1. Chapter 11
Inventory Management

2. Types of Inventories Raw materials & purchased parts Work in progress
Incoming students
Work in progress
Current students
Finished-goods inventories
(manufacturing firms) or merchandise (retail stores)
Graduating students
Replacement parts, tools, & supplies
Goods-in-transit to warehouses or customers
Students on leave

3. Functions of Inventory
To meet anticipated demand
To smooth production requirements
To decouple components of the production-distribution
To protect against stock-outs
To take advantage of order cycles
To help hedge against price increases or to take advantage of quantity discounts
To permit operations

4. Inventory performance measures and levers
Inventory level
Low or high
Customer service levels
Can you deliver what customer wants?
Right goods, right place, right time, right quantity
Inventory turnover
Cost of goods sold per year / average inventory investment
Inventory costs, more will come
Costs of ordering & carrying inventories
Decisions: Order size and time

5. Inventory Counting Systems
A physical count of items in inventory
Periodic/Cycle Counting System: Physical count of items made at periodic intervals
How much accuracy is needed?
When should cycle counting be performed?
Who should do it?
Continuous Counting System System that keeps track of
removals from inventory
continuously, thus monitoring
current levels of each item

6. Inventory Counting Systems (Cont’d)
Two-Bin System - Two containers of inventory; reorder when the first is empty
Universal Bar Code - Bar code printed on a label that has information about the item to which it is attached
RFID: Radio frequency identification

7. Key Inventory Terms
Lead time: time interval between ordering and receiving the order, denoted by LT
Holding (carrying) costs: cost to carry an item in inventory for a length of time, usually a year, denoted by H
Ordering costs: costs of ordering and receiving inventory, denoted by S
Shortage costs: costs when demand exceeds supply

8. Effective Inventory Management
A system to keep track of inventory
A reliable forecast of demand
Knowledge of lead times
Reasonable estimates of
Holding costs
Ordering costs
Shortage costs
A classification system

9. ABC Classification System
Classifying inventory according to some measure of importance and allocating control efforts accordingly.
Importance measure= price*annual sales
A - very important
B - mod. important
C - least important
Annual
$ volume
of items A B C
High
Low
Few
Many
Number of Items

10. Inventory Models Fixed Order Size - Variable Order Interval Models:
1. Economic Order Quantity, EOQ
2. Economic Production Quantity, EPQ
3. EOQ with quantity discounts
All units quantity discount
3.1. Constant holding cost
3.2. Proportional holding cost
4. Reorder point, ROP
Lead time service level
Fill rate
Fixed Order Interval - Variable Order Size Model
5. Fixed Order Interval model, FOI
Single Order Model
6. Newsboy model

11. 1. EOQ Model Assumptions: Only one product is involved
Annual demand requirements known
Demand is even throughout the year
Lead time does not vary
Each order is received in a single delivery
Infinite production capacity
There are no quantity discounts

12. Profile of Inventory Level Over Time
The Inventory Cycle
Profile of Inventory Level Over Time Q
Usage
rate
Quantity
on hand
Reorder
point
Time
Receive
order
Place
order
Receive
order
Place
order
Receive
order
Lead time

13. Average inventory held
Length of an inventory cycle
From one order to the next = Q/D
Inventory held over entire inventory cycle
Area under the inventory level = ½ Q (Q/D)
Average inventory held
= Inventory held over a cycle / cycle length
= Q/2

14. Total Cost
Annual
carrying
cost
ordering
Total cost = + Q 2 H D S
TC =

15. Figure 11-4

16. Cost Minimization Goal
Order Quantity (Q)
The Total-Cost Curve is U-Shaped
Ordering Costs QO
Annual Cost
(optimal order quantity)

17. Deriving the EOQ
Using calculus, we take the derivative of the total cost function and set the derivative (slope) equal to zero and solve for Q.
The total cost curve reaches its minimum where the carrying and ordering costs are equal.

18. EOQ example Demand, D = 12,000 computers per year.
Holding cost, H = 100 per item per year. Fixed cost, S = $4,000/order.
Find EOQ, Cycle Inventory, Optimal Reorder Interval and Optimal Ordering Frequency.
EOQ = , say 980 computers
Cycle inventory = EOQ/2 = 490 units
Optimal Reorder interval, T = year = 0.98 month
Optimal ordering frequency, n=12.24 orders per year.
Notes:

19. Optimal Quantity is robust

20. Total Costs with Purchasing Cost
Annual
carrying
cost
Purchasing
TC = + Q 2 H D S
ordering PD
Note that P is the price.

