Presentation on theme: "5.25.2 | 5.3 | 5.4 | 5.5 | 5.6 | 184.108.40.206.55.65.7 Probability Essentials n Concept of probability is quite intuitive; however, the rules of probability."— Presentation transcript:
| 5.3 | 5.4 | 5.5 | 5.6 | Probability Essentials n Concept of probability is quite intuitive; however, the rules of probability are not always intuitive or easy to master. n Mathematically, a probability is a number between 0 and 1 that measures the likelihood that some event will occur. –An event with probability zero cannot occur. –An event with probability 1 is certain to occur. –An event with probability greater than 0 and less than 1 involves uncertainty, but the closer its probability is to 1 the more likely it is to occur.
| 5.3 | 5.4 | 5.5 | 5.6 | Rule of Complement n The simplest probability rule involves the complement of an event. n If A is any event, then the complement of A, denoted by A c, is the event that A does not occur. n If the probability of A is P(A), then the probability of its complement, P(A c ), is P(A c )=1- P(A). n Equivalently, the probability of an event and the probability of its complement sum to 1.
| 5.3 | 5.4 | 5.5 | 5.6 | Addition Rule n We say that events are mutually exclusive if at most one of them can occur. That is, if one of them occurs, then none of the others can occur. n Events can also be exhaustive, which means that they exhaust all possibilities - one of these three events must occur. n Let A 1 through A n be any n events. Then the addition rule of probability involves the probability that at least one of these events will occur. P(at least one of A 1 through A n ) = P(A 1 ) + P(A 2 ) + + P(A n )
| 5.3 | 5.4 | 5.5 | 5.6 | Conditional Probability n Probabilities are always assessed relative to the information currently available. As new information becomes available, probabilities often change. n A formal way to revise probabilities on the basis of new information is to use conditional probabilities. n Let A and B be any events with probabilities P(A) and P(B). Typically the probability P(A) is assessed without knowledge of whether B does or does not occur. However if we are told B has occurred, the probability of A might change.
| 5.3 | 5.4 | 5.5 | 5.6 | Conditional Probability -- continued n The new probability of A is called the conditional probability of A given B. It is denoted P(A|B). –Note that there is uncertainty involving the event to the left of the vertical bar in this notation; we do not know whether it will occur or not. However, there is no uncertainty involving the event to the right of the vertical bar; we know that it has occurred. n The following formula conditional probability formula enables us to calculate P(A|B):
| 5.3 | 5.4 | 5.5 | 5.6 | Multiplication Rule n In the conditional probability rule the numerator is the probability that both A and B occur. It must be known in order to determine P(A|B). n However, in some applications P(A|B) and P(B) are known; in these cases we can multiply both side of the conditional probability formula by P(B) to obtain the multiplication rule. P(A and B) = P(A|B)P(B) n The conditional probability formula and the multiplication rule are both valid; in fact, they are equivalent.
| 5.3 | 5.4 | 5.5 | 5.6 | Assessing the Bendrix Situation n Now that we are familiar with the a number of probability rules we can put them to work in assessing the Bendrix situation. n To begin we will let A be the event that Bendrix meets its end-of-July deadline, and let B be the event that Bendrix receives the materials form its supplier by the middle of July. n The probabilities that we are best able to be assess on July 1 are probably P(B) and P(A|B).
| 5.3 | 5.4 | 5.5 | 5.6 | Assessing -- continued n They estimate a 2 in 3 chance of getting the materials on time; thus P(B)=2/3. n They also estimate that if they receive the materials on time then the chances of meeting the deadline are 3 out of 4. This is a conditional probability statement that P(A|B)=3/4. n We can use the multiplication rule to obtain: P(A and B) = P(A|B)P(B) = (3/4)(2/3) = 0.5 n There is a chance that Bendrix will gets its materials on time and meet its deadline.
