Managing Flow Variability: Safety Inventory

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Managing Flow Variability Safety Inventory
Forecasts Depend on (a) Historical Data and (b)
Market Intelligence. Demand Forecasts and
Forecast Errors Safety Inventory and Service
Level Optimal Service Level The Newsvendor
Problem Lead Time Demand Variability Pooling
Efficiency through Aggregation Shortening the
Forecast Horizon Levers for Reducing Safety
Inventory

2
Four Characteristics of Forecasts

• Forecasts are usually (always) inaccurate
  (wrong). Because of random noise.
• Forecasts should be accompanied by a measure of
  forecast error. A measure of forecast error
  (standard deviation) quantifies the managers
  degree of confidence in the forecast.
• Aggregate forecasts are more accurate than
  individual forecasts. Aggregate forecasts reduce
  the amount of variability relative to the
  aggregate mean demand. StdDev of sum of two
  variables is less than sum of StdDev of the two
  variables.
• Long-range forecasts are less accurate than
  short-range forecasts. Forecasts further into the
  future tends to be less accurate than those of
  more imminent events. As time passes, we get
  better information, and make better prediction.

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Demand During Lead Time is Variable N(µ,s)
Demand of sand during lead time has an average of
50 tons. Standard deviation of demand during lead
time is 5 tons Assuming that the management is
willing to accept a risk no more that 5.
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Forecasts should be accompanied by a measure of
forecast error
Forecast and a Measure of Forecast Error
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Demand During Lead Time
Inventory
Demand during LT
Time
Lead Time
6
ROP when Demand During Lead Time is Fixed
LT
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Demand During Lead Time is Variable
LT
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Demand During Lead Time is Variable
Inventory
Time
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Safety Stock
Safety stock reduces risk of stockout during lead
time
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Safety Stock
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Re-Order Point ROP
Demand during lead time has Normal distribution.
If we order when the inventory on hand is equal
to the average demand during the lead time then
there is 50 chance that the demand during lead
time is less than our inventory.
However, there is also 50 chance that the demand
during lead time is greater than our inventory,
and we will be out of stock for a while. We
usually do not like 50 probability of stock out
We can accept some risk of being out of stock,
but we usually like a risk of less than 50.
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Safety Stock and ROP
Service level
Probability of no stockout
ROP
Quantity
0
z
Each Normal variable x is associated with a
standard Normal Variable z
x is Normal (Average x , Standard Deviation x) ?
z is Normal (0,1)
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z Values
SL z value 0.9 1.28 0.95 1.65 0.99 2.33

• There is a table for z which tells us
• Given any probability of not exceeding z. What
  is the value of z
• Given any value for z. What is the probability
  of not exceeding z

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µ and s of Demand During Lead Time
Demand of sand during lead time has an average of
50 tons. Standard deviation of demand during lead
time is 5 tons. Assuming that the management is
willing to accept a risk no more that 5. Find
the re-order point. What is the service
level. Service level 1-risk of stockout 1-.05
.95 Find the z value such that the probability
of a standard normal variable being less than or
equal to z is .95 Go to normal table, look
inside the table. Find a probability close to
.95. Read its z from the corresponding row and
column.
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z Value using Table
Page 319 Normal table
0.05
z
Second digit after decimal
Z 1.65
Up to the first digit after decimal
1.6
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The standard Normal Distribution F(z)
F(z) Prob( N(0,1) lt z)
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Relationship between z and Normal Variable x
z (x-Average x)/(Standard Deviation of x) x
Average x z (Standard Deviation of x) µ
Average x s Standard Deviation of x
? x µ z s
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Relationship between z and Normal Variable ROP
LTD Lead Time Demand ROP Average LTD z
(Standard Deviation of LTD) ROP LTDzsLTD ?
ROP LTD Isafety
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Demand During Lead Time is Variable N(µ,s)
Demand of sand during lead time has an average of
50 tons. Standard deviation of demand during lead
time is 5 tons Assuming that the management is
willing to accept a risk no more that 5.
z 1.65
Compute safety stock Isafety zsLTD Isafety
1.64 (5) 8.2 ROP LTD Isafety ROP 50
1.64(5) 58.2
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Service Level for a given ROP

