1. 1 1 Random Variables A random variable is a numerical description of the A random variable is a numerical description of the outcome of an experiment. outcome of an experiment. A random variable is a numerical description of the A random variable is a numerical description of the outcome of an experiment. outcome of an experiment. A discrete random variable may assume either a A discrete random variable may assume either a finite number of values or an infinite sequence of finite number of values or an infinite sequence of values. values. A discrete random variable may assume either a A discrete random variable may assume either a finite number of values or an infinite sequence of finite number of values or an infinite sequence of values. values. A continuous random variable may assume any A continuous random variable may assume any numerical value in an interval or collection of numerical value in an interval or collection of intervals. intervals. A continuous random variable may assume any A continuous random variable may assume any numerical value in an interval or collection of numerical value in an interval or collection of intervals. intervals.

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3. 3 3 Random Variables Question Random Variable x Type Familysize x = Number of dependents reported on tax return reported on tax returnDiscrete Distance from home to store x = Distance in miles from home to the store site home to the store site Continuous Own dog or cat x = 1 if own no pet; = 2 if own dog(s) only; = 2 if own dog(s) only; = 3 if own cat(s) only; = 3 if own cat(s) only; = 4 if own dog(s) and cat(s) = 4 if own dog(s) and cat(s) Discrete

4. 4 4 The probability distribution for a random variable The probability distribution for a random variable describes how probabilities are distributed over describes how probabilities are distributed over the values of the random variable. the values of the random variable. The probability distribution for a random variable The probability distribution for a random variable describes how probabilities are distributed over describes how probabilities are distributed over the values of the random variable. the values of the random variable. We can describe a discrete probability distribution We can describe a discrete probability distribution with a table, graph, or formula. with a table, graph, or formula. We can describe a discrete probability distribution We can describe a discrete probability distribution with a table, graph, or formula. with a table, graph, or formula. Probability Distributions The probability distribution is defined by a The probability distribution is defined by a probability function, denoted by f ( x ), that provides probability function, denoted by f ( x ), that provides the probability for each value of the random variable. the probability for each value of the random variable. The probability distribution is defined by a The probability distribution is defined by a probability function, denoted by f ( x ), that provides probability function, denoted by f ( x ), that provides the probability for each value of the random variable. the probability for each value of the random variable. The required conditions for a discrete probability The required conditions for a discrete probability function are: function are: The required conditions for a discrete probability The required conditions for a discrete probability function are: function are: f ( x ) > 0  f ( x ) = 1

5. 5 5 Continuous Probability Distributions Continuous Probability Distributions f ( x ) x x Uniform Probability Distribution x f ( x ) Normal Probability Distribution x x f ( x ) Exponential Probability Distribution

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7. 7 7 Continuous Probability Distributions n A continuous random variable can assume any value in an interval on the real line or in a collection of intervals. n It is not possible to talk about the probability of the random variable assuming a particular value. n Instead, we talk about the probability of the random variable assuming a value within a given interval.

8. 8 8 Area Under the Curve The probability of the random variable assuming a value within some given interval from x 1 to x 2 is defined to be the area under the graph of the probability density function between x 1 and x 2. f ( x ) x x Uniform x1 x1x1 x1 x1 x1x1 x1 x2 x2x2 x2 x2 x2x2 x2 x Normal x1 x1x1 x1 x1 x1x1 x1 x2 x2x2 x2 x2 x2x2 x2 x1 x1x1 x1 x1 x1x1 x1 x2 x2x2 x2 x2 x2x2 x2 Exponential x x x1 x1x1 x1 x1 x1x1 x1 x2 x2x2 x2 x2 x2x2 x2

9. 9 9 Uniform Probability Distribution where: a = smallest value the variable can assume b = largest value the variable can assume b = largest value the variable can assume f ( x ) = 1/( b – a ) for a < x < b f ( x ) = 1/( b – a ) for a < x < b = 0 elsewhere = 0 elsewhere f ( x ) = 1/( b – a ) for a < x < b f ( x ) = 1/( b – a ) for a < x < b = 0 elsewhere = 0 elsewhere n A random variable is uniformly distributed whenever the probability is proportional to the interval’s length. Var( x ) = ( b - a ) 2 /12 E( x ) = ( a + b )/2