| 5.3 | 5.4 | 5.5 | 5.6 | Assessing -- continued n Other probabilities of interest exist in this example. n Let B c be the complement of B; it is the event that the materials from the supplier do not arrive on time. We know that P(B) = 1 - P(B c ) = 1/3 from the rule of complements. n Bendrix estimates that the chances of meeting the deadline are 1 out of 5 if the materials do not arrive on time, that is, P(A| B c ) = 1/5. The multiplication rule gives P(A and B c ) = P(A| B c )P(B c ) = (1/5)(1/3) =
| 5.3 | 5.4 | 5.5 | 5.6 | Assessing -- continued n In words, there is a 1 chance out of 15 that the materials will not arrive on time and Bendrix will meets its deadline. n The bottom line for Bendrix is whether it will meet its end-of-July deadline. After the middle of July the probability is either 3/4 or 1/5 because by this time they will know whether the materials have arrived on time. n But since it is July 1 the probability is P(A) - there is still uncertainty about whether B or B c will occur.
| 5.3 | 5.4 | 5.5 | 5.6 | Assessing -- continued n We can calculate P(A) from the probabilities we already know. Using the additive rule for mutually exclusive events we obtain P(A) =P(A and B) + P(A and B c ) = (1/2)+(1/15) = n In words, the chances are 17 out of 30 that Bendrix will meet its end-of-July deadline, given the information it has at the beginning of July.
| 5.3 | 5.4 | 5.5 | 5.6 | Probabilistic Independence n A concept that is closely tied to conditional probability is probabilistic independence. n There are situations unlike Bendix when P(A), P(A|B) and P(A| B c ) are not all different. They are situations where these probabilities are all equal. In this case we can say that events A and B are independent. n This does not mean they are mutually exclusive; it means that knowledge of one of the events is of no value when assessing the probability of the other event.
| 5.3 | 5.4 | 5.5 | 5.6 | Probabilistic Independence -- continued n The main advantage of knowing that two events are independent is that the multiplication rule simplifies to P(A and B) = P(A)P(B) n In order to determine if events are probabilistically independent we usually cannot use mathematical arguments; we must use empirical data to decide whether independence is reasonable.
Distribution of a Single Random Variable
| 5.3 | 5.4 | 5.5 | 5.6 | Background Information n An investor is concerned with the market return for the coming year, where the market return is defined as the percentage gain (or loss, if negative) over the year. n The investor believes there are five possible scenarios for the national economy in the coming year: rapid expansion, moderate expansion, no growth, moderate contraction, or serious contraction. n She estimates that the market returns for these scenarios are, respectively, 0.23, 0.18, 0.15, 0.09, and 0.03.
| 5.3 | 5.4 | 5.5 | 5.6 | Background Information -- continued n Also, she has assessed that the probabilities of these outcomes are 0.12, 0.40, 0.25, 0.15, and n We must use this information to describe the probability distribution of the market return.
| 5.3 | 5.4 | 5.5 | 5.6 | Type of Random Variables n A discrete random variable has only a finite number of possible values. n A continuous random variable has a continuum of possible values. n Mathematically, there is an important difference between discrete and continuos random variables. A proper treatment of continuos variables requires calculus. In this book we will only be dealing with discrete random variables.
| 5.3 | 5.4 | 5.5 | 5.6 | Discrete Random Variables n The properties of discrete random variables and their associated probability distributions are as follows: n Let X be a random variable and to specify the probability distribution of X we need to specify its possible values and their probabilities. This list of their probabilities sum to 1. n It is sometimes useful to calculate cumulative probabilities. A cumulative probability is the probability that the random variable is less than or equal to some particular values.
| 5.3 | 5.4 | 5.5 | 5.6 | Summarizing a Probability Distribution n A probability distribution can be summarized with two or three well-chosen numbers: –The mean, often called the expected value, is a weighted sum of the possibilities. It indicates the center of the probability distribution. –To measure the variability in a distribution, we calculate its variance or standard deviation. The variance is a weighted sum of the squared deviations of the possible values from the mean. As in the previous chapter the variance is represented in the squared units of X so a more natural measure of variability is the standard deviation.
