Inventory Class Notes
Why pay attention to Inventory
 May be the largest current asset
 A shortage may cause a stockout and an unhappy customer
 A shortage may result in a production stoppage
 Too much inventory can mean higher carrying costs and
negatively affect profits.
Inventory Functions
 Decoupling - When a manufacturing or other process
consists of two or more separate activities and where the
product of one activity may be a component of another
activity, it is important to have an inventory of these products
so that the manufacturing process can operate efficiently and
without delay.
 Storing resources - Farmers store hay
 Smoothing out irregularities in supply – Demand for many
products is seasonal. Manufactures must often produce more
products in the off-season and then store them until the
demand increases.
 Buying or producing in lots or batches – to reduce costs by
getting quantity discounts.
 Allows organizations to cope with perishable materials –
perishable goods.
 Storing Labour – oil producers
 Avoiding stockouts and shortages.
1
Two Inventory Decisions
1. How much to order
2. When to order
A major objective is to minimize total inventory costs.
Some of the most significant costs are:
1.
2.
3.
4.
The actual cost of the items.
The cost of Ordering.
The cost of Carrying (holding ) Inventory.
The cost of Stockouts.
Economic Order Quantity (EOQ) Model Assumptions
1. Demand D is known and occurs at a constant rate.
2. Lead Time is constant. Lead time is the time between the
placement of the order and the receipt of the order.
3. The order quantity Q is the same for each order. The
inventory level increases by Q units every time an order is
received.
4. The cost per order Co, is constant and does not depend on
quantity ordered.
5. The purchase cost per unit C, is constant and does not depend
on quantity ordered.
6. The inventory holding cost per unit per time period, Ch, is
constant. The total inventory holding cost depends on both
Ch and the size of the inventory.
7. Shortages such as stock-outs or backorders are not permitted.
8. The inventory position is reviewed continuously. As a result,
an order is placed as soon as an inventory position reaches
the reorder point.
2
See figure 6.2 on page 189 a description of inventory usage over
time.
Economic Order Quantity Model – EOQ
EOQ is a model designed to determine that particular quantity to
order that will minimize total inventory costs.
Types of Inventory Costs
Ordering Costs include all of the costs incurred into getting a unit
into the firm’s inventory and are expressed as a dollar cost per
order. See table 6.1 on page 188.
Holding costs (carrying costs) are the costs incurred because a
firm owns or maintains inventories. Some of the items included
are: interest expense, obsolescence, storage space, storage
administration, taxes, insurance, and theft. Holding costs can be
expressed as a percentage of average inventory value or as a cost
per unit per time period. See table 6.1 on page 188.
How Many to Order
2DCo
Q 
OPTIMAL UNITS PER ORDER
C
h
Q = number of pieces per order (cases, truckloads etc.)
EOQ = Q* = optimal # of pieces to order
D = annual demand in units
Co = ordering cost of each order
Ch = holding cost per unit per year
3
Graphically in figure 6.3 it shows that when EOQ assumptions are
met, total cost is minimized when Ordering costs are equal to
Holding costs.
Mathematically this is shown as:
𝑄
𝐷
𝐶ℎ = 𝐶𝑜
2
𝑄
Solving for Q we get:
2DCo
Q 
OPTIMAL UNITS PER ORDER
C
h
Sumco Pump Company Example
EOQ = Q* = ?
D = 1000 units
Co = $10/order
Ch = $0.50 per unit per year
Q  2(1000)(10) = √40,000 = 200 units
0.50
TC 
D
Co  Q C IS TOTAL COST
2 h
Q
1000
(10)  200 (0.5)  50 + 50 = 100
2
200
Since Annual Demand is 1000 units and there are 200 units per
order, then there must be 5 orders per year.
TC 
Average Inventory is
𝑸
𝟐
=
𝟐𝟎𝟎
𝟐
= 𝟏𝟎𝟎
Do problem 6-20
4
6-20. D = 100,000; Co = $10; Ch = $0.005
a. Q* 
2 100, 000 10 
 20, 000 number 6 screws
0.005
b. Number of orders per year 
D 100, 000

