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Chap 6

1. Amazon started as an online retailer without inventory but has become a leader in inventory management and warehousing. 2. Amazon uses technology and automation throughout its warehouses to efficiently pick, pack, and ship large volumes of customer orders while minimizing errors. 3. Orders are packed and shipped from the closest distribution center within 1-2 days to provide excellent customer service.
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0% found this document useful (0 votes)
47 views21 pages

Chap 6

1. Amazon started as an online retailer without inventory but has become a leader in inventory management and warehousing. 2. Amazon uses technology and automation throughout its warehouses to efficiently pick, pack, and ship large volumes of customer orders while minimizing errors. 3. Orders are packed and shipped from the closest distribution center within 1-2 days to provide excellent customer service.
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
You are on page 1/ 21

10/26/2023

Operations Management: Sustainability


and Supply Chain Management Outline (1 of 2)
Twelfth Edition
• Global Company Profile: Amazon.com
• The Importance of Inventory
Chapter 6
• Managing Inventory
Inventory Management
• Inventory Models
• Inventory Models for Independent Demand

1 2

Outline (2 of 2) Inventory Management at Amazon.Com (1 of 3)

• Probabilistic Models and Safety Stock • Amazon.com started as a “virtual” retailer – no inventory,
no warehouses, no overhead – just computers taking
• Single-Period Model
orders to be filled by others
• Fixed-Period (P) Systems
• Growth has forced Amazon.com to become a world
leader in warehousing and inventory management

3 4

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Inventory Management at Amazon.Com (2 of 3) Inventory Management at Amazon.Com (3 of 3)

1. Each order is assigned by computer to the closest 5. Crates arrive at central point where items are boxed and
distribution center that has the product(s) labeled with new bar code
2. A “flow meister” at each distribution center assigns work 6. Gift wrapping is done by hand at 30 packages per hour
crews
7. Completed boxes are packed, taped, weighed and
3. Technology helps workers pick the correct items from labeled before leaving warehouse in a truck
the shelves with almost no errors
8. Order arrives at customer within 1 - 2 days
4. Items are placed in crates on a conveyor, bar code
scanners scan each item 15 times to virtually eliminate
errors

5 6

Learning Objectives (1 of 2) Learning Objectives (2 of 2)

Conduct an ABC analysis Apply the production order quantity model


Explain and use cycle counting Explain and use the quantity discount model
Explain and use the EOQ model for independent inventory Understand service levels and probabilistic inventory
demand models
Compute a reorder point and explain safety stock

7 8

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Inventory Management Importance of Inventory

The objective of inventory management is to strike a • One of the most expensive assets of many companies
balance between inventory investment and customer representing as much as 50% of total invested capital
service
• Less inventory lowers costs but increases chances of
running out
• More inventory raises costs but always keeps customers
happy

9 10

Functions of Inventory Types of Inventory

1. To provide a selection of goods for anticipated demand • Raw material


and to separate the firm from fluctuations in demand – Purchased but not processed
2. To decouple or separate various parts of the production • Work-in-process (WIP)
process – Undergone some change but not completed
3. To take advantage of quantity discounts – A function of cycle time for a product
• Maintenance/repair/operating (MRO)
4. To hedge against inflation
– Necessary to keep machinery and processes productive
• Finished goods
– Completed product awaiting shipment

11 12

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Figure 12.1 The Material Flow Cycle Managing Inventory

1. How inventory items can be classified (ABC analysis)


2. How accurate inventory records can be maintained

13 14

ABC Analysis (1 of 5) ABC Analysis (2 of 5)

• Divides inventory into three classes based on annual Figure 12.2 Graphic Representation of ABC Analysis
dollar volume
– Class A - high annual dollar volume
– Class B - medium annual dollar volume
– Class C - low annual dollar volume
• Used to establish policies that focus on the few critical
parts and not the many trivial ones

15 16

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ABC Analysis (3 of 5) ABC Analysis (4 of 5)

ABC Calculation • Other criteria than annual dollar volume may be used
– High shortage or holding cost
– Anticipated engineering changes
– Delivery problems
– Quality problems

17 18

ABC Analysis (5 of 5) Record Accuracy (1 of 2)

• Policies employed may include • Accurate records are a critical


ingredient in production and
1. More emphasis on supplier development for A items inventory systems
2. Tighter physical inventory control for A items – Periodic systems require
3. More care in forecasting A items regular checks of inventory
▪ Two-bin system
– Perpetual inventory tracks
receipts and subtractions
on a continuing basis
▪ May be semi-automated

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19 20

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Record Accuracy (2 of 2) Cycle Counting

