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Probability Background 1 2 3 4 Random variables A random variable (r.v.) is a numerical outcome whose value depends on chance. We denote random variables with upper case letters, and their realized values with lower case letters. Random variables For example, suppose demand for sweaters next winter cannot be predicted with certainty, and we think it could be anywhere between 0 and 1000. Then we may denote the uncertain demand by random variable X, which takes on values in the set {0,1,…,1000}. If at the end of the season, the actual demand turns out to be 734 units, we say the value of X is x = 734. Probability distribution The expression {X ≤ x} is the uncertain event that the random variable X takes on a value less than or equal to x. The event is uncertain because whether it occurs or not depends on the value of X. Probability distribution The probability that the event occurs is denoted as P[X ≤ x]. As x varies, this probability defines a function: F(x) = P[X ≤ x] which is called the cumulative distribution function (c.d.f) of the random variable X. Probability distribution Sometimes, we write it as FX(x) to highlight that it is distribution function of X. The cumulative distribution of a random variable contains all information about it. Probability distribution A random variable X is called discrete if it can take on only a finite or countable number of values x1, x2,… with probabilities pi = P[X = xi] for i = 1, 2, … and Σi pi = 1. Probability distribution The function f(xi) = P[X = xi] for i = 1, 2, …, is called the probability mass function (p.m.f) of X. It is related to the cumulative distribution function as 𝐹(𝑥) = σ𝑥𝑖≤𝑥 𝑓 𝑥𝑖 Example of Discrete Random Variables L.L. BEAN EXAMPLE L.L. Bean Example • L.L.Bean is a large mail-order company that sells apparel. • One of the products L.L.Bean sells is ski jackets, • • for which the selling season is from November to February. The buyer at L.L.Bean currently purchases the entire season’s supply of ski jackets from the manufacturer before the start of the selling season. What is the probability that demand does not exceed 1,000 jackets? Copyright © 2016 Pearson Education, Inc. 13 – 14 L.L. Bean Example Demand Di Probability pi Cumulative Probability of Demand Being Di or Less (Pi) 400 0.01 0.01 500 0.02 0.03 600 0.04 0.07 700 0.08 0.15 800 0.09 0.24 900 0.11 0.35 1000 0.16 0.51 1100 0.20 0.71 1200 0.11 0.82 1300 0.10 0.92 1400 0.04 0.96 1500 0.02 0.98 1600 0.01 0.99 1700 0.01 1.00 Copyright © 2016 Pearson Education, Inc. 13 – 15 Probability distribution A random variable X is called continuous if it takes on a continuum of values x. In that case, it is improbable that it will take on any specific value x, i.e., P[X = x] = 0 for every x. Probability distribution Example: Suppose that you randomly ask this question: “Is it now exactly 12:30 PM?” What is the probability that you are answered “Yes”? Answer: 0 Comment: You are never exactly on time in your life. Probability distribution Then its cumulative distribution function is continuous in x. Often there exists a probability density function 𝑓(𝑥) such that 𝐹(𝑥) = 𝑥 ‫׬‬−∞ 𝑓 𝑢 𝑑𝑢 which is the area under the probability density function to the left of x. Probability density function of a triangular distribution Example of Continuous Random Variables TRIANGULAR DISTRIBUTION Triangular Distribution • Acknowledgement : The following notes are provided by Dr Nicola Ward Petty and Dr Shane Dye of Statistics Learning Centre: StatsLC.com Triangular Distribution • A triangular distribution is a continuous probability distribution with a probability density function shaped like a triangle. It is defined by three values: the minimum value a, the maximum value b, and the peak value c. Triangular Distribution • This is really handy as in a real-life situation we can often estimate the maximum and minimum values, and the most likely outcome, even if we don't know the mean and the standard deviation. Triangular Distribution • The triangular distribution has definite upper