10. 10  a b  a b Continuous Probability Distributions Continuous Probability Distributions  a b  a b x1x1x1x1 x2x2x2x2 x1x1x1x1 x1x1x1x1 P(x 1 ≤ x ≤ x 2 ) P(x ≤ x 1 ) P(x ≥ x 1 ) P(x ≥ x 1 )= 1- P(x<x 1 ) x2x2x2x2 x1x1x1x1 P(x 1 ≤ x ≤ x 2 )

11. 11 Example: Slater's Buffet Slater customers are charged for the amount of salad they take. Sampling suggests that the amount of salad taken is uniformly distributed between 5 ounces and 15 ounces. x = salad plate filling weight f(x)f(x) f(x)f(x) x x 1/10 Salad Weight (oz.) 5 5 10 15 f(x) = 1/10 for 5 < x < 15 f(x) = 1/10 for 5 < x < 15 = 0 elsewhere = 0 elsewhere f(x) = 1/10 for 5 < x < 15 f(x) = 1/10 for 5 < x < 15 = 0 elsewhere = 0 elsewhere E( x ) = ( a + b )/2 E( x ) = ( a + b )/2 = (5 + 15)/2 = (5 + 15)/2 = 10 = 10 E( x ) = ( a + b )/2 E( x ) = ( a + b )/2 = (5 + 15)/2 = (5 + 15)/2 = 10 = 10 Var( x ) = ( b - a ) 2 /12 Var( x ) = ( b - a ) 2 /12 = (15 – 5) 2 /12 = (15 – 5) 2 /12 = 8.33 = 8.33 Var( x ) = ( b - a ) 2 /12 Var( x ) = ( b - a ) 2 /12 = (15 – 5) 2 /12 = (15 – 5) 2 /12 = 8.33 = 8.33

12. 12 f(x)f(x) f(x)f(x) x x 1/10 Salad Weight (oz.) 5 5 10 15 P(12 < x < 15) = 1/10(3) =.3 What is the probability that a customer will take between 12 and 15 ounces of salad? Uniform Probability Distribution 12

13. 13 The Uniform Probability Distribution P(8< x < 12) = ? x x f ( x ) 55 1515 1212 1/101/10 88 P(8< x < 12) = (1/10)(12-8) =.4 x x f ( x ) 55 1515 1212 1/101/10 P(0< x < 12) = ? P(0< x < 12) = P(5<x < 12)= = (1/10)(12-5) =.7 P(0< x < 12) = P(5<x < 12)= = (1/10)(12-5) =.7 P(10< x < 22) = ? P(10< x < 22) = P(10<x < 15)= = (1/10)(15-10) =.5 P(10< x < 22) = P(10<x < 15)= = (1/10)(15-10) =.5

14. 14 Most computer languages include a functon that can be used to generate random numbers. In Excel, the RAND() function can be used to generate random numbers between 0 and 1. If we let x denote a random number generated using RAND(), then x is a continuous random variable with the following probability density function. f(x) = 1 for 0 ≤ x ≤ 1 f(x) = 0, elsewhere a) Graph the density function. Uniform Distribution Problem 4 f(x)f(x) f(x)f(x) x x 1 1 0 0 1 1 b) Compute P(.25 ≤ x ≤.75) c) Compute P( x ≤.3) d) Compute P( x >.6) e) Generate 100 uniform random variables between 20 and 80 using RAND() function. f) Compute Mean and StdDev for part (e)

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17. 17 The News Vendor Problem Swell Productions (The Retailer) is sponsoring an outdoor conclave for owners of collectible and classic Fords. The concession stand in the T-Bird area will sell clothing such as official Thunderbird racing jerseys. Suppose the probability of jerseys sales quantities is uniformly (and continuously) distributed between 100 and 400. Suppose sales price is $80 per jersey, purchase cost is $40, and unsold jerseys are returned to the manufacturer for $20 per unit. How many Jerseys Swell Production orders? 100400 Q Cu: Underage Cost Cu = 80-40 = 40 Co: Overage Cost Co = 40-20 = 20 SL* = Cu/(Cu+Co) SL* = 40/(40+20) = 2/3 (Q-100)/(400-100) = 2/3 Q= 300

