1
Inventory Management
and
Control
2
Inventory
Defined
• Inventory is the stock of any item or resource
held to meet future demand and can include: raw
materials, finished products, component parts,
supplies, and work-in-process
3
Inventory Classifications
Inventory
Process
stage
Raw Material
WIP
Finished Goods
Number
& Value
Demand
Type
A Items
B Items
C Items
Independent
Dependent
Other
Maintenance
Operating
4
Independent vs. Dependent Demand
Independent Demand (Demand for the final end-product
or demand not related to other items; demand created by
external customers)
Finished
product
A
B(4)
D(1)
Component parts
E(1
E(2)
)
Independent demand is uncertain
Dependent demand is certain
Dependent
C(2)
Demand
(Derived demand
for component
B(1)
E(3)
parts,
subassemblies,
raw materials, etcused to produce
final products)
5
Inventory Models
• Independent demand – finished goods, items that
are ready to be sold
– E.g. a computer
• Dependent demand – components of finished
products
– E.g. parts that make up the computer
6
Types of Inventories (1 of 2)
• Raw materials & purchased parts
• Partially completed goods called
work in progress
• Finished-goods inventories
(manufacturing firms)
or merchandise
(retail stores)
7
Types of Inventories (2 of 2)
• Replacement parts, tools, & supplies
• Goods-in-transit to warehouses or customers
8
The Material Flow Cycle (1 of 2)
9
The Material Flow Cycle (2 of 2)
Input
Wait
Time
Move
Time
Queue
Time
Setup
Time
Run
Time
Output
Cycle Time
Run time: Job is at machine and being worked on
Setup time: Job is at the work station, and the work station is
being "setup."
Queue time: Job is where it should be, but is not being
processed because other work precedes it.
Move time: The time a job spends in transit
Wait time: When one process is finished, but the job is waiting
to be moved to the next work area.
10
Performance Measures
• Inventory turnover (the ratio of annual
cost of goods sold to average inventory
investment)
• Days of inventory on hand (expected
number of days of sales that can be supplied
from existing inventory)
11
Functions of Inventory (1 of 2)
1. To “decouple” or separate various parts of the
production process, ie. to maintain
independence of operations
2. To meet unexpected demand & to provide high
levels of customer service
3. To smooth production requirements by meeting
seasonal or cyclical variations in demand
4. To protect against stock-outs
12
Functions of Inventory (2 of 2)
5. To provide a safeguard for variation in raw
material delivery time
6. To provide a stock of goods that will provide a
“selection” for customers
7. To take advantage of economic purchase-order
size
8. To take advantage of quantity discounts
9. To hedge against price increases
13
Disadvantages of Inventory
• Higher costs
– Item cost (if purchased)
– Holding (or carrying) cost
• Difficult to control
• Hides production problems
• May decrease flexibility
14
Inventory Costs
 Holding (or carrying) costs
 Costs for storage, handling, insurance, etc
 Setup (or production change) costs
 Costs to prepare a machine or process for
manufacturing an order, eg. arranging specific
equipment setups, etc
 Ordering costs (costs of replenishing inventory)
 Costs of placing an order and receiving goods
 Shortage costs
 Costs incurred when demand exceeds supply
15
Holding (Carrying) Costs
•
•
•
•
•
•
•
•
Obsolescence
Insurance
Extra staffing
Interest
Pilferage
Damage
Warehousing
Etc.
16
Inventory Holding Costs
(Approximate Ranges)
Category
Cost as a
% of Inventory Value
Housing costs (building rent, depreciation,
operating cost, taxes, insurance)
6%
(3 - 10%)
Material handling costs (equipment, lease or
depreciation, power, operating cost)
3%
(1 - 3.5%)
Labor cost from extra handling
3%
(3 - 5%)
Investment costs (borrowing costs, taxes,
and insurance on inventory)
11%
(6 - 24%)
Pilferage, scrap, and obsolescence
Overall carrying cost
3%
(2 - 5%)
26%
17
Ordering Costs
•
•
•
•
•
Supplies
Forms
Order processing
Clerical support
etc.
18
Setup Costs
•
•
•
•
Clean-up costs
Re-tooling costs
Adjustment costs
etc.
19
Shortage Costs
• Backordering cost
• Cost of lost sales
20
Inventory Control System
Defined
An inventory system is the set of policies and
controls that monitor levels of inventory and
determine what levels should be maintained,
when stock should be replenished and how large
orders should be
Answers questions as:
 When to order?
 How much to order?
21
Objective of Inventory Control
To achieve satisfactory levels of customer
service while keeping inventory costs within
reasonable bounds
Improve the Level of customer service
Reduce the Costs of ordering and carrying
inventory
22
Requirements of an 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
23
Inventory Counting (Control) Systems
• Periodic System
Physical count of items made at periodic intervals;
order is placed for a variable amount after fixed
passage of time.
• Perpetual (Continuous) Inventory System
System that keeps track
of removals from inventory
continuously, thus
monitoring current levels of
each item (constant amount is
ordered when inventory
declines to a predetermined level)
24
Inventory Models
 Single-Period Inventory Model
 One time purchasing decision (Examples: selling tshirts at a football game, newspapers, fresh
bakery products, fresh flowers)
 Seeks to balance the costs of inventory over stock
and under stock
 Multi-Period Inventory Models
 Fixed-Order Quantity Models
• Event triggered (Example: running out of stock)
 Fixed-Time Period Models
• Time triggered (Example: Monthly sales call by
25
Single-Period Inventory Model
26
Single-Period Inventory Model
• In a single-period model, items are received in the
beginning of a period and sold during the same
period. The unsold items are not carried over to
the next period.
• The unsold items may be a total waste, or sold at a
reduced price, or returned to the producer at some
price less than the original purchase price.
• The revenue generated by the unsold items is
called the salvage value.
3
(Newsboy Problem)
•Single period model: It is used to
handle ordering of perishables
(fresh fruits, flowers) and other
items with limited useful lives
(newspapers, spare parts for
