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Business Decision Models BU/EC275

Business Decision Models BU/EC275. Created by: Greg Overholt. Agenda. Waiting Lines Economic Analysis Simulation Crystal Ball / Arena Decision Analysis Payouts/Trees EVPI (Perfect Information) EVSI (Sample Info) Utility/Scoring. High-Level.

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Business Decision Models BU/EC275

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  1. Business Decision ModelsBU/EC275 Created by: Greg Overholt

  2. Agenda • Waiting Lines • Economic Analysis • Simulation • Crystal Ball / Arena • Decision Analysis • Payouts/Trees • EVPI (Perfect Information) • EVSI (Sample Info) • Utility/Scoring

  3. High-Level • Last term was all about deterministic models – when you have the same constraints and coefficients so you will get the same results. • This term is all about stochastic models – include some element of randomness

  4. Waiting Lines • 4 Types • Line Characteristics • Metrics • Calculations! Study of waiting line = Queuing Theory

  5. Overall Purpose

  6. 4 Types Autograph line (only 1 server) Eg: Customs (tons of servers and stations!) Old-School Cafeteria, where you went down the food line Bank Tellers (1 line, many servers)

  7. Characteristics! • Line structure: A/B/s - This format is called Kendall’s notation. • A: Arrival Distribution • B: Service Distribution • s: Number of Servers • A/B are typically either: • M: Markov’s dists: Poisson/Exponential Distribution • D: Constant (deterministic) • G: General Normal Distribution (mean / stdev) • Terms • Balking: when they come in and see the line so they don’t enter the queue • Reneging: when people leave the queue

  8. Poisson • Poisson is to describe the distribution of the probability that ‘x’ occurrences in an interval (number of successes). Discrete distribution = aka, the probability of each number is > 0 (vs continuous that is 0 for each individual possibility). =POISSON(x,λ,cumulative) If ‘cumulative’ = TRUE  less then If ‘cumul…’ = FALSE  equal to The rate / per hour (typically)

  9. Poisson Questions Students buy Tim Horton’s coffee at a rate of 100/hr (λ) Q1. What is the average time between arrivals? - 100 in an hour, so 60 mins/100 students = .6 mins per student Q2. What is prob that exactly 100 students come in the next hour? f(100) = 100100*e-100 / 100! f(100) = .0399 or 3.99% probability.

  10. Exponential • Exponential is the probability that the person will arrive/served within the first ‘x’ mins. • P(x < 2) = the probability that a person will arrive in 2 mins or less. (CUMULATIVE) =EXPONDIST(x,λ,cum) cumulative =TRUE  less then IF P > X, then take away the ‘1-’ (right tail test!)

  11. Exponential Question At the Tim Hortons, they take, on average, 1.5 mins to serve a customer. Q1. What is the service rate (per hour)? - We want this in hours, so 1.5 mins = .025 hours. - λ = 1/u = 1/.025 = 40 customers / hour. Q2. What is the prob that the service will take exactly 1.5 mins? - 0 (continuous probability, and the exact prob = 0) Q3. What is prob that service will take more then 3 mins? P(x > 3) = e -40(3/60) = .13533 = 13.53%

  12. Relationship between P and E • The relationship between the Poisson and exponential is very important. E.g. if the distribution for the time between arrivals is exponential with mean = B , then the distribution of the number of arrivals per unit of time is Poisson with mean = 1 / B • INVERSE RELATIONSHIP!!

  13. Waiting Line Inputs λ / μ = ρ (or U) = Utilization factor (probability that server is busy / prob that a customer has to wait) I = 1 – U: This is the probability that the server is idle == the probability that there are no customers in the queuing system (P0) If ρ = 60% (so busy for 60% of the time), I (idle %) would be 40% = the chance that the server is not serving a customer, and no one is in the queue.

  14. Terms / Things to Calculate Other Operating characteristics of the queuing system: • Po = probability the service facility is idle (pronounced “P-naught”) • Pn = probability of n units in system • Pw = probability an arriving unit must wait • Lq = average number of units in queue awaiting service • L = average number of units in system • Wq = average time a unit spends in queue awaiting service • W = average time a unit spends in system

  15. Example! Joe Ferris is a stock trader for the firm of Smith, Jones, Johnson, and Thomas, Inc. Stock transactions arrive at a mean rate of 20 per hour. Each order received by Joe requires an average of two minutes to process. (Assuming a M/M/1 system)

  16. Example! Arrival Rate Distribution Question: What is the probability that no orders are received within a 15-minute period? What is the mean number of orders in the 15 mins? (20 per hours = 1 every 3 mins, so average = 5 every 15 mins = λ Answer: P(x) = (λxe-λ)/x! P(0) = (50e-5)/0! = e-5 = .0067 = .67% chance =POISSON(0,5,false)

  17. Example! Arrival Rate Distribution Question: What is the probability that more than 6 orders arrive within a 15-minute period? Answer: P(x > 6) = 1 - [P(0) + P(1) + P(2) + P(3) + P(4) + P(5) + P(6)] = 1 - 0.762 = 0.238 = 23.8% chance that more then 6 orders arrive. =1-POISSON(6,5,true)

