1. Deterministic Inventory Models
Chapter 3
Deterministic Inventory Models

2. Inventory Inventory: A stock or store of goods.
Firms typically stock many items in inventory.
Many of the items a firm carries in inventory relate to the kind of business it engages in.

3. the reason to hold inventory can be:
Role of Inventory
the reason to hold inventory can be:
You bought a six-pack of soda, rather than a single bottle, because you don’t want to have to go to the store every time you want to drink a bottle of soda.
You bought a “family size” box of cereal, rather than a small box, because larger boxes are more cost-effective (cheaper per ounce) than smaller ones.

4. Examples
Manufacturing firms carry supplies of raw materials, purchased parts, finished items, spare parts, tools,....
Department stores carry clothing, furniture, stationery, appliances,...
Hospitals stock drugs, surgical supplies, life-monitoring equipment, sheets, pillow cases,...
Supermarkets stock fresh and canned foods, packaged and frozen foods, household supplies,...
Not all items in inventory are items to be sold.

5. The Nature and Importance of Inventories
Inventories are vita part of business
They are necessary for operations
They contribute to customer satisfaction
A typical firm probably has about 30% of its
current capital assets and perhaps as much as
90% of its working capital invested in inventory.
2. Firms that experience seasonal patterns in demand often build up inventories during off-season periods to meet overly high requirements during certain seasonal periods.
3. Manufacturing firms have used inventories as buffers between successive operations to maintain continuity of production that would otherwise be disrupted by events such as breakdowns of equipment and accidents that cause portion of operation to shut down temporarily.
4. Delayed deliveries and unexpected increase in demand increase the risk of shortages.

6. Inventory Level and Inventory Position
On-hand inventory (OH): number of units available in inventory
Backorders (BO): demands that have occurred but have not been satisfied yet
Usually OH and BO cannot both be non-zero
Inventory level (IL):
Note that

7. Inventory Level and Inventory Position
IL does not give us enough information to make ordering decisions
Need to consider inventory “in the pipeline”
Called on-order inventory (OO)
Make decisions based on inventory position (IP):
Usually make ordering decisions based on IP
Holding and stockout costs are based on IL
If lead time = 0, IL = IP

8. Inadequate Control of Inventories
Inadequate control of inventories can result in both under- and overstocking of items.
Understocking (too few) results in missed deliveries, lost sales, dissatisfied customers, and production bottlenecks (idle workers or machines).
Resulting underage cost.
Overstocking (too many) ties up funds that might be more productive elsewhere.
Resulting overage cost.
In other words, the manager tries to achieve a balance in stocking.
Goal: matching supply with demand!

9. Objective of Inventory Control
Two fundamental decisions:
When to order (timing)
How much to order (size)
Inventory management has two main concerns

10. EOQ Model Assumptions Demand is deterministic, constant
Rate D items/year
Stockouts not allowed
Costs:
Fixed cost S per order
Purchase cost c per unit ordered
Inventory holding cost h per unit per year
(No stockout penalty since stockouts not allowed)

11. Average Inventory (Q*/2)
EOQ Model
Inventory Level
Optimal Order Quantity (Q*)
Average Inventory (Q*/2)
Reorder Point (ROP)
Time
Lead Time

12. Inventory Costs Material Cost, C
Fixed Ordering cost, S: transportation cost, quotation, appraisal, preparing purchase order,
Holding or Carrying Cost, H, or h (as % of C): Interest, storage and handling, Tax, Insurance, shrinkage, pilferage, obsolescence, deterioration.

13. Average Inventory (Q*/2)
EOQ Model
Inventory Level
Optimal Order Quantity (Q*)
Average Inventory (Q*/2)
Reorder Point (ROP)
Time
Lead Time

14. Economic Order Quantity - EOQ
Annual
carrying
cost
TC = + Q 2 hC D S
ordering
Total cost is simple function of the lot size Q.

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

16. Deriving the EOQ
Using calculus, we take the derivative of the total cost function and set the derivative equal to zero and solve for Q. Total cost curve is convex i.e. curvature is upward so we obtain the minimizer.
T: Reorder interval length = EOQ/D.
n: Ordering frequency: number of orders per unit time = D/EOQ.
The total cost curve reaches its minimum where the inventory carrying and ordering costs are equal.

