Unformatted text preview: Goals of the Game
of the Game
• Experience the dynamics of a distribution
the dynamics of distribution The Distribution Game
The Distribution Game •
•
• Peter L. Jackson
Professor
School of O.R. and I.E.
2/4/2012 Industrial Data and Systems Analysis • system (demand, orders, lead time)
Discover rules for managing flow of
material
Optimize cost tradeoffs
Identify causes of inventory: pipeline
stock, cycle stock, safety stock
Compare different supply chain
configurations 2/4/2012 1 The Distribution Game
Distribution Game Industrial Data and Systems Analysis 2 Game Parameters
Parameters
Cost: $70 / unit
$70 unit
15 days transport + 1 day pick/pack/ship Supplier Order: $200
Hold:$0.21/$/yr Central
Warehouse
5 days transport + 1
day pick/pack/ship Retailer Retailer Order: $2.75
Hold:$0.25/$/yr Retailer Avg. demand: 2 /day
(Coeff. of Variation: 1) Customers
2/4/2012 Industrial Data and Systems Analysis 3 2/4/2012 Price: $100 / unit
$100 unit Industrial Data and Systems Analysis 4 Demonstration: with
Warehouse 2/4/2012 Demonstration: without
Warehouse 5 Industrial Data and Systems Analysis 2/4/2012 Industrial Data and Systems Analysis 6 Four Roles of Inventory
Roles of Inventory Four Roles of Inventory
Roles of Inventory Pipeline Stock
Stock
Cycle Stock
Stock Peter L. Jackson
Professor
School of O.R. and I.E.
Decoupling Stock
2/4/2012 Industrial Data and Systems Analysis 7 2/4/2012 Safety Stock
Stock Industrial Data and Systems Analysis 8 In your reading packet
In your reading packet Pipeline Stock
Stock Covers cycle stock
and decoupling stock Input rate, r Jackson, P.L. 2011. FillRate Driven
Safety Stock. ORIE 3120 Class Notes. Output rate, r Valueadded process time or transport time, T School of Operations Research and
Information Engineering, Cornell
University Ithaca NY
University, Ithaca, NY. Little’s Law: Average pipeline stock = rT Covers safety stock To reduce pipeline stock, reduce the duration of
th
valueadded activities 2/4/2012 Industrial Data and Systems Analysis 9 Decoupling Stock
Stock 2/4/2012 Industrial Data and Systems Analysis 10 A Useful Queueing Model
Useful Queueing Model
Jobs
arriving Waiting queue
1 Completed
jobs
departing Workstation • Decoupling stock: queues of work •
•
•
• between workstations caused by
variability in processing times 2/4/2012 Industrial Data and Systems Analysis 11 Focus on a single workstation
on single workstation
Jobs arrive at rate λ
Jobs can be completed at rate μ
can be completed at rate
λ < μ so the workstation is sometimes idle 2/4/2012 Industrial Data and Systems Analysis 12 No Decoupling Stock in
Deterministic Case Variability is the Enemy
is the Enemy • If the time between arrivals is constant
the time between arrivals is constant • Queues develop when the arrival rate temporarily exceeds the processing rate and the processing time is constant then
there will never be a queue
arrival busy arrival arrival arrival busy idle • Assume independent and identically
id • Utilization rate: fraction of time the distributed (i.i.d.) interarrival times, and
i.i.d. process times
i.i.d. process times workstation is busy; = busy/(busy+idle) 2/4/2012 Industrial Data and Systems Analysis idle 13 2/4/2012 Industrial Data and Systems Analysis Interpreting the Utilization
Rat
te How Shall We Measure
Variability? • Let ρ = λ/μ
• ρ is the utilization rate, the average fraction of 14 • Let ma = mean interarrival time (ma = 1/ λ)
mean interarrival time (m 1/