and lower limits, so we avoid unwanted extreme values. In addition the triangular distribution is a good model for skewed distributions. Triangular Distribution Triangular Distribution Triangular Distribution The probability density function of a triangular distribution is zero for values below a and values above b. It is piecewise linear 2 rising from 0 at a to 𝑏−𝑎 at c, then dropping down to 0 at b. The graph shows the probability density function of a triangle distribution with a = 1, b = 9 and c = 6. The peak is at c = 6 with a function value of 0.25. Triangular Distribution Triangular Distribution • Example 1: A burger franchise planning a new outlet in Auckland uses a triangular distribution to model the future weekly sales. They estimate that the minimum weekly sales is $1000 and the maximum is $6000. They also estimate that the most likely outcome is around $3000. Triangular Distribution • Example 1 (Continued): The graph of the probability density function reaches its maximum of 0.0004 at c = $3000. The graph of this probability density function is shown below. Triangular Distribution • Example 1 (Continued): Triangular Distribution • The probability density function is used to determine the probability that the random variable falls in some range. We want to determine the probability that the random variable is above a given value, below a given value, or between a pair of values. It is simply a matter of finding the area under the curve for the required interval. Triangular Distribution • Example 1 (Continued): A burger franchise planning a new outlet in Auckland wants to determine the probability the new outlet will have weekly sales of less than $2000. If the weekly sales are less than this, the outlet is unlikely to cover its costs. Triangular Distribution • Example 1 (Continued): So, they wish to calculate P(X < 2000). They use a triangular distribution to model the future weekly sales with a minimum value of a = $1000, and maximum value of b = $6000 and a peak value of c = $3000. • Answer: P(X < 2000) = 0.1 = 10% Triangular Distribution • Example 1 (Continued): Expected Value or Mean The expected value or the mean of a random variable X is the weighted average of all of its possible values, using probabilities as weights ෍ 𝑥𝑖 𝑓 𝑥𝑖 if 𝑋 is a discrete r. v. 𝐸 𝑋 = ∞ 𝑖 න 𝑢𝑓 𝑢 𝑑𝑢 if 𝑋 is a continuous r. v. −∞ We will also denote the mean of X by µX. Variance and Standard Deviation The variance of a random variable X is a measure of its variability from the mean. It is computed as the expected squared deviation of X from its mean 𝜇𝑋 and is denoted by 𝑉 𝑋 = 𝐸 𝑋 − 𝜇𝑋 2 . The square-root of the variance of X is called its standard deviation and is denoted by 𝜎𝑋 = 𝑉𝑋 = 𝐸 𝑋 − 𝜇𝑋 2 . L.L. Bean Example (Discrete) a) What is the expected demand? b)What is the standard deviation? Copyright © 2016 Pearson Education, Inc. 13 – 37 L.L. Bean Example (Discrete) DemandProb 400 0.01 D*Prob Deviance^2 4 3918.76 500 0.02 10 5533.52 600 0.04 24 7259.04 700 0.08 56 8502.08 800 0.09 72 4596.84 900 0.11 99 1746.36 1000 0.16 160 108.16 1100 0.2 220 1095.2 1200 0.11 132 3330.36 1300 0.1 130 7507.6 1400 0.04 56 5595.04 1500 0.02 30 4493.52 1600 0.01 16 3294.76 1700 0.01 17 4542.76 1026 61524 MEAN Copyright © 2016 Pearson Education, Inc. VAR 248.040319 STDEV 13 – 38 Triangular Distribution (Continuous) Discussion Problem W01-01-a) Variance = Risk in Finance • V[(value of) Stock 1] = 0.0449 • V[Stock 2] = 0.0069 • V[Stock 3] = 0.0011 assuming E[Stock 1] = E[Stock 2] = E[Stock 3] Which stock is best (of smallest variance) in risk? Covariance and Correlation Coefficient Suppose 𝑋1 and 𝑋2 are two random variables with means 𝜇1 and 𝜇2 and standard deviations 𝜎1 and 𝜎2 , respectively. The covariance of 𝑋1 and 𝑋2 is defined as the expected value of the product of their deviations from their respective means and is denoted by 𝐶𝑜𝑣 𝑋1 , 𝑋2 = 𝐸 𝑋1 − 𝜇1 𝑋2 − 𝜇2 . The correlation coefficient is then defined as 𝜌 = 𝐶𝑜𝑣𝜎 𝑋𝜎1,𝑋2 . 