17. Example 10.1
Demand for the Deskpro computer at Best Buy is 1000 units per month. Best
Buy incurs a fixed order placement, transportation, receiving cost of $4000 each
time an order is placed. Each computer costs Best Buy $500 and the retailer has
a holding cost of 20%. Evaluate the number of computers that the store manager
should order in each replenishment lot.
Demand, D = 12,000 computers per year
d = 1000 computers/month
Unit cost, C = $500
Holding cost fraction, h = 0.2
Fixed cost, S = $4,000/order
Q* = Sqrt[(2)(12000)(4000)/(0.2)(500)] = 980 computers
Cycle inventory = Q/2 = 490
Notes:

18. Example 1 (continued) Annual ordering and holding cost =
= (12000/980)(4000) + (980/2)(0.2)(500) = $97,980
Suppose lot size is reduced to Q=200, which would reduce flow time:
= (12000/200)(4000) + (200/2)(0.2)(500) = $250,000
To make it economically feasible to reduce lot size, the fixed cost associated with each lot would have to be reduced

19. Sensitivity to Q Suppose we cannot order Q* exactly
© 2014 Lawrence V. Snyder
Sensitivity to Q
Suppose we cannot order Q* exactly
e.g., must order in multiples of 10
or must order every month
How much more expensive is solution?
Theorem 3.2. For any Q > 0,

20. © 2014 Lawrence V. Snyder
Sensitivity to Q

21. Sensitivity to Q Expression grows slowly as Q moves away from Q*
© 2014 Lawrence V. Snyder
Sensitivity to Q
Expression grows slowly as Q moves away from Q*
Examples:
(We can double or halve Q and only increase cost by 25%)
In Fig. 3.3, cost curve is “flat” around minimum

22. © 2014 Lawrence V. Snyder
Example 3.2
Suppose Joe’s Corner Store orders 250 candy bars instead of Q* = 304
How much does cost increase?
Solution:
That is, using Q = 250 results in 1.9% increase in cost

23. Key Points from EOQ Model
In deciding the optimal lot size, the tradeoff is between setup (order) cost and holding cost.
If demand increases by a factor of 4, it is optimal to increase batch size by a factor of 2 and produce (order) twice as often.
If lot size is to be reduced, one has to reduce fixed order cost. To reduce lot size by a factor of 2, order cost has to be reduced by a factor of 4.
Notes:

24. Aggregating Multiple Products in a Single Order
Transportation is a significant contributor to the fixed cost per order
Can possibly combine shipments of different products from the same supplier
Can also have a single delivery coming from multiple suppliers or a single truck delivering to multiple retailers
Aggregating across products, retailers, or suppliers in a single order allows for a reduction in lot size for individual products because fixed ordering and transportation costs are now spread across multiple products, retailers, or suppliers

25. Lot Sizing with Multiple Products or Customers
In practice, the fixed ordering cost is dependent at least in part on the variety associated with an order of multiple models
A portion of the cost is related to transportation (independent of variety)
A portion of the cost is related to loading and receiving (not independent of variety)
Two scenarios:
Lots are ordered and delivered independently for each product
Lots are ordered and delivered jointly for all three models

26. Ex 3
Best Buy sells three models of computers, the Litepro, Medpro, and Heavypro.
Annual demands for the three products are DL=12000 for the Litepro, DM=1200
for the Medpro, and DH=120 for the Heavypro. Assume that each model costs Best
Buy $500. A fixed transportation cost of $4000 is incurred each time an order is
delivered. For each model ordered and delivered on the same truck, an additional
fixed cost of $1000 is incurred for receiving and storage. Best Buy incurs a holding
cost of 20%. Evaluate the lot sizes that the Best Buy manager should order if lots
for each product are ordered and delivered independently. Also evaluate the annual
cost of such a policy.
Notes:

27. Lot Sizing with Multiple Products
Demand per year
DL = 12,000; DM = 1,200; DH = 120
Common transportation cost, S = $4,000
Product specific order cost (i.e., receiving or loading cost for each product)
sL = $1,000; sM = $1,000; sH = $1,000
Holding cost, h = 0.2
Unit cost
CL = $500; CM = $500; CH = $500
Notes:

28. Delivery Options No Aggregation: Each product ordered separately
Complete Aggregation: All products delivered on each truck
Notes:

29. No Aggregation: Order Each Product Independently (Example 3)
Total cost = $155,140 (Ex 10.3)

30. Ex 10.4 Consider the Best Buy data in Example 3. The three
product managers have decided to aggregate and order all
three models each time they place an order. Evaluate the
optimal lot size for each model.
Notes:

31. Complete Aggregation: Order all products jointly
Total ordering cost S*=S+sL+sM+sH = $7,000
n: common ordering frequency
Annual ordering cost = n S*
Total holding cost:
(Recall: n=D/Q*)
Total inventory cost:

32. Aggregation: Order All Products Jointly
S* = S + sL + sM + sH = = $7000
n* = Sqrt[(DLhCL+ DMhCM+ DHhCH)/2S*]
= 9.75
QL = DL/n* = 12000/9.75 = 1230
QM = DM/n* = 1200/9.75 = 123
QH = DH/n* = 120/9.75 = 12.3
Cycle inventory = Q/2
Average flow time = (Q/2)/(weekly demand)

33. Complete Aggregation: Order All Products Jointly (Ex 4)
Annual order cost = 9.75 × $7,000 = $68,250
Annual total cost = $136,528 (Ex. 4)

34. Lessons from Aggregation
Aggregation allows firm to lower lot size without increasing cost
Complete aggregation is effective if product specific fixed cost is a small fraction of joint fixed cost

35. Power-of-Two Policies
Suppose we are given a base period
Week, day, work shift, etc.
Order interval T must be power-of-two multiple of base period
Place orders every 1 day, or 2 days, or 4 days, or…
Or every ½ day, or every ¼ day, or…
Key questions:
What order interval should we use?
How much more expensive than optimal T ?
© 2014 Lawrence V. Snyder

36. Why Power-of-Two? Simpler ordering schedule
Easier coordination in multi-stage systems
e.g., one-warehouse, multi-retailer (OWMR) problem
Retailers’ orders to central warehouse line up
Optimal policy for OWMR is not known
Power-of-two policies known to be close to optimal (Roundy 1985, Muckstadt and Roundy 1993)
More on this later in course
Mathematical convenience
425: “more later in course”
© 2014 Lawrence V. Snyder

37. Analysis Fixed base period TB
Reorder interval must be of the form for some k ∈ {…, –2, –1, 0, 1, 2, …}
Optimal order interval is
© 2014 Lawrence V. Snyder

38. Optimal Power-of-Two T
Let f (T) = EOQ cost as function of T:
f is convex
Therefore, optimal k is smallest k such that
425: fix: first <= should not have comma, second should have <==> …
© 2014 Lawrence V. Snyder

39. Optimal Power-of-Two T
Optimal order interval is T = TB2k, where k is smallest integer such that
© 2014 Lawrence V. Snyder

40. Error Bound
Theorem 3.3. If T is the optimal power-of-two interval and T* is the optimal order interval, then This holds for any TB.
Proof. We know that
© 2014 Lawrence V. Snyder

41. Error Bound (cont’d) We already know Therefore
Evaluate f(T) at endpoints:
© 2014 Lawrence V. Snyder

42. Error Bound (cont’d)
© 2014 Lawrence V. Snyder

43. Error Bound (cont’d) Since f is convex and , f(T) T optimal Po2 T
© 2014 Lawrence V. Snyder

44. Uniformly Distributed Optimal T
Theorem 3.4. If , then
Proof. Omitted.
© 2014 Lawrence V. Snyder

45. Example 3.4
Suppose Joe’s Corner Store must order in power-of-two multiples of 1 month
What is optimal order interval? What is cost error?
Solution: TB = 1/12 years
© 2014 Lawrence V. Snyder

46. Economies of Scale to Exploit Quantity Discounts
Price/Unit $3
Order quantity Unit Price
$3.00
$2.96
Over $2.92
$2.96
$2.92
5,000
10,000
Order Quantity

47. Economies of Scale to Exploit Quantity Discounts
Given a price schedule with quantity discounts, what is the optimal purchasing decision for a buyer seeking to maximize profits ?