• Incoming and outgoing record keeping must be accurate • Items are counted and records updated on a periodic
basis
• Stockrooms should be secure
• Often used with ABC analysis
• Necessary to make precise decisions about ordering,
scheduling, and shipping • Has several advantages

1. Eliminates shutdowns and interruptions


2. Eliminates annual inventory adjustment
3. Trained personnel audit inventory accuracy
4. Allows causes of errors to be identified and corrected
5. Maintains accurate inventory records

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21 22

Cycle Counting Example Control of Service Inventories

5,000 items in inventory, 500 A items, 1,750 B items, 2,750 • Can be a critical component of profitability
C items
• Losses may come from shrinkage or pilferage
Policy is to count A items every month (20 working days), B
• Applicable techniques include
items every quarter (60 days), and C items every six
months (120 days) 1. Good personnel selection, training, and discipline
Cycle Counting Number Of Items Counted 2. Tight control of incoming shipments
Item Class Quantity Policy Per Day
A 500 Each month 500 twentieths = 25 a day
3. Effective control of all goods leaving facility
500 20 = 25 day
B 1,750 Each quarter 1,750
1,750sixtieths
60 = 29= day
29 a day.

2,750 one hundred and


C 2,750 Every 6 months 120 = 23= 23
2,750twentieths daya day.

Blank Blank Blank 77 Total


day is 77
aday.

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Inventory Models (1 of 2) Inventory Models (2 of 2)

• Independent demand - the demand for item is • Holding costs - the costs of holding or “carrying”
independent of the demand for any other item in inventory over time
inventory
• Ordering cost - the costs of placing an order and
• Dependent demand - the demand for item is dependent receiving goods
upon the demand for some other item in the inventory
• Setup cost - cost to prepare a machine or process for
manufacturing an order
– May be highly correlated with setup time

25 26

Holding Costs (1 of 2) Holding Costs (2 of 2)


Table 12.1 Determining Inventory Holding Costs Holding costs vary considerably depending on the
Cost (And Range) As A
business, location, and interest rates. Generally greater
Percent Of Inventory than 15%, some high tech and fashion items have holding
Category Value
Housing costs (building rent or depreciation, operating 6% (3 - 10%)
costs greater than 40%.
costs, taxes, insurance)
Material handling costs (equipment lease or depreciation, 3% (1 - 3.5%)
power, operating cost)
Labor cost (receiving, warehousing, security) 3% (3 - 5%)
Investment costs (borrowing costs, taxes, and insurance 11% (6 - 24%)
on inventory)
Pilferage, space, and obsolescence (much higher in 3% (2 - 5%)
industries undergoing rapid change like tablets and smart
phones)
Overall carrying cost 26%

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Inventory Models for Independent Demand Basic EOQ Model

Need to determine when and how much to order Important assumptions


1. Basic economic order quantity (EOQ) model 1. Demand is known, constant, and independent
2. Production order quantity model 2. Lead time is known and constant
3. Quantity discount model 3. Receipt of inventory is instantaneous and complete
4. Quantity discounts are not possible
5. Only variable costs are setup (or ordering) and holding
6. Stockouts can be completely avoided

29 30

Figure 12.3 Inventory Usage over Time Minimizing Costs (1 of 6)

Objective is to minimize total costs

Figure 12.4 Costs as a Function of Order Quantity: Total costs

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Minimizing Costs (2 of 6) Minimizing Costs (3 of 6)

• By minimizing the sum of setup (or ordering) and holding Q = Number of units per order
costs, total costs are minimized Q* = Optimal number of units per order (EOQ)
• Optimal order size Q* will minimize total cost D = Annual demand in units for the inventory item

• A reduction in either cost reduces the total cost S = Setup or ordering cost for each order

• Optimal order quantity occurs when holding cost and H = Holding or carrying cost per unit per year
setup cost are equal Annual setup cost = (Number of orders placed per year) ×
(Setup or order cost per order)
 Annual demand   Setup or order 
=   
 Number of units in each order   cost per order 
D
=  S
Q 

33 34

Minimizing Costs (4 of 6) Minimizing Costs (5 of 6)


Q = Number of pieces per order Annual holding cost = (Average inventory level) ×
Q* = Optimal number of pieces per order (EOQ) (Holding cost per unit per year)
D = Annual demand in units for the inventory item  Order quantity 
=  (Holding cost per unit per year)
S = Setup or ordering cost for each order  2 
Q 
H = Holding or carrying cost per unit per year Annual holding cost =   H
2
Annual setup cost = (Number of orders placed per year) ×
(Setup or order cost per order)
 Annual demand   Setup or order 
=  
 Number of units in each order   cost per order 
D
Annual setup cost =   S
Q 