| 5.3 | 5.4 | 5.5 | 5.6 | MRETURN.XLS n This file contains the values and probabilities estimated by the investor in this example. n Mean, Probs, Returns, Var and Sqdevs have been specified as range names.
| 5.3 | 5.4 | 5.5 | 5.6 | Calculating Summary Measures n The summary measures for the probability distribution of the outcomes can be calculated as follows: –Mean return: =SUMPRODUCT(Returns,Probs) –Squared Deviations: =(C4-Mean)^2 –Variance: =SUMPRODUCT(SqDevs,Probs) –Standard Deviation: =SQRT(Var) n We see that the mean return is 15.3% and the standard deviation is 5.3%. What do these mean?
| 5.3 | 5.4 | 5.5 | 5.6 | Analyzing the Summary Measures n First, the mean or expected return does not imply that the most likely return is 15.3%, nor is this the value that the investor expects to occur. The value 15.3% is not even a possible market return. n We can understand these measures better in terms of long-run averages. n If we can see the coming year repeated many times, using the same probability distribution, then the average of these times would be close to 15.3% and their standard deviation would be 5.3%.
Derived Probability Distributions
| 5.3 | 5.4 | 5.5 | 5.6 | Background Information n A bookstore is planning on ordering a shipment of special edition Christmas calendars that they will sell for $15 a piece. n There will be only one order, so –if demand is less than the quantity ordered the excess calendars will be donated to a paper recycling company –if demand is greater than the quantity ordered, the excess demand will be lost and customers will take their business elsewhere n The bookstore estimates that the demand for calendars will be between 250 and 400.
| 5.3 | 5.4 | 5.5 | 5.6 | DERIVED.XLS n This file contains the probability distribution that the demand for calendars will follow. These estimates have been derived from subjective estimates and historical data. n If the bookstore decides to order 350 calendars, what is the probability distribution of units sold? What is the probability distribution of revenue?
| 5.3 | 5.4 | 5.5 | 5.6 | Derived Distributions of Units Sold and Revenue
| 5.3 | 5.4 | 5.5 | 5.6 | Solution n Let D, S,and R denote demand, units sold, and revenue. n The key to the solution is that each value of D directly determines the value of S, which in turn determines the value of R. n S is the smaller of D and the number ordered, 350, and R is $15 multiplied by the value of S. n Therefore we can derive the probability distributions of S and R with the following steps:
| 5.3 | 5.4 | 5.5 | 5.6 | Solution -- continued –Calculate Units sold : =MIN(B10,OnHand) –Calculate Revenue for each value of units sold: =UnitPrice*B20 –Transfer the Derived Probabilities for demand: =C10 –Calculate Means of demand, units sold, and revenue: =SUMPRODUCT(Revenues, DerivedProbs) –Calculate the Variances and Standard Deviations of demand, units sold and revenue. First, calculate the squared deviations of revenues from their mean in Column F, then calculate the sum of the products of these squared deviations and the revenue probabilities to obtain the variance of revenue. Finally, calculate the standard deviation as the square root of the variance.
| 5.3 | 5.4 | 5.5 | 5.6 | Summary Measures for Linear Functions n When one random variable is a linear function of another random variable X, there is a particularly simple way to calculate the summary measures of Y from the Summary measures of X. Y = a + bX for some constant a and b then: »mean: E(Y) = a + bE(X) »variance: Var(Y) = b 2 Var(X) »standard deviation: bStdev(X) n If Y is a constant multiple of X, that is a=0 then the mean and standard deviation of Y are this same multiple of the mean and standard deviation of X.