1 2 Covariance and Correlation Coefficient The value of the correlation coefficient is always between –1 and +1. A positive covariance or correlation coefficient implies that the two random variables tend to vary in the same direction (up or down). Similarly, negative covariance or correlation coefficient implies that on average they tend to move in the opposite direction. If 𝑋1 and 𝑋2 are independent then the two are uncorrelated, or 𝐶𝑜𝑣 𝑋1 , 𝑋2 = 0. Covariance and Correlation Coefficient Discussion Problem W01-01-b) Financial Hedging: to select a portfolio of common stocks giving precise mathematical meaning to the adage “Don't put all of your eggs in one basket.” Correlations of the values of Stocks 1, 2 and 3 are • ρ12 = 0.0032 • ρ13 = –0.0005 ρij = correlation of the values of Stocks i & j • ρ23 = 0.0006 Which pair of Stocks are best in hedging? Sums of Random Variables Consider two random variables X1 and X2. Then, it turns out that 𝐸 𝑋1 + 𝑋2 = 𝐸 𝑋1 + 𝐸 𝑋2 𝑉 𝑋1 + 𝑋2 = 𝑉 𝑋1 + 𝑉 𝑋2 + 2𝐶𝑜𝑣 𝑋1 , 𝑋2 Recall that if 𝑋1 and 𝑋2 are independent, then 𝐶𝑜𝑣 𝑋1 , 𝑋2 = 0. Sums of Random Variables It then follows that the expected value and variance of sums of independent random variables is equal to the sum of their expectations and variances, respectively; i.e., 𝐸 𝑋1 + 𝑋2 = 𝐸 𝑋1 + 𝐸 𝑋2 𝑉 𝑋1 + 𝑋2 = 𝑉 𝑋1 + 𝑉 𝑋2 Example of Discrete Random Variables DISCUSSION PROBLEM W01-02 L.L. Bean Example • L.L.Bean is a large mail-order company that • • sells apparel. One of the products L.L.Bean sells is ski jackets, for which the selling season is from November to February. The buyer at L.L.Bean currently purchases the entire season’s supply of ski jackets from the manufacturer before the start of the selling season. Copyright © 2016 Pearson Education, Inc. 13 – 48 L.L. Bean Example Demand Di (in hundreds) Probability pi Cumulative Probability of Demand Being Di or Less (Pi) 400 0.01 0.01 500 0.02 0.03 600 0.04 0.07 700 0.08 0.15 800 0.09 0.24 900 0.11 0.35 1000 0.16 0.51 1100 0.20 0.71 1200 0.11 0.82 1300 0.10 0.92 1400 0.04 0.96 1500 0.02 0.98 1600 0.01 0.99 1700 0.01 1.00 Copyright © 2016 Pearson Education, Inc. 13 – 49 Discussion Problem W01-02 L.L. Bean Example: a) What is the expected demand? b)What is the standard deviation? c) What is the probability that demand does not exceed 1,000 jackets? Copyright © 2016 Pearson Education, Inc. 13 – 50 Example of Continuous Random Variables (Triangular Distribution) DISCUSSION PROBLEM W01-03 ~ W01-06 Triangular Distribution • Acknowledgement : The following notes are provided by Dr Nicola Ward Petty and Dr Shane Dye of Statistics Learning Centre: StatsLC.com Triangular Distribution • A triangular distribution is a continuous probability distribution with a probability density function shaped like a triangle. It is defined by three values: the minimum value a, the maximum value b, and the peak value c. Triangular Distribution • This is really handy as in a real-life situation we can often estimate the maximum and minimum values, and the most likely outcome, even if we don't know the mean and standard deviation. Triangular Distribution • The triangular distribution has a definite upper and lower limit, so we avoid unwanted extreme values. In addition the triangular distribution is a good model for skewed distributions. Triangular Distribution Triangular Distribution The probability density function of a triangular distribution is zero for values below a and values above b. It is piecewise linear 2 rising from 0 at a to 𝑏−𝑎 at c, then dropping down to 0 at b. The graph below shows the probability density function of a triangle distribution with a = 1, b = 9 and c = 6. The peak is at c = 6 with a function value of 0.25. Triangular Distribution Triangular Distribution Triangular Distribution Triangular Distribution • Example 1: A burger franchise planning a new outlet in Auckland uses a triangular distribution to model the future weekly sales. They estimate that the minimum weekly sales is $1000 and the maximum is $6000. They also estimate that the most likely outcome is around $3000. Triangular Distribution • Example 1 (Continued): The graph of the probability density function reaches its maximum of 