specialized equipment).
27
28
Shortage cost (Cost of
Understocking)
• Shortage cost: generally, this cost
represents unrealized profit per unit
(Cu=Revenue per unit – Cost per
unit)
• If a shortage or stockout cost relates to
a spare part for a machine, then
shortage cost refers to the actual cost
of lost production.
29
Excess cost (Cost of Over
Stocking)
• Excess cost (Ce): difference
between purchase cost and salvage
value of items left over at the end
of a period.
• If there is a cost associated with
disposing of excess items, the
salvage cost will be negative.
30
Single Period Model
Given the costs of
overestimating/underestimating demand
and the probabilities of various demand
sizes the goal is to identify the order
quantity or stocking level that will
minimize the long-run excess
(overstock)or shortage costs
(understock).
31
Single-Period Models (Demand Distribution)
Demand may be discrete or continuous. The demand
of computer, newspaper, etc. is usually an integer.
Such a demand is discrete. On the other hand, the
demand of gasoline is not restricted to integers. Such a
demand is continuous. Often, the demand of
perishable food items such as fish or meat may also be
continuous.
• Consider an order quantity Q
• Let p
= probability (demand<Q)
= probability of not selling the Qth item.
• So, (1-p) = probability of selling the Qth item.
32
Single-Period Models (Discrete Demand)
• Expected loss from the Qth item = pce
• Expected profit from the Qth item = 1  p cs
• So, the Qth item should be ordered if pce  (1  p )cs
cs
or , p 
ce  cs
• Decision Rule (Discrete Demand):
– Order maximum quantity Q such that
cs
p
ce  cs
where p = probability (demand<Q)
33
Single-Period Model
Cs
P
Cs  Ce
The service level is the
probability that demand will
not exceed the stocking level.
The service level determines
the amount of stocking level
to keep.
Where :
Ce  Cost per unit of demand over estimated
Cs  Cost per unit of demand under estimated
P  Probabilit y that the unit will be sold
34
Optimal Stocking Level (Choosing optimum
Stocking level to minimize these costs is similar to
balancing a
Service level =
Cs
Cs + Ce
seesaw)
Cs = Shortage cost per unit
Ce = Excess cost per unit
Ce
Cs
Service Level
Quantity
So
Balance point
35
Service Level
Another way to define ‘Service
Level’ is:
• proportion of cycles in which no
stock-out occurs
36
Service Level
Order Cycle
1
2
3
4
5
6
7
8
9
10
Total
Demand
180
75
235
140
180
200
150
90
160
40
1450
Stock-Outs
0
0
45
0
0
10
0
0
0
0
55
Since there are two cycles out of ten in which a stockout
occurs, service level is 80%. This translates to a 96% fill rate.
There are a total of 1,450 units demand and 55 stockouts (which
means that 1,395 units of demand are satisfied).
37
Single Period Model
(Demand is represented by a discrete distribution)
• Unlike the continuous case where the optimal
solution is found by determining So which makes
the distribution function equal to the critical ratio
cs / (cs + ce), in the discrete case, the critical ratio
takes place between two values of F(So) or F(Q)
• The optimal So or Q corresponds to the higher
value of F( So) or F(Q).
(Note that, in the discrete case, the distribution
function increases by jumps)
SEE EXAMPLES 17 & 18 on page 576
38
Single-Period Models (Discrete Demand)
Example : Demand for cookies:
Demand
Probability of Demand
1,800 dozen
0.05
2,000
0.10
2,200
0.20
2,400
0.30
2,600
0.20
2,800
0.10
3,000
0,05
Selling price=$0.69, cost=$0.49, salvage value=$0.29
What is the optimal number of cookies to make?
c
39
Single-Period Models (Discrete Demand)
Cs= 0.69-0.49=$0.2, Ce= 0.49-0.29=$0.2
Order maximum quantity, Q such that
cs
p  Probabilit ydemand  Q  
 0.5
cs  ce
Demand, Q Probability(demand) Probability(demand<Q), p
1,800 dozen
0.05
0.05
2,000
0.10
0.15
2,200
0.20
0.35
2,400
0.30
0.65
2,600
0.20
0.85
2,800
0.10
0.95
3,000
0,05
1.00
40
Single-Period Models (Continous Demand)
• Often the demand is continuous. Even when
the demand is not continuous, continuous
distribution may be used because the
discrete distribution may be inconvenient.
• We shall discuss two distributions:
Uniform distribution
Normal distribution
41
Single-Period Models (Continuous Demand)
Example 2: The J&B Card Shop sells calendars. The oncea-year order for each year’s calendar arrives in
September. The calendars cost $1.50 and J&B sells them
for $3 each. At the end of July, J&B reduces the calendar
price to $1 and can sell all the surplus calendars at this
price. How many calendars should J&B order if the
September-to-July demand can be approximated by
a. uniform distribution between 150 and 850
42
Single-Period Models (Continuous Demand)
Overage cost
ce = Purchase price - Salvage value =1.5-1=$0.5
Underage cost
cs = Selling price - Purchase price =3-1.5=$1.5
43
Single-Period Models (Continuous Demand)
p
cs
ce  c s
=0.75
Now, find the Q so that p = probability(demand<Q) =0.75
Probability
Q* = a+p(b-a) =150+0.75(850-150)=675
Area
=
Area =
150
Demand
Q*
850
44
Single-Period Models (Continuous Demand)
Example 3: The J&B Card Shop sells calendars. The oncea-year order for each year’s calendar arrives in
September. The calendars cost $1.50 and J&B sells them
for $3 each. At the end of July, J&B reduces the calendar
price to $1 and can sell all the surplus calendars at this
price. How many calendars should J&B order if the
September-to-July demand can be approximated by
b. normal distribution with  = 500 and =120.
45
Single-Period Models (Continuous Demand)
Solution to Example 3: ce =$0.50, cu =$1.50 (see Example 2)
p
cs
1.50
=
= 0.75
ce  c s
1.50  0.50
46
Single-Period Models (Continuous Demand)
Now, find the Q so that p = 0.75
47
Single-Period Models
(Continuous Demand
Q= So = mean + zσ
= 500 + .68(120)
= 582
48
Single Period Example 15 (pg. 574)
Demand is uniformly distributed
•
•
•
•
•
Ce = $0.20 per unit
Cs = $0.60 per unit
Service level = Cs/(Cs+Ce) = .6/(.6+.2)
Service level = .75
Opt. Stock.Level=S0=300+.75(500-300)= 450 liters
Ce
Cs
Service Level = 75%
Quantity
Stockout risk = 1.00 – 0.75 = 0.25
49
Uniform Distribution
[Continuous Dist’n]
• A random variable X is uniformly distributed on the interval
(a,b), U(a,b), if its pdf and cdf are:
 1