Distribution of Two Random Variables: Scenario Approach
| 5.3 | 5.4 | 5.5 | 5.6 | Background Information n An investor plans to invest in General Motors (GM) stock and gold. n He assumes that the returns on these investments over the next year depend on the general state of the economy during the year. n He identifies four possible states of the economy: depression, recession, normal and boom. These four states have the following probabilities: 0.05, 0.30, 0.50, and 0.15.
| 5.3 | 5.4 | 5.5 | 5.6 | Background Information -- continued n The investor wants to analyze the joint distribution of returns on these two investments. n He also wants to analyze the distribution of a portfolio of investments in GM stock and gold.
| 5.3 | 5.4 | 5.5 | 5.6 | GMGOLD.XLS n This file contains the probabilities and estimated returns of the GM stock and the gold.
| 5.3 | 5.4 | 5.5 | 5.6 | Relating Two Random Variables n There are two methods for relating two random variables, the scenario approach and the joint probability approach. n The methods differ slightly in the way they assign probabilities to different outcomes. n Two summary measures, covariance and correlation, are used to measure the relationship between two variables in both methods.
| 5.3 | 5.4 | 5.5 | 5.6 | The Summary Measures n We have discussed summary measures with the same names, covariance and correlation, earlier. The summary measures we are looking at now go by the same name but are conceptually different. n In the past we have calculated them from data; here they are calculated from a probability distribution. n The random variables are X and Y and the probability that X and Y equal x i and y i is p(x i, y i ) is called a joint probability.
| 5.3 | 5.4 | 5.5 | 5.6 | Summary Measures -- continued n Although they are calculated differently, the interpretation is essentially the same as we previously discussed. n Each indicates the strength of a linear relationship between X and Y. If X and Y vary in the same direction then both measures are positive. If they vary in opposite directions then both measures are negative. n Covariance is more difficult to interpret because it depends on the units of measurement of X and Y. Correlation is always between -1 and +1.
| 5.3 | 5.4 | 5.5 | 5.6 | The Scenario Approach n The essence of the scenario approach in this example is that a given state of the economy determines both GM and gold returns, so that only four pairs of returns are possible. n These pairs are and 0.05, 0.10 and 0.20, 0.30 and -0.12, and 0.50 and Each pair has a joint probability. n To calculate means, variances and standard deviations, we treat GM and gold returns separately.
| 5.3 | 5.4 | 5.5 | 5.6 | Calculating Covariance and Correlation n We also need to calculate the covariance and correlation between the variables. To obtain these we use the following steps: 1Deviations between means: To calculate the covariance we need the sum of deviations from means, so we need to calculate these deviations with the formula =C4-GMMean in B14 and copy it down through B17. We also calculate this for gold. 2Covariance: Calculate the covariance between GM and gold returns in cell B23 with the formula =SUMPRODUCT(GMDevs,GoldDevs,Probs)
| 5.3 | 5.4 | 5.5 | 5.6 | Calculating Covariance and Correlation -- continued 3Correlation: Calculate the correlation between GM and gold returns in cell B24 with the formula =Covar/(GMStdv*GoldStedev) n The negative covariance indicates that GM and gold returns tend to vary in opposite directions, although it is difficult to judge the strength by the magnitude of the covariance. n The correlation of is also negative and indicates a moderately strong relationship. We cannot infer too much from this correlation though because the variables are not linear.
| 5.3 | 5.4 | 5.5 | 5.6 | Simulation n A simulation of GM and gold returns help explain the covariance and correlation. n There are two keys to this simulation: n First we must, simulate the states of the economy, not - at least not directly - the GM and gold returns. –We simulate this be entering a RAND function in A1 and then by entering the formulas VLOOKUP(A21,LTable,2) in B21 and VLOOKUP(A21,LTable,3) in C21. –This way uses the same random number, hence the same scenario, to generate both returns in a given row, and the effect is that only four pairs of returns are possible.
| 5.3 | 5.4 | 5.5 | 5.6 | Simulation -- continued n Second, once we have the simulated returns we can calculate the covariance and correlation of these numbers. –We calculate these in cells B8 and B9 with the formulas COVAR(SimGM,SimGOLD) and CORREL(SimGM,SimGold). These are built-in Excel functions. –A comparison of these summary measures with the previously calculated summary measures shows that there is reasonably good agreement between the covariance and correlation of the probability distribution and the measures based on the simulated values. The agreement is not perfect but will improve as more pairs are simulated.