48. Quantity Discounts
Pricing schedule has specified quantity break points q0, q1, …, qr, where q0 = 0
If an order is placed that is at least as large as qi but smaller than qi+1, then each unit has an average unit cost of Ci
The unit cost generally decreases as the quantity increases, i.e., C0>C1>…>Cr
The objective for the company (a retailer in our example) is to decide on a lot size that will minimize the sum of material, order, and holding costs

49. Quantity Discount Procedure (different from what is in the textbook)
Step 1: Calculate the EOQ for the lowest price. If it is feasible (i.e., this order quantity is in the range for that price), then stop. This is the optimal lot size. Calculate TC for this lot size.
Step 2: If the EOQ is not feasible, calculate the TC for this price and the smallest quantity for that price.
Step 3: Calculate the EOQ for the next lowest price. If it is feasible, stop and calculate the TC for that quantity and price.
Step 4: Compare the TC for Steps 2 and 3. Choose the quantity corresponding to the lowest TC.
Step 5: If the EOQ in Step 3 is not feasible, repeat Steps 2, 3, and 4 until a feasible EOQ is found.

50. Quantity Discounts: Example
Cost/Unit
Total Material Cost $3
$2.96
$2.92
Notes:
5,000
10,000
5,000
10,000
Order Quantity
Order Quantity

51. Quantity Discount: Example
Order quantity Unit Price
$3.00
$2.96
Over $2.92
q0 = 0, q1 = 5000, q2 = 10000
C0 = $3.00, C1 = $2.96, C2 = $2.92
D = units/year, S = $100/lot, h = 0.2

52. All-Unit Quantity Discount: Example
Step 1: Calculate Q2* = Sqrt[(2DS)/hC2]
= Sqrt[(2)(120000)(100)/(0.2)(2.92)] = 6410
Not feasible (6410 < 10001)
Calculate TC2 using C2 = $2.92 and q2 = 10001
TC2 = (120000/10001)(100)+(10001/2)(0.2)(2.92)+(120000)(2.92)
= $354,520
Step 2: Calculate Q1* = Sqrt[(2DS)/hC1]
=Sqrt[(2)(120000)(100)/(0.2)(2.96)] = 6367
Feasible (5000<6367<10000)  Stop
TC1 = (120000/6367)(100)+(6367/2)(0.2)(2.96)+(120000)(2.96)
= $358,969
TC2 < TC1  The optimal order quantity Q* is q2 = 10001

53. Why Quantity Discounts?
Coordination in the supply chain
Notes:

54. Coordination for Commodity Products
D = 120,000 bottles/year
SR = $100, hR = 0.2, CR = $3
SS = $250, hS = 0.2, CS = $2
Retailer’s optimal lot size = 6,324 bottles
Retailer cost = $3,795; Supplier cost = $6,009
Supply chain cost = $9,804
Notes:

55. Coordination for Commodity Products
What can the supplier do to decrease supply chain costs?
Coordinated lot size: 9,165; Retailer cost = $4,059; Supplier cost = $5,106; Supply chain cost = $9,165
Effective pricing schemes
quantity discount
Notes:

56. Quantity Discounts When Firm Has Market Power
The annual demand faced by a retailer is given by the demand
curve 360,000-60,000p, where p is the price at which the retailer
sells a type of product. The manufacturer incurs a production
cost of c=$2 per product. The manufacturer must decide on price
to charge the retailer, and the retailer in turn must decide on
the price (p) to charge the customer.
What are the optimal w for the manufacturer to charge the
retailer and the optimal retail price p for the retailer to charge
customer?
Notes:

57. Quantity Discounts When Firm Has Market Power
No inventory related costs
Demand curve
360, ,000p
What are the optimal prices and profits in the following situations?
The two stages make the pricing decision independently
Price = $5, Profit = $180,000, Demand = 60,000
The two stages coordinate the pricing decision
Price = $4, Profit = $240,000, Demand = 120,000
Notes:

58. Design a volume discount scheme that achieves the coordinated solution
Notes:

59. Production Order Quantity Model

61. Imax Inventory Level Production portion of cycle -u p - u Time
Supply Begins
Supply Ends
Demand portion of cycle with no supply

62. Production Portion of Cycle
POQ Cycles
Inventory Level
Imax=(p-u)*Q/p
=Q*·(1- u/p)
Production Portion of Cycle Q*
Time
Supply Begins
Supply Ends
Demand portion of cycle with no supply
Usage rate u is demand rate D in terms of the time measure same as production rate p.

