#### Transcript Slide 1

```Inventory (Chapter 16)
What is Inventory?
(to p2)
How Inventory works:
two ways of ordering
based on three elements
Inventory models
(to p3)
(to p4)
(to p5)
1
Inventory
- to study methods to deal with
“how much stock of items should be kept
on hands that would meet customer
demand”
Objectives are to determine:
a) how much to order, and
b) when to order
(to p1)
2
Two basic types of Inventory
Systems
1) continuous (fixed-order quantity)
• an order is placed for the same constant amount
when inventory decreases to a specified level, ie.
Re-order point
2) periodic (fixed-time)
• an order is placed for a variable amount after a
specified period of time
• used in smaller retail stores, drugstores, grocery
stores and offices
(to p1)
3
3 basic inventory elements
1. Carrying cost, Cc
•
Include facility operating costs, record
keeping, interest, etc.
2. Ordering cost, Co
•
Include purchase orders, shipping, handling,
inspection, etc.
3. Shortage (stock out) cost, Cs
•
Sometimes peanlties involved; if customer is
internal, work delays could result
(to p1)
4
Inventory models
Here, we only study the following three different
models:
1. Basic model
(to p6)
8th ed: 1, 15, 19,
9th ed: 1, 10, 13,
2. Model with “discount rate”
(to p23)
8th: 24, 26; 9th: 17, 18
3. Model with “re-order points”
(to p31)
8th: 36, 38; 9th: 26,28
5
1. Basic model
The basic model is known as:
“Economic Order Quantity” (EOQ) Models
Objective is to determine the optimal order
size that will minimize total inventory costs
(to p7)
How the objective is being achieved?
6
3 Basic EOQ models
Three models to be discussed:
1.
2.
3.
(to p8)
Basic EOQ model
EOQ model without instantaneous
(to p15)
receipt
(to p18)
EOQ model with shortages.
(to p5)
7
The Basic EOQ Model
• The optimal order size, Q,
is to minimize the sum of carrying costs and ordering costs.
• Assumptions and Restrictions:
- Demand is known with certainty and is relatively constant over time.
- No shortages are allowed.
- Lead time for the receipt of orders is constant. (will consider later)
- The order quantity is received all at once and instantaniously.
How to determine
the optimal value
Q *?
(to p10)
8
Determine of Q
We try to
•
•
•
Find the total cost that need to spend for keeping inventory
on hands
= total ordering + stock on hands
Determine its optimal solution by finding its first derivative
with respect to Q
How to get these values?
1. Find out the total carrying cost
2. Find out the total ordering cost
3. Total cost = 1 + 2
4. d (Total cost) /d Q = 0, and find Q*
(to p10)
9
The Basic EOQ Model
We assumed that, we will only keep half the inventory over a year then
The total carry cost/yr = Cc x (Q/2).
Then , Total cost =
TC  C D  C Q
Q
2
o
c
Total order cost = Co x (D/Q)
Finding optimal Q*
(to p11)
10
The Basic EOQ Model
• EOQ occurs where total cost curve is at minimum value and carrying cost equals
ordering cost:
TC  C D  C Q
Q
2
o
min
Then,
c
Q * 2C D
C
o
*
(How to obtain this?)
c
•Where is Q* located in our model?
(to p12)
11
The Basic EOQ Model
• Total annual inventory cost is sum of ordering and carrying cost:
To order inventory
To keep inventory
TC  C D  C Q
Q
2
o
c
Figure 16.5 The EOQ cost model
Try to get this value
Examples
(to p13)
12
The Basic EOQ Model
Example
Consider the following:
Model parameters : Cc  \$0.75, Co  \$150, D  10,000yd
Optimal order size : Q** 2CoD  2(150)(10,000)  2,000 yd
Cc
(0.75)
Total annual inventory cost : TC min  Co D  Cc Qopt  (150) 10,000  (0.75) (2,000)  \$1,500
Q*
2
2,000
2
Number of orders per year : D  10,000  5
Q * 2,000
Order cycle time  311 days  311  62.2 store days
5
D / Q*
No. of working days/yr
Note: You should pay attention that
all measurement units must be the same
(to p14)
Consider the same example, with yearly
13
The Basic EOQ Model
EOQ Analysis with monthly time frame
Model parameters : Cc  \$0.0625 per yd per month, Co  \$150 per order, D  833.3 yd per month
Optimal order size : Q*  2CoD  2(150)(833.3)  2,000 yd
Cc
(0.0625)
Total monthly inventory cost : TC min  Co D  Cc Q *  (150) (833.3)  (0.0625) (2,000)  \$125 per month
2
2,000
2
Q*
Total annual inventory cost  (\$125)(12)  \$1,500
(unit be based on yearly)
12 months a year
14
(to p7)
The EOQ Model with Noninstantaneous Receipt
The order quantity is received gradually over time and inventory is drawn on
at the same time it is being replenished.