, a xb
f ( x)   b  a
0,
otherwise
xa
0,
x a
F ( x)  
, a xb
b  a
xb
1,
• Properties
– P(x1 < X < x2) is proportional to the length of the interval [F(x2) –
F(x1) = (x2-x1)/(b-a)]
– E(X) = (a+b)/2
V(X) = (b-a)2/12
• U(0,1) provides the means to generate random numbers, from
which random variates can be generated.
49
50
Poisson Distribution
[Discrete Dist’n]
• Poisson distribution describes many random processes quite
well and is mathematically quite simple.
– where a > 0, pdf and cdf are:
 e a a x

p( x)   x! , x  0,1,...
0,
otherwise
e a a i
F ( x)  
i!
i 0
x
– E(X) = a = V(X)
50
51
Normal Distribution
[Continuous
Dist’n]
• A normally distributed random variable X has the pdf:
 1  x   2 
1
f ( x) 
exp  
 ,    x  
2

 2
 
 
    
– Mean:
2

0
– Variance:
– Denoted as X ~ N(,2)
• Special properties:
.
– symmetric about .
– The maximum value of the pdf occurs at x = ; the mean and mode are
equal.
51
52
Normal Distribution
[Continuous
Dist’n]
• Evaluating the distribution:
– Use numerical methods (no closed form)
– Independent of  and , using the standard normal distribution:
Z ~ N(0,1)
– Transformation of variables: let Z = (X - ) / ,
x 

F ( x )  P  X  x   P Z 

 

