ISOM 2700: Operations Management
Session 16: Inventory Management II
Dongwook Shin
Dept. ISOM, HKUST Business School
Newsvendor Model Review
• Underage cost Cu
• Marginal benefit when an additional unit is sold
• Overage cost Co
• Marginal cost when an additional unit is not sold
Pr Demand ≤
n∗
Cu
=
= Cri8cal Frac8le
Cu +Co
• When demand is N(μ, σ!)
n∗ =μ+z∗ ×σ
Cu
∗
z =normsinv
Cu +Co
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Newsvendor Model: Mr. Choi’s Problem
• Mr. Choi operates a news-stand and sells SCMP
• Orders copies of the newspaper from the publisher at a cost of
$4 per copy
• Sells at a retailing price of $7
• How many copies should he order every day?
The demand can be described
with a normal distribution with
mean 100 and standard
deviation 12
Critical Fractile = 0.4286
z = -0.18
Optimal Quantity = 100 +
(-0.18) 12 = 98
2
Service Level Requirement
• Service level is the probability of no shortage
• What is the order quantity n that guarantees the service
level of S?
Pr Demand ≤ n = S
• When demand is N(μ, σ2 )
n=μ+zσ
z=normsinv(S)
3
Newsvendor Model: Mr. Choi’s Problem
• How much should Mr. Choi order if he wants to ensure a
minimum service level of 70%?
Mr. Choi needs to increase the order quantity because the service level at
the optimal order quantity is only 42.86%. In particular, Mr. Choi should
order 𝑛 = 𝜇 + 𝑧𝜎 = 107 (using a round-up rule), where 𝑧 =
𝑛𝑜𝑟𝑚𝑠𝑖𝑛𝑣 0.7 = 0.54
• What if Mr. Choi wants to ensure a minimum service
level of 10%?
Mr. Choi can order n* = 98 because the optimal order quantity achieves
service level of 42.86% > 10%
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Learning Objectives: Session 16
• Economic Order Quantity (EOQ) model
• Lead time and re-order point
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Example: Sweater
• M&S has a stable demand for a line of sweater it offers.
Each week there is a demand for 100 sweaters. M&S
incurs a fixed cost of $5000 every time it places an
order. The marginal cost of a sweater is $400, and the
shop’s cost of capital is approximately 1% per week
• How often should M&S order?
• How many units should be
ordered each time?
6
Inventory Costs
• Fixed ordering (setup) costs
• Handling charges, preparing purchase order
• Supplier selection, negotiations
• Freight and insurance
• Inventory carrying costs
• Insurance cost
• Maintenance cost
• Opportunity cost of alternative investment
• Underage costs
• Loss of profit
• Loss of good will or reputation (hard to quantify)
• Overage costs
• Leftover items are costly
7
The Inventory “Saw-Tooth” Pattern
Average
Inventory (Q/2)
Inventory
Q
D
Time
Q/D
Shipment
arrives
Shipment
arrives
• Q = Quantity in each order (what we need to choose)
• D = Demand Rate
• Q/D = Time between shipments
• D/Q = Order frequency per unit time
8
Tradeoffs
Inventory level
Q = 100
Total demand = 5200/year
# of orders/year = 52
Average inventory = 50
Time
1 week
Total demand = 5200/year
# of orders/year = 104
Average inventory = 25
Q = 50
Time
0.5 week
9
Economic Order Quantity (EOQ) Model:
How Much to Order?
Given:
D - Demand per unit of time (year, week, …)
S - Setup or Order Cost ($/setup; $/order)
H - Marginal holding cost ($/per unit per unit of time)
Decide: Order quantity Q
Inventory Level
Q
Average
Inventory (Q/2)
Costs
•
•
•
Holding cost = HQ/2 per year
Ordering cost = SD/Q per year
Total cost = HQ/2 + SD/Q per year
Time
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EOQ Model: How Much to Order?
Annual Cost
lC
a
t
o
T
rve
u
C
ost
ng
i
d
l
Ho
ve
r
u
tC
s
o
C
Order (Setup) Cost Curve
Optimal Order
Quantity (Q*)
Order Quantity
Q* balances the setup costs with inventory holding costs
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Solution to EOQ Model
DS QH
Total cost = TC(Q) =
+
Q
2
d TC(Q)
DS H
=−
+
2
2
dQ
Q
2DS
Q∗ =
H
Economic
Order
Quantity
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Characteristics of EOQ
2DS
∗
Q =
H
• The optimal batch size trades off setup and holding costs
• At Q=Q*, setup cost per unit time = holding cost per unit time
• Square-root relationship between Q* and (D, S):
• If demand increases by a factor of 4, the optimal batch size
increases by a factor of 2 and order twice as often
• To reduce batch size by a factor of 2, setup cost has to be
reduced by a factor of 4
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EOQ Model Summary
Given D (demand rate), S (setup cost), and H (holding cost):
Time between two orders = Q/D
Order frequency per unit time = D/Q
Holding cost per unit time = HQ/2
Order or setup cost per unit time = SD/Q
Total cost per unit time = HQ/2 + SD/Q
Optimal order quantity Q∗ =
2DS
H
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Example: Sweater
• M&S has a stable demand for a line of sweater it offers.