2 = variance of interarrival time
• Let σ a
• Let ca = σa / ma, the coefficient of
the coefficient of time the workstation is in use
• 1 ρ is the probability you will find the queue
empty and the workstation idle if you observe it
at random time
at a random time
• μ λ is the average rate at which queues
decrease, once a queue has formed
• ρ/(1 ρ)= λ/(μ λ) a unitless measure of the (lack
of) ability of workstation to work off queues
2/4/2012 Industrial Data and Systems Analysis variation of interarrival time
• ca is a unitless measure of variability
unit
measure of variability
• ca ≥ 1 corresponds to high variability • ca = 1 is characteristic of the exponential
is characteristic of the exponential distribution • ca ≤ 1 for our examples
15 2/4/2012 Industrial Data and Systems Analysis 16 Variability
Variability Variability of Effective Process
Ti
Time • c2a is called the squared coefficient of • Process times are usually not highly variation (SCV) of the interarrival process variable on their own • Or, “the arrival SCV” • Time to make a part if everything goes well doesn’t change much from piece to piece
doesn’t change much from piece to piece
• Breakdowns, rework, setups make effective
process time more variable • Similarly, let me = mean effective processing time (me = 1/ μ)
• Let σ2e = variance of effective processing
variance of effective processing
time
• Why “effective”? Recall our adjustments
effective Recall our adjustments
to the throughput rate (speed loss,
breakdowns, etc.)
2/4/2012 Industrial Data and Systems Analysis • Let ce = σe / me, the coefficient of variation of effective process time
• c2e is called the process SCV
th
SCV
2 is not easy to estimate
• Unfortunately, c e
17 2/4/2012 • Let Wq = expected waiting time in queue
expected waiting time in queue
• Then, a reasonable approximation is • Let Lq = expected number of units in
expected number of units in queue
• Little’s Law: Lq = λWq
• So, a reasonable approximation is c 2 + c 2 ρ Wq = a e 2 1 − ρ me c 2 + c 2 ρ 2 Lq = a e 2 1 − ρ • Suppose interarrival times and effective
interarrival times and effective process times are exponentially
distributed (c
distributed (ca = ce = 1). Then,
1). Then, ρ
λme mean arrivals during service time
Wq = 1 − ρ me = μ − λ = rate at which queue is reduced rate which Industrial Data and Systems Analysis 18 Decoupling Stock: Expected
Queue Length Kingman’s Approximation
Approximation 2/4/2012 Industrial Data and Systems Analysis 19 • What happens as ρ approaches 1?
2/4/2012 Industrial Data and Systems Analysis 20 The Impact of Utilization and
Variability on Decoupling Stock
St • Decrease utilization
utilization Decoupling Stock:
Expected Queue Length Kingman's Approximation to the G/G/1 Queue
th G/G/1 • I.e. increase μ relative to λ (throughput improvements)
• Reduce λ (the DGR)
• Buy extra capacity (extra machines to
increase
increase μ) 100
80
60
40
20 • Reduce variability 0
0 0.2 0.4 0.6 0.8 • Increase MTTF, reduce MTTR
MTTF reduce MTTR
• Reduce or eliminate rework
• Reduce or eliminate setup times
or eliminate setup times 1 Utilization Rate
High Variability 2/4/2012 Medium Variability Low Variability 21 Industrial Data and Systems Analysis Series of Queues
of Queues
1 2 Industrial Data and Systems Analysis 22 • Throughput rate = min(λ,μ)
• If λ>μ then this workstation is a bottleneck and it 3 will have to work overtime to catch up (or λ will
have to be reduced)
have to be reduced)
• In general, we will assume λ<μ (i.e. ρ<1).