• SL Prob (LTD ROP)
• LTD is normally distributed ? LTD N(LTD, sLTD
  ).
• ROP LTD zsLTD ? ROP LTD Isafety ? I
  safety z sLTD

At GE Lightings Paris warehouse, LTD 20,000,
sLTD 5,000 The warehouse re-orders whenever ROP
24,000 I safety ROP LTD 24,000 20,000
4,000 I safety z sLTD ? z I safety / sLTD
4,000 / 5,000 0.8 SL Prob (Z 0.8) from
Appendix II ? SL 0.7881
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Excel Given z, Compute Probability
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Excel Given Probability, Compute z
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µ and s of demand per period and fixed LT
Demand of sand has an average of 50 tons per
week. Standard deviation of the weekly demand is
3 tons. Lead time is 2 weeks. Assuming that the
management is willing to accept a risk no more
that 10. Compute the Reorder Point
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µ and s of demand per period and fixed LT
R demand rate per period (a random variable) R
Average demand rate per period sR Standard
deviation of the demand rate per period
L Lead time (a constant number of periods)
LTD demand during the lead time (a random
variable) LTD Average demand during the lead
time sLTD Standard deviation of the demand
during lead time
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µ and s of demand per period and fixed LT
A random variable R N(R, sR) repeats itself L
times during the lead time. The summation of
these L random variables R, is a random variable
LTD
If we have a random variable LTD which is equal
to summation of L random variables R LTD
R1R2R3.RL Then there is a relationship
between mean and standard deviation of the two
random variables
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µ and s of demand per period and fixed LT
Demand of sand has an average of 50 tons per
week. Standard deviation of the weekly demand is
3 tons. Lead time is 2 weeks. Assuming that the
management is willing to accept a risk no more
that 10.
z 1.28, R 50, sR 3, L 2
Isafety zsLTD 1.28(4.24) 5.43 ROP 100
5.43
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Lead Time Variable, Demand fixed
Demand of sand is fixed and is 50 tons per
week. The average lead time is 2 weeks. Standard
deviation of lead time is 0.5 week. Assuming that
the management is willing to accept a risk no
more that 10. Compute ROP and Isafety.
28
µ and s of lead time and fixed Demand per period
L lead time (a random variable) L Average lead
time sL Standard deviation of the lead time
R Demand per period (a constant value)
LTD demand during the lead time (a random
variable) LTD Average demand during the lead
time sLTD Standard deviation of the demand
during lead time
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µ and s of demand per period and fixed LT
A random variable L N(L, sL) is multiplied by a
constant R and generates the random variable LTD.
If we have a random variable LTD which is
equal to a constant R times a random variables
L LTD RL Then there is a relationship between
mean and standard deviation of the two random
variables
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Variable R fixed L.Variable L fixed R
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Lead Time Variable, Demand fixed
Demand of sand is fixed and is 50 tons per
week. The average lead time is 2 weeks. Standard
deviation of lead time is 0.5 week. Assuming that
the management is willing to accept a risk no
more that 10. Compute ROP and Isafety.
z 1.28, L 2 weeks, sL 0.5 week, R 50 per
week
Isafety zsLTD 1.28(25) 32 ROP 100 32
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Both Demand and Lead Time are Variable
R demand rate per period R Average demand
rate sR Standard deviation of demand L lead
time L Average lead time sL Standard deviation
of the lead time LTD demand during the lead
time (a random variable) LTD Average demand
during the lead time sLTD Standard deviation of
the demand during lead time
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Optimal Service Level The Newsvendor Problem
How do we choose what level of service a firm
should offer?
Cost of Holding Extra Inventory
Improved Service
Optimal Service Level under uncertainty
The Newsvendor Problem
The decision maker balances the expected costs of
ordering too much with the expected costs of
