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1. HARAMAYA UNIVERSITY
POST GRADUATE PROGRAM DIRECTORATE
College: Natural and Computational Sciences
Department: Mathematics
Program: MSc. In Operation Research
PRESENTATION ASSIGNMENT OF OPERATION RESEARCH II
Prepared by: - Rauf Dina Abrahim ID No: PGP/650/14
Submitted to: - Sileshi D. (Ph.D.)
December, 2022 GC.
Haramaya University, Ethiopia

2. 1. DETERMINISTIC EOQ INVENTORY MODELS
1.1 Introduction to Basic Inventory Models
The purpose of inventory theory is to determine rules that management can use to
minimize the costs associated with maintaining inventory and meeting customer
demand
Costs Involved in Inventory Models
 Ordering and Setup Cost.
 Unit Purchasing Cost.
 Holding or Carrying Cost.
 Stock out or Shortage Cost.
Assumptions of EOQ Models
 Repetitive Ordering.
 Constant Demand.
 Constant Lead Time.
 Continuous Ordering.

3. 1.2 The Basic Economic Order Quantity Model
Assumptions of the Basic EOQ Model
For the basic EOQ model to hold, certain assumptions are required (for
the sake of definiteness, we assume that the unit of time is one year):
 Demand is deterministic and occurs at a constant rate.
 In an order of any size (say q units is placed, an ordering
and setup cost K is incurred.
 The lead time for each order is zero.
 No shortages are allowed.
 The cost per unit-year of holding inventory is h

4. Derivation of Basic EOQ Model
 We begin by making some simple observations.
 We should never place an order when I, the inventory level, is greater than zero; if
we place an order then the we are incurring an unnecessary holding cost.
 If I=0, we must place an order to prevent a shortage.
 Each time an order is placed (when I=0) we should order the same quantity, q.
 We now determine the value of q that minimizes annual cost (call it q*).
 Let TC(q) be the total annual cost incurred if q units are ordered each time that I=0.
Note that
 TC(q)= annual cost of placing orders + annual purchasing cost + annual holding a
cost …. (1)

5. Cont...
To do this let us consider the following notations:
 K = Cost of Placing an order
 h = holding cost
 D = Annual Demand
q = Order Quantity
p = purchasing cost
 I = Re-Order Point (level of inventory at which order is placed)
d = Demand Rate (say, daily demand)

6. cont...
So, Number of orders placed in that year =
𝐷
𝑞
and length of each cycle is
𝑞
𝐷
(from fig. 1).
 And Annual Ordering cost =
𝐾𝐷
𝑞
………………. (1)
 Average inventory = [Area of the triangle (Figure-1)/Length of the cycle] =
1
2
𝑞𝑇
𝑇
=
𝑞
2
;
 Or, Average inventory = [(Maximum Inventory + Minimum Inventory)/2] = (q+0)/2 =
𝑞
2
.
 So, Annual Carrying Cost =
ℎ𝑞
2
………………….. (2)
 And purchasing cost =𝑃𝐷 ……………………………………….. (3)
Hence from I-III we obtained
 TC(q)= annual cost of placing orders + annual purchasing cost + annual holding a cost.
 𝑇𝐶 𝑞 =
ℎ𝑞
2
+ 𝑝𝐷 +
𝐾𝐷
𝑞
………………. (4)
 And TC* = Minimum Inventory Cost.

7. Cont.
Figure-1: A simple inventory model based on fixed demand and fixed lead time.

8. Cont.…
 To find the value of q that minimizes, 𝑇𝐶 𝑞 we set, 𝑇𝐶′ 𝑞 = 0 equal to zero. These
yields
𝑇𝐶′ (𝑞) = −
𝐾𝐷
𝑞2 +
ℎ
2
= 0 …………………… (5)
Thus from (3) we obtained 𝑞 = ±
2𝐾𝐷
ℎ
 The economic order quantity, or EOQ, minimizes TC(q).
That is 𝑞∗
=
2𝐾𝐷
ℎ
1
2
………………………………… (6)
Thus, q* does indeed minimize total annual cost.
 The figure on the next page confirms the fact that at q*, the annual holding and
ordering costs are the same.

9. Cont..
From Figure-2 it is evident at EOQ; Ordering Cost = Holding Cost
That is
𝐾𝐷
𝑞∗
=
ℎ𝑞∗
2
…………………... (7)
From this equation (7), we can say that: 𝑞 ∗2
=
2𝐾𝐷
ℎ
.
So, 𝑞∗
=
2𝐾𝐷
ℎ
1
2
Figure 2 Trade-Off between Holding Cost and Ordering Cost.