( x ) /
1 z2 / 2

e
dz

2

( x ) /

 ( z )dz   ( x  )
, where ( z )  
z

1 t 2 / 2
e
dt
2
52
53
Normal Distribution
[Continuous
Dist’n]
• Example: The time required to load an oceangoing vessel, X, is
distributed as N(12,4)
– The probability that the vessel is loaded in less than 10 hours:
 10  12 
F (10)  
   (1)  0.1587
 2 
• Using the symmetry property, (1) is the complement of  (-1)
53
54
Single Period Model
(Demand is represented by a continous distribution)
Our college basketball team is playing in a tournament
game this weekend. Based on our past experience we
sell on average 2,400 shirts with a standard deviation
of 350 and we can assume that demand for shirts is
approximately normally distributed. We make $10
on every shirt we sell at the game, but lose $5 on
every shirt not sold. What is the optimal stocking
level for shirts?
So =mean + zσ
Cs = $10 and Ce = $5; P ≤ $10 / ($10 + $5) = .667
Z.667 = .432
therefore we need 2,400 + .432(350) = 2,551 shirts
55
Multi-Period Inventory Models
Fixed-Order Quantity Models (Types of)
Economic Order Quantity Model (EOQ)
Economic Production Order Quantity (Economic
Lot Size) Model (EPQ)
Economic Order Quantity Model with Quantity
Discounts
Fixed Time Period (Fixed Order Interval) Models
56
Fixed Order Quantity Models:
Economic Order Quantity Model
57
Economic Order Quantity Model
Assumptions (1 of 2):
• Demand for the product is known with certainty, it is
constant and uniform throughout the period
• Lead time (time from ordering to receipt) is known and
constant
• Price per unit of product is constant (no quantity
discounts). So it is not included in the total cost.
• Inventory holding cost is based on average inventory
58
Economic Order Quantity Model
Assumptions (2 of 2):
• Ordering or setup costs are constant
• All demands for the product will be satisfied (no
backorders are allowed)
• No stockouts (shortages) are allowed
• The order quantity is received all at once.
(Instantaneous receipt of material in a single
lot)
The goal is to calculate the order quantitiy that
minimizes total cost
59
Basic Fixed-Order Quantity Model and
Reorder Point Behavior
1. You receive an order quantity Q.
Number
of units
on hand
(Inv.
Level)
Q
Q
4. The cycle then repeats.
Q
R
2. You start using
them up over time.
L
R = Reorder point
Q = Economic order quantity
L = Lead time
Time
L
3. When you reach down to
a level of inventory of R,
you place your next Q
sized order.
60
EOQ Model
Inventory Level
Order
Quantity
(Q)
Average
Inventory
(Q/2)
Demand
rate
Reorder
Point
(ROP)
Order placed
Lead Time
Order received
Time
61
EOQ Cost Model: How Much to Order?
By adding the holding and ordering costs together, we
determine the total cost curve, which in turn is used to find
the optimal order quantity that minimizes total costs
Annual
cost ($)
Total Cost
Slope = 0
HQ
Carrying Cost =
2
Minimum
total cost
SD
Ordering Cost = Q
Optimal order
Qopt
Order Quantity, Q
62
Why Holding Costs Increase?
• More units must be stored if more are
ordered
Purchase Order
Description
Qty.
Microwave
1
Order quantity
Purchase Order
Description
Qty.
Microwave
1000
Order quantity
63
Why Ordering Costs Decrease ?
Cost is spread over more units
Example: You need 1000 microwave ovens
1000 Order (Postage $ 0.33)
1 Order (Postage $330)
Purchase Order
Description
Qty.
Microwave
1000
PurchaseOrder
Order
Purchase
PurchaseOrder
OrderQty.
Description
Purchase
Description Qty.
Qty.
Description
Microwave Qty. 11
Description
Microwave
Microwave
Microwave
11
Order quantity
64
Basic Fixed-Order Quantity (EOQ)
TC=Total annual
cost
Model Formula
D =Annual demand
Total
Annual =
Cost
Annual
Annual
Annual
Purchase + Ordering + Holding
Cost
Cost
Cost
D
Q
TC = DC + S + H
Q
2
C =Cost per unit
Q =Order quantity
S =Cost of placing
an order or setup
cost
R =Reorder point
L =Lead time
H=Annual holding
and storage cost
per unit of inventory
65
EOQ Cost Model
Using calculus, we take the first derivative of the total cost
function with respect to Q, and set the derivative (slope) equal to
zero, solving for the optimized (cost minimized) value of Qopt
SD
Annual ordering cost =
Q
SD
HQ
TC =
+
HQ
Q
2
Annual carrying cost =
2
SD
H
TC
=- 2 +
Q
2
SD
HQ
Q
Total cost =
+
Q
2
SD
H
0 =- 2 +
Q
2
Deriving Qopt
Qopt =
2SD
H
Proving equality
of costs at
optimal point
SD HQ
=
Q
2
Q2
Qopt =
2S D
=
H
2SD
H
66
Deriving the EOQ
1) How much to order?
Q OPT =
2DS
=
H
2(Annual D em and)(Order or Setup Cost)
Annual Holding Cost
2) When to order?
_
We also need a
reorder point to tell
us when to place an
order
R eo rd er p o in t, R = d L
