CHAPTER 3 ECONOMIC ORDER QUANTITY (EOQ) FOR ITEMS WITH IMPERFECT
3.3 Mathematical Model
Figure 3.1 shows how the inventory behaves with a buyer. It should be noted here that Salameh and Jaber (2000) suggested this behavior. The screening and consumption of the inventory continues until time Ο1, after which all the defectives (B1) are withdrawn from inventory as a single batch and are sold to the secondary market. The consumption process continues at the demand rate until the end of cycle time T. Due to inspection error, some of the items used to fulfill the demand would be defective. These defective items are later returned to the inventory and are shown in Figure 3.1 as B2. The position of B2 is chosen to ensure the disposal of two defective batches (B1 and B2) at the same time. The cycle inventory is kept at a minimum level this way. It may be beneficial to sell the defective lot in two batches (π΅2β² and π΅2β²β²) in a cycle, as shown in Figure 3.2, where the holding cost of the returned inventory is less;
however, implicitly, this may incur additional cost such as transportation (fixed and variable cost), which makes the savings in inventory less than the additional costs incurred. So, in this
Q
B2
Inventory level
Z2
Z1
B1
Time
Ο1 t2
T
T
79 chapter, it is assumed here that the two types of defective items (screened and misclassified) are sold at the same time, as shown in Figure 3.1. The model in Figure 3.2 may be addressed in a technical note sometime in the near future.
Figure 3.2 Selling the defective items twice in a cycle
To avoid shortages, it is assumed that the number of nondefective items is at least equal to the adjusted demand, that is the sum of the actual demand (D) and items that are replaced for the ones returned (Ξ³m2 D) from the market over T. Thus
π β π(1β πΎ)π1β ππΎ(1β π2)β₯ π·π+πΎπ2π π(1β πΎ)β ππ1(1β πΎ)β₯ π·π
π(1β πΎ)(1β π1)β₯ π·π
So, for the limiting case, the cycle length (which includes the screening time) can be written as π= π(1β πΎ)(1β π1)
π· (3.1)
Inventory level
Z2
Z1
B1
Ο1 t2
T
π΅2β² π΅2β²β²
Time
80 It should be noted that the above expression is unaffected by the Type II error and reduces to the cycle length in Salameh and Jaber (2000) if the Type I error becomes zero.
Consider now the different cases of misclassifications that an inspection process can have.
There are four possibilities in such an inspection process. Those are: Case (1) A nondefective item is classified as nondefective; Case (2) A nondefective item is classified as defective; Case (3) A defective item is classified as nondefective and Case (4) A defective item is classified as defective. This scenario is depicted in Figure 3.3 below. The number of items going into different categories following these cases is given by:
Case (1): π(1β πΎ)(1β π1) Case (2): π(1β πΎ)π1 Case (3): ππΎπ2
Case (4): ππΎ(1β π2)
Now B1 and B2 are given by π΅1 =π(1β πΎ)π1+ππΎ(1β π2) π΅2 = ππΎπ2
Figure 3.3 Four possibilities in the inspection process
Inspection/Decision Picking an Item for Inspection
Nondefective (1 β Ξ³)
Defective (Ξ³)
Classify an item as defective (1 β m2)
Classify a bad item as Nondefective Classify an item as Nondefective (1 β m1)
Classify a good item as defective
81 The items in batch B2 are returned from the market at the rate ππΎπ2/π and are taken from the inventory with batch B1. Therefore the revenue from salvaging B = B1 + B2 items within a cycle are given by:
π 1 =π£(π΅1+π΅2) =ππ(1β πΎ)π1+πππΎ(1β π2) +πππΎπ2 or
π 1 =ππ(1β πΎ)π1+πππΎ
The revenue from selling the good items is computed as π 2 = π 1π(1β πΎ)(1β π1) +π 1ππΎπ2
So, the total revenue is given as π =π 1+π 2
π =π 1π(1β πΎ)(1β π1) +π 1ππΎπ2+ππ(1β πΎ)π1+πππΎ (3.2)
Consider now the different costs of the inventory system. The procurement cost per cycle is
ππΆ =π΄π+π1π (3.3)
where π1 is the variable cost. The screening cost per cycle is the sum of the costs of inspection and misclassifications which is given by
πΌπΆ(π) =ππ+ππ(1β πΎ)ππ1+ππππΎπ2 (3.4)
The holding cost per cycle is the cost of carrying the (i) nondefective lot, (ii) defective lot and (iii) returned lot. So, from Figure 3.1 the holding cost for a cycle can be written as
π»πΆ(π) =β οΏ½(π β π1)π1
2 +π1π1+π2π‘2
2 οΏ½+β οΏ½π΅2π 2 οΏ½
where Z represents the stock level before and after the screening/inspection process. Replacing B2 with QπΎm2, Z2 with Z1 β B1, Z1 with Q β DΟ1 and Ο1 with D/x, it can be written as
82 π»πΆ(π) =βππ2
2 οΏ½2 π₯ β
π· π₯2+π2
π· οΏ½+β οΏ½ππΎπ2π
2 οΏ½ (3.5)
where π = 1βπ·π₯β(π1β πΎ) +πΎ(π1+π2)
Therefore the total cost per cycle is given by summing up Eq. (3.3), Eq. (3.4) and Eq. (3.5), i.e.