  18. Example! Service Rate Distribution Question: What is the mean service rate per hour? Answer: Since Joe Ferris can process an order in an average time of 2 minutes (.033 hours) then the mean service rate, μ, is μ = 1/(mean service time), or 60/2 = 30 / hr

  19. Example! Service Time Distribution Question: What % of orders will take less than one minute to process? Answer: Since units are expressed in hrs, P(T ≤ 1 minute) = P(T ≤ 1/60 hour) (μ = 30 per hour – from previous slide) Using exponential distribution, P(T ≤ t) = 1 - e-μt P(T<1/60) = 1 - e-30(1/60) = 1 - .6065 = .3935 = 39.35% = EXPONDIST(1/60, 30,1) The value gotten from e-μt is the area to the right, but here we want the area to the left (so 1 – value)

  20. Calculate System variables NOTE: L > Lq and W > Wq. If you calculate something otherwise, re-check your work!

  21. Average Time in System Question: What is average time an order must wait from time Joe receives order until it is finished being processed (i.e. its turnaround time)? Answer: This is an M/M/1 queue with λ = 20 per hour & μ = 30 per hour. Average time an order waits in the system is: W = 1/(μ - λ) = 1/(30 - 20) = 1/10 hour or 6 minutes

  22. Example! Average Length of Queue Question: What is the average # of orders Joe has waiting to be processed? Answer: Average # of orders waiting in queue is: Lq = λ2/[μ(μ - λ)] = (20)2/[(30)(30-20)] = 400/300 = 4/3 = on average there are 1.33 orders waiting in the queue.

  23. Example! Utilization Factor Question: What percentage of time is Joe processing orders? Answer: % of time Joe is processing orders is equivalent to utilization factor, λ/μ. Thus, % of time he is processing orders is: λ/μ = 20/30 = 2/3 or 66.67% of the time.

  24. Example 2 (M/M/2) Smith, Jones, Johnson, & Thomas, Inc. has begun a major advertising campaign which it believes will increase its business 50%. To handle the increased volume, the company has hired an additional floor trader, Sue Hanson, who works at the same speed as Joe Ferris (2 minute or 30 per hour). Recall λ = 20 per hour & μ = 30 per hour with just Joe (M/M/1) > Note that the new arrival rate of orders, λ, is 50% higher than that of problem (A). Thus, λ = 1.5(20) = 30 per hour.

  25. Example 2 (M/M/2) Sufficient Service Rate Question: Why will Joe Ferris alone not be able to handle the increase in orders? Answer: Since Joe Ferris processes orders at a mean rate of μ = 30 per hour, then λ = μ = 30 and utilization factor, λ/μ = 1. This implies queue of orders will grow infinitely large. Hence, Joe alone cannot handle this increase in demand. CANNOT HAVE λ being = or > then μ - can’t have people arriving a rate faster then you can serve them – queue will grow infinitely!

  26. Example 2 (M/M/2) Probability of no Units in System Question: What is probability that neither Joe nor Sue will be working on an order at any point in time? Answer: Given λ = 30, μ = 30, k = 2 & [λ/μ] = 1, probability that neither Joe nor Sue will be working is: A = 33% chance that neither Joe nor Sue will be working. Summation only applies to term directly to right

  27. Example 2 (M/M/2) Average Time in System Question: What is average time in the system (turnaround time) for an order with both Joe & Sue working? Answer: Average turnaround time is average waiting time in system (not just the line!!) = W.

  28. Big Problem!! • What is the ideal number of servers?? • Balance speed and cost (both service cost and waiting cost!!) • Need to compare TOTAL costs across the different scenarios.

  29. How? • Change Queue Priority • Change Queue Style (1 line reduces wait time vs multiple lines) • Improve Service Rate • More channels / servers • Faster channels (technology / self-serve) • EXAM: Will ask you to do an ‘economic analysis’ of 2 scenarios that have different number of server! (compare total costs!)

  30. Economic Analysis • The advertising campaign of SJJT, Inc. was so successful that business actually doubled. The mean rate of stock orders arriving is now 40/hr and the company must decide how many traders to employ. Each trader hired can process an order in an average time of 2 min. • Based on a number of factors the brokerage firm has determined the average waiting cost/minute for an order to be $0.50. Traders hired will earn $20/hr in wages & benefits. Using this information compare the total hourly cost of having 2 traders with that of having 3 traders.

  31. Economic Analysis Economic Analysis of Waiting Lines Total Hourly Cost = Cost of Service + Cost of Waiting = (Total salary cost per hour) + (Total hourly cost for orders in the system) = ($20 per hour) x (Number of traders) + ($30 waiting cost per hour) x (Number of Orders per hour) x (Average wait per order) = 20k + 30 x lW but L = lW  Little’s Flow Equations = 20k + 30L

  32. Economic Analysis Thus, L must be determined for k = 2 traders and for k = 3 traders with l = 40/hr and m = 30/hr (since the average service time is 2 minutes (1/30 hr). TOTAL COST = 20k + 30L  WE DON’T  WE HAVE  WE DON’T  WE HAVE

  33. Economic AnalysisCost of Two Servers Probability of no Units in System 20% chance that there will be no units in the system.