_
d = average daily demand (constant)
L = Lead time (constant)
67
EOQ Model Equations
2 ×D ×S
H
Expected Number of Orders = N = D
Q*
Optimal Order Quantity = Q* =
Expected Time Between Orders
d =
D
Working Days / Year
ROP = d × L
=T =
Working Days / Year
N
68
EOQ Example 1 (1 of 3)
Given the information below, what are the EOQ and
reorder point?
Annual Demand = 1,000 units
Days per year considered in average daily demand = 365
Cost to place an order = $10
Holding cost per unit per year = $2.50
Lead time = 7 days
Cost per unit = $15
69
EOQ Example 1(2 of 3)
Q O PT =
d =
2D S
=
H
2(1,000 )(10)
= 89.443 units or 90 u n its
2.50
1,000 units / year
= 2.74 units / day
365 days / year
_
R eo rd er p o in t, R = d L = 2 .7 4 u n its / d ay (7 d ays) = 1 9 .1 8 o r 2 0 u n its
In summary, you place an optimal order of 90 units. In the
course of using the units to meet demand, place the next
order of 90 units when you only have 20 units left.
70
EOQ Example I(3 of 3)
Orders per year = D/Qopt
= 1000/90
= 11 orders/year
TCmin =
TCmin =
SD
Q
+
(10)(1,000)
90
TCmin = $ 111 + $111 = 22 $
Order cycle time= 365/(D/Qopt)
= 365/11
= 33.1days
HQ
2
+
(2,5)(90)
2
71
EOQ Example 2(1 of 2)
Determine the economic order quantity
and the reorder point given the following…
Annual Demand = 10,000 units
Days per year considered in average daily demand =
365
Cost to place an order = $10
Holding cost per unit per year = 10% of cost per unit
Lead time = 10 days
Cost per unit = $15
72
EOQ Example 2(2 of 2)
Q OPT =
2D S
=
H
2 (1 0 ,0 0 0 )(1 0 )
= 3 6 5 .1 4 8 u n its, o r 3 6 6 u n its
1 .5 0
10,000 units / year
d=
= 27.397 units / day
365 days / year
_
R = d L = 27.397 units / day (10 days) = 273.97 or 274 u n its
Place an order for 366 units. When in the course of
using the inventory you are left with only 274 units,
place the next order of 366 units.
73
EOQ Example 3
H = $0.75 per yard
Qopt =
2SD
H
Qopt =
2(150)(10,000)
(0.75)
Qopt = 2,000 yards
S = $150
D = 10,000 yards
SD
HQ
TCmin =
+
Q
2
TCmin
(150)(10,000) (0.75)(2,000)
=
+
2,000
2
TCmin = $750 + $750 = $1,500
Orders per year = D/Qopt Order cycle time =311 days/(D/Qopt)
= 10,000/2,000
= 311/5
= 5 orders/year
= 62.2 store days
74
When to Reorder with EOQ Ordering ?
• Reorder Point – is the level of inventory at which a
new order is placed
ROP = d . L
• Safety Stock - Stock that is held in excess of
expected demand due to variable demand rate
and/or lead time.
• Service Level - Probability that demand will not
exceed supply during lead time (probability that
inventory available during the lead time will meet
the demand) 1 - Probability of stockout
75
Reorder Point Example
Demand = 10,000 yards/year
Store open 311 days/year
Daily demand = 10,000 / 311 = 32.154 yards/day
Lead time = L = 10 days
R = dL = (32.154)(10) = 321.54 yards
76
Determinants of the Reorder Point
•
•
•
•
The rate of demand
The lead time
Demand and/or lead time variability
Stockout risk (safety stock)
77
Probabilistic Models
Answer how much & when to order
Allow demand and lead time to vary
 Follows normal distribution
 Other EOQ assumptions apply
Consider service level & safety stock
 Service level = 1 - Probability of stockout
 Higher service level means more safety stock
 More safety stock means higher ROP
78
Quantity
Safety Stock
Maximum probable demand
during lead time
Expected demand
during lead time
ROP
Safety stock
Safety stock reduces risk of
stockout during lead time
LT
Time
79
Reorder Point With Variable Demand
Inventory level
Q
Reorder
point, R
0
LT
LT
Time
80
Inventory level
Reorder Point with a Safety Stock
Q
Reorder
point, R
Safety Stock
0
LT
LT
Time
81
Reorder Point With Variable Demand and
Constant Lead Time
R = dL + zd L
where
d = average daily demand
L = lead time
d = the standard deviation of daily demand
z = number of standard deviations
corresponding to the service level
probability
zd L = safety stock
82
Reorder Point for Service Level
Probability of
meeting demand during
lead time = service level
Probability of
a stockout
Safety stock
zd L
dL
Expected Demand
R
The reorder point based on a normal distribution of LT demand
83
Reorder Point for Variable Demand
(Example)
The carpet store wants a reorder point with a
95% service level and a 5% stockout probability
d = 30 yards per day, (demand is normally distributed)
d = 5 yards per day
L= 10 days
For a 95% service level, z = 1.65
R = dL + z d L
Safety stock = z d L
= 30(10) + (1.65)(5)( 10)
= (1.65)(5)( 10)
= 326.1 yards
= 26.1 yards
84
Shortages and Service Levels
It is also important to specify:
1) Expected number of units short per order cycle
E(n) =E(z) σdLT
where E(z) is standardized number of units short obtained
from Table 12.3, pg. 569.
2) Expected number of units short per year
E(N) =E (n) (D/Q)
3) Annual Service Level
SLannual = 1- E(N)/D that is percentage of demand filled
directly from inventory, known also as FILL RATE.
85
Example 10 (pg. 568)– shortages and service
levels
 Suppose standard deviation of lead time demand is known to be
20 units. Lead time demand is approximately normal.
(a) For lead time service level of 90 percent, determine the
expected number of units short for any other cycle.
(b) What lead time service level would imply an expected
shortage of 2 units?
85
86
Answer – shortage and service levels
(a) For lead time service level of 90 percent, determine the expected
number of units short for any other cycle.
 σdLT= 20 units
 lead time service level is 0.90 from z table (lead time), E(z)= 0.048
(page 569, table 12.3)
 E(n) =E(z) σdLT = (0.048) (20)= 0.96 or about 1 unit.
(b) What lead time service level would an expected shortage of 2
units imply?
 E(n) = 2
 E(n) =E(z) σdLT or E(z) = E(n) / σdLT =(2)/(20)= 0.100 from the
table, lead time service level is 81.06 percent or 81.7%
86
87
Shortages and Service Levels
• Expected number of units short per year
See example 11, page 568
• Annual Service Level
See example 12, page 570
Note that annual service level will usually
be grater than the cycle service level
88
Fixed Order Quantity Models:
-Noninstantaneous ReceiptProduction Order Quantity
(Economic Lot Size)
Model
89
Production Order Quantity Model
Production done in batches or lots
Capacity to produce a part exceeds that part’s
usage or demand rate
Allows partial receipt of material
 Other EOQ assumptions apply
Suited for production environment
 Material produced, used immediately
 Provides production lot size
Lower holding cost than EOQ model
Answers how much to order and when to order
90
POQ Model Inventory Levels (1 of 2)
Inventory Level
Maximum
inventory
level
Demand portion of cycle with
no supply
Production portion of
cycle
Supply Supply
Begins Ends
Time
91
POQ Model Inventory Levels (2 of 2)
Inventory Level
Inventory level with no demand
Production
Portion of
Cycle
Q*
Supply Supply
Begins Ends
Max. Inventory
Q/p·(p- u)
Average
inventory
Q/2(1- u/p)
Demand portion of
cycle with no supply
Time
92
POQ Model Equations
Maximum inventory level
Setup Cost
Holding Cost
=
D
Q
= Q*
(
1 -
u
p
)
* S
= 1/2 * H * Q
( )
1-
u
p
D = Demand per year
S = Setup cost
H = Holding cost
d = Demand per day
p = Production per day
93
Production Order Quantity Example
(1 of 2)
H = $0.75 per yard
S = $150
u = 10,000/311 = 32.2 yards per day
2SD
POQopt =
H 1- u
p
SD HQ
u
TC = Q + 2 1 - p
D = 10,000 yards
p = 150 yards per day
2(150)(10,000)
=
32.2
0.75 1 150
= 2,256.8 yards
= $1,329
2,256.8
Q
Production run =
=
= 15.05 days per order
150
p
94
Production Quantity Example
(2 of 2)
H = $0.75 per yard
S = $150
u= 10,000/311 = 32.2 yards per day
D = 10,000 yards
p = 150 yards per day
2CoD
10,000
2(150)(10,000)
D
Number of production runs =
=
= 4.43 runs/year
2,256.8 = 2,256.8 yards
Q
Qopt =
=
32.2
Cc 1 - d
0.75 1 150
p
u
32.2
Maximum inventory level = Q 1 = 2,256.8 1 p
150
CoD CcQ
d
= 1,772 yards
TC = Q + 2 1 - p = $1,329
2,256.8
Q
Production run =
=
= 15.05 days per order
150
p
95
Fixed-Order Quantity Models:
Economic Order Quantity Model
with Quantity Discounts
96
Quantity Discount Model
• Answers how much to order & when to order
• Allows quantity discounts
– Price per unit decreases as order quantity
increases
– Other EOQ assumptions apply
• Trade-off is between lower price & increased
holding cost
Total cost with purchasing cost
SD
iP Q
TC =
+
+ PD
Q
2
Where P: Unit Price
97
Cost
Total Costs with Purchasing Cost
Adding Purchasing cost TC with PD
doesn’t change EOQ
TC without PD
PD
0
EOQ
Quantity
98
Quantity Discount Models
• There are two general cases of quantity
discount models:
1.Carrying costs are constant (e.g. $2 per
unit).
2.Carrying costs are stated as a percentage off
purchase price (20% of unit price)
99
1) Total Cost with Constant Carrying Costs
(Compute the Common Optimal Order Quantity
Total Cost
TCa
TCb
Decreasing
Price
TCc
CC a,b,c
OC
EOQ
Quantity
2) Total Cost with Variable Carrying Cost
(Compute Optimal Order Quantity for each price
range)
Based on the same assumptions as the EOQ model,
the price-break model has a similar Qopt formula:
2DS
2(Annual Demand)(Or der or Setup Cost)
Q OPT =
=
iC
Annual Holding Cost
i = percentage of unit cost attributed to carrying inventory