• In a series of workstations with ρi <1 for each
station i, the throughput rate at each station will
th th
be λ
• The bottleneck will be the workstation with the
highest utilization, ρi.. Since the throughput is the
same at each station (λ), the bottleneck will be
the workstation, i, with the lowest processing
rate, μi another
• Variability in departure times from one
station becomes variability in arrival
times at the next station Industrial Data and Systems Analysis 2/4/2012 Interpreting Utilization
Utilization • Output of one workstation is input to 2/4/2012 How to Reduce Decoupling
St
Stock? 23 2/4/2012 Industrial Data and Systems Analysis 24 Data We Assume We Know
Data We Assume We Know
ma=1/ λ c2a =σ 2a λ 2
2 me1=1/ μ1 c2 e1 • Input Data
Data λ
1 λ Example =σ 2 e1 μ1 2 3 me2=1/ μ2 c2 e2 =σ 2 e2 μ2 2 Station
Time per piece (seconds)
Coefficient of variation 0 (Interarrivals)
40
0.25 1 2
25
1.25 3
38
0.75 37
0.75 me3=1/ μ3 c2 e3 =σ 2e3 μ32 • Given the input rate, λ, the arrival SCV, c2a, the processing rates, μi , and the
process SCV’s
process SCV’s, c2ei, for each station i, we
for each station we
would like to estimate the expected queue
length in front of each station 2/4/2012 Industrial Data and Systems Analysis 25 2/4/2012 Industrial Data and Systems Analysis Estimating Departure
Variability
bilit Two Formulas to Remember
Formulas to Remember • Let c2d denote the departure SCV from a 26 • Kingman’s Approximation for Queue queue (the squared coefficient of variation
of interdeparture times)
• A good simple approximation is
good simple approximation is Length 2
2
cd = ρ 2 ce2 + (1 − ρ 2 )ca • Approximation for Departure Time Variability
bilit • If utilization is high, the process SCV 2
2
cd = ρ 2 ce2 + (1 − ρ 2 )ca dominates (and is transmitted to the next
dominates (and is transmitted to the next
station)
• If utilization is low, the arrival SCV
dominates
2/4/2012 Industrial Data and Systems Analysis 2 ca + ce2 ρ 2 Lq = 2 1 − ρ • These two formulas are sufficient to estimate queue lengths at each station
estimate queue lengths at each station 27 2/4/2012 Industrial Data and Systems Analysis 28 Solution
Solution
Station
Time per piece (seconds)
Coefficient of variation
of variation Station
Process Rates (pieces per second)
Utilization
SCV
Departure SCV Performance View
View
0 (Interarrivals)
40
0.25 1
25
1.25 38
0.75 37
0.75 0 (Interarrivals)
0.025 1
0.0400
0.6250
1.5625
0.6484 2
0.0263
0.9500
0.5625
0.5709 3
0.0270
0.9250
0.5625
0.5637 Variability Term
Utilization Term
Capacity Term
Expected Waiting Time in Queue
Decoupling Stock c 2 + c 2 ρ Wq = a e 2 1 − ρ me 2/4/2012 0.0625
0.0625 2 3 Total Expected Process Time
Expected Process Time
Total Expected Time in Queue
Queue Factor • Most of the time a unit spends in factory is 0.8125
0.6055
0.5667
1.6667 19.0000 12.3333
25.0000 38.0000 37.0000
33.8542 437.1484 258.5993
0.8464 10.9287
6.4650 Lq = λWq
2
2
cd = ρ 2 ce2 + (1 − ρ 2 )ca Industrial Data and Systems Analysis 100.0000
729.6019
8.2960 29 Cycle Stock
Stock queue time
• One example: Total processing time for a
ti
small metal part: 55 seconds; Total order
flow time: 55 days
flow time: 55 days
• Lots of opportunities for improvement!
2/4/2012 30 Industrial Data and Systems Analysis Cycle Stock
Stock
Cumulative
Units Order
Size
Throughput
rate Average
Cycle
Stock • Cycle stock: inventory in the system Average
Queue
Time because of batching operations
because of batching operations Time
2/4/2012 Industrial Data and Systems Analysis 31 2/4/2012 Industrial Data and Systems Analysis 32 Cycle Stock Calculation
Cycle Stock Calculation Reasons for Batching
for Batching • If Q is the order (batch) size, then the average inventory on hand will be Q/2
• By Little’s Law, the average queue time
due to batching = Q/(2*Throughput rate)
• To reduce queue time (and cycle stock),
reduce Q
• But there is always a reason for batching 2/4/2012 Industrial Data and Systems Analysis 33 A Simple Model for Economic
Order Quantit
tity 34 • The average number of orders placed per year is dollars per order placed.
