normal distribution is the normal distribution with mean m = 0 and standard deviation s = 1. The graph of its probability density resembles a bell (see the green curve). f(x) =
17. Simulation About 68% of values drawn from a normal distribution are within 1 standard deviation away from the mean; about 95% of the values are within two standard deviations and about 99.7% lie within 3 standard deviations. There is positive probability weight on all values, both positive and negative, but the weight becomes negligible as values become very positive or very negative.
18. Simulation One convenient feature of the normal distribution is that a linear function of a normal random variable is another normal random variable. For any number m and any positive number s > 0, if the random variable X has the standard normal distribution, then the linear combination m + sX is a normal random variable with mean m and standard deviation s. Here are graphs of normal probability densities for various means m and variances s2.
19. Repeated Risk Repeated Risk
20. Repeated Risk Overview Repeated Risk differs from one-time risk because of the central limit theorem. Many independent risks average to a certainty, so each should be accepted if it has positive expected value.
21. Repeated Risk Risk management depends on whether an investor will face a risk just once, like a person one year from living off retirement savings, or repeatedly, like a young person 40 years from living off retirement savings. Consider an investment (gamble) where you receive $11 if a coin comes up heads, and you loose $10 if the coin comes up tails. Accepting that gamble just once involves risk. There is a 50% chance you will loose. But that risk of loss becomes negligible (it is eventually much less than the risk of dying in a accident today) as the gamble is repeated. Simulate the results of 10, 100, and 1000 coin flips for someone with an initial net worth of $500,000.
22. Repeated Risk Download Excel 2003 workbook http://faculty.pepperdine.edu/jburke2/ba452/PowerP2/Example 1.xls
23. Repeated Risk Simulate each coin flip by a standard continuous uniform random variable, with values between 0 and 1. “Heads” (net gain +$11.00) is simulated by the variable having a value between 0 and 0.5. “Tales” (net gain -$10.00) by a value between 0.5 and 1. Since the random variable is continuous, the probability that its value is exactly 0.5 is zero, so the intervals can overlap at the value 0.5.
24. Repeated Risk The net gain from each flip is computed by the function VLOOKUP. RAND() refers to the standard continuous uniform variable. $A$8:$C$9 refers to the dimensions of the probability table with A8 the upper left cell, and C9 the lower right cell. The 3 specifies that the third column of the table (Net Gain) is returned as the result of the coin flip. Repeat that formula for each row.
25. Repeated Risk Add the net gain from the initial coin flip to the initial net worth. Repeat for flips 2, …, 10. Hit F9 to recalculate a simulation. Do 20 simulations, and compute the percentage of times the 10 coin flips increases net worth. According to the law of large numbers, you will most likely get an increase in net worth more than 50% of the time. (I computed an increase 70% of the time.)
26. Repeated Risk Do similar simulations for 100 flips and 1000 flips. Use the Excel hide command to view your results. 1000 flips has the highest likelihood of increasing worth.
27. Risk Analysis Risk Analysis
28. Risk Analysis Overview Risk Analysis predicts the outcome of an uncertain decision. The simplest case concerns finding best-case and worst-case outcomes. Complicated cases concern expected value and confidence intervals.
29. Risk Analysis Hewlett Packard makes computer equipment. Hewlett Packard’s product design group developed a prototype for a new portable printer. Market and financial research has accurately predicted the following parameters (constants): Selling price = $249 per unit. Administrative cost = $400,000. Advertising cost = $600,000. However, there is uncertainty about three other terms affecting profit, which are considered to be random variables. The mean (expected value) of direct labor cost on each unit = $45. The mean of the parts cost on each unit = $90. The mean of the first year demand = 15,000.
30. Risk Analysis What-if analysis generates values for the probabilistic inputs then computes the resulting profit. Let c1 = direct labor cost on each unit. Let c2 = parts cost on each unit. Let x = first-year demand. Hence, profit = P = (249 - c1 – c2)x – 1,000,000. If all probabilistic inputs equaled their means, then profit = p = (249-Ec1-Ec2)Ex – 1,000,000 = (249-45-90)15,000 – 1,000,000 = $710,000 Assume random variables c1, c2, and x are independent. Hence, Ep= E{(249 - c1 – c2)x – 1,000,000} = E{(249 - c1 – c2)x} – 1,000,000 = E(249 - c1 – c2)Ex – 1,000,000 = (249 - Ec1 – Ec2)Ex – 1,000,000 = $710,000. Profit is at its mean when variables are at their means.
