CHAPTER 4 ECONOMIC ORDER QUANTITY MODEL FOR ITEMS WITH IMPERFECT
5.3 Model Extensions
119
π=οΏ½β 2π΄π£π·
π£2οΏ½1βπ·ποΏ½ββππ =1βπ ππ +π· βππ =1βπ£1π οΏ½ππ 2EοΏ½ππ 2οΏ½οΏ½1+Eπ₯οΏ½πΎπ οΏ½οΏ½+ππ Eπ[ππ ]οΏ½1βEοΏ½πΎπ οΏ½βπ·π₯οΏ½οΏ½+2βππ =1βπ£1π ππ (1βE[ππ ])οΏ½1βπ·ποΏ½ (5.12) The denominator of Eq. (5.10) is positive since D < P, 0 < E[ππ ] < 1, and E[πΎπ ] +π·π₯ < 1.
To validate this approximation, one thousand examples were tried. That is, the exact and approximate production quantities were computed through Eq. (5.10) and Eq. (5.12) respectively. The difference between the costs calculated with these production quantities, was almost zero in one thousand cases. Thus, a solution procedure for the last two mechanisms would be as follows:
1. Estimate an approximate production quantity, using Eq. (5.12).
2. Estimate an approximate multiplier for supplier s using Eq. (5.11).
3. Determine integer values of the multipliers as βπΎπ β and βπΎπ β
4. Determine an exact production quantity for each combination of the multipliers from step 3, using Eq. (5.10).
5. Determine an annual cost using Eq. (5.9), for each combination from step 3, using Q from step 4.
6. Determine an optimal annual cost of the supply chain as the minimum of the costs from step 4. This will indicate the optimal production quantity and the optimal set of multipliers.
120 while some defective items as non-defectives. In other words, they will attribute a percentage of defective to each supplier, different from the actual one. Thus, the defective items of type s classified by the inspection process would be
ππ β² =πππ (1βE[πΎπ ])E[π1] +πππ E[πΎπ ](1βE[π2])
=πππ {(1βE[πΎπ ])E[π1] + E[πΎπ ](1βE[π2])} =πππ E[ππ ]
Thus, the fraction accommodating the leftovers of type s in a cycle, will be given as E[ππ π] = 1β(E[πmax]βE[ππ ])
where E[πmax] = max{E[ππ ], π = 1,2, β¦ ,π} and Eq. (5.1) can be written as
ππ =πππ {1β(E[πmax]βE[ππ ])} =πππ E[ππ π] (5.13)
So, the vendorβs total cost of the raw materials can now be written as
πΆπ£π(π) =βππ =1οΏ½(ππ£π +ππ )πππ E[ππ π] +βπ£1π 2π2οΏ½ππ 2EοΏ½ππ π2 οΏ½(1+E[ππ₯ π ])+οΏ½ππ E[ππ π π]οΏ½ οΏ½1β E[ππ ]βπ·π₯οΏ½οΏ½+ βπ£1π πππ (1βE[ππ π])οΏ½E[π]βπποΏ½οΏ½
(5.14)
The defective raw material misclassified by an inspector ends up making a defective product. This is assumed to cost the vendor an extra ππE[π2]π βππ =1E[πΎπ ]. This may be taken as a goodwill cost or warranty cost. The loss due to misclassifying nondefective raw material is neglected for simplicity here. The rest of the model remains the same as in section 5.2.