| 5.3 | 5.4 | 5.5 | 5.6 | Simulation of GM and Gold Returns
| 5.3 | 5.4 | 5.5 | 5.6 | Portfolio Analysis n The final part of this example is to analyze a portfolio consisting of GM stock and gold. n We assume that the investor has $10,000 and puts some fraction of this in GM stock and the rest in gold. n The key to the analysis is that there are only four possible scenarios -- that is, there are only four possible portfolio returns. n In this case we calculate the entire portfolio return distribution and summary measures in the usual way.
| 5.3 | 5.4 | 5.5 | 5.6 | Portfolio Analysis -- continued n One thing of interest is to see how the expected portfolio return and standard deviation of portfolio return change as the amount the investor puts into GM stock changes. –To do this we use a data table or mean and stdev of portfolio return as a function of GM investment. n A graph of these measures show that the expected portfolio return steadily increases as more and more is put into GM.
| 5.3 | 5.4 | 5.5 | 5.6 | Portfolio Analysis -- continued n However, we must note that the standard deviation, often used as a measure of risk, first decreases, then increases. n This means there is trade-off between expected return and risk (as measured by the standard deviation). n The investor could obtain a higher expected return by putting more of his money into GM; but past a fraction of approximately 0.4, the risk also increases.
| 5.3 | 5.4 | 5.5 | 5.6 | Distribution of Portfolio Return
Distribution of Two Random Variables: Joint Probability Approach
| 5.3 | 5.4 | 5.5 | 5.6 | SUBS.XLS n The left and top margins of the table show the possible values of demand for the products. n A company sells two products, product 1 and 2, that tend to be substitutes for each one another.The company has assessed the joint probability distribution of demand for the two products during the coming months. n This joint distribution appears in the Demand sheet of this file.
| 5.3 | 5.4 | 5.5 | 5.6 | SUBS.XLSSUBS.XLS -- continued n Demand for product 1 (D 1 ) can range from 100 to 400 (in increments of 100) and demand for product 2(D 2 ) can range from (in increments of 50). n Each possible value of D 1 can occur for each possible value of D 2 with the joint probability given in the table. n Given this joint probability distribution, describe more fully the probabilistic structure of demands for the two products.
| 5.3 | 5.4 | 5.5 | 5.6 | Joint Probability Approach n In this example we use an alternative method for specifying probability distribution. n A joint probability distribution, specified by all probabilities of the form p(x, y), indicates that X and Y are related and also how each of X and Y is distributed in its own right. n The joint probability of X and Y determines the marginal distributions of both X and Y, where each marginal distribution is the probability distribution of a single random variable.
| 5.3 | 5.4 | 5.5 | 5.6 | Joint Probability Approach -- continued n The joint distribution also determines the conditional distributions of X given Y, and of Y given X.
| 5.3 | 5.4 | 5.5 | 5.6 | Marginal Distributions n We begin by finding the marginal distributions of demands for each product. n These are the row and column sums of the joint probabilities. n The marginal distributions indicate that in-between values of the demands for each product are most likely, whereas extreme values in either direction are less likely. n These distributions tell us nothing of the relationship between the demands for the products.
| 5.3 | 5.4 | 5.5 | 5.6 | Conditional Distributions n A better way to do learn about this relationship is to calculate the conditional distributions of the demands. n We begin with with the conditional distribution for D 1 given D 2. n To calculate we create a new table. In each row of the table we fix the value of D 2 at the value given in column B. We can then calculate the conditional probabilities of the values of D 1.
| 5.3 | 5.4 | 5.5 | 5.6 | Conditional Distributions -- continued n This is the joint probability divided by the marginal probability of the D 2. They can be calculated all at once by entering the formula =C5/$G5. n We also check that each row of the table is a probability in its own right by summing across the rows. These sums should equal one. n Similarly the conditional distributions of the D 2 given the D 1 can be calculated in another table by entering the formula =C5/C$10. Each column sum should equal one.