5
Q 20, 000
Total ordering cost = 5($10) = $50 per year
c. Average inventory 
Q 20, 000

 10, 000 units
2
2
Total holding cost = 10,000(0.005) = $50 per year
When to Order
Reorder Point (ROP)
ROP = (demand per day) X (lead time in days)
Example
D = Annual Demand = 8000 units
d = Daily demand = 40 units
Q = Order Quantity = 400 units
L = Lead Time = 3 days
ROP = (demand per day) X (lead time in days) = D x L
ROP = (40) X (3) = 120
Do 6-27
5
Production Run Model
 Eliminating the Constant Supply Rate Assumption in EOQ
 Assuming Production rate exceeds demand rate
 During the production run excess inventory is built up.
When the production run ends, inventory is gradually
reduced until the next production run is started.
Where:
Q = number of pieces per order, or production run
Ch= holding cost per unit of inventory per year
Cs = setup cost (replaces ordering cost)
D = Annual Demand in units
d= daily demand rate
p = daily production rate
t = length of production run in days
𝑸∗ = √
𝑪
𝟐𝑫𝑪𝒔
𝒅
(𝟏−
)
𝒉
𝒑
= Optimal Production Quantity
6
Brown Manufacturing
Annual Demand = D = 10,000 units
Setup Costs = Cs = $100
Carrying Cost = Ch = $0.50 per unit per year
Daily Production Rate = p = 80 units daily
Daily Demand rate = d = 60 units daily
𝑸∗
∗
= √
𝑪
𝑸 = √
𝟐𝑫𝑪𝒔
𝒅
(𝟏−
)
𝒉
𝒑
= Optimal Production Quantity
𝟐(𝟏𝟎,𝟎𝟎𝟎)(𝟏𝟎𝟎)
𝟎.𝟓(𝟏−
𝟔𝟎
)
𝟖𝟎
= 4000 units
Length of the production cycle =
𝐷
𝑄
𝑝
Production runs per year = 𝑄 =
=
4000
80
10,000
4000
𝑑
= 50 𝑑𝑎𝑦𝑠
= 2.5
Maximum Inventory Level = 𝑄(1 − 𝑝)
Students Do 6-28
7
6-27. a. EOQ 
2  2,500 18.75
2 DCo

 250 units .
Ch
1.5
b. Average inventory 
Annual holding cost 
Q 250

 125
2
2
Q
Ch  125 1.5   $187.50
2
c. Number of orders per year 
Annual ordering cost

D 2,500

 10
Q
250
D
Co  10 18.75   187.5
Q
d. Total cost = $187.5 + $187.5 + 2500(15) = $37,875
e. With 250 days per year, and 10 orders per year, the number of days between orders
= 250/10 = 25 days.
f. ROP = d × L = (2500 units per year/250 days per year) × 2 = 20 units
6-28. a. daily demand = 2500/250 = 10 units per day
b. Q* 
2  2,500  25
2 DCs

 324.92
 d
 10 
1.48 1  
Ch 1  
p
 50 

c. 324.92/50 = 6.5 days. Inventory sold = (10 units/day)(6.5 days)
= 65 units.
d. Maximum inventory level = Q(1 – d/p)
= 324.92(1 – 10/ 50)
= 259.94
Average inventory = 0.5(Maximum inventory level)
= 0.5 (259.94) = 129.97
Annual holding cost = (average inventory) Ch = 129.97 (1.48) = $192.35
e. Number of production runs = D/Q =2500/324.92 = 7.694
Annual setup cost = (D/Q)Cs = 7.694(25) = $192.35
f. Including the cost of production, the annual cost is
$192.35 + $192.35 + 2,500(14.80) = $37,384.71
g. ROP = d × L = 10 × 0.5 = 5 units
8
Quantity Discount Models
1. For each discount price(C), compute EOQ = √
2𝐷𝐶𝑜
𝐼𝐶
2. If EOQ < minimum for discount, adjust the quantity to Q =
minimum for discount.
3. For each EOQ or adjusted Q, compute total cost
𝑇𝐶 =
𝐷
𝑄
𝐶𝑜 + 𝐶ℎ + 𝐷𝐶
𝑄
2
4.
Choose the lowest cost quantity.
Quantity Discounts
D = 5000
Category
1
2
3
Co = 49
Order Size
0 to 999
1000 to 1999
2000 and over
Ch = 20% of unit cost
Unit Cost
$5.00
$4.80
$4.75
Q
1
2DCo
 2(5000)49  700
IC
(0.20)(5.00)
Q 
2
2DCo
 2(5000)49  714
(0.20)(4.80)
IC
Q 
3
2DCo
 2(5000)49  718
IC
(0.20)(4.75)
9
Since Q2 and Q3 are not high enough to qualify for
the discount they must be adjusted upward to the
nearest quantity that would allow the product to be
purchased at the discount price.
Therefore Q2 = 1000 and Q3 = 2000
In the original EOQ model the annual purchase cost
was not included because it was constant and never
affected by the inventory order policy. However,
with discounts, the annual purchase cost (DC) is
included in the equation for total cost.
TC  D Co  Q C  DC
2 h
Q
 700 
(
5000
)
(1.00)  (5000)(5.00)  25,700
TC 
(49)  