• Let h denote the cost of holding one unit in
denote the cost of holding one unit in
inventory for one year.
• Let λ denote the demand rate, measured in units
per year. If you want to measure demand on a
shorter term basis, such as units per month, then
you must change the units of h , the holding
cost, to be on the same time basis.
th
ti
• Let Q denote the order size, in units. This is the
decision variable we are trying to choose.
Industrial Data and Systems Analysis Industrial Data and Systems Analysis The Basic Tradeoff
Basic Tradeoff • Let K denote the fixed order cost, measured in 2/4/2012 2/4/2012 •
• •
• 35 λ / Q.
The average annual cost of placing orders is
K*
K* λ / Q.
Assuming that demand occurs constantly and
continuously throughout the year at rate
continuously throughout the year at rate λ , the
the
average inventory is given by Q/2.
The average annual inventory cost is h*Q/2.
Let f(Q) denote the total average annual cost of
placing orders and holding inventory. Then,
f(Q) = K* λ / Q + h*Q/2. 2/4/2012 Industrial Data and Systems Analysis 36 The EOQ
The EOQ Safety Stock
Stock • Solution: the optimal economic order
the optimal economic order quantity (EOQ) is given by
Q* = 2λ K
h • Insight: to reduce cycle stock, reduce K,
to reduce cycle stock, reduce K, • Safety stock: the expected net inventory
th the fixed cost of the operation
• What are the benefits to reducing cycle
What are the benefits to reducing cycle
stock?
2/4/2012 (on hand less backorders) at the time of
delivery of replenishment order
delivery of a replenishment order 37 Industrial Data and Systems Analysis Order Points and Lead Time
Points and Lead Time 2/4/2012 38 Industrial Data and Systems Analysis Order Point, Order Quantity
Point, Order Quantity
Inventory
Level Order More! Q
R L Q L Q
L B Time • ‘Inventory Position’ = on hand Stockout! 2/4/2012 Industrial Data and Systems Analysis backorders + on order
on order
• When inventory position falls below R,
order
order Q additional units
additional units Delivery 39 2/4/2012 Industrial Data and Systems Analysis 40 Safety Stock
Safety Stock Analysis
• Mark cycle from order placement times
cycle from order placement times
• Total demand that occurs in one cycle = Q
• Let B denote the number of customer
denote the number of customer Inventory
Level Q
R L Q L Q backorders that occur during one cycle
• Assume there is at most one order
there is at most one order
outstanding (ok if Q is large relative to Lμ)
• The number of customer backorders is
The number of customer backorders is
unaffected by Q L B Time • Safety stock: expected net inventory (on hand – backorders) just before a delivery
• Assume daily demand is normally
distributed (mean variance
distributed (mean μ, variance σ2)
2/4/2012 Industrial Data and Systems Analysis 41 Order Point Problem
Point Problem 42 • Order lead time is L days
lead time is
days
• Let DL be demand over L days
• Backorders exist if lead time demand exceeds that can be satisfied without a stockout
• Let f denote the fill rate:
f = (QE[B]) / Q
• Suppose we want to achieve some minimum
fill rate
fill rate, fmin : choose R to ensure that
choose
to ensure that
(QE[B]) / Q ≥ fmin
• That is, choose R to satisfy
is choose
to satisfy
E[B] = Q(1 fmin)
Industrial Data and Systems Analysis Industrial Data and Systems Analysis Order Point Analysis