| 5.3 | 5.4 | 5.5 | 5.6 | Summary Measures n Various summary measures can now be calculated. –Expected values: The expected demands follow from the marginal distributions and are calculated in cells B32 and C32 by these formulas =SUMPRODUCT(Demands1,Prob1) and =SUMPRODUCT(Demands2,Prob2). –Variances and standard deviations:These are also calculated from the marginal distributions in the usual way. We first find squared deviations from the means and calculate the weighted sum of these squared deviations.
| 5.3 | 5.4 | 5.5 | 5.6 | Summary Measures -- continued –Covariance and correlation: The formulas are the same as before but we proceed differently. We now form a complete table of products of deviations from the means by using the formula =(C$4-MeanDem1)*($B5- MeanDem2) in C37 and copying it to C37:F41. Then we calculate the covariance in cell B47 with the formula =SUMPRODUCT(ProdDevsDem,JtProbs). Finally, we calculate the correlation in B48 with the formula =CovarDem/(StdevDem1*StDevDem2).
| 5.3 | 5.4 | 5.5 | 5.6 | Analysis n The best way to see the joint behavior of D 2 and D 1 is to look in the conditional probability tables. n For example: Compare the probabilities in the conditional distribution table of D 1, given D 2. The value of D 2 increases, while the probabilities for D 1 tend to shift to the left. In other words, as the demand for product two increases the demand for product 1 tends to decrease. n This can be seen more clearly in the following graph.
| 5.3 | 5.4 | 5.5 | 5.6 | Conditional Distributions of Demand 1 Given Demand 2
| 5.3 | 5.4 | 5.5 | 5.6 | Analysis -- continued n The graph shows that when D 2 is large D 1 tends to be small, although again this is a tendency not a perfect relationship. n When we say that two products are substitutes for one another, this is the type of behavior we imply. n By symmetry, the conditional distribution of D 2 given D 1 shows the same type of behavior. n This is shown in the next graph.
| 5.3 | 5.4 | 5.5 | 5.6 | Conditional Distributions of Demand 2 Given Demand 1
| 5.3 | 5.4 | 5.5 | 5.6 | Conclusions n The information in these graphs is confirmed - to some extent - by the covariance and correlation between the demands for the products. n In particular, their negative values indicate that the demands for the products move in opposite directions. n Also the small correlation indicates that the relationship between these demands is far from perfect. There is still a reasonably good chance that when D 1 is large D 2 will be large, and when D 1 is small D 2 will be small.
| 5.3 | 5.4 | 5.5 | 5.6 | Assessing Joint Probability Distributions n Using the joint probability approach can often times be quite difficult especially when there are many possible values for each of the random variables n One approach is to proceed backwards from the way we proceeded in this example. n Instead of specifying the joint probabilities and then deriving the marginal and conditional distributions, we can specify either the marginal or conditional probabilities and use theses to calculate the joint possibilities.
| 5.3 | 5.4 | 5.5 | 5.6 | Assessing Joint Probability Distributions -- continued n The advantage of this procedure is that it is probably easier and more intuitive for a business manager. n He gets ore control over the relationship between the two random variables, as determined by the conditional probabilities he assesses.
| 5.3 | 5.4 | 5.5 | 5.6 | JTPROBS.XLS n An file shows an example of the indirect method of assessing joint probabilities.
| 5.3 | 5.4 | 5.5 | 5.6 | JTPROBS.XLS n The shaded regions of the spreadsheet represent probabilities assessed directly, and the joint probabilities are calculated from these. n The formula used in C20 is =C11*C$6 and then this is copied to C20:F24. n The associated graph on the next slide appears to be consistent with the meaning of substitute products.
| 5.3 | 5.4 | 5.5 | 5.6 | Conditional Distributions of Demand 2 Given Demand 1
Independent Random Variables
| 5.3 | 5.4 | 5.5 | 5.6 | Background Information n A distributor of parts keeps track of the inventory of each part type at the end of every week. n If the inventory of a given part is at or below a certain value called the reorder point, the distributor places an order for the part. n The amount ordered is a constant called the order quantity.