ordering too little to determine the optimal
order quantity.
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Optimal Service Level The Newsvendor Problem
Cost 1800, Sales Price 2500, Salvage Price
1700 Underage Cost 2500-1800 700, Overage
Cost 1800-1700 100
What is probability of demand to be equal to
130? What is probability of demand to be less
than or equal to 140? What is probability of
demand to be greater than 140? What is
probability of demand to be equal to 133?
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Optimal Service Level The Newsvendor Problem
What is probability of demand to be equal to
116? What is probability of demand to be less
than or equal to 160? What is probability of
demand to be greater than 116? What is
probability of demand to be equal to 13.3?
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Optimal Service Level The Newsvendor Problem
What is probability of demand to be equal to
130? What is probability of demand to be less
than or equal to 140? What is probability of
demand to be greater than 140? What is
probability of demand to be equal to 133?
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Compute the Average Demand
Average Demand 1000.02 1100.051200.08
1300.091400.11 1500.16 1600.20 1700.15
1800.08 1900.052000.01 Average Demand
151.6
How many units should I have to sell 151.6 units
(on average)? How many units do I sell (on
average) if I have 100 units?
38
Suppose I have ordered 140 Unities. On average,
how many of them are sold? In other words, what
is the expected value of the number of sold
units?
When I can sell all 140 units? I can sell all
140 units if ? R 140 Prob(R 140) 0.76 The the
expected number of units sold for this part-
is (0.76)(140) 106.4 Also, there is 0.02
probability that I sell 100 units? 2 units Also,
there is 0.05 probability that I sell 110
units?5.5 Also, there is 0.08 probability that I
sell 120 units? 9.6 Also, there is 0.09
probability that I sell 130 units? 11.7 106.4 2
5.5 9.6 11.7 135.2
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Suppose I have ordered 140 Unities. On average,
how many of them are salvaged? In other words,
what is the expected value of the number of sold
units?
0.02 probability that I sell 100 units. In that
case 40 units are salvaged ? 0.02(40) .8 0.05
probability to sell 110 ? 30 salvage? 0.05(30)
1.5 0.08 probability to sell 120 ? 30 salvage?
0.08(20) 1.6 0.09 probability to sell 130 ? 30
salvage? 0.09(10) 0.9 0.8 1.5 1.6 0.9
4.8 Total number Solved 135.2 _at_ 700
94640 Total number Salvaged 4.8 _at_ -100
-480 Expected Profit 94640 480 94,160
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Cumulative Probabilities
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Number of Units Sold, Salvages
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Total Revenue for Different Ordering Policies
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Analytical Solution for the Optimal Service Level
Net Marginal Benefit Net Marginal Cost
MB p c MC c - v
MB 2,500 - 1,800 700 MC 1,800 - 1,700
100
Suppose I have ordered Q units. What is the
expected cost of ordering one more units? What is
the expected benefit of ordering one more
units? If I have ordered one unit more than Q
units, the probability of not selling that extra
unit is if the demand is less than or equal to Q.
Since we have P( R Q). The expected marginal
cost MC P( R Q) If I have ordered one unit
more than Q units, the probability of selling
that extra unit is if the demand is greater than
Q. We know that P(RgtQ) 1- P( R Q). The
expected marginal benefit MB 1-Prob.( R Q)
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Analytical Solution for the Optimal Service Level
As long as expected marginal cost is less than
expected marginal profit we buy the next unit. We
stop as soon as Expected marginal cost
Expected marginal profit
MCProb(R Q) MB 1 Prob(R Q)
If we assume demand is normally distributed, What
quantity corresponds to this service level ?
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Analytical Solution for the Optimal Service Level
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Aggregate Forecast is More Accurate than
Individual Forecasts
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Physical Centralization