Point Analysis • ‘Fill rate’ is the expected fraction of demand
rate is the expected fraction of demand 2/4/2012 2/4/2012 the order point:
B = max(DLR,0)
• Assume daily demands are independent and
daily demands are independent and
identically distributed
• Recall: variance of sum of independent random variables is the sum of the variances
variables is the sum of the variances • Then, DL is normally distributed: N(Lμ,Lσ2) 43 2/4/2012 Industrial Data and Systems Analysis 44 ‘Normalize’ the Demand
the Demand ‘Normalize’ the Reorder Point
the Reorder Point • Let Z = (DL  Lμ)/L1/2σ
(D
• Then Z has a standard normal • Let k = (R  Lμ)/L1/2σ
)
• Then, E[B] = L1/2σE[max(Zk,0)]
• Choose k to satisfy
E[max(Zk,0)] = Q(1 fmin)/(L1/2σ)
• After some calculus:
E[max(Zk,0)] = φ(k)  k(1Φ(k)) distribution: Z~N(0,1) where
• φ(k) is the standard normal density function and
• Φ(k) is the standard normal cumulative distribution
function
function 4 3 3.5 2 2.5 1 1.5 0.5 0 1 0.5 2 1.5 3 2.5 4 2/4/2012 3.5 Standard normal
probability density
function Industrial Data and Systems Analysis 45 Order Point Solution
Technique 46 • μ = 15 units per day; σ2 = 16
• Q = 45 units; L = 2 days
• What value of R will ensure that the fill φ(k)  k(1Φ(k)) = Q(1 fmin)/(L1/2σ)
• In Microsoft Excel:
Microsoft Excel: rate is at least 98%? (ie. fmin=0.98)
• Calculate r.h.s. of equation:
Q(1 fmin)/(L1/2σ)= 45 * (0.02) / (2 1/2 *4)
= 0.159
• Search for solution to:
φ(k)  k(1Φ(k)) = 0.159 φ(k)=EXP((k^2)/2)/ SQRT(2*PI())
Φ(k)=NORMSDIST(k)
• Call the solution, k*
• Value of R that ensures fill rate ≥ fmin:
of
that ensures fill rate
R = Lμ + k*L1/2σ
Industrial Data and Systems Analysis Industrial Data and Systems Analysis Example • Search for k that solves:
for that solves: 2/4/2012 2/4/2012 47 2/4/2012 Industrial Data and Systems Analysis 48 Solution
Solution Formulas to Remember
to Remember 0.4
0.35 • Find k to satisfy 0.3
0.25
l.h.s. 0.2 φ(k)  k(1Φ(k)) = Q(1 fmin)/(L1/2σ)
k(1
Q(1
• Reorder point is given by
R = Lμ + k*L1/2σ
• Safety stock is given by
SS = k*L1/2σ r.h.s. 0.15
0.1
0.05
2 1.8 1.6 1.4 1.2 1 0.8 0.6 0.4 0 0.2 0
k • k* = 0.63. Now, solve for R: R = Lμ + k*L1/2σ =2*15 + 0.63 * (2 1/2 *4)
k
0.63 (2
4)
= 30 + 3.56 = 33.56
• Order point should be at least 33.56 units (34 if
whole units must be ordered)
whole units must be ordered)
2/4/2012 Industrial Data and Systems Analysis Std. dev. of lead time demand
49 Observations Concerning FillRate Driven Safety Stock
St 2/4/2012 Industrial Data and Systems Analysis 50 Four Roles of Inventory
Roles of Inventory • Safety stock is proportional to the
stock is proportional to the standard deviation of lead time demand
• Safety stock is proportional to the square
stock is proportional to the square
root of the lead time
• Only the tail of the probability distribution
the tail of the probability distribution
is used
• Safety stock depends inversely on the
stock depends inversely on the
order quantity, Q!
2/4/2012 Industrial Data and Systems Analysis Pipeline Stock
Stock
Cycle Stock
Stock Decoupling Stock
51 2/4/2012 Safety Stock
Stock Industrial Data and Systems Analysis 52 ...
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 Spring '09
 JACKSON
 Systems Analysis, Normal Distribution, Variance, industrial data, Industrial Data and Systems Analysis

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