1 700
 2 
 1000 
(0.96)  (5000)(4.80)  24725
TC  (5000) (49)  

2 1000
 2 
 2000 
(0.95)  (5000)(4.75)  24,822.50
TC  (5000) (49)  

3 2000
 2 
10
6-38. D = 1,000; unit cost = $50; Co = $40;
Ch = 0.25 × unit cost
Q
2 1, 000  40 
 80
0.25  50 
With discount, unit cost = (1 – 0.03) × $50 = $48.50
Qd* 
2 1, 000  40 
 81.22
0.25  48.50
which should be adjusted to minimum orderable quantity (i.e., 200).
Original total cost  1, 000  50 
1, 000
80
 40   0.25  50
80
2
= $51,000
Discount cost  1, 000  48.50 
1, 000
200
 40 
 0.25  48.50  $49,912.50
200
2
Therefore, North Manufacturing should take the discount.
Safety Stock
Safety Stock is additional inventory carried by some businesses in
order to avoid stockout situations during times when either demand
or lead time is unusually high.
Two Important Factors are:
Stockout Cost
Holding Cost
Stockout cost usually involves lost sales and lost goodwill.
How to determine the level of safety Stock – Service Level.
Service level is the percentage of time the customers’ demands are
met. When demand during lead time is normally distributed, we
can determine the safety stock level using the normal distribution.
11
Where ROP = (Avg. demand during lead time) + ZσdLT
Where
Z is the number of standard deviations for a given service level.
σdLT is the standard deviation of demand during lead time.
Hinsdale Company – SKU A3378
Demand during lead time is normally distributed
The mean µ = 350
The standard Deviation σ = 10
95% service level is desired. Therefore Z = 1.65
Since ROP = (Avg. demand during lead time) + ZσdLT, then
ROP = 350 + 1.65(10) = 366.5 or about 367 units.
Since demand during lead time is 350 then the safety sock level is
17 units.
Do Problem 39
6-39.  = 60;  = 7
Safety stock for 90% service level
= Z (at 0.90) = 7 × 1.28 = 8.96  9
6-40. Ch = $0.50;  = 600;  = 7
Safety stock for 90% service level  9
Carrying cost = 9 × 0.5 = $4.50
Safety stock for 95% service level = 7 × 1.65  12
Carrying cost = $6.00
Safety stock for 98% service level = 7 × 2.05  15
Carrying cost = $7.50
12
Omit 206 -209
Single Period Inventory Models
Some products are either of no value or a much reduced value if
they are not sold during the period for which they were produced
or manufactured. For example: Newspapers, Magazines and
Perishable foods
Marginal Analysis With Discrete Distributions
Marginal Profit (MP) the additional profit from one additional unit
Marginal Loss (ML) is the loss that occurs when a unit is stocked
but not sold.
The decision to stock is based on probability. We will decide to
stock an additional unit only if the expected marginal profit from
that unit exceeds the expected marginal loss.
Let P = the probability of selling one additional unit.
Then (1-p) is the probability the unit will not sell.
Therefore, Expected Marginal Profit = P(MP) and
Expected Marginal Loss = (1-P)ML. The decision to stock then is
