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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

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 manager’s 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.

Demand During Lead Time is Variable N(μ,σ)

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%.

Forecasts should be accompanied by a measure of forecast error

Forecast and a Measure of Forecast Error

A large demand

during lead time

Average demand

during lead time

ROP

Safety stock

Time

LT

Safety Stock

Safety stock reduces risk of

stockout during lead time

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%.

stockout

Average

demand

Safety

stock

z-scale

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)

Risk of a

stockout

Probability of

no stockout

ROP

Quantity

Average

demand

Safety

stock

0

z

z-scale

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 anyprobability of not exceeding z. What is the value of z
- Given anyvalue forz. What is the probability of not exceeding z

μand σ 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.

95% Probability

The table will give you z

Probability

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

Risk of a

stockout

Probability of

no stockout

ROP

Quantity

Average

demand

Safety

stock

0

z

z-scale

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

σ = Standard Deviation of x

x = μ +z σ

Risk of a

stakeout

Probability of

no stockout

ROP

Quantity

Average

demand

Safety

stock

0

z

z-scale

Relationship between z and NormalVariable ROP

LTD = Lead Time Demand

ROP = Average LTD +z (Standard Deviation of LTD)

ROP = LTD+zσLTD ROP = LTD + Isafety

Demand During Lead Time is Variable N(μ,σ)

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 = zσLTD

Isafety = 1.64 (5) = 8.2

ROP = LTD + Isafety

ROP = 50 + 1.64(5) = 58.2

Service Level for a given ROP

- SL= Prob (LTD ≤ ROP)
- LTD is normally distributed LTD = N(LTD, sLTD).
- ROP = LTD + zσLTD ROP = LTD + Isafety I safety = z sLTD

- At GE Lighting’s 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

μ and σ 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

μ and σof demand per period and fixed LT

R: demand rate perperiod(a random variable)

R: Average demand rate perperiod

σR:Standard deviation of the demand rate perperiod

L: Lead time (a constant number of periods)

LTD: demand during the lead time (a random variable)

LTD: Average demand during the lead time

σLTD:Standard deviation of the demand during lead time

μ and σ ofdemand per period and fixed LT

A random variable R:N(R, σR) 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 = R1+R2+R3+…….+RL

Then there is a relationship between mean and standard deviation of the two random variables

μ and σ 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, σR = 3, L = 2

Isafety = zσLTD = 1.28(4.24) = 5.43

ROP= 100 + 5.43

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.

R

L

μ and σ oflead time and fixed Demand per period

L: lead time (a random variable)

L: Average lead time

σL: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

σLTD:Standard deviation of the demand during lead time

R

L

μ and σ of demand per period and fixed LT

A random variable L:N(L, σL) is multipliedby a constant R and generates the random variable LTD.

If we have a random variable LTD which is equal to a constant Rtimes a random variables L

LTD = RL

Then there is a relationship between mean and standard deviation of the two random variables

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, σL = 0.5 week, R = 50 per week

Isafety = zσLTD = 1.28(25) = 32

ROP= 100 + 32

Both Demand and Lead Time are Variable

R: demand rate per period

R: Average demand rate

σR:Standard deviation of demand

L: lead time

L: Average lead time

σL:Standard deviation of the lead time

LTD: demand during the lead time (a random variable)

LTD: Average demand during the lead time

σLTD:Standard deviation of the demand during lead time

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.

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?

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?

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?

Compute the Average Demand

Average Demand =

+100×0.02 +110×0.05+120×0.08 +130×0.09+140×0.11 +150×0.16

+160×0.20 +170×0.15 +180×0.08 +190×0.05+200×0.01

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?

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 units5.5

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

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 @ 700 = 94640

Total number Salvaged 4.8 @ -100 = -480

Expected Profit = 94640 – 480 = 94,160

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(R>Q) = 1- P( R ≤ Q).

The expected marginal benefit = MB× [1-Prob.( R ≤ Q)]

In acontinuous model: SL* = Prob(R ≤ Q*) =

Analytical Solution for the Optimal Service LevelAs 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

MC×Prob(R ≤ Q*) ≥ MB× [1 – Prob(R ≤ Q*)]

If we assume demand is normally distributed, What quantity corresponds to this service level ?

Probability Less than Upper Bound is 0.87493

0.4

0.35

0.3

0.25

Density

0.2

z = 1.15

0.15

0.1

0.05

0

-4

-3

-2

-1

0

1

2

3

4

Critical Value (z)

Analytical Solution for the Optimal Service LevelPhysical 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, σLTD )LTD2: N(LTD, σLTD )

- Both warehouses have identical service levels
- To provide desired SL, each location must carry Isafety = zσLTD
- z is determined by the desired service level
- The total safety inventory in the decentralized system is

Independent Lead time demands at two locations

- Decrease in safety inventory by a factor of

LTDC = LTD1 + LTD2 LTDC = LTD + LTD = 2 LTD

Centralization reduced the safety inventory by a factor of 1/√2

- 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 = 1.65×5000= 8250

independent Lead time demands at N locations

Independent demand in N locations: Total safety inventory to provide a specific SL increases not by N but by √N

- Centralization of N locations:

If centralization of stocks reduces inventory, why doesn’t 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.

Dependent Demand

- Does centralization offer similar benefits when demands in multiple locations are correlated?
- LTD1 and LTD2are statistically identically distributed but correlated with a correlation coefficient of ρ .

No Correlation: ρ close to 0

+ Correlation, + Perfect Correlation

Positive Correlation: ρ close to 1

Perfect Positive Correlation: ρ = +1

Negative Correlation: ρ close to -1

Perfect Negative Correlation: ρ = -1

The safety inventory in the two-location decentralized system is larger than in the centralized system by a factor of

Correlation- If demand is positively fully correlated, ρ = 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

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

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 A
- Exceeds Available stock

- Location B
- Less than 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.

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.

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).

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.

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.

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

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