| 5.3 | 5.4 | 5.5 | 5.6 | Assumptions n The ordering lead time is negligible. n Sales are lost if customer demand during any week is greater than that weeks beginning inventory; that is, there is no backlogging of demand. n Customer demands for a given part type in different weeks are independent random variables. n The marginal distribution of weekly demand for a given part type is the same each week.
| 5.3 | 5.4 | 5.5 | 5.6 | INVNTORY.XLS n This file contains the plant managers estimated data in the shaded area for a particular part type. n She wants to calculate the mean revenue in each of the first weeks, given that the initial inventory at the beginning of week 1 is 250, the value shown in cell B12.
| 5.3 | 5.4 | 5.5 | 5.6 | Independent & Dependent n When random variables are independent, any information about the values of any of the random variables is worthless in predicting any of the others. n Random variables in real world applications are not usually independent; they are usually related in some way, in which case they are dependent. n However, we often make an assumption of independence in mathematical models to simplify the analysis.
| 5.3 | 5.4 | 5.5 | 5.6 | Joint Distribution of Demands n Due to the assumption that weekly demands are independent and have the same distribution, all the manager needs to assess is a single weekly distribution of demand. n To obtain these joint probabilities, we calculate products of marginals by entering the formula =VLOOKUP(C$20,DistTable,2)*VLOOKUP($B21,DistTable,2) in cell C21 and copying it into range C21:G25. n This formula simply multiplies the two marginal probabilities corresponding to the demands in the top and left margins of the joint probability table.
| 5.3 | 5.4 | 5.5 | 5.6 | Joint Distribution of Demand -- continued n To check we calculate the row and column sums. The column sums agree with the probabilities in the distribution of demand in each week, as they should. n The calculations are shown on the next slide.
| 5.3 | 5.4 | 5.5 | 5.6 | Calculations
| 5.3 | 5.4 | 5.5 | 5.6 | Calculating Mean Revenue - Week 1 n The mean revenue calculation portion of this example is more complex. This is particularly true for the revenue in week 2 because it depends on the demands of both periods. n Beginning with the revenue in week 1, the revenue is the unit price multiplied by the smaller of the on-hand inventory and demand in week 1. n We use this formula to calculate revenue for each value of demand in week 1: =UnitPrice*MIN(C29,InitInv)
| 5.3 | 5.4 | 5.5 | 5.6 | Calculating Mean Revenue - Week 1 -- continued n Given these values we can calculate the mean revenue in week 1 by entering the formula =SUMPRODUCT(Revenues1,Probs1).
| 5.3 | 5.4 | 5.5 | 5.6 | Calculating Mean Revenue - Week 2 n The revenue in week 2 is the unit price multiplied by the number of units sold in week 2, and this latter quantity is the smaller of the beginning inventory in week 2 and the demand in week 2. n The complex portion is that the inventory in week 2 depends on the demand in week1, because this demand determines how much (if any) is left at the end of week 1 and whether an order was placed. n The analysis breaks down into three cases in which I 0, D 1 and RP denote beginning inventory in week 1, the demand in week 1, and the reorder point.
| 5.3 | 5.4 | 5.5 | 5.6 | Calculating Mean Revenue - Week 2 -- continued n One of following occurs: –If I 0 -D 1 < 0, the ending inventory in week 1 is 0, and an order is placed. This brings the beginning inventory in week 2 up to the 400 units (the order quantity). –If 0< I 0 -D 1 < = RP, then positive inventory is on hand at the end of week 1, but demand in week 1 is large enough to trigger an order. Therefore, beginning inventory in week 2 is I 0 -D 1 plus the order quantity. –If I 0 -D 1 > RP, no order is triggered, so the beginning inventory in week 2 equals the ending inventory in week 1, I 0 -D 1.
| 5.3 | 5.4 | 5.5 | 5.6 | Calculating Mean Revenue - Week 2 -- continued n Putting all this together we can calculate the revenue in week 2 for each combination of week 1 and week 2 demands. n The following formula should be copied into the range C37:G41 =UnitPrice*MIN($B37,IF(InitInv- C$36<=0,OrderQuan, IF(IntiInv-C$36<=ReorderPt,InitInv- C$36+OrderQuan,InitInv-C$36))) n This formula is complex but it simply implements the logic on the previous slide with nested IF functions.