31. Risk Analysis If Hewlett Packard were neutral toward risk, then all they would care about is expected profit. (In the same way, gamblers seek wagers with positive expected profit. In particular, gamblers do not fixate on the probability of a loss.) In that case, no further analysis is required, Ep = $710,000. Risk analysis is required when, rather than being neutral, Hewlett Packard is adverse toward risk. Simulations are needed when problems are too complex to be computed exactly. Some problems are so complex, the expected value cannot be computed exactly. The Hewlett Packard problem allows the expected value to be computed exactly, but not other statistics, such as the probability of a loss.
32. Risk Analysis Hewlett Packard believes possible values of c1 (direct labor cost for each unit) range from $43 to $47 per unit. Hewlett Packard believes possible values of c2 (parts cost for each unit) range from $80 to $100 per unit. Hewlett Packard believes possible values of x (first-year demand) range from 1,500 to 28,000. In the best-case scenario, cost is low, demand is high, and profit = p = (249-43–80)28000 – 1,000,000 = $2,591,000. In the worst-case scenario, cost is high, demand is low, and profit = p = (249-47–100)1500 – 1,000,000 = -$847,000. (Negative profit is a loss.)
33. Risk Analysis On the one hand, if Hewlett Packard were perfectly pessimistic, then all they would care about is the worst-case scenario. On the other hand, if Hewlett Packard were perfectly optimistic, then all they would care about is the best-case scenario. For all cases in between, Hewlett Packard cares about the probabilities of alternative scenarios. For that, they first need to assess probabilities for all random variables. Download Excel 2003 workbook template http://faculty.pepperdine.edu/jburke2/ba452/PowerP2/Template.xls That template facilitates simulations with a mixture of inputs of random variables with one of three types of distributions: a general finite distribution, a continuous uniform distribution, and a general normal distribution.
34. Risk Analysis
35. Risk Analysis Hewlett Packard believes possible values of c1 (direct labor cost for each unit) have the following finite probability distribution. (Hence, adapt the general distribution part of the template.) Direct labor cost Probability c1 = $43 0.1 c1 = $44 0.2 c1 = $45 0.4 c1 = $46 0.2 c1 = $47 0.1 Prob( c1 = $47 ) = 0.1 = 1.0 – 0.9
36. Risk Analysis Hewlett Packard believes possible values of c2 (parts cost for each unit) depend on the general economy, the overall demand for parts, and the pricing policy of Hewlett Packard’s parts suppliers. Specifically, they believe c2 has a continuous uniform distribution that ranges from $80 to $100 per unit. (Hence, adapt the general distribution part of the template.)
37. Risk Analysis Hewlett Packard reassesses its beliefs about possible values of x (first-year demand). Rather than ranging from 1,500 to 28,000, they now believe that demand can possibly have any value. Specifically, Hewlett Packard believes x has a normal distribution with a mean m = 15,000 and standard deviation s = 4,500. (Hence, adapt the normal distribution part of the template.)
38. Risk Analysis Under Simulation Trials, copy formulas for the three variables: C1, C2, X. Define profit = (249 – C1 – C2)*X – 1,000,000.
39. Risk Analysis Drag down row 21 to row 520 for 500 trials.
40. Risk Analysis For this particular problem, we already precisely computed the mean for the probability distribution of profit, Ep= $710,000. The purpose of the simulation is to estimate other features of the probability distribution of profit. Compute Summary Statistics for the 500 simulation trials.
41. Risk Analysis The central limit theorem predicts the averages of simulated profit from many repeated simulations are most likely close their expected value, Ep= $710,000. Since $756,242 on the previous slide is far from $710,000, more simulations are needed. Try 5,000 simulations. Even then, profit seems far from $710,000, so try more, …
42. Risk Analysis Recall: The simulation for Hewlett Packard assumed that demand x has a normal distribution with a mean m = 15,000 and standard deviation s = 4,500. There is no upper limit to profits in the best-case scenario of the true probability distribution of profit because there is no upper limit to demand. The maximum profits in the 500 scenario trials serve as a practical upper bound to profits. There is no lower limit to profits in the worst-case scenario of the true probability distribution because demand x can be negative. A more practical worst-case scenario is to set demand x = 0, which makes profit p = -$1,000,000
43. Risk Analysis The true probability distribution of profit is not normal, so we do not know with 95% probability that profits are within 2 standard deviations of the mean. We can use the simulation to estimate the probabilities of being within two standard deviations of the mean. The true mean is $710,000. The estimated standard deviation is about $520,000. So profits are within 2 estimated standard deviations ($1,040,000) when $710,000 – $1,040,000 < p < $710,000 + $1,040,000 – $330,000 < p < $1,750,000 For 5,000 trials, the estimated probability that profits are – $330,000 < p < $1,750,000 is 1 – 0.0498, which is about 95%.
44. BA 452 Quantitative Analysis End of Lesson C.1