C = cost per unit
Since “C” changes for each price-break, the formula above
will have to be used with each price-break cost value
100
101
Quantity Discount – How Much to Order?
102
Price-Break Example 1 (1 of 3)
ORDER SIZE
0 - 99
100 - 199
200+
PRICE
$10
8 (d1)
6 (d2)
For this problem holding cost is given as a constant value, not
as a percentage of price, so the optimal order quantity is the
same for each of the price ranges. (see the figure 12.9)
103
Price Break Example 1 (2 of 3)
TC = ($10 )
TC (d1 = $8 )
Inventory cost ($)
TC (d2 = $6 )
Carrying cost
Ordering cost
Q(d1 ) = 100 Qopt
Q(d2 ) = 200
104
Price Break Example 1 (3 of 3)
TC = ($10 )
TC (d1 = $8 )
Inventory cost ($)
TC (d2 = $6 )
Carrying cost
Ordering cost
Q(d1 ) = 100 Qopt
Q(d2 ) = 200
The lowest total cost is at the second price break
105
Price Break Example 2
QUANTITY
1 - 49
50 - 89
90+
Qopt =
PRICE
$1,400
1,100
900
2SD
=
H
S = $2,500
H = $190 per computer
D = 200
2(2500)(200)
= 72.5 PCs
190
For Q = 72.5
H Qopt
SD
TC =
+
2 + PD = $233,784
Qopt
For Q = 90
HQ
SD
TC =
+ 2 + PD = $194,105
Q
106
Price-Break Example 3
(1 of 4)
A company has a chance to reduce their inventory
ordering costs by placing larger quantity orders using the
price-break order quantity schedule below. What should
their optimal order quantity be if this company purchases
this single inventory item with an e-mail ordering cost of
$4, a carrying cost with a rate of 2% of the unit price, and
an annual demand of 10,000 units?
Order Quantity(units) Price/unit($)
0 to 2,499
$1.20
2,500 to 3,999 1.00
4,000 or more .98
107
Price-Break Example (2 of 4)
First, plug data into formula for each price-break value of “C”
Annual Demand (D)= 10,000 units
Cost to place an order (S)= $4
Carrying cost % of total cost (i)= 2%
Cost per unit (C) = $1.20, $1.00, $0.98
Next, determine if the computed Qopt values are feasible or not
Interval from 4000 & more, the Q OPT =
Qopt value is not feasible
Interval from 2500-3999, the
Qopt value is not feasible
Interval from 0 to 2499, the
Qopt value is feasible
Q OPT =
Q OPT =
2DS
=
iC
2DS
=
iC
2DS
=
iC
2(10,000)(4)
= 2,020 units
0.02(0.98)
2(10,000)(4)
= 2,000 units
0.02(1.00)
2(10,000)( 4)
= 1,826 units
0.02(1.20)
108
Price-Break Example 2 (3 of 4)
Since the feasible solution occurred in the first pricebreak, it means that all the other true Qopt values occur
at the beginnings of each price-break interval. Why?
Because the total annual cost function is
a “u” shaped function
Total
annual
costs
So the candidates
for the pricebreaks are 1826,
2500, and 4000
units
0
1826
2500
4000
Order Quantity
109
Price-Break Example 2 (4 of 4)
Next, we plug the true Qopt values into the total cost annual cost
function to determine the total cost under each price-break
D
Q
TC = DC +
S+
iC
Q
2
TC(0-2499)=(10000*1.20)+(10000/1826)*4+(1826/2)(0.02*1.20)
= $12,043.82
TC(2500-3999)= $10,041
TC(4000&more)= $9,949.20
Finally, we select the least costly Qopt, which in this
problem occurs in the 4000 & more interval. In summary,
our optimal order quantity is 4000 units
110
Multi-period Inventory Models:
Fixed Time Period
(Fixed-Order- Interval)
Models
111
Fixed-Order-Interval Model
Orders are placed at fixed time intervals
Order quantity for next interval? (inventory is
brought up to target amount, amount ordered
varies)
Suppliers might encourage fixed intervals
Requires only periodic checks of inventory
levels (no continous monitoring is required)
Risk of stockout between intervals
112
Inventory Level in a Fixed Period
System
Various amounts (Qi) are ordered at regular time intervals
(p) based on the quantity necessary to bring inventory up
to target maximum
Target maximum
Q1
Q4
Q2
d Inventory
Q3
p
p
p
Time
113
Fixed-Interval Benefits
Tight control of inventory items
Items from same supplier may yield savings
in:
 Ordering
 Packing
 Shipping costs
May be practical when inventories cannot
be closely monitored
114
Fixed-Interval Disadvantages
 Requires a larger safety stock
 Increases carrying cost
 Costs of periodic reviews
115
Fixed-Time Period Model with Safety Stock
Formula
q = Average demand + Safety stock – Inventory currently on hand
q = d(T + L) + Z  T + L - I
Where :
q = quantitiy to be ordered
T = the number of days between reviews
L = lead time in days
d = forecast average daily demand
z = the number of standard deviations for a specified service probabilit y
 T + L = standard deviation of demand over the review and lead time
I = current inventory level (includes items on order)
116
Fixed-Time Period Model:
Determining the Value of T+L
 T+ L =
 