10. Cont..
Example 1
Braneast Airlines
Braneast Airlines uses 500 taillights per year. Each time an order for taillights is placed,
an ordering cost of $5 is incurred. Each light costs 40¢, and the holding cost is 8¢/light/year.
Assume that demand occurs at a constant rate and shortages are not allowed. What is the
EOQ? How many orders will be placed each year? How much time will elapse between
the placement of orders?
Solution
We are given that K=$5, h=$0.08/light/year, and D=500 lights/year. The EOQ is
𝑞∗ =
2 × 5 × 500
0.08
1
2
= 250

11. cont.
Hence, the airline should place an order for 250 taillights each time that inventory reaches
zero. 𝑂𝑟𝑑𝑒𝑟 =
𝐷
𝑞∗
=
500
250
= 2 𝑜𝑟𝑑𝑒𝑟
The time between placement (or arrival) of orders is simply the length of a cycle. Since the
length of each cycle is
𝑞∗
𝐷
, the time between orders will be
𝑞∗
𝐷
=
250
500
= 0.5𝑦𝑒𝑎𝑟

12. 1.3 Computing the Optimal Order Quantity When Quantity Discounts Are
Allowed
 Up to now, we have assumed that the annual purchase cost does not depend on the order size.
In real life, however, suppliers often reduce the unit purchasing price for large orders. Such price
reductions are referred to as quantity discounts.
 If we let q be the quantity ordered each time an order is placed, the general quantity discount
model analyzed in this section may be described as follows:
 If 𝑞 < 𝑏1, 𝑝1 𝑑𝑜𝑙𝑙𝑎𝑟𝑠.
 If 𝑏1 ≤ 𝑞 < 𝑏2 𝑒𝑎𝑐ℎ 𝑖𝑡𝑒𝑚 𝑐𝑜𝑠𝑡 𝑝2 𝑑𝑜𝑙𝑙𝑎𝑟𝑠.
 If 𝑏𝑘−2 ≤ 𝑞 < 𝑏𝑘−1 𝑒𝑎𝑐ℎ 𝑖𝑡𝑒𝑚 𝑐𝑜𝑠𝑡 𝑝𝑘−1 𝑑𝑜𝑙𝑙𝑎𝑟𝑠.
 If 𝑏𝑘−1 ≤ 𝑞 < 𝑏𝑘 = ∞ 𝑒𝑎𝑐ℎ 𝑖𝑡𝑒𝑚 𝑐𝑜𝑠𝑡 𝑝𝑘 𝑑𝑜𝑙𝑙𝑎𝑟𝑠.
 Since 𝑏1, 𝑏2, 𝑏3, … , 𝑏𝑘−1 are points where a price change occurs, we refer to them as price
break points

13. cont..
 Since larger order quantities should be associated with lower prices, we have 𝒑𝒌 <
𝒑𝒌−𝟏 < 𝒑𝒌−𝟐 < … < 𝒑𝟏.
 Now to find the order quantity minimizing total annual costs, we need the following
definitions:
 𝐸𝑂𝑄𝑖 = quantity that minimizes total annual cost if, for any order quantity, the
purchasing cost of the item is 𝑝𝑖.
 If 𝑇𝐶𝑖 𝑞 = total annual cost if each order is q units at a price 𝑝𝑖.
 𝐸𝑂𝑄𝑖 is admissible if 𝑏𝑖−1 ≤ 𝐸𝑂𝑄𝑖 < 𝑏𝑖.
 𝑇𝐶 𝑞 =actual annual cost of q items is ordered each time an order is placed.
 In general, the value of q minimizing 𝑇𝐶 𝑞 can be either a break point or some 𝐸𝑂𝑄𝑖.
 The following observations are helpful in determining the point that minimizes 𝑇𝐶 𝑞 .

14. cont..
 . Where 𝑇𝐶 𝑞 =
ℎ𝑞
2
+ 𝑝𝐷 +
𝐾𝐷
𝑞
……………. (8)
 For any value of 𝑞,
𝑻𝑪𝒌(𝒒) < 𝑻𝑪𝒌−𝟏(𝒒) < 𝑻𝑪𝒌−𝟐(𝒒) < … < 𝑻𝑪𝟏(𝒒)
 If 𝐸𝑂𝑄𝑖 is admissible, then minimum cost for 𝑏𝑖−1 ≤ 𝑞 < 𝑏𝑖 occurs for 𝑞 =
𝐸𝑂𝑄𝑖. If 𝐸𝑂𝑄𝑖 < 𝑏𝑖, the minimum cost for 𝑏𝑖−1 ≤ 𝑞 < 𝑏𝑖 occurs for 𝑞 = 𝑏𝑖.
 If 𝑞 = 𝐸𝑂𝑄𝑖 for d=h/p using (5) we obtained 𝐸𝑂𝑄𝑖 =
2𝐾𝐷
𝑝𝑖𝑑
1
2
………………. (9)

15. Example
A local accounting firm in Smalltown orders boxes of floppy disks (10 disks to a box)
from a store in Megalopolis. The per-box price charged by the store depends on the
number of boxes purchased (see Table 1). The accounting firm uses 10,000 disks per
year. The cost of placing an order is assumed to be $100. The only holding cost is the
opportunity cost of capital, which is assumed to be 20% per year. For this example,
𝑏1 =100, 𝑏2 =300, 𝑝1 =$50.00, 𝑝2 =$49.00, and 𝑝3 =$48.50

