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MGTSC 352. Lecture 23: Inventory Management Big Blue Congestion Management Introduction: Asgard Bank example Simulating a queue Types of congested systems, queueing template. Pg. 161. BigBluePills, Inc. Expensive drug treatments Perishable – last only 3 months Order once every 3 months
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MGTSC 352 Lecture 23: Inventory Management Big Blue Congestion Management Introduction: Asgard Bank example Simulating a queue Types of congested systems, queueing template
Pg. 161 BigBluePills, Inc. • Expensive drug treatments • Perishable – last only 3 months • Order once every 3 months • Regular cost: $400 per treatment • If demand > order size, place rush order • Rush cost: $1,000 per treatment • Price to patient: $650 • How much should they order?
Single period models • Perishable product • Past demand data • Must decide how much to order before knowing actual demand for the period • Must live with the consequences • We’ve seen this before: it’s called the “newsvendor problem”
Solution 1 • Average demand = 18, so … • … let’s order 18 each quarter • Profit = 18 (650 – 400) = $4,500 • Right? • Q < D lose ? / unit • Q > D lose ? / unit • Do these cancel out on average?
Tradeoffs • Q > D wasted product • Lose $400 • On product bought but not sold • Q < D loss on Sale • Lose $1000 - $650 = $350 • On Rush order (no lost sales) • If order = average demand each outcome will occur half the time. • Is that what you want?
Solution 2: simulation Solution 1 predicted this profit with Q = 18
The Flaw of Averages When input is uncertain... output given average input may not equal the average output
Huh? • Average BigBluePill demand = 18 • Profit for Q = 18, given D = 18: $4,500 • Average profit with Q = 18: $1,740 • Less than half • Optimal Q = 20, with avg. profit: $1,786 • Using avg. demand(ignoring variability) • Seriously overestimates profit • Results in a suboptimal decision
How bad can it get? • What if rush cost is $1,800(instead of $1,000) • The “averaging analyst” will still recommend Q = 18 and estimate P = $4,500. • The actual profit with Q = 18 will be -$565. • Using Q = 20 generates P = $1,217 • Using average inputs is a bad idea. “How bad” will depend on data.
In general Profit(AVERAGE(Demand1, Demand2, …, Demandn)) AVERAGE(Profit(Demand1), Profit(Demand2), …, Profit(Demandn))
Simple example of the flaw of averages: • A drunk on a highway • Random walk
Consider the drunk’s condition • The AVERAGE location of the drunk • Middle of the road • The outcome at the middle of the road ALIVE • What do you think the average outcome for the drunk is? DEAD Average inputs do not result in average outputs.
Congestion Asgard Bank ATM Pg. 168
pg. 168 Asgard Bank: Times Between Arrivals(pg. 173)
Asgard Bank: Arrival Rate • Given: avg. time between arrivals = 1.00 minute • average arrival rate per hour= = ?
Asgard Bank: Service Rate • Given: avg. service time = 0.95 minutes • average service rate per hour(if working continuously)= = • Note: • the service rate is not the rate at which customers are served • it’s the rate at which customers could be served, if there were enough customers • service rate = capacity of a server
Asgard Bank – Collecting Data • Real system: • Record arrival time, service start, service end • Compute inter-arrival times, service times, waiting time, time in system • Simulating the system: • Simulate inter-arrival times, service times • Compute arrival time, service start, service end, time in system, waiting time
Given:Average inter-arrival time = 1.00 min.Average service time = 0.95 min. Next customer arrives and begins service Customer leaves Customer arrives and begins service Why Do Customers Wait? 0.95 time What’s missing from this picture?
What’s missing from this Picture? VARIABILITY!
Including Randomness: Simulation • Service times: Normal distribution, mean = 57/3600 of an hour stdev = 10/3600 of an hour MAX(NORMINV(RAND(),57/3600,10/3600),0) • Inter-arrival times: Exponential distribution, mean = 1/ 60 of an hour. (1/60)*LN(RAND()) To Excel …
71 arrivals Simulated Lunch Hour 1:
50 arrivals Simulated Lunch Hour 2:
Unused capacity Simulated Lunch Hour 3:
Causes of Congestion • Higher than average number of arrivals • Lower than average service capacity • Lost capacity due to timing Lesson: For a service where customers arrive randomly, it is not a good idea to operate the system close to its average capacity
Anatomy of a Congested System (pg. 172) waiting room = queue parallel servers potential customers
Notation: M/M/s/K/N • Inter-arrival time distribution • Service time distribution • M = exponential distribution • G = general distribution • s = number of servers • K = max. # of customers allowed in system • being served + waiting • N = population size • K and N are left out when they are infinite
Analyzing a Congested System (pg. 174) Inputs System Description Model of the System Measures of Quality of Service Outputs Measures important to Servers
· m (mu) = service rate (per server per time unit) · l (lambda) = arrival rate (per time unit) · s = number of servers · C = maximum number that can wait in line = queue capacity · M = number of potential customers · s (sigma sub S) = standard deviation of service times S System Description
· Wq = average time in queue · W = average time in system · Lq = average number of customers in queue · L = average number of customers in system · SL = service level = fraction of customers that wait less than some given amount of time · PrBalk = fraction of customers that balk (do not enter system) · PrWait = fraction of customers that wait Measures of Quality of Service
· r (rho) = utilization = fraction of time each server is busy Measures Important to Servers
And now the formulas for the simplest case: M / M / 1 Or would you prefer an Excel template?
Template.xls • Does calculations for • M/M/s • M/M/s/s+C • M/M/s//M • M/G/1 • Want to know more? Go to http://www.bus.ualberta.ca/aingolfsson/qtp/ • Asgard Bank Data • Model: M/G/1 • Arrival rate: 1 per minute • Average service time: 57/60 min. • St. dev of service time: 10/60 min.
Asgard Conclusions • The ATM is busy 95% of the time. • Average queue length = 9.3 people • Average no. in the system = 10.25(waiting, or using the ATM) • Average wait = 9.3 minutes • What if the arrival rate changes to … • 1.05 / min.? • 1.06 / min.?