0.0004 at c = $3000. The graph of this probability density function is shown below. Triangular Distribution • Example 1 (Continued): Triangular Distribution • The probability density function is used to determine the probability that the random variable falls in some range. We want to determine the probability that the random variable is above a given value, below a given value, or between a pair of values. It is simply a matter of finding the area under the curve for the required interval. Triangular Distribution • W01-03-a): A burger franchise planning a new outlet in Auckland wants to determine the probability the new outlet will have weekly sales of less than $2000. If the weekly sales are less than this the outlet is unlikely to cover its costs. Triangular Distribution • W01-03-a) (Continued): So, they wish to calculate P(X < 2000). They use a triangular distribution to model the future weekly sales with a minimum value of a = $1000, and maximum value of b = $6000 and a peak value of c = $3000. • Answer: P(X < 2000) = 0.1 = 10% Triangular Distribution • W01-03-a) (Continued): Triangular Distribution • W01-03-b) Assume that weekly sales is independent across weeks. What is the expected sales for 3 weeks? • W01-03-c) Assume that weekly sales is independent across weeks. What is the standard deviation of the sales for 3 weeks? Triangular Distribution a= 1000 mean = 3333.333 mean*3= c= 3000 var = var*3= b= 6000 1055556 10000 3166667 Triangular Distribution • Example 2: Voting for the student representative on a school’s Board of Trustees has closed but the votes have not been counted. Simon Pegg (a candidate) thinks about how many votes he thinks he will get. He thinks the most likely value is around 550, but he could get as many as 900 or as few as 200. Triangular Distribution • Example 2 (Continued): Simon models the number of votes he may have received as a triangular distribution with minimum value a = 200, maximum value b = 900 and peak value c = 550. The graph of the probability density function reaches its maximum of 0.002857 at c = 550 and is shown below. Triangular Distribution • Example 2 (Continued): Triangular Distribution • W01-04 Voting for the student representative on a school’s Board of Trustees has closed but the votes have not been counted. Candidate Simon Pegg wants to determine the probability that he received more than 450 votes. This means Simon wants to determine P(X > 450). Triangular Distribution • W01-04 (Continued): Triangular Distribution • W01-04 (Continued): a= 200 250 0.002041 0.255102 c= 550 350 0.002857 0.744898 b= 900 Triangular Distribution • Example 3: • W01-05 Find P(6.5 < X < 8) for a triangular distribution with minimum value a = 1, maximum value b = 9 and peak value c = 6. • Answer: P(6.5 < X < 8) = ? a= 1 1 c= 6 2.5 b= 9 3 0.083333 0.041667 0.21875 0.208333 0.260417 0.25 Triangular Distribution • W01-05 (Continued) Triangular Distribution • W01-06 Find P(1.2 < X < 2.6) for a triangular distribution with minimum value a = 0.8, maximum value b = 2.8 and peak value c = 2.0. • Answer: P(1.2 < X < 2.6) = ? a= 0.8 0.266667 0.708333333 0.025 c= 2 0.666667 1 1 0.25 b= 2.8 0.8 1.2 0.8 0.2 Triangular Distribution • W01-06 (Continued) TIME SERIES Tahoe Salt Consider the demand for rock salt used primarily to melt snow. This salt is produced by a firm called Tahoe Salt, which sells its salt through a variety of independent retailers around the Lake Tahoe area of the Sierra Nevada Mountains. In the past, Tahoe Salt has relied on estimates of demand from a sample of its retailers, but the company has noticed that these retailers always overestimate their purchases, leaving Tahoe (and even some retailers) stuck with excess inventory. Copyright © 2016 Pearson Education, Inc. 7 – 81 Tahoe Salt After meeting with its retailers, Tahoe has decided to produce a collaborative forecast. Tahoe Salt wants to work with the retailers to create a more accurate forecast based on the actual retail sales of their salt. Quarterly retail demand data for the past three years are shown in Table 7-1 and