18. 18 The expected number of participants in a conference is uniformly distributed between 100 and 700. The participants spend one night in the hotel and the cost is paid by the conference. The hotel has offered a rate of $200 per room if a block of rooms is reserved (non-refundable) in advance. The rate in the conference day is $300. All rooms will be single occupied. How many rooms should we reserve in the non-refundable block to minimize our expected total cost. 100 700 The News Vendor Problem Service level (Probability of demand not exceeding what we have ordered) SL* = Cu/(Cu+Co) Co: Overage cost Co = 200. Cu: Underage cost Cu = 300-200 = 100

19. 19 The News Vendor Problem a=100 b=700 ? B-1=600 1/600 0.3333 SL* = Cu/(Cu+Co) SL* = 100/(100+200) = 1/3 SL* = (Q-a)/(b-a) = (Q-100)/600 = 1/3 Q= 300 Uniform Random Variable Generation X= a+(b-a)Rand() X= 100+600Rand()

20. 20 Simulation of Project Management Network

21. 21 Simulation of Project Management Network

22. 22 Simulation of Project Management Network

23. 23 Normal Probability Distribution n The normal probability distribution is the most important distribution for describing a continuous random variable. n It is widely used in statistical inference. n It has been used in a wide variety of applications including: including: Heights of people Heights of people Rainfall amounts Rainfall amounts Test scores Test scores Scientific measurements Scientific measurements n Abraham de Moivre, a French mathematician, published The Doctrine of Chances in 1733. n He derived the normal distribution.

24. 24 Normal Probability Distribution  = 3.14159 e = 2.71828 The distribution is symmetric; its skewness measure is zero. The highest point on the normal curve is at the mean, which is also the median and mode. mean, which is also the median and mode. The distribution is symmetric; its skewness measure is zero. The highest point on the normal curve is at the mean, which is also the median and mode. mean, which is also the median and mode. Standard Deviation  Mean  x

25. 25 Normal Probability Distribution The entire family of normal probability distributions is defined by its mean  and its standard deviation . The entire family of normal probability distributions is defined by its mean  and its standard deviation . The standard deviation determines the width of the curve: larger values result in wider, flatter curves. The standard deviation determines the width of the curve: larger values result in wider, flatter curves. The mean can be any numerical value: negative, zero, or positive. The mean can be any numerical value: negative, zero, or positive.

26. 26 Normal Probability Distribution Probabilities for the normal random variable are given by areas under the curve. The total area under the curve is 1 (.5 to the left of the mean and.5 to the right). x  – 3   – 1   – 2   + 1   + 2   + 3  68.26% 95.44% 99.72%

27. 27 Standard Normal Probability Distribution A random variable having a normal distribution with a mean of 0 and a standard deviation of 1 is said to have a standard normal probability distribution. The letter z is used to designate the standard normal random variable. A random variable having a normal distribution with a mean of 0 and a standard deviation of 1 is said to have a standard normal probability distribution. The letter z is used to designate the standard normal random variable.  0 z

28. 28 Standard Normal Probability Distribution We can think of z as a measure of the number of standard deviations x is from . is used to compute the z value is used to compute the z value given a cumulative probability. given a cumulative probability. is used to compute the z value is used to compute the z value given a cumulative probability. given a cumulative probability. NORMSINVNORMSINV NORM.S.INV is used to compute the cumulative is used to compute the cumulative probability given a z value. probability given a z value. is used to compute the cumulative is used to compute the cumulative probability given a z value. probability given a z value. NORMSDISTNORMSDIST NORM.S.DIST The “S” in the function names reminds us that they relate to the standard normal probability distribution.