21. Total Costs with PD Cost Adding Purchasing cost doesn’t change EOQ
TC with PD
TC without PD PD
Quantity
Adding Purchasing cost doesn’t change EOQ

22. 2. Economic Production Quantity
Production done in batches or lots
Capacity to produce a part exceeds the part’s usage or demand rate
Assumptions of EPQ are similar to EOQ except orders are received incrementally during production
This corresponds to producing for an order with finite production capacity

23. Economic Production Quantity Assumptions
Only one item is involved
Annual demand is known
Usage rate is constant
Usage occurs continually
Production rate p is constant
Lead time does not vary
No quantity discounts

24. Economic Production Quantity
Usage
Production
& Usage
Production
& Usage
Usage
Inventory Level

25. Average inventory held
(Q/p)(p-D)
p-D D
Q/p
Time
Q/D
Average inventory held=(1/2)(Q/p)(p-D)
Total cost=(1/2)(Q/p)(p-D)H+(D/Q)S

26. EPQ example
Demand, D = 12,000 computers per year. p=20,000 per year. Holding cost, H = 100 per item per year. Fixed cost, S = $4,000/order.
Find EPQ.
EPQ = EOQ*sqrt(p/(p-D))
=979.79*sqrt(20/8)=1549 computers
Notes:

27. 3. All unit quantity discount
Cost/Unit
Order Quantity
5,000
10,000
Two versions
Constant H
Proportional H $3
$2.96
$2.92

28. 3.1.Total Cost with Constant Carrying Costs
TCb
Total Cost
Decreasing
Price
TCc
Annual demand*discount
EOQ
Quantity

29. 3.1.Total Cost with Constant Carrying Costs
TCb
Total Cost
Decreasing
Price
TCc
Annual demand*discount
EOQ
Quantity

30. Price a > Price b > Price c
Example Scenario 1
Price a > Price b > Price c a b c
Total Cost
TCa
TCb
TCc
Q*=EOQ
Quantity

31. Price a > Price b > Price c
Example Scenario 2
Price a > Price b > Price c a b c
Total Cost
TCa
TCb
TCc
EOQ Q*
Quantity

32. Price a > Price b > Price c
Example Scenario 3
Price a > Price b > Price c a b c
Total Cost
TCa
TCb
TCc
EOQ Q*
Quantity

33. Price a > Price b > Price c
Example Scenario 4
Price a > Price b > Price c a b c
Total Cost
TCc
TCa
TCb
Q*=EOQ
Quantity

34. 3.1. Finding Q with all units discount with constant holding cost
Note all the price ranges have the same EOQ.
Stop if EOQ=Q1 is in the lowest cost range (highest quantity range), otherwise continue towards quantity break points which give lower costs
Quantity
Total Cost 1 2

35. All-units quantity discounts Constant holding cost
A popular shoe store sells 8000 pairs per year. The fixed cost of ordering shoes from the distribution center is $15 and holding costs are taken as $12.5 per shoe per year. The per unit purchase costs from the distribution center is given as
C3=60, if 0 < Q < 50
C2=55, if 50 <= Q < 150
C1=50, if 150 <= Q
where Q is the order size. Determine the optimal order quantity.

36. Solution There are three ranges for lot sizes in this problem:
(0, q2=50),
(q2=50, q1=150)
(q1=150,infinite).
Holding costs in all there ranges of shoe prices
are given as H=12.5,
EOQ is not feasible in the lowest price range because
138.6 < 150.
The order quantity q1=150 is a candidate with cost
TC(150)=8000(50)+8000(15)/150+(12.5)(150)/2
=401,900
Let us go to a higher cost level of (q2=50, q1=150).
EOQ=138.6 is in the appropriate range, so it is another candidate with cost
TC(138.6)=8000(55)+8000(15)/138.6+(12.5)(138.6)/2
=441,732
Since TC(150) < TC(132.1), Q=150 is the optimal solution.
Remark: In these computations, we do not need to compute TC(50), why? Because TC(50) >= TC(132.1).