πΆπ= ππΆ+πΌπΆ+π»πΆ or
πΆπ(π) =π΄π+π1π+ππ+ππ(1β πΎ)ππ1+ππππΎπ2+βπ
2 οΏ½οΏ½2 π₯ β
π· π₯2 +π2
π· οΏ½ π2+ππΎπ2ποΏ½
The term π2 is simplified in Appendix 2. Figure 3.1 depicts the behavior of different types of inventory in the order cycle. The triangle at the bottom represents the defective lot that is returned by the market and is accumulated into the salvaged lot.
The total profit per cycle can now be written as the difference between the total revenue and total cost per cycle, that is
ππ(π) =π 1π(1β πΎ)(1β π1) +π ππΎπ2+ππ(1β πΎ)π1+πππΎ β οΏ½π΄π+π1π+ππ+ππ(1β πΎ)ππ1+ππππΎπ2 +βπ
2 οΏ½οΏ½2 π₯ β
π· π₯2+π2
π· οΏ½ π2+ππΎπ2ποΏ½οΏ½(3.6) Since πΎ, π1 and π2 are assumed in this thesis to be independent and identically distributed random variables with probability density functions π1(πΎ), π2(π1) and π3(π2), respectively. The expected total profit can be written as
E[ππ(π)] =π 1π(1βE[πΎ])(1βE[π1]) +π πE[π2]E[πΎ] +ππ(1βE[πΎ])E[π1] +ππE[πΎ]β π΄πβ π1π β ππ β ππ(1βE[πΎ])πE[π1]β πππE[π2]E[πΎ]
ββπ
2 οΏ½οΏ½2 π₯ β
π·
π₯2+E[π2]
π· οΏ½ π2+πE[π2]E[π]E[πΎ]οΏ½
(3.7)
Now from Eq. (3.1), the expected cycle length would be:
E[π] =π(1βE[πΎ])(1βE[π1])
π· (3.8)
83 Maddah and Jaber (2008b) corrected the approach in the Salameh and Jaber (2000) model to determine the annual profit. They suggested using renewal reward theorem. Using this new approach, the expected annual profit for our model, is written as
E[πππ(π)] = E[ππ(π)]
E[π]
or
E[πππ(π)] =π 1π·+ π π·E[πΎ]E[π2]
(1βE[πΎ])(1βE[π1]) +
π£π·E[π1] (1βE[π1]) +
π£π·E[πΎ]
(1βE[πΎ])(1βE[π1])β
π· οΏ½π΄ππ +π1+π+ππ(1βE[πΎ])E[π1] +ππE[πΎ]E[π2] +βπ
2 οΏ½οΏ½2 π₯ β π·
π₯2+ E[π2] π· οΏ½ ποΏ½οΏ½
(1βE[πΎ])(1βE[π1]) ββππE[πΎ]E[π2]
2
(3.9)
It should be noted that Eq. (3.9) converges to the expected annual profit equation in Maddah and Jaber (2008b) once the value of errors goes to zero. It can be demonstrated that this expected annual profit follows a concave function. The first derivative of Eq. (3.9) is given by
d
dπE[πππ(π)] =βπ· οΏ½β π΄ππ2 +βπ
2 οΏ½οΏ½2 π₯ β π·
π₯2+ E[π2] π· οΏ½οΏ½οΏ½
(1βE[πΎ])(1βE[π1]) ββE[πΎ]E[π2] 2 The second derivative of Eq. (3.9) is
d2
dπ2E[πππ(π)] = β 2π΄ππ·
π3(1βE[πΎ])(1βE[π1])
Since π΄π> 0, D> 0, 0 < E[πΎ] < 1, and 0< E[m1] < 1, then d2
dπ2E[πππ(π)]< 0 for every π > 0, suggesting that the annual profit in Eq. (3.9) is concave. This fact is also demonstrated graphically in the next section. Thus, the optimal order size that represents the maximum annual profit, is determined by setting the first derivative equal to zero and solving for Q to get
84
πβ =οΏ½ 2π΄ππ·
βπE[πΎ]E[π2](1βE[πΎ])(1βE[π1]) +βππ· οΏ½2 π₯ β π·
π₯2+ E[π2] π· οΏ½
(3.10)
It should be noticed that the denominator in Eq. (3.10) remains always positive, since x > D, 0 , E[πΎ]< 1, 0< E[π1]< 1, βπ > 0, and E[π2] > 0 from Appendix 2. Note also that when Ξ³ = m1 = m2
= 0, Eq. (3.10) reduces to the traditional EOQ formula.