29. 29 n Excel Formula Worksheet Using Excel to Compute Standard Normal Probabilities AB 1 2 3 P(z < 1.00)=NORM.S.DIST(1,1) 4 P(0.00< z < 1.00) 5 P(0.00< z < 1.25) 6 P(-1.00< z < 1.00) 7 P(z > 1.58) 8 P(z < -0.50) 9 Probabilities: Standard Normal Distribution 0.8413 =NORM.S.DIST(1,1)-NORM.S.DIST(0,1) =NORM.S.DIST(1.25)-NORM.S.DIST(0,1) 0.3944 =NORM.S.DIST(1,1)-NORM.S.DIST(-1,1) 0.6827 =1-NORM.S.DIST(1.58,1) 0.0571 =NORM.S.DIST(-0.5,1) 0.3085 0.3413

30. 30 n Excel Formula Worksheet Using Excel to Compute Standard Normal Probabilities AB 1 2 3 z value with.10 in upper tail=NORM.S.INV(0.9) 4 z value with.025 in upper tail ) 5 z value with.025 in lower tail =NORM.S.INV(0.025) 6 Findingz Values, Given Probabilities 1.28 =NORM.S.INV(0.975) 1.96 =NORM.S.INV(0.025) -1.96

31. 31 Standard Normal Probability Distribution Pep Zone sells auto parts and supplies including a popular multi-grade motor oil. When the stock of this oil drops to 20 gallons, a replenishment order is placed. The store manager is concerned that sales are being lost due to stockouts while waiting for a replenishment order. Pep Zone sells auto parts and supplies including a popular multi-grade motor oil. When the stock of this oil drops to 20 gallons, a replenishment order is placed. The store manager is concerned that sales are being lost due to stockouts while waiting for a replenishment order. It has been determined that demand during replenishment lead-time is normally distributed with a mean of 15 gallons and a standard deviation of 6 gallons It has been determined that demand during replenishment lead-time is normally distributed with a mean of 15 gallons and a standard deviation of 6 gallons The manager would like to know the probability of a stockout during replenishment lead-time. In other words, what is the probability that demand during lead-time will exceed 20 gallons? The manager would like to know the probability of a stockout during replenishment lead-time. In other words, what is the probability that demand during lead-time will exceed 20 gallons? P ( x ≥ 20) = ? P ( x ≥ 20) = ?

32. 32 z = ( x -  )/  z = ( x -  )/  = (20 - 15)/6 = (20 - 15)/6 =.833333 =.833333 z = ( x -  )/  z = ( x -  )/  = (20 - 15)/6 = (20 - 15)/6 =.833333 =.833333 n Solving for the Stockout Probability Step 1: Convert x to the standard normal distribution. Step 2: Find the area under the standard normal curve to the left of z =.83. curve to the left of z =.83. Step 2: Find the area under the standard normal curve to the left of z =.83. curve to the left of z =.83. NORM.S.DIST(0.83333,1) NORM.S.DIST(0.83333,1) Standard Normal Probability Distribution.7977.7977

33. 33 P ( z >.83) = 1 – P ( z.83) = 1 – P ( z <.83) = 1-.7977 = 1-.7977 =.2023 =.2023 P ( z >.83) = 1 – P ( z.83) = 1 – P ( z <.83) = 1-.7977 = 1-.7977 =.2023 =.2023 n Solving for the Stockout Probability Step 3: Compute the area under the standard normal curve to the right of z =.83. curve to the right of z =.83. Step 3: Compute the area under the standard normal curve to the right of z =.83. curve to the right of z =.83. Probability of a stockout of a stockoutProbability P ( x > 20) Standard Normal Probability Distribution

34. 34 n Solving for the Stockout Probability 0.83 Area =.7967 Area = 1 -.7977 =.2023 =.2023 z Standard Normal Probability Distribution

35. 35 n Standard Normal Probability Distribution Standard Normal Probability Distribution If the manager of Pep Zone wants the probability If the manager of Pep Zone wants the probability of a stockout during replenishment lead-time to be no more than.05, what should the reorder point be? --------------------------------------------------------------- ---------------------------------------------------------------