16. cont..
Each time an order is placed for disks, how many boxes of disks should be ordered?
How many orders will be placed annually? What is the total annual cost of meeting the
accounting firm’s disk needs?
Solution
Given b1 = 100, b2 = 300, p1 =$50.00, p2 = $49.00, p3 = $48.50, d=0.2, K=$100 and
D = 1000 boxes per year. We first determine the best order quantity for p3 = $48.50 and
300 ≤ q.
𝐸𝑂𝑄3 =
2𝐾𝐷
𝑝𝑖𝑑
1
2
=
2(100)1000
0.2(48.50)
1
2
= 143.59.
Since EOQ3 < 300, EOQ3 is not admissible. Therefore q ≥ 300, TC3(q) is minimized by
𝑞3
∗
= 300.
We next consider p2=$49.00 and 100 ≤ q < 300. Then
𝐸𝑂𝑄2 =
2𝐾𝐷
𝑝𝑖𝑑
1
2
=
2(100)(1000)
9.8
1
2
= 142.86

17. Cont...
Since 100 ≤ 𝐸𝑂𝑄2 < 300, 𝑡ℎ𝑢𝑠 𝐸𝑂𝑄2 is admissible, and for a price 𝑝2 =$49.00, the best we
can do is to choose 𝑞2
∗
=142.86.
Since 𝑞2
∗
is admissible, p1 =%50.00 and 0 ≤ q < 100 cannot yield the order quantity minimizing
TC(q). Thus, either 𝑞2
∗
or 𝑞3
∗
will minimize TC(q).
To determine which of these order quantities minimizes TC(q), we must find the smaller of
𝑇𝐶3(300) and 𝑇𝐶2(142.86). For 𝑞3
∗
the annual holding cost is $9.70.
Thus, for𝑞3
∗
,
Annual ordering cost = 100(1000/300) = $333.33.
Annual purchasing cost = 1000(48.50) = $48,500.
Annual holding cost = (½) (300) (9.7) = $1455.
𝑇𝐶3 300 =$50, 288.33.

18. Cont...
For 𝑞2
∗
the annual holding cost is $9.80. Thus, for 𝑞2
∗
,
Annual ordering cost = 100(1000/142.86) = $699.99.
Annual purchasing cost = 1000(49) = $49,000.
Annual holding cost = (½) (142.86) (9.8) = $700.01.
𝑇𝐶2 142.86 =$50, 400.
Thus 𝑞3
∗
= 300 will minimize TC(q).
Our analysis shows that each time an order is placed, 300 boxes of disks should
be ordered. Then 3.33 orders are placed each year.

19. 1.4 The Continuous Rate EOQ Model
Many goods are produced internally rather than purchased from an outside supplier.
If a company meets demand by making its own products, the continuous rate EOQ
model will be more realistic than the traditional EOQ model.
The continuous rate EOQ model assumes that a firm can produce a good at a rate of r
units per time period. This means that during any time period of length, t, the firm can
produce rt units.
We define :-
 q = number of units produced during each production run.
 K = cost of setting up a production run.
 H = cost of holding one unit in inventory for one year.
 D = annual demand for the product.
 r = rate at which firm can make product per unit time (r>D)

20. Cont...
 Assuming that a production run begins at time 0, the variation of inventory over time is
 described by Figure 3. At the beginning of a production run, we are producing at a rate of 𝑟
units per year, and demand is occurring at a rate of 𝐷 units per year. Thus, until 𝑞 units are
produced, inventory increases at a rate of 𝑟 − 𝐷 𝑢𝑛𝑖𝑡𝑠 per year.
 The variation of inventory over time is shown.
 𝑂𝑝𝑡𝑖𝑚𝑎𝑙 𝑟𝑢𝑛 𝑠𝑖𝑧𝑒 =
2𝐾𝐷𝑟
ℎ(𝑟−𝐷)
1
2
= 𝐸𝑂𝑄
𝑟
𝑟−𝐷
1
2
…………………… (7)
 As r increases, production occurs at a more rapid rate. Hence, for large r, the rate model
should approach the instantaneous delivery situation of the EOQ model.

21. Cont...
Figure 3 Variation of Inventory for Continuous Rate EOQ Model

22. Example Macho Auto Company
Macho Auto Company needs to produce 10,000 car chassis per year. Each is valued at
$2,000. The plant has the capacity to produce 25,000 chassis per year. It costs $200 to set
up a production run, and the annual holding cost is 25¢, per dollar of inventory. Determine the optimal production run
size. How many production runs should be made each year?
Solution:
We are given that;
r =25,000 chassis per year.
D=10,000 chassis per year.
h=0.25($2,000)/chassis/year =$500/chassis/year.
K=$200 per production run.
From (7),Optimal run size =
2𝐾𝐷𝑟
ℎ(𝑟−𝐷)
1
2
=
2(200)(10000)(25000)
500(25000−10000)
1
2
= 115.47.
Also
10000
115.47
= 86.60 production runs will be made each year.

23. Thank-you!!!