63. POQ Model Equations D= Demand S = Setup cost H = Holding cost
p = Production rate
u= Usage rate

64. POQ Model Equations D= Demand S = Setup cost H = Holding cost
p = Production rate
u= Usage rate

65. POQ Example 1
You’re a production planner for Stanley Tools. Stanley Tools faces an annual demand of 30,000 units of screw. The production rate is 300 units per day with 300 working days a year. Production setup cost is $150 per order. Carrying cost is $1.50 per screw driver. What is the optimal lot size?
D= Demand = 30,000/yr
S= Setup cost = $150
H = Holding cost = $1.5/unit/yr
p = Production per day = 300/day
u=usage rate per day=30,000/300=100/day

66. POQ Example 1
Optimal lot size:

67. POQ Example 1

68. 3.3 Periodic Review: The Wagner-Whitin Model
© 2014 Lawrence V. Snyder

69. Problem Statement The Wagner-Whitin Model Similarities to EOQ
Wagner and Whitin (1958)
Similarities to EOQ
Fixed order cost
Stockouts not allowed
Choose order quantity to minimize total cost
Differences from EOQ
Periodic review
Demand may change over time
© 2014 Lawrence V. Snyder

70. Problem Statement (cont’d)
T-period, finite-horizon model
Lead time = 0
May not be optimal to order in every period
ZIO is optimal (we will show)
Therefore, need to choose how many whole periods’ of demand to order each time we order
In each period, decide whether to order
And if so, how much
© 2014 Lawrence V. Snyder

71. Notation dt = demand in period t K = fixed cost
h = holding cost per item per period
(not per year)
Assume both are >0
Ignore purchase cost c
© 2014 Lawrence V. Snyder

72. Sequence of Events In each period:
Replenishment order (if any) is placed and received instantly
Demand occurs and is satisfied from inventory
Holding costs are assessed based on on-hand inventory
Important to specify sequence of events in periodic- review models
Assume IL = 0 at time 0
© 2014 Lawrence V. Snyder

73. ZIO Property (cont’d) If items had been ordered in t instead, then
Holding cost would decrease
Fixed cost would stay the same
Contradicts assumption that original policy was optimal
Corollary. Always order integer number of periods’ of demand
In t, order dt or dt+dt+1 or dt+dt+1+dt+2 or …
Therefore just need to decide which periods to order in
© 2014 Lawrence V. Snyder

74. DP Recursion determine next period s in which to order
(s = T+1 ⇒ this is last order)
evaluate given choice of s
cost in t, …, s – 1
fixed cost (in t)
holding cost (in t + 1, …, s – 1)
cost for next and all subsequent orders
© 2014 Lawrence V. Snyder

75. DP Recursion θt θt+1 θt+2 θt+3 θT–1 θT θT+1 … t t+1 t+2 t+3 T–1 T T+1
order in t : K
order again in t + 1: + θt+1
order again in t + 2: + hdt+1 + θt+2
order again in t + 3: + h(dt+1 + 2dt+2) + θt+3 …
order again in T – 1: + h(dt+1 + 2dt+2 + … + (T – 2 – t)dT–2) + θT–1
order again in T : + h(dt+1 + 2dt+2 + … + (T – 2 – t)dT–2 + (T – 1 – t)dT–1) + θT
don’t order again: + h(dt+1 + 2dt+2 + … + (T – 2 – t)dT–2 + (T – 1 – t)dT–1 + (T – t)dT ) + θT+1
© 2014 Lawrence V. Snyder

76. Algorithm 3.1 (Wagner-Whitin)
Set θT+1 ← 0 and t ←T
Compute θt using and set s(t) = minimizer
Set t ← t – 1
If t = 0, STOP
Otherwise, go to 2
© 2014 Lawrence V. Snyder

77. Shortest Path Representation (cont’d)
425: D should be lower-case throughout
Figure 3.9
© 2014 Lawrence V. Snyder

78. Example 3.9 Garden center sells bags of compost K = 500, h = 2, T = 4
(d1, …, d4) = (90, 120, 80, 70)
Find optimal order quantities and cost
Solution:
© 2014 Lawrence V. Snyder

79. Example 3.9 (cont’d)
© 2014 Lawrence V. Snyder

80. Example 3.9 (cont’d) Order in periods 1 and s(1) = 3
Q1 = d1+ d2 = 210
Q3 = d3+ d4 = 150
Total cost = θ1 = 1380
© 2014 Lawrence V. Snyder