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Minimizing Costs (6 of 6) An EOQ Example (1 of 6)

Optimal order quantity is found when annual setup cost Determine optimal number of needles to order
equals annual holding cost
D = 1,000 units
Solving for Q* 2DS = Q 2H S = $10 per order
2DS
Q =2 H = $.50 per unit per year
H
D Q  Q* =
2DS 2 DS
 S =  H Q* =
Q  2 H H

2(1,000)(10)
Q* = = 40,000 = 200 units
0.50

37 38

An EOQ Example (2 of 6) An EOQ Example (3 of 6)


Determine expected number of orders Determine optimal time between orders

D = 1,000 units D = 1,000 units

S = $10 per order S = $10 per order

I = $.50 per unit per year I = $.50 per unit per year
Q* =200 units
Q* =200 units
N = 5 orders/year
Demand D
Expected number of = N = =
Order quantity Q  Expected time between orders = T =
Number of working days per year
orders Expected number of orders
1,000
N= = 5 orders per year 250
200 T= = 50 days between orders
5
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39 40

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An EOQ Example (4 of 6) The EOQ Model


Determine the total annual cost When including actual cost of material P
D = 1,000 units Q* = 200 units Total annual cost = Setup cost + Holding cost + Product
S = $10 per order N = 5 orders/year cost
H = $.50 per unit per year T = 50 days
D Q
Total annual cost = Setup cost + Holding cost TC = S + H + PD
Q 2
D Q
TC = S+ H
Q 2
1,000 200
= ( $20 ) + ( $.50 )
200 2
= (5)($10) + (100)($.50)
= $50 + $50 = $100
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41 42

Robust Model An EOQ Example (5 of 6)

• The EOQ model is robust Determine optimal number of needles to order

• It works even if all parameters and assumptions are D = 1,000 units Q1,000 = 200 units
not met S = $10 per order T = 50 days
• The total cost curve is relatively flat in the area of the H = $.50 per unit per year Q1,500 = 244.9 units
EOQ N = 5 orders/year

Ordering old Q* Ordering new Q*


D Q
TC = S+ H 1,500 244.9
Q 2 = ($10) + ($.50)
244.9 2
1,500 200
= ($10) + ($.50) = 6.125($10) + 122.45($.50)
200 2
= $75 + $50 = $125 = $61.25 + $61.22 = $122.47
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43 44

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An EOQ Example (6 of 6) Reorder Points

Only 2% less than the total cost of $125 when the order • EOQ answers the “how much” question
quantity was 200
• The reorder point (ROP) tells “when” to order
• Lead time (L) is the time between placing and receiving
an order
 Demand  Lead time for a new 
ROP =  
 per day  order in days 

ROP = d x L
D
d=
Number of working days in a year

45 46

Reorder Point Curve Reorder Point Example


Figure 12.5 The Reorder Point (ROP) Demand = 8,000 iPhones per year
250 working day year
Lead time for orders is 3 working days, may take 4

D
d=
Number of working days in a year
= 8,000 / 250 = 32 units
ROP = d x L
= 32 units per day × 3 days = 96 units
= 32 units per day × 4 days = 128 units
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Production Order Quantity Model (1 of 5) Production Order Quantity Model (2 of 5)

1. Used when inventory builds up over a period of time Q = Number of units per order p = Daily production
after an order is placed H = Holding cost per unit per year d = Daily demand/usage rate
2. Used when units are produced and sold simultaneously t = Length of the production run in
days

( Annual
holding cost ) (perHolding
unit per year )
Figure 12.6 Change in Inventory Levels over Time for the
Production Model
inventory = ( Average inventory level )  cost

( Annuallevel
inventory
) = (Maximum inventory level ) / 2

( Maximum
inventory level
) (
= Total produced during
the production run
) (− Total used during
the production run
)
= pt − dt
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49 50

Production Order Quantity Model (3 of 5) Production Order Quantity Model (4 of 5)

( Maximum
)(= Total produced during
) ( − Total used during
) Setup cost = ( D / Q )S

( )
inventory level the production run the production run 1

HQ 1 −
d
= pt − dt Holding cost =
2  p
However, Q = total produced = pt ; thus t = Q/p
D 1   d 
 Maximum  Q  Q   d
S =
  p 
HQ 1 −
 inventory level  = p  p  − d  p  = Q  1 − p  Q 2
       
2 DS 2 DS
 d  
2
Maximum inventory level Q Q = Q *p =
  d    d 
Holding cost = (H ) =
1 −  p  H 
2 
H 1−  
  p  H 1−  
2   p 
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51 52

