Koç University Graduate School of Business
MBA Program
OPSM 501: Operations Management
Week 12:
Inventory Management
Order-up-to model
Zeynep Aksin
zaksin@ku.edu.tr
Levers for Managing Inventories
 Theoretical Inventory Ith=R x Tth
–
–
–
–
Reduce critical activity times
Eliminate non-value added work
Move work from critical to non-critical
Redesign process to replace serial with parallel
processing
 Cycle inventory
– Average inventory per cycle=Q/2
– Reduce set-up to reduce cycle inventory
Levers for Managing Inventories
 Seasonal Inventory
– Use pricing and incentive tactics to smooth demand
– Increase resource flexibility
 Safety inventory-this is next!
A Multi-Period Inventory Model
 Often, there are multiple reorder opportunities
 Consider a central distribution facility which orders from
a manufacturer and delivers to retailers. The distributor
periodically places orders to replenish its inventory
4
Set Up: Simple Supply Chain
orders
Supply
Pipeline
stock
On-hand
inventory
Inventory position
 Three key questions:
– How often to review?
– When to place an order?
– How much to order?
5
Timing in the order up-to model



Time is divided into periods of equal length, e.g., one hour, one month.
During a period the following sequence of events occurs:
– A replenishment order can be submitted.
– Inventory is received.
– Random demand occurs.
Lead times:
– An order is received after a fixed number of periods, called the lead time.
– Let l represent the length of the lead time.
An example with l =
1
13-6
Order up-to model vs. newsvendor model
 Both models have uncertain future demand, but
there are differences…
Inventory obsolescence
Number of
replenishments
Demand occurs during
replenishment
Newsvendor
Order up-to
After one period
Never
One (maybe two or three
with some reactive
capacity)
Unlimited
No
Yes
– Newsvendor applies to short life cycle products with
uncertain demand and the order up-to applies to long
life cycle products with uncertain, but stable, demand.
13-7
Periodic Review, Order-up-to Policy
Inventory Position = Quantity + Quantity
on hand on order
S - Base stock level/Order-up-to Point
p - Review period; l - Replenishment lead time
 - Demand per unit time
ss - Safety stock
Ordering Rule:
Place an order every p periods so as to bring your
inventory position to the Base Stock Level, S.
8
Periodic review with no demand variability
Inventory position
Inventory Level
On-hand inventory
(p+l)
p
l
0
p
l
2p
p+l
3p
2p+l
4p
3p+l
time
9
Periodic review with no demand variability
Order Quantity, Q = p
Average Cycle stock = Q/2 = p / 2
Pipeline stock =  l
Order-up-to point, S =  (p + l)
10
Why hold Safety Inventory?
 Demand uncertainty
 Supply uncertainty
Measures of product availability
– Product fill rate (f): fraction of demand that is
satisfied from product in inventory
– Cycle service level (CSL): fraction of
replenishment cycles that end with all the
customer demand being met
11
Periodic review with variable demand
Order-up-to point (S) =  (p+l) + Safety Stock (ss)
Average Order Quantity (Q) = p
Average Pipeline stock =  l
Average Cycle stock = Q/2 = p / 2
Safety Stock = ss = ?
12
Determination of the Safety Stock
Inventory position
Inventory Level
On-hand inventory
p+l+ss
p+ss
l+ss
ss
0
p
l
2p
p+l
3p
2p+l
4p
3p+l
13
time
Probabilistic Models
Key idea:
Order-up-to target covers demand over time
period of p+l
X
 ( p  l)
+ (ss  z
p l )
=S
14
Designing for a target CSL
Safety Stock (ss) = z
p l
Choosing z:
a=CSL= P(demand during p+l <= S)
z= Fs-1(CSL)
15
Example #1
Given:
Solve:
p = 2 weeks
l = 3 weeks
 = 1.5 units per week
2 = 4 units per week
Target service level, CSL=95%
F ( z )  0.95
Safety stock = ss  z
so from table, z = 1.64
p  l  1.64  2 5  7.36
Base stock level = S  p  l  z
p  l  1.5  2  1.5  3  7.36  14.86
Average on-hand inv= 1.5x2/2 + safety stock=1.5+7.36=8.86
16
Example #2
Given:
Solve:
p = 2 weeks
l = 1 week
 = 1.5 units per week
2 = 4 units per week
Target service level, CSL=95%
F z   0.95
Safety stock = ss  z
so from table, z = 1.64
p  l  1.64  2 3  5.70
17
Base stock level =
S  l  p  z p  l  1.5  2  1.5 1  5.70  10.2
Computer example continued
 Suppose we do not know the base-stock level S
 We know the company uses a periodic-review,
order-up-to policy
 From company data we know that average onhand inventory is 12.6 units
 What service level is the store providing?
