1.21k likes | 1.33k Vues
This presentation by Stephen Duarte delves into the critical aspects of queuing theory and stochastic models in decision-making. It covers an array of topics including economic analysis, simulation techniques like Monte Carlo and Arena, and decision analysis frameworks such as Expected Value of Perfect Information (EVPI) and Expected Value of Sample Information (EVSI). The focus is on understanding waiting lines, arrival and service distributions, metrics for performance analysis, and the implications of randomness in operational efficiency. Attendees are encouraged to engage and ask questions.
E N D
Business Decision Models BU/EC275 Created by: Greg Overholt Presented by: Stephen Duarte bu275sos@hotmail.com
Agenda • Waiting Lines • Economic Analysis • Simulation • Monte Carlo / Arena • Decision Analysis • Payouts/Trees • EVPI (Perfect Information) • EVSI (Sample Info) • Utility/Scoring • Remember to ask questions!! Thumbs up/down?
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
Waiting Lines • 4 Types • Line Characteristics • Metrics • Calculations! Study of waiting line = Queuing Theory
4 Types Autograph line (only 1 server) Old-School Cafeteria, where you went down the food line
Bank Tellers (1 line, many servers) Eg: Departures (tons of servers and stations!)
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 • Terms • Balking: when they come in and see the line so they don’t enter the queue • Reneging: when people leave the queue
Poisson • the distribution of the probability that ‘x’ occurrences in an interval (number of successes). • Discrete distribution = aka, the probability of each number is > 0 • PROBABILITY FOR A SPECIFIC VALUE The rate / per hour (typically)
Poisson • Excel =POISSON(x,λ,cumulative) If ‘cumulative’ = TRUE less then If ‘cumul…’ = FALSE equal to
Poisson Questions Students arrive at Tim Horton’s at a rate of 10/hr (λ) Q1. What is the average time between arrivals? (What is µ?) - 10 in an hour, so 60 mins/10 students = 6 mins per student Inverse Relationship! Example Q2. What is prob that exactly 5 students come in the next hour? f(5) = 105*e-10 / 5! f(5) = .0378 or 3.78% probability
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) IF P > X, then take away the ‘1-’ (right tail test!)
Excel =EXPONDIST(x,λ,cum) cumulative =TRUE less then
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%
Poisson vs Exponential Poisson P(x=xo)=POISSON(x,λ,false) P(x≤xo)=POISSON(x,λ,true) P(x>xo)=1 - POISSON(x,λ,true) Exponential P(x=xo)= EXPONDIST(x,λ,false) = 0 P(x≤xo)= EXPONDIST(x,λ,true) P(x>xo)= 1 - EXPONDIST(x,λ,true) cumulative =TRUE less then
Waiting Line Inputs NOW WE SEPARATE λ AND µ
λ / μ = U = Utilization factor • probability that server is busy • probability that a customer has to wait • I = 1 – U : Idle time • the probability that the server is idle • the probability that there are no customers in the queuing system (P0) • If U = 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.
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
Example! Joe Ferris is a Sub Maker at MR Sub in the terrace. Sub orders arrive at a mean rate of 20 per hour. Each order received by Joe requires an average of two minutes to Make the sub. (Assuming a M/M/1 system)
Example! Arrival Rate Distribution Question: What is the probability that no orders are received within a 15-minute period? Poisson because you are looking for a specific value 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)
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)
Example! Service Rate Distribution Question: What is the average service rate per hour? (What is μ?) Answer: Since Joe Ferris can process an order in an average time of 2 minutes (.033 hours) We know that: Average Service Time = 1/ μ, and we are looking for average service rate (which is just μ). So: Average Service Time = 0.033 hours = 1/ μ 0.033 = 1/ μ -> μ = 1/0.033 Therefore, μ = 30 / hr
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)
Calculate System variables NOTE: L > Lq and W > Wq. If you calculate something otherwise, re-check your work!
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
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.
Example! Utilization Factor Question: What percentage of time is Joe making subs? 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.
Example 2 (M/M/2) Mr Sub 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 sub maker, 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.
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 that Joe is working all the time, he is unable to take breaks, in order to comply with labour standards, we cannot do this CANNOT HAVE λ being = or > then μ - can’t have people arriving a rate faster then you can serve them – queue will grow infinitely!
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
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.
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.
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!)
Economic Analysis • The advertising campaign of Mr Sub. was so successful that business actually doubled. The mean rate of sub orders arriving is now 40/hr and the company must decide how many sub makers to employ. Each sub maker hired can process an order in an average time of 2 min. • Based on a number of factors the sub shop has determined the average waiting cost/minute for an order to be $0.50. Sub makers hired will earn $20/hr in wages & benefits. Using this information compare the total hourly cost of having 2 sub makers with that of having 3 sub makers.
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 (sub makers) + ($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
Economic Analysis Thus, L must be determined for k = 2 sub makers and for k = 3 2 sub makers 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
Economic AnalysisCost of Two Servers Probability of no Units in System 20% chance that there will be no units in the system.
Economic AnalysisCost of Two Servers Cost of System
Economic AnalysisCost of Three Servers Probability of no Units in System 25% chance that there will be no units in the system.
Economic AnalysisCost of Three Servers Cost of System
Sub Makers Sub Makers
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.
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
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
What Program to use!! • Excel: when time isn’t a factor, and probabilities are known! • 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
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 4 to 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 4 to 6 clientsSo, RN’s 0 0.3 = less then 40.3 1 = 4 to 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.
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 = 4 to 6 clients Expense Probability: If has less then 4 clients, then take RN*$2000, and add that to $3,000. If has 4 to 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
Monte Carlo • 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. • Initial seed value is the number you begin with