T+ L
i 1
di

2
Since each day is independent and  d is constant,
 T+ L =
(T + L) d 2
The standard deviation of a sequence of random
events equals the square root of the sum of the
variances
117
Order Quantity for a
Periodic Inventory System
where
Q = d(tb + L) + zd
d
T
L
d
T+L -I
= average demand rate
= the fixed time between orders
= lead time
= standard deviation of demand
zd T + L = safety stock
I = inventory level
z = the number of standard deviations
for a specified service level
118
Fixed-Period Model with Variable
Demand (Example 1)
d
d
T
L
I
z
= 6 bottles per day
= 1.2 bottles
= 60 days
= 5 days
= 8 bottles
= 1.65 (for a 95% service level)
Q = d(T + L) + zd
T+L -I
= (6)(60 + 5) + (1.65)(1.2)
= 397.96 bottles
60 + 5 - 8
119
Fixed-Time Period Model with
Variable Demand (Example 2)(1 of 3)
Given the information below, how many units
should be ordered?
Average daily demand for a product is 20 units.
The review period is 30 days, and lead time is
10 days. Management has set a policy of
satisfying 96 percent of demand from items in
stock. At the beginning of the review period
there are 200 units in inventory. The standard
deviation of daily demand is 4 units.
120
Fixed-Time Period Model with Variable
Demand (Example 2)(2 of 3)
 T+ L =
(T + L) d =
2
 30 + 10  4  2 = 25.298
So, by looking at the value from the Table, we have a
probability of 0.9599, which is given by a z = 1.75
121
Fixed-Time Period Model with Variable
Demand (Example 2) (3 of 3)
q = d(T + L) + Z  T + L - I
q = 20(30 + 10) + (1.75)(25. 298) - 200
q = 800  44.272 - 200 = 644.272, or 645 units
So, to satisfy 96 percent of the demand,
you should place an order of 645 units at
this review period
122
ABC Classification System
• Demand volume and value of items vary
• Items kept in inventory are not of equal
importance in terms of:
–
dollars invested
–
profit potential
–
sales or usage volume
–
stock-out penalties
123
ABC Classification System
Classifying inventory according to some
measure of importance and allocating control
efforts accordingly.
A - very important
B - mod. important
C - least important
High
A
Annual
$ value
of items
B
C
Low
Low
High
Percentage of Items
124
ABC Analysis
Classify inventory into 3 categories typically
on the basis of the dollar value to the firm
$ volume = Annual demand x Unit cost
A class, B class, C class Policies based on
ABC analysis
– Develop class A suppliers more carefully
– Give tighter physical control of A items
– Forecast A items more carefully
125
Classifying Items as ABC
Class
A
B
C
% Annual $ Usage
100
80
60
% $ Vol
70-80
15
5-10
A
40
B
20
C
0
0
50
100
% of Inventory Items
% Items
5-15
30
50-60
126
ABC Classification
PART
UNIT COST
ANNUAL USAGE
1
2
3
4
5
6
7
8
9
10
$ 60
350
30
80
30
20
10
320
510
20
90
40
130
60
100
180
170
50
60
120
127
ABC Classification
PART
9
8
2
1
4
3
6
5
10
7
TOTAL
PART
VALUE
$30,600
1
16,000
2
14,000
3
5,400
4
4,800
5
3,900
3,600
6
3,000
7
2,400
8
1,700
9
$85,400
10
% OF TOTAL % OF TOTAL
UNIT
ANNUAL
USAGE
VALUECOSTQUANTITY
% CUMMULATIVE
35.9
$ 60
18.7
350
16.4
30
6.3
5.680
4.630
4.220
3.510
2.8
320
2.0
510
20
6.0
5.0
4.0
9.0
6.0
10.0
18.0
13.0
12.0
17.0
90
40
130
60
100
180
170
50
60
120
6.0
11.0
15.0
24.0
30.0
40.0
58.0
71.0
83.0
100.0
128
ABC Classification
PART
9
8
2
1
4
3
6
5
10
7
TOTAL
PART
VALUE
$30,600
1
16,000
2
14,000
3
5,400
4
4,800
5
3,900
3,600
6
3,000
7
2,400
8
1,700
9
$85,400
10
% OF TOTAL % OF TOTAL
UNIT
ANNUAL
USAGE
VALUECOSTQUANTITY
% CUMMULATIVE
35.9
$ 60
18.7
350
16.4
30
6.3
5.680
4.630
4.220
3.510
2.8
320
2.0
510
20
6.0
5.0
4.0
9.0
6.0
10.0
18.0
13.0
12.0
17.0
90
A
40
130
60
B
100
180
170
C
50
60
120
6.0
11.0
15.0
24.0
30.0
40.0
58.0
71.0
83.0
100.0
129
ABC Classification
PART
TOTAL
PART
VALUE
9 $30,600
1
8
16,000
2
2
14,000
3
1 CLASS
5,400
4
4
4,800
A3,900
5
3
B3,600
6
6
C3,000
5
7
10
2,400
8
7
1,700
9
$85,400
10
% OF TOTAL % OF TOTAL
UNIT
ANNUAL
USAGE
VALUECOSTQUANTITY
% CUMMULATIVE
35.9
6.0
$ 60
18.7
5.0
350
16.4 % OF TOTAL
4.0
30
6.3
ITEMS
VALUE9.0
5.680
6.0
9, 8, 2 4.630
71.010.0
1, 4, 3 4.220
16.518.0
6, 5, 10,3.5
7
12.513.0
10
2.8
12.0
320
2.0
17.0
510
20
6.0
90
11.0
A
40
15.0
% OF TOTAL
130
24.0
QUANTITY
60
B 15.030.0
100
40.0
180 25.058.0
60.071.0
170
C
83.0
50
100.0
60
120
130
ABC Classification
C
100 –
B
% of Value
80 –
60 –
A
40 –
20 –
0 |–
0
|
20
|
40
|
60
% of Quantity
|
80
|
100
131
Inventory Accuracy and
Cycle Counting
• Inventory accuracy refers to how well the
inventory records agree with physical count.
• Cycle counting refers to Physical Count of
items in inventory.
• Used often with ABC classification
– While A items are counted most often (e.g.,
daily), C items are counted the least frequently.
132
Last Words
Inventories have certain functions.
But too much inventory
- Tends to hide problems
- Costly to maintain
So it is desired
• Reduce lot sizes
• Reduce safety stocks