18
Example #3
12.6= (1.5 x 2)/2 + z x 2 x
23
z =2.48
F(2.48)=0.993
19
Computer store: determining policy
parameters
 Store wants to re-evaluate order frequency
 Retain service level of 95%
 Apple charges a fixed fee of $25 for shipping
and handling of order
 Store’s order processing cost is $15
 The model being considered has wholesale
price of $3000
 Holding cost rate is estimated to be 20%
20
Example #4
 Compute EOQ
–
–
–
–
h= (3000x0.2)/52weeks/yr=$11.5
K=15+25=40
Q*=3.2
p = Q*/=3.2/1.5=2.15
21
Delayed Product Differentiation
 Products start off undifferentiated; at some
point, product variety explodes
 Trade-off between product variety vs. inventory
and service levels
 Design the product so the point of
differentiation is delayed as much as possible
 Don’t commit to FGI early on
22
DPD- Standardization
 Using common components or processes
– Reduces complexity of manufacturing
– Increases “flexibility” of work-in-process
– Improves service level
 Examples:
– standardizing head driver board & print mechanism
interface in b&w and color printers
– generic printer for Mac and Windows users
23
DPD-Modular Design
 Decomposing the complete product into
submodules that are easily assembled; delay
assembly of product specific modules
– Can increase no. of modules
– Same benefits as standardization
 Examples:
– Power supply module in the HP Deskjet printer
– Inserting plastic color panel to generic products
– Channel assembly in PC industry
24
DPD – Process Restructuring
Postponing (if necessary) reverse operations
 Operation divided into two steps, first step
common to all products
 Reverse the order of two operations with first
operation common to all products
 Example:
– Benetton (dye & knit  knit & dye)
25
1.
Isolate the variable functions/features of the product in one (or a few)
physical components.
26
2.
Minimize the variable cost of the differentiating components of the product.
30%
N=10 variants of product
25%
Reduction in
Total Inventory
20%
(includes
pipeline, cycle,
and safety
inventories)
15%
N=8
N=6
N=4
10%
N=2
5%
0
-5%
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
Cost of Differentiating Components as Fraction of Total
Assumptions: Fill rate = 0.98
Review period = 1 week
Order lead time = 6 weeks
Coefficient of variation of demand = 1.0
27
3. Ensure that production/supply-chain precedence requirements allow the
differentiating elements of the product to be added last.
28
When is DPD appropriate?







High uncertainty in demand mix
Long lead times
Short product life cycle
High inventory /stock out costs
Not too costly/time consuming to customize
High value to core component
Low variable cost of differentiating
components
29
Announcement 1
 Next week field trip to Mercedes coach plant
 Departure from campus 8:10-visit starts @ 9:30
 Bus info: MUHAMMET GÜLER 05392294281
34 ZP 4194
 Intermediate stop at Ataturk Oto Sanayi: 8:35
 HOŞDERE OTOBÜS FABRİKASI-check web
site for directions
 Sanayi Mah. Mercedes Bulvarı No. 5, 34500
Esenyurt / İstanbul
Tel: (0 212) 622 70 00 Pbx
Fax: (0 212) 622 84 00
Announcement 2
 Read the Temsa case before the trip
 Bonus assignment-can be done in pairs (5%):
 Take notes-ask questions-take photos if allowed
to
– Strategy: Comment on the 4 product attributes for
Mercedes: PQTV
– Process documentation: Provide a high level
process flow chart
– Process Selection: Analyze volume, variety level
and its fit with the type of process (position the plant
on the product-process matrix based on this analysis)
Announcement 3
 Last session: will play the beer game in-class
– Need to read the handout that I will distribute before
coming to class
– Need to be on time since we will start at 11:00 sharpaim for arrival at 10:45.
 Final exam on January 4 @ 10:00
 Room will be announced