Order Quantities when
Demand is Approximately
Level
Chapter 5
Inventory Management
Dr. Ron Tibben-Lembke
Inventory Costs
Costs associated with inventory:

Cost of the products

Cost of ordering

Cost of hanging onto it

Cost of having too much / disposal

Cost of not having enough (shortage)
Shrinkage Costs

How much is stolen?




2% for discount, dept. stores, hardware,
convenience, sporting goods
3% for toys & hobbies
1.5% for all else
Where does the missing stuff go?




Employees: 44.5%
Shoplifters: 32.7%
Administrative / paperwork error: 17.5%
Vendor fraud: 5.1%
Inventory Holding Costs
Category
% of Value
Housing (building) cost
6%
Material handling
3%
Labor cost
3%
Opportunity/investment
Pilferage/scrap/obsolescence
Total Holding Cost
11%
3%
26%
ABC Analysis

Divides on-hand inventory into 3 classes


Basis is usually annual $ volume


A class, B class, C class
$ volume = Annual demand x Unit cost
Policies based on ABC analysis
Develop class A suppliers more
 Give tighter physical control of A items
 Forecast A items more carefully

Classifying Items
as ABC
% Annual $ Usage
100
80
60
A
40
B
20
C
0
0
50
100
% of Inventory Items
150
ABC Classification Solution
Stock #
Vol.
206
105
019
144
207
26,000
200
2,000
20,000
7,000
Total
Cost
$ Vol.
$ 36 $936,000
600 120,000
55 110,000
4
80,000
10
70,000
1,316,000
%
ABC
ABC Classification Solution
Stock #
Vol.
206
105
019
144
207
26,000
200
2,000
20,000
7,000
Total
Cost
$ Vol.
$ 36 $936,000
600 120,000
55 110,000
4
80,000
10
70,000
%
71.1
9.1
8.4
6.1
5.3
1,316,000 100.0
ABC
A
A
B
B
C
Economic Order Quantity
Assumptions

Demand rate is known and constant

No order lead time

Shortages are not allowed

Costs:
A - setup cost per order
 v - unit cost
 r - holding cost per unit time

EOQ
Inventory
Level
Q*
Optimal
Order
Quantity
Decrease Due to
Constant Demand
Time
EOQ
Inventory
Level
Q*
Optimal
Order
Quantity
Instantaneous
Receipt of Optimal
Order Quantity
Time
EOQ
Inventory
Level
Q*
Reorder
Point
(ROP)
Time
Lead Time
EOQ
Inventory
Level
Q*
Average
Inventory Q/2
Reorder
Point
(ROP)
Time
Lead Time
Total Costs

Average Inventory = Q/2

Annual Holding costs = rv * Q/2

# Orders per year = D / Q

Annual Ordering Costs = A * D/Q

Annual Total Costs = Holding + Ordering
Q
D
TC (Q)  vr *  A *
2
Q
How Much to Order?
Annual Cost
Holding Cost
= H * Q/2
Order Quantity
How Much to Order?
Annual Cost
Ordering Cost
= A * D/Q
Holding Cost
= H * Q/2
Order Quantity
How Much to Order?
Annual Cost
Total Cost
= Holding + Ordering
Order Quantity
How Much to Order?
Total Cost
= Holding + Ordering
Annual Cost
Optimal Q
Order Quantity
Optimal Quantity
Total Costs =
Q
D
vr *  A *
2
Q
Optimal Quantity
Total Costs =
Q
D
vr *  A *
2
Q
Take derivative
with respect to Q =
vr
D
 A* 2
2
Q
Optimal Quantity
Total Costs =
Q
D
vr *  A *
2
Q
Take derivative
with respect to Q =
vr
D
 A* 2  0
2
Q
Set equal
to zero
Optimal Quantity
Total Costs =
Q
D
vr *  A *
2
Q
Take derivative
with respect to Q =
vr
D
 A* 2  0
2
Q
Solve for Q:
vr DA
 2
2 Q
Set equal
to zero
Optimal Quantity
Total Costs =
Q
D
vr *  A *
2
Q
Take derivative
with respect to Q =
vr
D
 A* 2  0
2
Q
Solve for Q:
vr DA
 2
2 Q
2 AS
Q 
vr
2
Set equal
to zero
Optimal Quantity
Total Costs =
Q
D
vr *  A *
2
Q
Take derivative
with respect to Q =
vr
D
 A* 2  0
2
Q
Set equal
to zero
Solve for Q:
vr DA
 2
2 Q
2 AS
Q 
vr
2
2 AS
Q
vr
Sensitivity