| 5.3 | 5.4 | 5.5 | 5.6 | Calculating Mean Revenue - Week 2 -- continued n Next, we calculate the mean revenue in week 2 in the usual way, as a sum of products of possible revenues and their probabilities with the formula: =SUMPRODUCT(Revenues2,JtProbs) n The one advantage to doing all this work is that now we can change any of the inputs in the shaded cells and the mean revenues will be recalculated automatically.
Weighted Sums of Random Variables
| 5.3 | 5.4 | 5.5 | 5.6 | INVEST.XLS n An investor has $100,000 to invest, and she would like to invest it in a portfolio of eight stocks. n She has gathered historical data on the returns of these stocks and has used the data to estimate means, standard deviations and correlations for the stock returns. n This file contains summary measures obtained from historical data. She believes they are also relevant for future returns.
| 5.3 | 5.4 | 5.5 | 5.6 | INVEST.XLSINVEST.XLS -- continued n The investor would like to analyze a portfolio of these stocks using certain investment amounts. n What is the mean annual return from this portfolio? What are its variance and standard deviation?
| 5.3 | 5.4 | 5.5 | 5.6 | Input Data
| 5.3 | 5.4 | 5.5 | 5.6 | Solution n This is a typical weighted sum model. n The random variables, the Xs are the annual returns from the stocks; the weights, the as are the dollar amounts invested in the stocks; and the summary measures of the Xs are given in rows 12,13 and of the input data. n We can obtain the mean return from the portfolio in cell B49 by using the formula =SUMPRODUCT(Weights, Means)
| 5.3 | 5.4 | 5.5 | 5.6 | Solution -- continued n We are not quite ready to calculate the variance of the portfolio return. n The reason why is because we do not currently know the Var(Y). But these are related to standard deviations and correlation by Var(X i ) = (Stdev(X i )) 2 Cov(X i, X j ) = StDev(X i ) X Stdev(X j ) X Corr(X i,X j )
| 5.3 | 5.4 | 5.5 | 5.6 | Solution -- continued n To calculate these in Excel, it is useful to create a column of standard deviations in Column L by using Excels TRANSPOSE function. n To do this highlight the range L12:L19 and type the formula =TRANSPOSE(Stedevs) and press Ctrl- Shift-Enter. n Next we form a table of variances and covariances of the Xs i the range B28:I35, using the formula =$L12*B$13*B17 in cell B28 and copying it to the range.
| 5.3 | 5.4 | 5.5 | 5.6 | Solution -- continued n Finally, we need to calculate the portfolio variance in cell B50. n To do this we form a table of terms needed and then sum these terms as in the following steps. –Row of weights Enter the weights in row 38 by highlighting the range B38:I18 and typing the formula =Weights and pressing Ctrl-Enter. –Column of weights Enter these same weights as a column in the range A39:A46 by highlighting the range, typing the formula =TRANSPOSE(Weights) and pressing Ctrl-Shift- Enter.
| 5.3 | 5.4 | 5.5 | 5.6 | Solution -- continued –Table of terms Now use these weights and the covariances to fill in the table of terms required for the portfolio variance. To do so, enter the formula =$A39*B28*B$38 in cell B39 and copy it to the range B39:I46. –Portfolio variance and standard deviation Calculate the portfolio variance in cell B50 with the formula =SUM(PortVarTerms). Then calculate the standard deviation of the portfolio return in cell B51 as the square root of the varince. n The results are shown in the calculation on the next slide.
| 5.3 | 5.4 | 5.5 | 5.6 |
| 5.3 | 5.4 | 5.5 | 5.6 | Solution -- continued n The standard deviation of approximately $11,200 is sizable. The standard deviation is th a measure of the portfolios risk. n Investors always want a large mean return, but they also want low risk. n They realize though that often time the only way to obtain high men returns is to assume more risk.