(to p16)
Example: Let p = production, d = demand, then the total cost (TC) =
Figure 16.6 The EOQ model with noninstantaneous order receipt
Always greater than 0
why?
15
The EOQ Model with Noninstantaneous Receipt
Model Formulation
p  daily rate at which the order is received over time ( or production rate)
d  daily rate at which inventory is demanded

Q
Total annual inventory cost :TC  Co D  Cc 1 d 
Q
2  p
Optimal order size : Q* 
Assuming placing an order/yr
2CoD
Cc(1 d / p)
(to p17)
Example
16
The EOQ Model with Noninstantaneous Receipt
Example
Let,
Co  \$150, Cc  \$0.75 per unit,
D  10,000 yd per year,  10,000/311  32.2 yd per day, p  150 yd per day
Optimal order size : Q* 
2CoD
2(150)(10,000)

 2,256.8 yd


 32.2 
d
0.751 

Cc 1  
150 
p



Total minimum annual inventory cost : TC min  Co
Production run length 
D
Q *  d 
(10,000)
(2,256.8)  32.2 
 Cc
1    (150)
 (.075)
1 
  \$1,329

Q*
2  p
(2,256.8)
2
150 

Q * 2,256.8

 15.05 days
p
150
Number of orders per year (production runs) 
D 10,000

 4.43 runs
Q * 2,256.8

d
 32.2 
Maximum inventory level  Q * 1    2,256.81 
  1,772 yd
p
150




( to p7)
17
The EOQ Model with Shortages
Here, we allow Q being shortage, so that we could borrow or replenish the stocks
later
Max level of inventory
• In the EOQ model wth shortages, the assumption that shortages cannot exist is relaxed.
• Assumed that unmet demand can be backordered with all demand eventually satisfied.
Shortage = S/Q
Shortage
What we needed
On hand = (Q-S)/Q
Total cost is
t1 + t2 = S/Q + (Q-S)/Q = 1
18
(to p19)
The EOQ Model with Shortages
½*base* height = ½ * (Q-S) * (Q-S)/Q
= ½ * (Q-S)2 /Q
Area = ½ * (S/Q) * S
= ½ * S2 /Q
Total cost =
(to p20)
• In the EOQ model wth shortages, the assumption that shortages cannot exist is relaxed.
• Assumed that unmet demand can be backordered with all demand eventually satisfied.
Shortage = S/Q
Shortage
What we needed
On hand = (Q-S)/Q
19
The EOQ Model with Shortages
Total cost  Total shortage costs  total carrying costs  total ordering cost
Total inventory cost : TC  Cs
Optimal order quantity : Q* 

S2
(Q  S ) 2
D
 Cc
 Co
2Q
2Q
Q
2CoD  Cs  Cc 


Cc  Cs 
Cc 

 Cc  Cs 
Shortage level : S *  Q * 
Example
(to p21)
20
The EOQ Model with Shortages
Example
Let,
Co  \$150, Cc  \$0.75 per yd, Cs  \$2 per yd, D  10,000 yd




Optimal order quantity : Q*  2CoD  Cs  Cc   2(150)(10,000)  2  0.75   2,345.2 yd
Cc  Cs 
0.75
2 





Shortage level : S*  Q *  Cc   2,345.2 0.75   639.6 yd
 Cc  Cs 
 2  0.75 
2
2
2
2
Total inventory cost :TC  Cs S  Cc (Q * S*)  Co D  (2)(639.6)  (0.75)(1,705.6)  (150)(10,000)
2,345.2
2Q *
2Q *
Q * 2(2,345.2)
2(2,345.2)
 \$174.44  465.16  639.60
 \$1,279.20
(to p22)
21
The EOQ Model with Shortages
Number of orders  D  10,000  4.26 orders per year
Q 2,345.2
Maximum inventory level  Q  S  2,345.2  639.6  1,705.6 yd
Time between orders  t 
= Q/D
days per year
311

 73.0 days between orders
number of orders 4.26
Q  S 2,345.2 - 639.6

 0.171 or 53.2 days
D
10,000
Time during which there is a shortage  t2  S  639.6  0.064 year or 19.9 days
D 10,000
Time during which inventory is on hand  t1 
(to p7)
22
2. Model with “discount rate”
Price discounts are often offered if a
predetermined number of units is ordered
or when ordering materials in high volume.
How do we decide if we should order more
to take advantage of the discount being
offered?