36. 36 n Solving for the Reorder Point 0 Area =.9500 Area =.0500 z z.05 Standard Normal Probability Distribution This is done using =NORM.S.INV(0.95) 1.645

37. 37 n Solving for the Reorder Point Step 2: Convert z.05 to the corresponding value of x. Step 2: Convert z.05 to the corresponding value of x. x =  + z.05  x =  + z.05   = 15 + 1.645(6) = 24.87 or 25 = 24.87 or 25 x =  + z.05  x =  + z.05   = 15 + 1.645(6) = 24.87 or 25 = 24.87 or 25 A reorder point of 25 gallons will place the probability A reorder point of 25 gallons will place the probability of a stockout during leadtime at (slightly less than).05. of a stockout during leadtime at (slightly less than).05. Standard Normal Probability Distribution

38. 38 Normal Probability Distribution n Solving for the Reorder Point 15 x 24.87 Probability of a stockout during replenishment lead-time =.05 Probability of a stockout during replenishment lead-time =.05 Probability of no stockout during replenishment lead-time =.95 Probability of no stockout during replenishment lead-time =.95

39. 39 n Solving for the Reorder Point By raising the reorder point from 20 gallons to By raising the reorder point from 20 gallons to 25 gallons on hand, the probability of a stockout decreases from about.20 to.05. This is a significant decrease in the chance that This is a significant decrease in the chance that Pep Zone will be out of stock and unable to meet a customer’s desire to make a purchase. Standard Normal Probability Distribution

40. 40 Using Excel to Compute Normal Probabilities n Excel has two functions for computing cumulative probabilities and x values for any normal distribution: NORM.DIST is used to compute the cumulative NORM.DIST is used to compute the cumulative probability given an x value. probability given an x value. NORM.DIST is used to compute the cumulative NORM.DIST is used to compute the cumulative probability given an x value. probability given an x value. NORM.INV is used to compute the x value given NORM.INV is used to compute the x value given a cumulative probability. a cumulative probability. NORM.INV is used to compute the x value given NORM.INV is used to compute the x value given a cumulative probability. a cumulative probability.

41. 41 n Excel Formula Worksheet Using Excel to Compute Normal Probabilities AB 1 2 3 P(x > 20) =1-NORM.DIST(20,15,6,1) 4 5 6 7x value with.05 in upper tail 8 Probabilities: Normal Distribution Findingx Values, Given Probabilities =0.2023 =NORM.INV(0.95,15,6) =24.87 Note: P( x > 20) =.2023 here using Excel, while our previous manual approach using the z table yielded.2033 due to our rounding of the z value.

42. 42 Reorder Point If average demand during the lead time is 200 and standard deviation of demand during lead time is 25. At what level of inventory we should order such that with 90% confidence we will not have stockout. =NORM.S.INV(0.9) = 1.2816 (X-  )/σ =1.2816 X=  +1.2816σ X= 200 +1.2816(25) 200+32=NORM.INV(0.9,200,25) 232.0388 232.0388

43. 43 Average lead time demand is 20,000 units. Standard deviation of lead time demand is 5,000 units. The warehouse currently orders a 14-day supply, 28,000 units, each time the inventory level drops to 24,000 units. What is the probability that the demand during the period exceeds inventory? X= 24000  = 20000 σ = 5000 =NORM.DIST(24000,20000,5000,1) = = 0.788145 In 78.81 % of the order cycles, the warehouse will not have a stockout. Risk = 21.19%. Service Level

44. 44 Optimal Service Level: The News Vendor Problem An electronics superstore is carrying a 60” LEDTV for the upcoming Christmas holiday sales. Each TV can be sold at $2,500. The store can purchase each unit for $1,800. Any unsold TVs can be salvaged, through end of year sales, for $1,700. The retailer estimates that the demand for this TV will be Normally distributed with mean of 150 and standard deviation of 15. How many units should they order? Note: If they order 150, they will be out of stock 50% of the time. Which service level is optimal? 80%, 90%, 95%, 99%?? Cost =1800, Sales Price = 2500, Salvage Value = 1700 Underage Cost = Marginal Benefit = p-c = 2500-1800 = 700 Overage Cost = Marginal Cost = c-v = 1800-1700 = 100 Optimal Service Level = SL* = P( LTD ≤ ROP) = Cu/(Cu+Co) Or in NVP Terminology SL* = P( R ≤ Q*) = Cu/(Cu+Co)