Suppose we do not order optimal EOQ, but
order Q instead, and Q is p percent larger

Q = (1+p) * EOQ

Percentage Cost Penalty given by:
 p2 

PCP  50
1 p 

EOQ = 100, Q = 150, so p = 0.5
50*(0.25/1.5) = 8.33 a 8.33% cost increase
Figure 5.3 Sensitivity
Percentage Cost Penalty using Q different from the EOQ
30
25
20
PCP
15
10
5
0
-0.6
-0.4
-0.2
0
-5
p
0.2
0.4
0.6
A Question:

If the EOQ is based on so many
horrible assumptions that are never
really true, why is it the most
commonly used ordering policy?
Benefits of EOQ



Profit function is very shallow
Even if conditions don’t hold
perfectly, profits are close to optimal
Estimated parameters will not throw
you off very far
Tabular Aid 5.1







For A = $3.20 and r = 0.24%
Calculate Dv =total $ usage (or sales)
Find where Dv fits in the table
Use that number of months of supply
D = 200, v = $16, Dv=$3,200
From table, buy 1 month’s worth
Q = D/12 = 200/12 = 16.7 = 17
How do you get a table?

Decide which T values you want to
consider: 1 month, etc.

Use same v and r values for whole table

For each neighboring set of T’s, put them
into
288 A
Dv 
T1T2 r
How do you get a table?

For example, A = $3.20, r = 0.24

To find the breakpoint between 0.25 and 0.5

Dv = 288 * 3.2 / (0.25 * 0.5 * 0.24)

= 921.6 / 0.03 = 30,720

So if Dv is less than this, use 0.25, more
than that, use 0.5

Find 0.5 and 0.75 breakpoint:

Dv = 288 * 3.2/(0.5 * 0.75 * 0.24) = 10,2240
Why care about a table?




Some simple calculations to get set up
No thinking to figure out lot sizes
Every product with the same ordering cost
and holding cost rate can use it
Real benefit - simplified ordering


Every product ordered every 1 or 2 weeks, or
every 1, 2, 3, 4, 6, 12 months
Order multiple products on same schedule:



Get volume discounts from suppliers
Save on shipping costs
Savings outweigh small increase from non-EOQ orders
Uncoordinated Orders
Time
Simultaneous Orders
Time
Same T = number months supply allows firm to order at
same time, saving freight and ordering expenses
Adjusted some T’s, changed order times
Offset Orders
Same T = number months supply allows firm to control
maximum inventory level by coordinating replenishments
With different T, no consistency
Quantity Discounts

How does this all change if price
changes depending on order size?

Explicitly consider price:
2 AD
Q
vr
Discount Example
D = 10,000
A = $20
Price
v = 5.00
4.50
3.90
Quantity
Q < 500
501-999
Q >= 1000
r = 20%
EOQ
633
666
716
Discount Pricing
Total Cost
Price 1
Price 2
Price 3
X 633
X 666
X 716
500
1,000
Order Size
Discount Pricing
Total Cost
Price 1
Price 2
Price 3
X 633
X 666
X 716
500
1,000
Order Size
Discount Example
Order 666 at a time:
Hold 666/2 * 4.50 * 0.2= $299.70
Order 10,000/666 * 20 = $300.00
Mat’l 10,000*4.50 = $45,000.00 45,599.70
Order 1,000 at a time:
Hold 1,000/2 * 3.90 * 0.2=$390.00
Order 10,000/1,000 * 20 = $200.00
Mat’l 10,000*3.90 = $39,000.00 39,590.00
Discount Model
1.Compute EOQ for each price
2.Is EOQ ‘realizeable’? (is Q in range?)
If EOQ is too large, use lowest
possible value. If too small, ignore.
3.Compute total cost for this quantity
4.Select quantity/price with lowest total
cost.
Adding Lead Time

Use same order size

Order before inventory depleted

R = DL where:
2DA
Q
vr
D = annual demand rate
 L = lead time in years