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Production Order Quantity Example Production Order Quantity Model (5 of 5)

D = 1,000 units Note:


S = $10 D 1,000
d = 4 = =
Number of days the plant is in operation 250
H = $0.50 per unit per year * 2 DS
Q =
p = 8 units per day
p H
1− d p ( )
When annual data are used the equation becomes:
d = 4 units per day *= 2(1,000)(10)
Qp

0.50 1−( 4 8)  * 2 DS
Qp =

=
20,000
=
 Annual demand rate 
80,000 H 1 − 
0.50(1 2)
 Annual production rate 
= 282.8 hubcaps, or 283 hubcaps
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53 54

Quantity Discount Models (1 of 4) Quantity Discount Models (2 of 4)

• Reduced prices are often available when larger quantities Total annual cost = Setup cost + Holding cost + Product cost
are purchased D Q
TC = S+ IP + PD
• Trade-off is between reduced product cost and increased Q 2
holding cost
where Q = Quantity ordered P = Price per unit
Table 12.2 A Quantity Discount Schedule D = Annual demand in units I = Holding cost per unit per
S = Ordering or setup cost per year expressed as a
order percent of price P
Price Range Quantity Ordered Price Per Unit P
Initial price 0 to 119 $100 2DS
Q* =
Discount price 1 200 to 1,499 $ 98 IP
Discount price 2 1,500 and over $ 96 Because unit price varies, holding cost is expressed as a percent (I) of
unit price (P)
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55 56

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Quantity Discount Models (3 of 4) Quantity Discount Models (4 of 4)

Steps in analyzing a quantity discount Figure 12.7 EOQs and Possible Best Order Quantities for the Quantity
Discount Problem with Three Prices in Table 12.2 (see slide 55)

1. Starting with the lowest possible purchase price,


calculate Q* until the first feasible EOQ is found. This is
a possible best order quantity, along with all price-break
quantities for all lower prices.
2. Calculate the total annual cost for each possible order
quantity determined in Step 1. Select the quantity that
gives the lowest total cost.

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57 58

Quantity Discount Example (1 of 2) Quantity Discount Example (2 of 2)

Calculate Q* for every discount Table 12.3 Total Cost Computations for Chris Beehner Electronics
2DS
starting with the lowest price Q* =
IP
2(5,200)($200) Annual Annual Annual
Q$96 * = = 278 drones / order Order Unit Ordering Holding Product Total Annual
(.28)($96) Quantity Price Cost Cost Cost Cost
275 $98 $3,782 $3,773 $509,600 $517,155
Infeasible – calculate Q* for next-higher price 1,500 $96 $693 $20,160 $499,200 $520,053

2(5,200)($200)
Q$98 * = = 275 drones / order Choose the price and quantity that gives the lowest total cost
(.28)($98) Buy 275 drones at $98 per unit
Feasible

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Quantity Discount Variations Probabilistic Models and Safety Stock

• All-units discount is the most popular form • Used when demand is not constant or certain
• Incremental quantity discounts apply only to those • Use safety stock to achieve a desired service level and
units purchased beyond the price break quantity avoid stockouts
• Fixed fees may encourage larger purchases
ROP = d × L + ss
• Aggregation over items or time
• Truckload discounts, buy-one-get-one-free offers,
Annual stockout costs = The sum of the units short for each
one-time-only sales
demand level × The probability of that demand level × The
stockout cost/unit × The number of orders per year

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61 62

Safety Stock Example (1 of 2) Safety Stock Example (2 of 2)

ROP = 50 units Safety Additional Total


Stock Holding Cost Stockout Cost Cost
Orders per year = 6
20 20(20)($5)
times $5 == $100
$100 $0 $100
Stockout cost = $40 per frame 10 (10)(.1)
10 = ($40)(6)
times 0.1 = $240
times $40 times 6 = $240. $290
10(10)($5)
times $5 == $ .50
$50

Carrying cost = $5 per frame per year 0 $0


10(10)(.2)($40)(6) + (20)(.1)($40)(6)
times 0.2 times $40 = 0.1
times 6 + 20 times $960 $960
times $40 times 6 = $960.
Number Of Units Probability
30 .2
40 .2
A safety stock of 20 frames gives the lowest total cost
ROP → 50 .3
ROP = 50 + 20 = 70 frames
60 .2
70 .1
Blank 1.0

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Probabilistic Demand (1 of 3) Probabilistic Demand (2 of 3)


Figure 12.8 Probabilistic Demand for a Hospital Item Use prescribed service levels to set safety stock when the
cost of stockouts cannot be determined