37. Example: Q1 feasible stop
3.2. Summary of finding Q with all units discount with proportional holding cost
Note each price range has a different EOQ.
Stop if Q1=“EOQ of the lowest price” feasible
Otherwise continue towards higher costs until an EOQ becomes feasible.
In each price range, evaluate the lowest cost.
Lowest cost is either at an EOQ or price break quantity
Pick the minimum cost among all evaluated
Total Cost
Example: Q1 feasible stop 1
Quantity

38. 3.2. Finding Q with all units discount with proportional holding cost
Total Cost 2
Example: Q1 infeasible, Q2 feasible, Break point 1 is selected since TC1 < TC2 1’
Quantity

39. 3.2. Finding Q with all units discount with proportional holding cost
Stop if 1 is feasible, otherwise continue towards higher costs until a EOQ becomes feasible. Evaluate cost at all alternatives
Total Cost 1’ 2
Quantity

40. All-units quantity discounts Proportional holding cost
A popular shoe store sells 8000 pairs per year. The fixed cost of ordering shoes from the distribution center is $15 and holding costs are taken as 25% of the shoe costs. The per unit purchase costs from the distribution center is given as
C3=60, if 0 < Q < 50
C2=55, if 50 <= Q < 150
C1=50, if 150 <= Q
where Q is the order size. Determine the optimal order quantity.

41. Solution There are three ranges for lot sizes in this problem:
(0, q2=50),
(q2=50, q1=150)
(q1=150,infinite).
Holding costs in there ranges of shoe prices
are given as
H3=(0.25)60=15,
H2 =(0.25)55=13.75
H1 =(0.25)50=12.5.
EOQ1 is not feasible because < 150.
The order quantity q1=150 is a candidate with cost
TC(150)=8000(50)+8000(15)/150+(0.25)(50)(150)/2
=401,900
Let us go to a higher cost level of (q2=50, q1=150).
EOQ2=132.1 is in the appropriate range, so it is another candidate with cost
TC(132.1)=8000(55)+8000(15)/132.1+(0.25)(55)(132.1)/2
=441,800
Since TC(150) < TC(132.1), Q=150 is the optimal solution.
We do not need to compute TC(50) or EOQ3, why?

42. Types of inventories (stocks) by function
Deterministic demand case
Anticipation stock
For known future demand
Cycle stock
For convenience, some operations are performed occasionally and stock is used at other times
Why to buy eggs in boxes of 12?
Pipeline stock or Work in Process
Stock in transfer, transformation. Necessary for operations.
Students in the class
Stochastic demand case
Safety stock
Stock against demand variations

43. 4. When to Reorder with EOQ Ordering
Reorder Point - When the quantity on hand of an item drops to this amount, the item is reordered. We call it ROP.
Safety Stock - Stock that is held in excess of expected demand due to variable demand rate and/or lead time. We call it ss.
(lead time) Service Level - Probability that demand will not exceed supply during lead time. We call this cycle service level, CSL.

44. Optimal Safety Inventory Levels
An inventory cycle Q
ROP
time
Lead Times
Shortage

45. Safety Stock Quantity Maximum probable demand Expected demandLT
Time
Expected demand
during lead time
Maximum probable demand
ROP
Quantity
Safety stock

46. Inventory and Demand during Lead Time
ROP
Inventory=
ROP-DLT
ROP
Upside
down
Inventory
DLT: Demand
During LT LT
Demand
During LT

47. Shortage and Demand during Lead Time
Demand
During LT LT
Shortage
DLT: Demand During LT
ROP
Shortage=
DLT-ROP
Upside
down
ROP

48. Cycle Service Level
Cycle service level: percentage of cycles with shortage

49. DLT : Demand during lead time LT and demand may be uncertain.

50. Reorder Point ROP = E(DLT) + z σDLT ROP Risk of a stockout
Service level
Probability of
no stockout
Expected
demand
Safety
stock z
Quantity
z-scale
ROP = E(DLT) + z σDLT

51. Finding safety stock from cycle service level CSL
Note we use normal density for the demand during lead time
The excel function normsinv has default values of 0 and 1 for the mean and standard deviation. Defaults are used unless these values are specified.