Each week there is a demand for 100 sweaters. M&S
incurs a fixed cost of $5000 every time it places an
order. The marginal cost of a sweater is $400, and the
shop’s cost of capital is approximately 1% per week
D = 100/week
S = $5000
H = $4/week$unit
Optimal order size Q = QEOQ = 500
Time between orders = Q/D = 5 weeks
Weekly ordering cost = (D/Q)S = $1000
Weekly holding cost = (Q/2)H = $1000
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Example: Training Flight Attendant
Ace Airline has a need for 100 new flight attendants per month as
replacements for those who leave
Trainees are put through a two-month school. The fixed cost of running
one session of this school is $150,000. Any number of sessions can be
run during the year but must be scheduled so that the airline always has
enough flight attendants
The cost of having excess attendants is simply
the salary that they receive, which is $15,000
per month. How many sessions of the school
should Ace Airline run each year, and how many
flight attendants should be in each session?
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Example: Training Flight Attendant
D = 1200/year, S =$150,000, H = $180,000/year2employee
Optimal session size Q = 44.7≈45
Number of sessions per year = 26.7
Time between sessions = 0.0375 years ≈ 2 weeks
Annual training cost = 4,000,000
Annual holding cost = 4,050,000
Average waiting time after training until job assignment =
Q/2D = 6.8 days ≈ 1 week (Use Little’s law)
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Example: Training Flight Attendant
• On any given day, how many trainees do you expect to find in
school? Does that depend on the class size?
Average inventory = Average flow rate x Average flow time = (100
per month) x (2 months) = 200
• We have implicitly assumed that Ace Airline starts paying the salary
of $15,000 per month only at the end of the two-month school.
Such a practice drew significant complaints from the trainees. Ace
decided to change its practice and pay the trainees during the
training session as well. How would the new policy change Ace's
class size?
Additional cost per year = ($15,000/month) x (1200 employees/year)
x (2 months) = $36,000,000/year
This cost is independent of the class size Q*, and therefore, the
optimal class size is not affected by the new policy
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Extension: Per-unit Purchasing Cost
Given:
D - Demand per unit of time (year)
S - Setup cost ($/order)
H - Marginal holding cost ($/year)
C – Purchasing cost ($/unit)
Decide: Order quantity Q
Costs
•
•
•
•
Holding cost = HQ/2 per year
Ordering cost = SD/Q per year
Purchasing cost = DC
Total cost = DC+HQ/2 + SD/Q per year
EOQ is not
affected by the
purchasing cost
2DS
∗
Q =
H
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Learning Objectives: Session 16
• Economic order quantity (EOQ) model
• Lead time and re-order point
20
Lead Time in EOQ model
• EOQ formula determines how much to order
• Lead time determines when to order
• (M&S Store case revisited) It takes 1 week from the time
M&S places an order to the time the order is received.
This means the lead time is 1 week. If demand is stable
as before, when should the store place an order?
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M&S Store: When to Order?
• Given the one-week lead time, M&S should place an
order when the on-hand inventory drops to 100, the
demand during the lead time
• The Re-Order Point (ROP) equals 100
On-hand inventory
Q=500
Lead time = 1 week
LT demand = 100
ROP = 100
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EOQ Model with Lead time
Inventory Level
Average
Inventory
(Q*/2)
Optimal
Order
Quantity
(Q*)
Re-Order
Point
ROP= D x L
Lead Time
Time
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EOQ Example
George Heinrich uses 18,000 units per year of a certain component.
The per unit cost of this component is $500. Each order costs George
$100. He operates 360 days per year and has found that an order
must be placed with his supplier 4 days before he can expect to
receive that order. The holding cost is $5 per unit per year. Find
• Economic order quantity
𝑄∗
=
2×18000×100
= 848.5 𝑢𝑛𝑖𝑡𝑠
5
• Annual holding cost and annual ordering cost
𝑄∗𝐻
𝐷𝑆
= $2121.3,
= $2121.3
2
𝑄∗
• Reordering point
18000
𝑅𝑂𝑃 =
×4 = 200 𝑢𝑛𝑖𝑡𝑠
360
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Takeaways
• An EOQ model is applicable when
• Demand does not change significantly from one ordering
period to another
• Example: Long lifecycle stable products such as groceries,
automobile components, chemicals, heavy industrial
equipment, etc
• Video: EOQ problem walkthrough
• https://www.youtube.com/watch?v=JCt1IVSjsuM&feature=yo
utu.be
• Next class: Inventory Management III
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