charted in Figure 7-1. Copyright © 2016 Pearson Education, Inc. 7 – 82 Tahoe Salt Year Quarter Period, t Demand, Dt 1 2 1 8,000 1 3 2 13,000 1 4 3 23,000 2 1 4 34,000 2 2 5 10,000 2 3 6 18,000 2 4 7 23,000 3 1 8 38,000 3 2 9 12,000 3 3 10 13,000 3 4 11 32,000 4 1 12 41,000 TABLE 7-1 Copyright © 2016 Pearson Education, Inc. 7 – 83 Tahoe Salt FIGURE 7-1 In Figure 7-1, observe that demand for salt is seasonal, increasing from the second quarter of a given year to the first quarter of the following year. The second quarter of each year has the lowest demand. Each cycle lasts four quarters, and the demand pattern repeats every year. Copyright © 2016 Pearson Education, Inc. 7 – 84 Tahoe Salt FIGURE 7-1 There is also a growth trend in the demand, with sales growing over the past three years. The company estimates that growth will continue in the coming year at historical rates. Copyright © 2016 Pearson Education, Inc. 7 – 85 Time Series Plots • A time series or time sequence is a data set in which the observations are recorded in the order in which they occur. • A time series plot is a graph in which the vertical axis denotes the observed value of the variable and the horizontal axis denotes the time. • When measurements are plotted as a time series, we often see •Trends (Moving Average) •Cycles = seasonality Time Series Plots •A moving average (MA) is a widely used indicator in technical analysis that helps smooth out demand action by filtering out the “noise” from random demand fluctuations. It is a trend-following, or lagging, indicator because it is based on past prices. 4-period moving average Period t Demand Dt Moving Average 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 8,000 13,000 23,000 34,000 10,000 18,000 23,000 38,000 12,000 13,000 32,000 41,000 19,500 20,000 21,250 21,250 22,250 22,750 21,500 23,750 24,500 Forecast Ft Error Et 19,500 20,000 21,250 21,250 22,250 22,750 21,500 23,750 24,500 24,500 24,500 24,500 9,500 2,000 -1,750 -16,750 10,250 9,750 -10,500 -17,250 Forecast by 4-period moving average 45,000 40,000 35,000 30,000 25,000 20,000 15,000 10,000 5,000 0 0 2 4 6 8 10 12 14 Regression (Simple linear) regression is a statistical method that allows us to summarize and study relationships between two continuous (quantitative) variables: • One variable, denoted x, is regarded as the predictor variable. • The other variable, denoted y, is regarded as the response variable. Regression Since we are interested in summarizing the trend between two quantitative variables, the natural question arises — "what is the best fitting line?" You were probably shown a scatter plot of (x, y) data and were asked to draw the "most appropriate" line through the data. Regression You can try it now on a set of heights (x) and weights (y) of 10 students. Look at the plot below. The red line best summarizes the trend Y between height and weight. X 63 127 (inch) (lb) 64 66 69 69 71 71 72 73 75 121 142 157 162 156 169 165 181 208 Regression Coefficients Intercept X Variable 1 -266.5 Trend = 6.138 Discussion Problem W01-07 Perform Regression on the moving averages to see the trend in Tahoe Salt Example. (Use Excel | Data Analysis | Regression) • Answer: Demand increases by 554 every quarter Coefficients Intercept 17427.78 X Variable 1 554.1667 4-period moving average Period t Demand Dt Moving Average 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 8,000 13,000 23,000 34,000 10,000 18,000 23,000 38,000 12,000 13,000 32,000 41,000 19,500 20,000 21,250 21,250 22,250 22,750 21,500 23,750 24,500 Forecast Ft Error Et 19,500 20,000 21,250 21,250 22,250 22,750 21,500 23,750 24,500 24,500 24,500 24,500 9,500 2,000 -1,750 -16,750 10,250 9,750 -10,500 -17,250 PERT Application of beta distribution Program Evaluation and Review Technique (www.se.cuhk.edu.hk/~seem3530/files/ProjMgt-IA-PERT.ppt) SHOPPING MALL RENOVATION Estimation of the activity duration Example: An activity was performed 40 times in the past, requiring a time between 10 to 70 hours. The figure below shows the frequency distribution. PERT SEEM 3530 97 Estimation