only if P(MP) ≥ (1-P)ML. Solving for P, we get
𝑃 ≥
𝑀𝐿
𝑀𝐿 + 𝑀𝑃
Steps of Marginal Analysis with Discrete Distribution
13
𝑀𝐿
 Determine the value of 𝑃 ≥
.
𝑀𝐿+𝑀𝑃
 Construct a probability table with a cumulative probability
column.
 Keep ordering inventory as long as the probability of selling
𝑀𝐿
at least one additional unit is greater than
.
𝑀𝐿+𝑀𝑃
Café du Donut
MP = 2
ML = 4
Daily Sales
4
5
6
7
8
9
10
Step 1.
Prob. Of Demand Cumulative Prob.
0.05
P(x ≥ 4) = 1.00
0.15
P(x ≥ 5) = 0.95
0.15
P(x ≥ 6) = 0.80
0.20
P(x ≥ 7) = 0.65
0.25
P(x ≥ 8) = 0.45
0.10
P(x ≥ 9) = 0.20
0.10
P(x ≥ 10) = 0.10
1.00
𝑃 ≥
𝑀𝐿
𝑀𝐿+𝑀𝑃
𝑃 ≥
4
4+2
=
4
6
= 0.67
Therefore keep stocking units as long as there is at least a .67
chance of selling the unit.
Step 2. Add a cumulative probability column to the table.
Step 3. Order 6 cartons
14
Do problem 6-53
6-53.
Demand Probability P(Demand  _____)
50
0.05
1.00
75
0.10
0.95
100
0.20
0.85
125
0.30
0.65
150
0.20
0.35
175
0.10
0.15
200
0.05
0.05
a. ML = 20; MP = 80 – 20 = 60
ML/(ML + MP) = 20/(20 + 60) = 0.25
Stock 150
b. ML = 35; MP = 80 – 35 = 45
ML/(ML + MP) = 35/(35 + 45) = 0.4375
Stock 125
c.
Demand Probability P(Demand  _____)
50
0.25
1.00
75
0.25
0.75
100
0.25
0.50
125
0.25
0.25
ML = 20; MP = 100 – 20 = 80
ML/(ML + MP) = 20/(20 + 80) = 0.20
Stock 125
15
Marginal Analysis with the normal distribution
Need to know µ, σ, ML, MP
Newspaper Example – Joe’s Newsstand
µ = 60
σ = 10
Step 1. 𝑃 ≥
𝑀𝐿
𝑀𝐿+𝑀𝑃
ML = 0.20
𝑃 ≥
.20
.20+.30
MP = 0.30
=
.20
.50
= 0.40
Step 2. Use the normal curve to highlight the area of interest and
determine the z value.
Step 3. Solve for X using 𝑍 =
𝑋−𝜇
𝜎
where 𝑋 = 𝜇 + 𝑍𝜎
𝑋 = 60 + 0.25(10) = 62.5 𝑛𝑒𝑤𝑠𝑝𝑎𝑝𝑒𝑟𝑠
Do problem 6-55
6-55. ML = 35; MP = 15
ML/(ML + MP) = 35/(35 + 15) = 0.70
Using the normal distribution table, the z-value associated with the upper 70% of the
normal distribution is between –0.52 and –0.53. We will choose –0.53 since P(Z  –
0.53) is slightly more than 0.70.
The quantity to stock is
X* =  + z = 45,000 – 0.53(4,450) = 42,641.5.
16
Victorian Bakery specializes in making cheesecakes that are then
sold to restaurants. The cheesecakes can be produced for $3 each
and sold for $7 each. Any cheesecakes left at the end of the day
are sold to a retirement home for $1 each. The demand for the
cheesecakes is normally distributed with a mean of 62 and a
standard deviation of 15. Using marginal analysis, recommend the
level of demand the bakery should plan to meet in order to
maximize the expected value of profit.
17
µ = 62
σ = 15
Step 1. 𝑃 ≥
𝑀𝐿
𝑀𝐿+𝑀𝑃
ML = 2
𝑃 ≥
2
4+2
MP = 4
=
2
6
= 0.3333
Step 2. Use the normal curve to highlight the area of interest and
determine the z value.
Step 3. Solve for X using 𝑍 =
𝑋−𝜇
𝜎
where 𝑋 = 𝜇 + 𝑍𝜎
𝑋 = 62 + 0.44(15) = 68.6 𝑐ℎ𝑒𝑒𝑠𝑒𝑐𝑎𝑘𝑒𝑠
18
ABC Analysis
Inventory
Group
Dollar Usage
(%)
Inventory
Items (%)
A
B
C
70
20
10
10
20
70
Use
Quantitative
Control?
YES
SOMETIMES
NO
19