45. 45 Optimal Service Level: The Newsvendor Problem Underage cost = Marginal Benefit =Cu = 2500-1800 = 700 Overage Cost = Marginal Cost = Co = 1800-1700 = 100 SL* = Cu/(Cu+Co) SL* = 700/800 = 0.875 1.15 Probability of excess inventory 0.875 Probability of shortage0.125 LTD =N(150,15) ROP = LTD + Isafety = LTD + zσ LTD = 150+1.15(15) Isafety = 17.25 = 18 ROP = 168 Risk = 12.5%

46. 46 The News Vendor Problem The expected number of participants in a conference is normally distributed with mean of 500 and standard deviation of 150. The participants spend one night in the hotel and the cost is paid by the conference. The hotel has offered a rate of $200 per room if a block of rooms is reserved (non-refundable) in advance. The rate in the conference day is $300. All rooms will be single occupied. How many rooms should we reserve in the non-refundable block to minimize our expected total cost. Service level (Probability of demand not exceeding what we have ordered) SL* = Cu/(Cu+Co) Co = 200, Cu = 300-200 = 100 SL* = Cu/(Cu+Co) = 100/(100+200) = 1/3 =NORM.INV(1/3,500,150)=436

47. 47 Daily demand for your merchandise has mean of 20 and standard deviation of 5. Sales price is $100 per unit of product. You have decided to close this business line in 64 days. Your supplier has also decided to close this line immediately, but has agreed to provide your last order at a cost of $60 per unit. Any unsold product will be disposed at cost of $10 per unit. How many units do you order Problem Game- The News Vendor Problem

48. 48 SL = Cu/(Cu+Co) = 40/(40+70) = 0.3636. Due to high overage cost, SL*< 50%. Z(0.3636) = ？ The optimal Q = LTD + z σ LTD =NORM.S.INV(0.3636) = -0.34885 Q = 1280 -0.34885(40) ROP = 1280-13.9541 ROP = 1266.0459 =NORM.INV(probability, mean, standard_dev) =NORM.INV(0.3636,1280,40=1266.0459 Problem Game- The News Vendor Problem

49. 49 100000 in One Stock or 10000 in Each of 10 stocks Suppose there are 10 stocks and with high probability they all have Normal pdf return with mean of 5% and standard deviation of 5%. These stocks are your only options and no more information is available. You have to invest $100,000. What do you do? =NORM.S.INV(rand()) =z = 0.441475 X= µ +  z = X = 5%+ 0.441475(5%) X= 5%+2.21% = 7.21% =NORM.INV(probability, mu, sigma) =NORM.INV(probability, 5%, 5%) =NORM.INV(rand(), 5%, 5%) = -8.1%

50. 50 100,000 vs. 10(10,000) investment in N(5%, 5%)

51. 51 Risk Aversion Individual

52. 52 Exponential Probability Distribution n The exponential random variables can be used to describe: Time between vehicle arrivals at a toll booth Time between vehicle arrivals at a toll booth Distance between major defects in a highway Distance between major defects in a highway Time required to complete a questionnaire Time required to complete a questionnaire time it takes to complete a task. time it takes to complete a task. n In waiting line applications, the exponential distribution is often used for interarrival time and service times. n A property of the exponential distribution is that the mean and standard deviation are equal. n The exponential distribution is skewed to the right. Its skewness measure is 2.

53. 53 n Density Function Exponential Probability Distribution for x > 0 n Cumulative Probabilities P(x≥x 0 )= 1- P(x≤ x 0 ) = e –x0/  = EXP(-x 0 /  ) e = 2.71828  = expected or mean in terms of time (in Poisson Distribution it was number of occurrences over a period of time x 0 = some specific value of x

54. 54 Exponential Probability Distribution n Example: Al’s Full-Service Pump The time between arrivals of cars at Al’s full- The time between arrivals of cars at Al’s full- service gas pump follows an exponential probability distribution with a mean time between arrivals of 3 minutes. Al would like to know the probability that the time between two successive arrivals will be 2 minutes or less.