of the activity duration Example: An activity was performed 40 times in the past, requiring a time between 10 to 70 hours. The figure below shows the frequency distribution. PERT SEEM 3530 98 Estimation of the activity duration The probability distribution of the activity is approximated by a probability frequency distribution. PERT SEEM 3530 99 Estimation of the activity duration In project scheduling, we usually use a beta distribution to represent the time needed for each activity. PERT SEEM 3530 100 Estimation of the activity duration • Three key values we use in the time estimate for each activity: a = optimistic time, which means that there is little chance that the activity can be completed before this time; c = most likely time, which will be required if the execution is normal; b = pessimistic time, which means that there is little chance that the activity will take longer. PERT SEEM 3530 101 Estimation of Mean and SD • The expected or mean time is given by: µ = (a + 4c + b)/6 The variance is: 𝜎 2 = (b-a) 2/36 ▪ The standard deviation is (b - a)/6 PERT SEEM 3530 102 Estimation of Mean and SD For our example (Figure 7-3), we have a = 10, b = 70, c = 35. Therefore µ = 36.6, and 𝜎 2 =100. PERT SEEM 3530 103 Estimation of Mean and SD Beta-distribution a PERT c b Expected task time: 𝝁 = 𝒂+𝟒𝒄+𝒃 𝟔 Standard deviation: 𝒃−𝒂 𝟔 SEEM 3530 𝝈= , V = 𝝈𝟐 = 𝒃−𝒂 𝟔 104 𝟐 Example: Shopping Mall Renovation Activity A: Prepare initial design B: Identify new potential clients C: Develop prospectus for tenants D: Prepare final design E: Obtain planning permission F: Obtain finance from bank G: Select contractor H: Construction I: Finalize tenant contracts J: Tenants move in PERT SEEM 3530 IP a 1 4 A 2 A 1 D 1 E 1 D 2 G, F 10 B, C, E 6 I, H 1 m 3 5 3 8 2 3 4 17 13 2 b 5 12 10 9 3 5 6 18 14 3 105 Expected Activity Time and SD Act A B C D E F G H I J PERT a 1 4 2 1 1 1 2 10 6 1 m 3 5 3 8 2 3 4 17 13 2 b 5 12 10 9 3 5 6 18 14 3 SEEM 3530 µ 3 6 4 7 2 3 4 16 12 2 1+ 4 3 + 5 2 t= =3 6 0.44 1.78 1.78  = (12−4 ) = 1.78 6 1.78 0.11 0.44 0.44 1.78 1.78 0.11 2 2 106 Discussion Problem W02-01: Expected Activity Time and SD? Act A B C D E F G H I J PERT a 1 4 2 1 1 1 2 10 6 1 m 3 5 3 8 2 3 4 17 13 2 b 5 12 10 9 3 5 6 18 14 3 SEEM 3530 D ? V ? D= 1+ 4 3 + 5 =3 6 12− 4 = 1.78 ) 6 2 V =( 107 The PERT Approach ▪ Now assume that the activity times are independent random variables. ▪ Further, assume that there are n activities in the project, k of which make the longest path in the mean time. Denote the activity times of the activities in the path by the random variables di with mean E(di) and variances V(di), for i = 1, 2, …, k. PERT SEEM 3530 108 The PERT Approach (cont’d) ▪ Then, the total length of the path is a random variable X = d1 + d2 +,…, + dk ▪ The mean length, E(X), and its variance, V(X): E(X)= E(d1)+E(d2)+,…, +E(dk) V(X)= V(d1)+V(d2)+,…, +V(dk) PERT SEEM 3530 109 Discussion Problem W02-02: a) The mean time of path A-D-E-F-H-J? b) The standard deviation of the time of the path? I,12 B,6 1 C,4 J,2 E,2 End F,3 A,3 PERT D,7 G,4 SEEM 3530 H,16 110 Important Continuous Distributions Normal Distribution Undoubtedly, the most widely used model for the distribution of a random variable is a normal distribution. • Central limit theorem • Gaussian distribution Important Continuous Distributions Normal Distribution Important Continuous Distributions Normal Distribution Important Continuous Distributions Normal Distribution Important Continuous Distributions Normal Distribution Important Continuous Distributions Normal Distribution Normal Distribution Important Continuous Distributions Normal Distribution Important Continuous Distributions Normal Distribution Normal Distribution Normal Distribution Probability Background 1 PERT Application of beta distribution Program Evaluation and Review Technique (www.se.cuhk.edu.hk/~seem3530/files/ProjMgt-IA-PERT.ppt) SHOPPING MALL RENOVATION Estimation of the activity duration Example: An activity was performed 40 times in the past, requiring a time between 10 to 70 hours. The figure below shows the frequency