(to p24)
23
Quantity Discounts with Constant Carrying Costs
Analysis Approach
Solution method:
1. We first determine the optimal order size, Q*
2. We then compare with any lower total cost with a discount price
and accept the one has the minimum total cost
Example
(to p25)
24
Quantity Discounts with Constant Carrying Costs Example
(1 of 2)
Consider the following example:
The following discount schedule is offered, which size of order should we subscript?
Co  \$2,500, Cc  \$190 per unit, D  200
Quantity
1- 49
50 – 89
90 +
Then, we have
Q* 
Price
\$1,400
1,100
900
Falls in this section,
Now we compare the TC of this, to
The next discount class, 90+
2CoD
2(2,500)(200)

 72.5
Cc
190
(to p26)
25
Quantity Discounts with Constant Carrying Costs Example
(2 of 2)
- Compute total cost at eligible discount price (\$1,100):
CoD
Q*
 Cc
 PD
Q*
2
(2,500)(200)
(72.5)

 (190)
 (1,100)(200)  \$233,784
(72.5)
2
TC* 
- Compare with total cost of with order size of 90 and price of \$900:
CoD
Q*
TC* 
 Cc
 PD
Q*
2
(2,500)(200) (190)(90)


 (900)(200)  \$194,105
(90)
2
- Because \$194,105 < \$233,784, maximum discount price should be taken and 90 units ordered.
Do we always take the offer?
(to p27)
26
No
• It depends on the carrying cost
• Example:
(to p28)
• Carrying Costs as a Percentage of Price
27
Quantity Discounts with Carrying Costs as a Percentage of
Price
Example
(1 of 2)
- Consider the same example, but with 15% of TC as carrying cost
- Then, the carrying cost for each category is as follows:
- Data:
Co = \$2,500, D = 200 computers per year
Quantity
Price (TC)
Carrying Cost
0 - 49
\$1,400
1,400(.15) = \$210
50 - 89
1,100
90 +
900
1,100(.15) = 165
900(.15) = 135
(to p29)
Chapter 16 - Inventory
Management
28
Quantity Discounts with Carrying Costs as a Percentage of
Price
Example
(2 of 2)
- Compute optimum order size for purchase price without discount and C c = \$210: Q*  2CoD  2(2,500)(200)  69
Cc
210
- Compute new order size:
Q* 
2(2,500)(200)
 77.8
165
- Compute minimum total cost: TC 
(note: 69 falls onto 50-89 category)
CoD
Q*
(2,500)(200)
(77.8)
 Cc
 PD 
165
 (1,100)(200)  \$232,845
Q*
2
77.8
2
- Compare with cost, discount price of \$900, order quantity of 90: TC  (2,500)(200)  (135)(90)  (900)(200)  \$191,630
90
2
- Optimal order size computed as follows:
Q* 
2(2,500)(200)
 86.1
135
(less than 90 as needed)
- Since this order size is less than 90 units , it is not feasible,thus optimal order size is 77.8 units.
(to p5)
29
Quantity Discount Model Solution with QM for Windows
Exhibit 16.4
30
3. Model with “re-order points”
• The reorder point is the inventory level at which a new order is placed.
• Order must be made while there is enough stock in place to cover demand during lead time.
• Formulation: R = dL, where d = demand rate per time period,
Then R = dL = (10,000/311)(10) = 321.54
Working days/yr
What would
happen?
(to p32)
31
Reorder Point
• Inventory level might be depleted at slower or faster rate during lead time.
• When demand is uncertain, safety stock is added as a hedge against stockout.
Two possible scenarios
No Safety
stocks!
Safety stock!
We should then ensure
Safety stock is secured!
How?
(to p33)
32
Determining Safety Stocks Using Service Levels
• We apply the Z test to secure its safety level,
R  d L  (Zd L )
Safety stock
Reorder point
Average sample demand
How these values are represented in the diagram of normal distribution?
33
(to p34)
Reorder Point with Variable Demand
R  d L  Zd L
where
R  reorder point
d  average daily demand
d  the standard deviation of daily demand
Z  number of standard deviations corresponding to service level probabilit y
Zd L  safety stock
Example
(to p35)
34
Reorder Point with Variable Demand
Example
Example:
determine reorder point and safety stock for service level of 95%.
d  30 yd per day, L  10 days,   5 yd per day
d
For 95% service level, Z  1.65 (Appendix A)
R  d L  Z L  30(10)  (1.65)(5)( 10 )  300  26.1 326.1 yd
d
Safety stock is second term in reorder point formula : 26.1.
(to p5)
35
Determining the Reorder Point with Excel
Exhibit 16.5
36
```