• Physical Centralization the firm consolidates
  all its warehouses in one location from which is
  can serve all customers.
• Example Two warehouses. Demand in the two ware
  houses are independent.
• Both warehouses have the same distribution for
  their lead time demand.
• LTD1 N(LTD, sLTD )
  LTD2 N(LTD, sLTD )

Both warehouses have identical service levels To
provide desired SL, each location must carry
Isafety zsLTD z is determined by the desired
service level The total safety inventory in the
decentralized system is
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Independent Lead time demands at two locations
LTDC LTD1 LTD2 ? LTDC LTD LTD 2 LTD
?
Centralization reduced the safety inventory by a
factor of 1/v2

GE lighting operating 7 warehouses. A warehouse
with average lead time demand of 20,000 units
with a standard deviation of 5,000 units and a
95 service level needs to carry a safety
inventory of Isafety 1.655000 8250
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independent Lead time demands at N locations
Centralization of N locations

• If centralization of stocks reduces inventory,
  why doesnt everybody do it?
• Longer response time
• Higher shipping cost
• Less understanding of customer needs
• Less understanding of cultural, linguistics, and
  regulatory barriers
• These disadvantages my reduce the demand.

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Dependent Demand

• Does centralization offer similar benefits when
  demands in multiple locations are correlated?
• LTD1 and LTD2 are statistically identically
  distributed but correlated with a correlation
  coefficient of ? .

No Correlation ? close to 0
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Correlation, Perfect Correlation
Positive Correlation ? close to 1
Perfect Positive Correlation ? 1
Negative Correlation ? close to -1
Perfect Negative Correlation ? -1
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Correlation
If demand is positively fully correlated, ? 1,
centralization offers no benefits in the
reduction of safety inventory Benefits of
centralization increases as the demand on the two
locations become negatively correlated. The best
case is ? -1, where we do not need safety
inventory at all
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Principle of Aggregation and Pooling Inventory

• Inventory benefits ? due to principle of
  aggregation.
• Statistics Standard deviation of sum of random
  variables is less than the sum of the individual
  standard deviations.
• Physical consolidation is not essential, as long
  as available inventory is shared among various
  locations ? Pooling Inventory
• Virtual Centralization
• Specialization
• Component Commonality
• Delayed Differentiation
• Product Substitution

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Virtual Centralization
Virtual Centralization inventory pooling in a
network of locations is facilitated using
information regarding availability of goods and
subsequent transshipment of goods between
locations to satisfy demand.
Location B Less than Available stock

• Location A
• Exceeds Available stock

1. Information about product demand and
availability must be available at both
locations 2. Shipping the product from one
location to a customer at another location must
be fast and cost effective
Pooling is achieved by keeping the inventories at
decentralized locations.
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Specialization, Substitution

• Demand for both products exist in both locations.
  But a large portion of demand for P1 is in
  location A, while a large portion of demand for
  P2 is in location B.

Location B Product P2
Location A Product P1
Both locations keep average inventory. Safety
inventory is kept only in the specialized
warehouse
One other possibility to deal with variability is
product substitution.
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Component Commonality

• Up to now we have discussed aggregating demand
  across various geographic locations, either
  physical or virtual
• Aggregating demand across various products has
  the same benefits.
• Computer manufacturers offer a wide range of
  models, but few components, CPU, RMA, HD, CD/DVD
  drive, are used across product lines.
• Replace Make-to-stock with make Make-to-Order
• Commonality MTO
• Commonality Safety inventory of the common
  components much less than safety inventory of
  unique components stored separately.
• MTO Inventory cost is computed in terms of WIP
  cost not in terms of finished good cost (which
  is higher).

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Postponement (Delayed Differentiation)

• Forecasting Characteristic Forecasts further
  into the future tends to be less accurate than
  those of more imminent events.
• Since shorter-range forecasts are more accurate,
  operational decisions will be more effective if
  supply is postponed closer to the point of actual
  demand.
• Two Alternative processes (each activity takes
  one week)
• Alternative A (1) Coloring the fabric, (2)
  assembling T-shirts
• Alternative B (1) Assembling T-shirts, (2)
  coloring the fabric
• No changes in flow time. Alternative B postponed
  the color difference until one week closer to the
  time of sale. Takes advantage of the forecasting
  characteristic short-Range forecast more
  accurate.

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Postponement (Delayed Differentiation)

• Two advantages Taking advantage of two demand
  forecasting characteristics
• Commonality Advantage At week 0 Instead of
  forecast for each individual item, we forecast
  for aggregates item uncolored T-shirt. Forecast
  for aggregate demand is more accurate than
  forecast for individual item. It is easier to
  more accurately forecast total demand for
  different colored T-shirts for next week than the
  week after the next.
• Postponement Advantage Instead of forecasting
  for each individual items two weeks ahead, we do
  it at week 1. Shorter rang forecasts are more
  accurate. It is easier to more accurately
  forecast demand for different colored T-shirts
  for next week than the week after the next.

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Lessons Learned

• Levels for Reducing Safety Capacity
• Reduce demand variability through improved
  forecasting
• Reduce replenishment lead time
• Reduce variability in replenishment lead time
• Pool safety inventory for multiple locations or
  products
• Exploit product substitution
• Use common components
• Postpone product-differentiation processing until
  closer to the point of actual demand