distribution. PERT SEEM 3530 3 Estimation of the activity duration Example: An activity was performed 40 times in the past, requiring a time between 10 to 70 hours. The figure below shows the frequency distribution. PERT SEEM 3530 4 Estimation of the activity duration The probability distribution of the activity is approximated by a probability frequency distribution. PERT SEEM 3530 5 Estimation of the activity duration In project scheduling, we usually use a beta distribution to represent the time needed for each activity. PERT SEEM 3530 6 Estimation of the activity duration • Three key values we use in the time estimate for each activity: a = optimistic time, which means that there is little chance that the activity can be completed before this time; c = most likely time, which will be required if the execution is normal; b = pessimistic time, which means that there is little chance that the activity will take longer. PERT SEEM 3530 7 Estimation of Mean and SD • The expected or mean time is given by: µ = (a + 4c + b)/6 The variance is: 𝜎 2 = (b-a) 2/36 ▪ The standard deviation is (b - a)/6 PERT SEEM 3530 8 Estimation of Mean and SD For our example (Figure 7-3), we have a = 10, b = 70, c = 35. Therefore µ = 36.6, and 𝜎 2 =100. PERT SEEM 3530 9 Estimation of Mean and SD Beta-distribution a PERT c b Expected task time: 𝝁 = 𝒂+𝟒𝒄+𝒃 𝟔 Standard deviation: 𝒃−𝒂 𝟔 SEEM 3530 𝝈= , V = 𝝈𝟐 = 𝒃−𝒂 𝟔 10 𝟐 Example: Shopping Mall Renovation Activity A: Prepare initial design B: Identify new potential clients C: Develop prospectus for tenants D: Prepare final design E: Obtain planning permission F: Obtain finance from bank G: Select contractor H: Construction I: Finalize tenant contracts J: Tenants move in PERT SEEM 3530 IP a 1 4 A 2 A 1 D 1 E 1 D 2 G, F 10 B, C, E 6 I, H 1 c 3 5 3 8 2 3 4 17 13 2 b 5 12 10 9 3 5 6 18 14 3 11 Expected Activity Time and SD Act A B C D E F G H I J PERT a 1 4 2 1 1 1 2 10 6 1 c 3 5 3 8 2 3 4 17 13 2 b 5 12 10 9 3 5 6 18 14 3 SEEM 3530 µ 3 6 4 7 2 3 4 16 12 2 1+ 4 3 + 5 2 t= =3 6 0.44 1.78 1.78  = (12−4 ) = 1.78 6 1.78 0.11 0.44 0.44 1.78 1.78 0.11 2 2 12 Discussion Problem W02-01: Expected Activity Time and SD? Act A B C D E F G H I J PERT a 1 4 2 1 1 1 2 10 6 1 c 3 5 3 8 2 3 4 17 13 2 b 5 12 10 9 3 5 6 18 14 3 SEEM 3530 µ ? 2 ? D= 1+ 4 3 + 5 =3 6 12− 4 = 1.78 ) 6 2 V =( 13 The PERT Approach ▪ Now assume that the activity times are independent random variables. ▪ Further, assume that there are n activities in the project, k of which make the longest path in the mean time. Denote the activity times of the activities in the path by the random variables di with mean E(di) and variances V(di), for i = 1, 2, …, k. PERT SEEM 3530 14 The PERT Approach (cont’d) ▪ Then, the total length of the path is a random variable X = d1 + d2 +,…, + dk ▪ The mean length, E(X), and its variance, V(X): E(X)= E(d1)+E(d2)+,…, +E(dk) V(X)= V(d1)+V(d2)+,…, +V(dk) ▪ We call the longest path as critical path. (See the flow chart on the next page, which illustrates the precedence relations between activities (IP in the table of activities).) PERT SEEM 3530 15 Discussion Problem W02-02: a) The mean time of path A-D-E-F-H-J? b) The standard deviation of the time of the path? I,12 B,6 1 C,4 J,2 E,2 End F,3 A,3 PERT D,7 G,4 SEEM 3530 H,16 16 Important Continuous Distributions Normal Distribution Undoubtedly, the most widely used model for the distribution of a random variable is a normal distribution. • Central limit theorem • Gaussian distribution Important Continuous Distributions Normal Distribution Important Continuous Distributions Normal Distribution Important Continuous Distributions Normal Distribution Important Continuous Distributions Normal Distribution Important Continuous Distributions Normal Distribution Normal Distribution Important Continuous Distributions Normal Distribution Important Continuous Distributions Normal Distribution Normal Distribution Normal Distribution PERT Application of normal distribution Program Evaluation and Review Technique (www.se.cuhk.edu.hk/~seem3530/files/ProjMgt-IA-PERT.ppt) SHOPPING MALL