52. Example: Safety inventory vs. Lead time variability
D = 2,500/day; D = 500
L = 7 days; CSL = 0.90
Normsinv(0.9)=1.3, either from table or Excel
If LT=0, DLT=sqrt(7)*500=1323
ss=1323*normsinv(0.9)=1719.8
ROP=(D)(L)+ss=
If LT=1, DLT=sqrt(7*500* *2500*1)=2828
ss=2828*normsinv(0.9)=3676
ROP=(D)(L)+ss=

53. Expected shortage per cycle
First let us study shortage during the lead time
Ex:

54. Expected shortage per cycle
If we assume that DLT is normal,

55. Example: Finding expected shortage per cycle
Suppose that the demand during lead time has expected value 100 and stdev 30, find the expected shortage if ROP=120.
z=( )/30=0.66.
E(z)=0.153 from Table 11-3.
Expected shortage = 30*0.153=4.59

56. Fill rate Fill rate is the percentage of demand filled from the stock
In a cycle
Fill rate = 1-(Expected shortage during LT) / Q
For normally distributed demand

57. Example: computing the fill rate
Suppose that the demand during lead time has expected value 100 and stdev 30, find the expected shortage if ROP=120. Compute the fill rate if order sizes are 200. Compute the annual expected shortages if there are 12 order cycles per year.
Expected shortage per cycle=4.59 from the last example.
Fill rate = /20=0.7705
Annual expected shortage=12*4.59=55.08.

58. Determinants of the Reorder Point
The rate of demand
The lead time
Demand and/or lead time variability
Stockout risk (safety stock)

59. 5. Fixed-Order-Interval Model
Orders are placed at fixed time intervals
Order quantity for next interval?
Suppliers might encourage fixed intervals
May require only periodic checks of inventory levels
Items from same supplier may yield savings in:
Ordering
Packing
Shipping costs
May be practical when inventories cannot be closely monitored

60. FOI compared against variable order interval models
A single order must cover the demand until the next order arrives. Exposure to random demand during not only lead time but also before.
Requires higher safety stock than variable order interval models.
May provide savings in set up / ordering costs.

61. 6. How many to order for the winter? Parkas at L.L. Bean
Notes: Mention that L.L. Grain is a mail order company deciding on the number of units of a Fall jacket to order. An estimate of demand using past information and expertise of buyers is given here. What should the appropriate order quantity be? In general distribution may not be known. Discuss methodology used in the Matching supply and demand article as a possibility in deciding on demand uncertainty (distribution).

62. Parkas at L.L. Bean Cost per parka = c = $45
Sale price per parka = p = $100
Discount price per parka = $50
Holding and transportation cost = $10
Salvage value per parka = s = $40
Profit from selling parka = p-c = = $55
Underage cost=$55
Cost of overstocking = c-s = = $5
Overage cost=$5
Notes: What information is required to make the ordering decision? Stress cost of understocking and overstocking. How to evaluate these costs for this example?

63. 6. Single Period Model
Single period model: model for ordering of perishables and other items with limited useful lives
Shortage cost: generally the unrealized profits per unit, $55 for L.L.Bean. We call this underage.
Excess cost: difference between purchase cost and salvage value of items left over at the end of a period, $5 for L.L.Bean. We call this overage.

64. Parkas at L.L. Bean Expected demand = 10 (‘00) parkas
Expected profit from ordering 10 (‘00) parkas = $499
The underage and overage probabilities after ordering 1100 parkas
P(D>1100):Probability of underage
P(D<1100):Probability of overage
Notes: Show detailed calculation of expected profit.

65. Optimal level of product availability
p = sale price; s = outlet or salvage price; c = purchase price
CSL = Probability that demand will be at or below reorder point
At optimal order size,
Expected Marginal Benefit from raising order size =
=P(Demand is above stock)*(Profit from sales)=(1-CSL*)(p - c)
Expected Marginal Cost =
=P(Demand is below stock)*(Loss from discounting)=CSL*(c - s).
Let Co= c-s; Cu=p-c, then the optimality condition is
(1-CSL*)Cu = CSL* Co,
CSL* = Cu / (Cu + Co)
Notes:

66. Parkas at L.L. Bean
Notes: Discuss marginal benefit and marginal cost of each jacket. We keep increasing order size as long as expected benefit exceeds expected cost.

67. Order Quantity for a Single Order
Co = Cost of overstocking = $5
Cu = Cost of understocking = $55
Q* = Optimal order size
Notes: Explain how formula is derived using decision tree.

68. Optimal Order Quantity without Normal Demands
0.917
Notes:
Optimal Order Quantity = 13(‘00)

69. Operations Strategy Too much inventory Wise strategy
Tends to hide problems
Easier to live with problems than to eliminate them
Costly to maintain
Wise strategy
Reduce lot sizes
Reduce set ups
Reduce safety stock
Aggregate negatively correlated demands
Remember component commonality
Delayed postponement