55. 55 x x f(x)f(x) f(x)f(x).1.3.4.2 0 1 2 3 4 5 6 7 8 9 10 Time Between Successive Arrivals (mins.) Exponential Probability Distribution P ( x < 2) = 1 - 2.71828 -2/3 = 1 – EXP(-2/3) = P ( x < 2) = 1 - 2.71828 -2/3 = 1 – EXP(-2/3) = n Example: Al’s Full-Service Pump 1-0.5134 = 0.4866 1-0.5134 = 0.4866

56. 56 n Excel Formula Worksheet Using Excel to Compute Exponential Probabilities AB 1 2 3 P(x < 2)=EXPON.DIST(2,1/3,1) 4 Probabilities: Exponential Distribution AB 1 2 3 P(x < 2)0.4866 4 Probabilities: Exponential Distribution =EXPON.DIST(2,1/3,0)= ??

57. 57 Exponential Probability Distribution Average trade time in Ameritrade is one second. Ameritrade has promised its customers if trade time exceeds 5 second it is free (a $10.99 cost saving. The same promises have been practiced by Damion Pizza (A free regular pizza) and Wells Fargo ($5 if waiting time exceeds 5 minutes). There are 150,000 average daily trade. What is the cost to Ameritrade” P(x≥ X0) = e -X0/  = e -5/1 = 0.006738 Probability of of not meeting the promise is 0.6738% 0.006738*150,000* = 1011 orders @10.99 per order = 10.99*1011 = $11111 per day What was the cost if they had improved their service level by 50% that is to make it free for transactions exceeding 2.5 secs. e -2.5/1 = 0.082085 8.2085%*150,000*10.99 = $135317 per day We cut the promised time by half, our cost increased 12 times.

58. 58 Exponential Probability Distribution In a single phase single server service process and exponentially distributed interarrival time and service times, the actual total time that a customer spends in the process is also exponentially. Suppose total time the customers spend in a pharmacy is exponentially distributed with mean of 15 minutes. The pharmacy has promised to fill all prescriptions in 30 minutes. What percentage of the customers cannot be served within this time limit? P(x≥30) = EXP(-30/15) = 0.1353 13.53% of customers will wait more than 30 minutes. = P(x≤30) = EXPON.DIST(30,1/15,1) = P(x≤30) = 0.864665 P(x ≥ 30) = 1- P(x≤30) = 1- 0.864665 = 0.1353

59. 59 Exponential Probability Distribution 90% of customers are served in less than what time limit? 1-e -X0/  = 0.9 Find X0 e -X0/  = 0.1  e -X0/15 = 0.1 ln (0.1) = -X0/15 X0 = -15ln(0.1) X0 = -15(-2.30259) = 34.5388 More than 90% of customers are served in less than 35 mins. What is the probability of 3 customers arriving in 30 minutes One customer arrives per 15 minutes.

60. 60 Exponential Probability Distribution 90% of customers are served in less than what time limit?

61. 61 Exponential Probability Distribution SOLVER

62. 62 Exponential Random Variable P(x ≤ X0) = 1-e (-X0/µ) P(x ≤ X0) = rand() = 1-e (-X0/µ) 1-rand() = e (-X0/µ) 1-rand() by itself is a rand() rand() = e (-X0)/µ) e (-X0/µ) = rand() X0= -µrand() x= -µrand()

63. 63 The Poisson distribution provides an appropriate description of the number of occurrences per interval The Poisson distribution provides an appropriate description of the number of occurrences per interval The exponential distribution provides an appropriate description of the length of the interval between occurrences The exponential distribution provides an appropriate description of the length of the interval between occurrences One customer arrives per 15 minutes. The average number of customers arriving in 30 mins is 2. This is Poisson distribution. =POISSON.DIST(3,2,1) =0.857123