RENOVATION (REVISITED) The PERT Approach ▪ The mean project length, E(X), and its variance, V(X): E(X)= E(d1) + E(d2) + … + E(dk) V(X)= V(d1) + V(d2) + … + V(dk) ▪ Assumption: ▪ Activity times are independent random variables. ▪ The project duration (=sum of times of activity on a critical path) is normally distributed. ▪ Based on the Central Limit Theorem, which states that the distribution of the sum of independent random variables is approximately normal when the number of terms in the sum is sufficiently large. PERT SEEM 3530 29 The PERT Approach (cont’d) ▪ Using a normal distribution, the probability of completing the project in not more than some given time T: X-E(X) T -E(X) T -E(X) P(X  T) = P( ------------  ------------- ) = P(Z  ----------) V(X)1/2 V(X)1/2 V(X)1/2 where Z is the standard normal deviate with mean 0 and variance 1. • The probability for P(Z < ), given any , can be found using normal distribution tables. PERT SEEM 3530 30 PERT SEEM 3530 31 Example: Shopping Mall Renovation Activity A: Prepare initial design B: Identify new potential clients C: Develop prospectus for tenants D: Prepare final design E: Obtain planning permission F: Obtain finance from bank G: Select contractor H: Construction I: Finalize tenant contracts J: Tenants move in PERT IP a 1 4 A 2 A 1 D 1 E 1 D 2 G, F 10 B, C, E 6 I, H 1 SEEM 3530 c 3 5 3 8 2 3 4 17 13 2 b 5 12 10 9 3 5 6 18 14 3 32 Example: Issues to Address 1. Schedule the project. 2. What is the probability of completing the project in 36 weeks? PERT SEEM 3530 33 Expected Activity Time and SD Act A B C D E F G H I J PERT a 1 4 2 1 1 1 2 10 6 1 c 3 5 3 8 2 3 4 17 13 2 b 5 12 10 9 3 5 6 18 14 3 µ 3 6 4 7 2 3 4 16 12 2 SEEM 3530 1+ 4 3 + 5 2 t= =3 6 0.44 1.78 1.78  = (12−4 ) = 1.78 6 1.78 0.11 0.44 0.44 1.78 1.78 0.11 2 2 34 CPM with Expected Activity Times I,12 B,6 1 C,4 J,2 E,2 End F,3 A,3 PERT D,7 G,4 SEEM 3530 H,16 35 Critical Path and Expected Time 1. Critical path: A-D-E-F-H-J. 2. Expected Completion time: 33 weeks 3. What is the probability to complete the project within 36 weeks? -- Use the critical path to assess the probability PERT SEEM 3530 36 Discussion Problem W02-03 The longest path in the mean time is A-D-EF-H-J. Assume that the length of the longest path is the whole project duration. What is the probability to complete the project within 36 weeks? Use the longest path to assess the probability PERT SEEM 3530 37 Probability Assessment Expected project completion time: Sum of the expected activity times along the critical path. Used to obtain probability of project  = 3+7+2+3+16+2 = 33 completion Variance of project-completion time Sum of the variances along the critical path. 2 = 0.44+1.78+0.11+0.44+1.78+0.11= 4.66  = 2.15 PERT SEEM 3530 38 Assessment by Normal Distribution P(X  36) = ? Assume X ~ N(33, 2.152) Normal Distribution  = 2.15 -  36 - 33 T = = 1.4 z = .  2.15 Standardized Normal Distribution  = 33 36 PERT P(Z  1.4) = ?  =1 z X SEEM 3530  = 0 1.4 z Z 39 Obtain the Probability Standardized Normal Probability Table (Portion) Z .00 .01 .02 P(Z
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RUNNING HEAD: Engineering Statics

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Engineering Statics
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DISCUSSION PROBLEM W01-01-a
Stock 3 is the best because it has the smallest variance of 0.0011. In risk finance, the lower the
variance value the better the capital in terms of risk.
DISCUSSION PROBLEM W01-01-b
Stock 1 and 3 are best in hedging since they have a negative correlation, which helps in mitigating
losses in that as the value of one stock increases, the amount of the other decreases.
DISCUSSION PROBLEM W01-02
a. The expected demand calculated as follows:
𝐷𝑒𝑚𝑎𝑛𝑑 = ∑ 𝐷𝑖 𝑃𝑖 𝑥 100
∑ 𝐷𝑖 𝑃𝑖 = (4𝑥0.01) + (5𝑥0.02) + (6𝑥0.04) + (7𝑥0.08) + (8𝑥0.09) + (9𝑥0.11) +
(10𝑥0.16) + (11𝑥0.20) + (12𝑥0.11) + (13𝑥0.10) + (14𝑥0.04) + (15𝑥0.020) +
(16𝑥0.01) + (17𝑥0.01) = 10.26.
𝐷𝑒𝑚𝑎𝑛𝑑 (𝑀𝑒𝑎𝑛) = 10.26 x 100= 1026 jackets
b. The standard deviation c...


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