ROP = demand during lead time + Z dLT

Where Z = Number of standard deviations


 dLT = Standard deviation of demand during lead time

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65 66

Probabilistic Demand (3 of 3) Probabilistic Example (1 of 4)


 = Average demand = 350 kits

 dLT = Standard deviation of demand during lead time =


10 kits

Stockout policy = 5% (service level = 95%)

Using Appendix I, for an area under the curve of 95%, the Z = 1.645

Safety stock = Z dLT = 1.645(10) = 16.5 kits

Reorder point = Expected demand during lead time + Safety stock


= 350 kits + 16.5 kits of safety stock
= 366.5 or 367 kits
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Other Probabilistic Models (1 of 4) Other Probabilistic Models (2 of 4)

• When data on demand during lead time is not available, Demand is variable and lead time is constant
there are other models available
ROP = (Average daily demand × Lead
1. When demand is variable and lead time is constant time in days) + Z dLT
2. When lead time is variable and demand is constant
3. When both demand and lead time are variable
where  dLT =  d Lead time
 d = Standard deviation of demand per day

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69 70

Probabilistic Example (2 of 4) Other Probabilistic Models (3 of 4)


Average daily demand (normally distributed) = 15 Lead time is variable and demand is constant
Lead time in days (constant) = 2
ROP = (Daily demand × Average lead time
Standard deviation of daily demand = 5 in days) + Z × (Daily demand) ×  LT
Service level = 90% where  LT = Standard deviation of lead time in days
Z for 90% = 1.28 From Appendix I

ROP = (15 units  2 days ) + Z dLT


= 30 + 1.28 ( 5 ) ( 2)
= 30 + 9.02 = 39.02  39

Safety stock is about 9 computers


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Probabilistic Example (3 of 4) Other Probabilistic Models (4 of 4)

Daily demand (constant) = 10 Both demand and lead time are variable
Average lead time = 6 days ROP = (Average daily demand
× Average lead time) + Z dLT
Standard deviation of lead time =  LT = 1

Service level = 98%, so Z (from Appendix I) = 2.055 where  d = Standard deviation of demand per day
ROP = (10 units × 6 days) + 2.055(10 units)(1)  LT = Standard deviation of lead time in days

= 60 + 20.55 = 80.55 ( Average lead time   d 2 )


 dLT =
Reorder point is about 81 cameras
+ Average daily demand) 2 2LT

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73 74

Probabilistic Example (4 of 4) Single-Period Model


Average daily demand (normally distributed) = 150 • Only one order is placed for a product
Standard deviation =  d = 16 • Units have little or no value at the end of the sales period
Average lead time 5 days (normally distributed)
Cs = Cost of shortage = Sales price/unit – Cost/unit
Standard deviation =  LT = 1 day
Co = Cost of overage = Cost/unit – Salvage value
Service level = 95%, so Z = 1.645 (from Appendix I)

ROP = (150 packs  5 days) + 1.645 dLT


Cs
 dLT = ( 5 days 16 ) + (150
2 2
 12 ) = ( 5  256 ) + ( 22,500 1) Service level =
Cs + Co
= (1,280 ) + ( 22,500 ) = 23,780  154
ROP = (150  5) + 1.645(154)  750 + 253 = 1,003 packs
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Single-Period Example (1 of 2) Single-Period Example (2 of 2)


Average demand =  = 120 papers / day From Appendix I, for the area .579, Z  .20

Standard deviation =  = 15 papers The optimal stocking level

Cs = cost of shortage = $1.25 – $.70 = $.55 = 120 copies + (.20) ( )


Co = cost of overage = $.70 – $.30 = $.40
= 120 + (.20)(15) = 120 + 3 = 123 papers
Cs
Service level =
Cs + Co The stockout risk = 1 − Service level
.55
=
.55 + .40 = 1 − .579 = .422 = 42.2%
.55
= =.579
.95
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77 78

Fixed-Period (P) Systems (1 of 3) Fixed-Period (P) Systems (2 of 3)

• Fixed-quantity models require continuous monitoring • Inventory counted only at end of period
using perpetual inventory systems
• Order brings inventory up to target level
• In fixed-period systems orders placed at the end of a – Only relevant costs are ordering and holding
fixed period – Lead times are known and constant
• Periodic review, P system – Items are independent of one another

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Fixed-Period (P) Systems (3 of 3) Fixed-Period Systems


Figure 12.9 Inventory Level in a Fixed-Period (P) System • Inventory is only counted at each review period
• May be scheduled at convenient times
• Appropriate in routine situations
• May result in stockouts between periods
• May require increased safety stock

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