64. 64 Examples of Poisson distributed random variables: Examples of Poisson distributed random variables: the number of knotholes in 14 linear feet of the number of knotholes in 14 linear feet of pine board pine board the number of knotholes in 14 linear feet of the number of knotholes in 14 linear feet of pine board pine board the number of vehicles arriving at a toll the number of vehicles arriving at a toll booth in one hour booth in one hour the number of vehicles arriving at a toll the number of vehicles arriving at a toll booth in one hour booth in one hour Poisson Probability Distribution Bell Labs used the Poisson distribution to model Bell Labs used the Poisson distribution to model the arrival of phone calls. the arrival of phone calls. Bell Labs used the Poisson distribution to model Bell Labs used the Poisson distribution to model the arrival of phone calls. the arrival of phone calls.

65. 65 A Poisson distributed random variable is often A Poisson distributed random variable is often useful in estimating the number of occurrences useful in estimating the number of occurrences over a specified interval of time or space over a specified interval of time or space A Poisson distributed random variable is often A Poisson distributed random variable is often useful in estimating the number of occurrences useful in estimating the number of occurrences over a specified interval of time or space over a specified interval of time or space It is a discrete random variable that may assume It is a discrete random variable that may assume an infinite sequence of values (x = 0, 1, 2,... ). an infinite sequence of values (x = 0, 1, 2,... ). It is a discrete random variable that may assume It is a discrete random variable that may assume an infinite sequence of values (x = 0, 1, 2,... ). an infinite sequence of values (x = 0, 1, 2,... ). Poisson Probability Distribution The occurrence or nonoccurrence in any interval is independent of the occurrence or interval is independent of the occurrence or nonoccurrence in any other interval. nonoccurrence in any other interval. The occurrence or nonoccurrence in any interval is independent of the occurrence or interval is independent of the occurrence or nonoccurrence in any other interval. nonoccurrence in any other interval. The probability of an occurrence is the same for any two intervals of equal length. for any two intervals of equal length. The probability of an occurrence is the same for any two intervals of equal length. for any two intervals of equal length.

66. 66 n Poisson Probability Function Poisson Probability Distribution where: where: x = the number of occurrences in an interval x = the number of occurrences in an interval f(x) = the probability of x occurrences in an interval f(x) = the probability of x occurrences in an interval  = mean number of occurrences in an interval  = mean number of occurrences in an interval e = 2.71828 e = 2.71828 x ! = x ( x – 1)( x – 2)... (2)(1) x ! = x ( x – 1)( x – 2)... (2)(1)

67. 67 Poisson Probability Distribution n Poisson Probability Function In practical applications, x will eventually become In practical applications, x will eventually become large enough so that f ( x ) is approximately zero large enough so that f ( x ) is approximately zero and the probability of any larger values of x and the probability of any larger values of x becomes negligible. becomes negligible. In practical applications, x will eventually become In practical applications, x will eventually become large enough so that f ( x ) is approximately zero large enough so that f ( x ) is approximately zero and the probability of any larger values of x and the probability of any larger values of x becomes negligible. becomes negligible. Since there is no stated upper limit for the number Since there is no stated upper limit for the number of occurrences, the probability function f ( x ) is of occurrences, the probability function f ( x ) is applicable for values x = 0, 1, 2, … without limit. applicable for values x = 0, 1, 2, … without limit. Since there is no stated upper limit for the number Since there is no stated upper limit for the number of occurrences, the probability function f ( x ) is of occurrences, the probability function f ( x ) is applicable for values x = 0, 1, 2, … without limit. applicable for values x = 0, 1, 2, … without limit.

68. 68

69. 69 Simulation of Break-Even Analysis Fixed Cost =INT(A$3+(A$4-A$3)*RAND()) Variable Cost=INT(B$3+(B$4-B$3)*RAND()) Sales Price=INT($C$3+$C$4*NORM.S.INV(RAND())) Sales =-INT($D$3*LN(RAND()))

70. 70 Simulation of Break-Even Analysis

71. 71 End Continous PD and Poisson