  34. Economic AnalysisCost of Two Servers Cost of System

  35. Economic AnalysisCost of Three Servers Probability of no Units in System 25% chance that there will be no units in the system.

  36. Economic AnalysisCost of Three Servers Cost of System

  37. Simulation The process of designing a mathematical or logical model of a real system and then conducting computer-based experiments with the model to describe, explain, and predict the behavior of the real system. Advantages: • Leads to a better understanding of the real system • Simulation is far more general then mathematical models • Simulation answers ‘what-if’ questions Disadvantages: • No guarantee that it will provide good answers • Time consuming • The simulation technique still lacks a standardized approach • Unlike analytical techniques, it is NOT an optimizing technique.

  38. SIMULATION Simulation is the most widely used Decision Analysis technique Simulation is NOT an optimizing technique, but a descriptive tool of the system under various conditions

  39. Types of Simulations Static vs Dynamic • Does time have a role in model? Continuous change vs Discrete change • Can system change continuously or only at discrete points in time Deterministic vs Probabilistic • Is everything certain, or is there uncertainty? • Use Monte Carlo technique for uncertainty Most operational models are: • Dynamic, Discrete change, Probabilistic

  40. Equation for profit model • Without simulation, you can get worst / avg / best case results.. But need a formula to figure out profit: Let P = the unit price. Marketing has set this already: P = $40:00. Let Q = the quantity to produce (i.e. the answer to the question we’ve posed). Let D = the quantity sold in the market (i.e. the Demand). Let FC = the fixed costs (depending on which machine tool we end up buying), and Let UC = the unit cost (the planning numbers above). • PROFIT = P * [min(Q,D)] - (UC * Q) - FC

  41. What Program to use!! • Excel: when time isn’t a factor, and probabilities are known! • Crystal Ball: Crystal Ball automatically calculates thousands of different "what if" cases, saving the inputs and results of each calculation as individual scenarios. Analysis of these scenarios reveals to you the range of possible outcomes, their probability of occurring, which input has the most effect on your model and where you should focus your efforts. • Arena: Arena is most effective when modeling and analyzing business, service, or manufacturing processes or flows. Use when things are very dynamic, changes with time, and multiple steps

  42. Probabilistic ‘Excel’ Style question:

  43. Exam Example • A consultant knows that the monthly costs incurred is uniformly distributed in the range ($3000, $5000) if he has less than 4 clients per month. If he has 4 to 6 clients that month, his expenditure is either $5000 or $6000 [both are equally likely]. He never has more than 6 clients a month and 30% of the time he has less than 4 clients a month. His monthly revenue from clients is either $4000 or $7,000. The probability that his revenue is $4000 is twice the probability that his revenue is $7000. Expense Probability: If has less then 4 clients, then take RN*$2000, and add that to $3,000. If has 6 clients, then 50% will be $5K, 50% will be 6K, so if 0.0  0.5, then take $5K, if above 0.5 then use $6K Client Probability:30% = less then 4,70% will have 6 clientsSo, RN’s 0  0.3 = less then 40.3  1 = 6 clients Revenue Probability: $4K is twice as probable.. So 66% chance of being 4K, 33% chance of being 7K. So if 0.0  0.66 go with $4K, if above .66 go with $7K.

  44. no $6K $4K .35*2K + 3K = 3,700 yes $7K .27*2K + 3K = 3,540 yes $4K no $5K $4K Client Probability:30% = less then 4,70% will have 6 clientsSo, RN’s 0  0.3 = less then 40.3  1 = 6 clients Expense Probability: If has less then 4 clients, then take RN*$2000, and add that to $3,000. If has 6 clients, then 50% will be $5K, 50% will be 6K, so if 0.0  0.5, then take $5K, if above 0.5 then use $6K Revenue Probability: $4K is twice as probable.. So 66% chance of being 4K, 33% chance of being 7K. So if 0.0  0.66 go with $4K, if above .66 go with $7K. -$2K $3.4K $0.46K $1K

  45. Crystal Ball • The Monte Carlo technique is defined as a technique for selecting numbers randomly from a probability distribution for use in a trial (computer run) of a simulation model. • Crystal Ball is a risk analysis and forecasting program that uses Monte Carlo simulation to provide a statistical range of results.

  46. Crystal Ball Previous Final Exam Question: What is initial seed value? - the starting random number generated that you want to use in your sequence. So if you have a seed value of 100, then your test will use the 100th, 101st, 102nd random number generated until you don’t need anymore.

  47. Crystal Ball

  48. Crystal Ball • Overall: Spreadsheets do not account for the variability of the real world. • CB allows you to you range of inputs to explore the range of possible outcomes AND the probability of their occurrence.

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