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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 - PowerPoint PPT Presentation
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MGTSC 352
Lecture 23: Inventory Management
Big Blue
Congestion ManagementIntroduction: Asgard Bank example
Simulating a queueTypes of congested systems, queueing template
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?
Pg. 161
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”
Five Years of Demand DataAverage 18.00
Stdev 6.66 min 5.00
max 34.00
Category bin Frequency<=5 5 1
6 - 10 10 311 - 15 15 316 - 20 20 521 - 25 25 626 - 30 30 1> 30 1
01234567
<=56 - 10
11 - 1516 - 2021 - 2526 - 30> 30
Quarterly demand
Frequency
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
-$10,000
-$8,000
-$6,000
-$4,000
-$2,000
$-
$2,000
$4,000
$6,000
$8,000
5 10 15 20 25 30 35 40 45
Order size
Average profit
5th percentile
95th percentile
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
Asgard Bank: Times Between Arrivals
(pg. 173)
0
10
20
30
40
50
60
70
0.00 - 0.250.25 - 0.500.50 - 0.750.75 - 1.001.00 - 1.251.25 - 1.501.50 - 1.751.75 - 2.002.00 - 2.252.25 - 2.502.50 - 2.752.75 - 3.003.00 - 3.253.25 - 3.503.50 - 3.753.75 - 4.00
> 4.00
Time between Arrivals (min.)
Frequency
Average = 1.00 min.St. dev. = 0.92 min.Arrival rate = λ = ?
pg. 168
Asgard Bank: Arrival Rate
• Given: avg. time between arrivals = 1.00 minute• average arrival rate per hour
= λ = ?
Asgard Bank: Service Times
0
10
20
30
40
50
60
70
80
0.00 - 0.100.20 - 0.300.40 - 0.500.60 - 0.700.80 - 0.901.00 - 1.101.20 - 1.301.40 - 1.501.60 - 1.701.80 - 1.902.00 - 2.102.20 - 2.302.40 - 2.502.60 - 2.702.80 - 2.903.00 - 3.103.20 - 3.303.40 - 3.503.60 - 3.703.80 - 3.90Duration of Service (min.)
Frequency
Average = 0.95 min. (57 sec.)St. dev. = 0.17 min. (10 sec.)
service rate = μ = ?
Asgard Bank: Service Rate
• Given: avg. service time = 0.95 minutes• average service rate per hour
(if working continuously)= μ =
• 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
Customer numberArrival timeInter-arrival timeService startsService endsService timeTime in systemWaiting time1 0.00 0.00 0.50 0.50 0.50 0.002 2.00 2.00 2.00 2.25 0.25 0.25 0.003 2.00 0.00 2.25 3.00 0.75 1.00 0.254 2.75 0.75 3.00 3.25 0.25 0.50 0.255 3.25 0.50 3.25 3.75 0.50 0.50 0.006 3.80 0.55 3.80 4.00 0.20 0.20 0.007 3.90 0.10 4.00 4.20 0.20 0.30 0.108 3.90 0.00 4.20 4.50 0.30 0.60 0.309 3.90 0.00 4.50 5.00 0.50 1.10 0.60
10 5.20 1.30 5.20 5.50 0.30 0.30 0.00
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
Why Do Customers Wait?
Given:Average inter-arrival time = 1.00 min.Average service time = 0.95 min.
0.95
Customer arrives and begins service
Customer leavestimeNext customer arrives
and begins service
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 …
0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00
1
11
21
31
41
51
61
Customer number
Time (hours)
Waiting time
Service time
Press F9 to recalculate
Simulated Lunch Hour 1:
71 arrivals
0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00
1
11
21
31
41
51
61
Customer number
Time (hours)
Waiting time
Service time
Press F9 to recalculate
Simulated Lunch Hour 2:
50 arrivals
0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00
1
11
21
31
41
51
61
Customer number
Time (hours)
Waiting time
Service time
Press F9 to recalculate
Simulated Lunch Hour 3:
Unused capacity
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
potential customers parallel servers
Notation: M/M/s/K/N
1. Inter-arrival time distribution2. Service time distribution
• M = exponential distribution• G = general distribution
3. s = number of servers4. K = max. # of customers allowed in system
• being served + waiting
5. N = population size• K and N are left out when they are infinite
Types of Congested Systems We Will Analyze (pg. 173)
Type Service time dist’n
Inter-arrival time
dist’n
# of servers
# that can wait
# of potential
customers
M/M/s Exponential exponential s
M/M/s/s+C exponential exponential s C
Finite Population
exponential exponential s M
M/G/1 general exponential 1
Analyzing a Congested System (pg. 174)
System Description
Measures of Quality of Service
Measures important to Servers
Model of the System
Inputs
Outputs
System Description
λ μ (mu) = service rate (per server per time unit) (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
Measures of Quality of Service
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 Important to Servers
(rho) = utilization = fraction of time each server is busy
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.?