5.3.2 Learning in Vendorβs Production Process
In this section, it is assumed that the vendorβs production process follows Wright (1936) learning curve. That is, vendor produces the final product at an increasing production rate which is consumed at a constant rate. Let us assume that Tpi, Tdi and Ti are the production time, depletion time and the cycle time, respectively, in any cycle, as shown in Figure 5.3. The process produces a fixed quantity Q and builds up a maximum inventory Zi, in each cycle i. The level of inventory in each cycle can be expressed as a function of time as
121 Ξ¦π(π‘) =οΏ½π(π‘)β π·π‘ 0 <π‘ <πππ
π·ππ β π·π‘ πππ <π‘ <ππ (5.15)
Let us now assume that b is the learning exponent, while 0 β€ bi< 1 is the learning exponent in cycle i of production. Faster learning is associated with higher values of b. The production time in a cycle i is written as
πππ =β«(πβ1)πππ π1π₯βππππ₯ or
πππ =π1π1βπποΏ½π1βππ(1βπβ(πβ1)1βπποΏ½
π) (5.16)
Figure 5.3 Vendorβs inventory of the final product with learning in ith cycle
where T1 = 1/P is the time to assemble (produce) the first unit on the learning curve. To understand the changing learning rate, assume that the vendor produces x units in one cycle and y units in the next. If T1 and T2 are times to produce the first unit in the two cycles, the time to produce the yth unit can be written as
ππ¦ = π1(π₯+π¦)βπ (5.17)
and
Time Zi
Tdi
Tpi
Ti
Inventory level
122 ππ¦ = π2π¦βπ2 or
ππ¦ = π1(π₯+ 1)βππ¦βπ2 (5.18)
Equating expressions (5.17) and (5.18), it can be written as π2 = π[log(π₯+π¦)βlog(π₯+1)]
log(π¦)
In case of producing a fixed quantity Q in each cycle, the new learning rate can be written as ππ =π[log(ππ)βlog{(πβ1)π+1}]
log(π) (5.19)
Now, the average inventory of finished products in a cycle i can be written as
β« Ξ¦0ππ π(π‘)ππ‘= β«0πππ(π β π·π‘)ππ‘+πππ2ππ After simplification, it can be written as
β« Ξ¦0ππ π(π‘)ππ‘= π2π·2βπ1π2βππ(1βποΏ½π1βππβ(πβ1)1βπποΏ½
π)(2βππ) (5.20)
So, the vendorβs total cost of the finished products in cycle i would be πΆπ£π(π) =π΄π£+ππ1π1βπποΏ½π(1βπ1βππβ(πβ1)1βπποΏ½
π) +βπ£2οΏ½2π·π2βπ1π2βππ(1βποΏ½π1βππβ(πβ1)1βπποΏ½
π)(2βππ) οΏ½
Inventory buildup behaves linearly when learning in production is fast. Therefore, using Eq. (5.2), the vendorβs total cost of the raw material, in cycle i would be
πΆπ£π(π)=βππ =1οΏ½(ππ£π +ππ )πππ E[ππ ] +βπ£1π πππ (1βE[ππ ])οΏ½E[π]βπποΏ½+ βπ£1π 2π2οΏ½ππ 2EοΏ½ππ 2οΏ½(1+E[πΎπ₯ π ])+πβππππ E[ππ ]π(1βπ1οΏ½π1βππβ(πβ1)1βπποΏ½
π) οΏ½1βE[πΎπ ]βπ·π₯οΏ½οΏ½οΏ½
So, the vendorβs total cost in a cycle would be
123 πΆπ£(π)=π΄π£+
β οΏ½(ππ£π +ππ )πππ E[ππ ] +βπ£1π 2π2οΏ½ππ 2EοΏ½ππ 2οΏ½(1+E[πΎπ₯ π ])+πβππππ E[ππ ]π(1βπ1οΏ½π1βππβ(πβ1)1βπποΏ½
π) οΏ½1βE[πΎπ ]β
ππ =1
π·
π₯οΏ½οΏ½+βπ£1π πππ (1βE[ππ ])οΏ½E[π]βπποΏ½οΏ½+ππ1π1βπποΏ½π(1βπ1βππβ(πβ1)1βπποΏ½
π) +
βπ£2οΏ½π2π·2β π1π2βππ(1βποΏ½π1βππβ(πβ1)1βπποΏ½
π)(2βππ) οΏ½ and the vendorβs annual cost would be E[ππΆππ£(π)]=π΄π£π·
π +βπ£2οΏ½π2 βπ1π·π1βππ(1βποΏ½π1βππβ(πβ1)1βπποΏ½
π)(2βππ) οΏ½+π· βππ =1οΏ½βπ£1π 2ποΏ½ππ 2EοΏ½ππ 2οΏ½(1+E[πΎπ₯ π ])+
πβππππ E[ππ ]π1οΏ½π1βππβ(πβ1)1βπποΏ½
(1βππ) οΏ½1βE[πΎπ ]βπ·π₯οΏ½οΏ½+ (ππ£π +ππ )ππ E[ππ ]οΏ½+
βππ =1βπ£1π πππ (1βE[ππ ])οΏ½1βπ·ποΏ½+ππ·π1πβπποΏ½π(1βπ1βππβ(πβ1)1βπποΏ½
π)
(5.21)
Using Eq. (5.5), the total annual cost of the supply chain for the equal cycle time can be written as
E[ππΆπ(π)]=(π΄π£+βππ =1π΄π )π·
π +βπ£2οΏ½π2βπ1π·π1βππ(1βποΏ½π1βππβ(πβ1)1βπποΏ½
π)(2βππ) οΏ½+ π· β οΏ½βπ£1π 2ποΏ½ππ 2EοΏ½ππ 2οΏ½(1+E[πΎπ₯ π ])+πβππππ E[ππ ]π(1βπ1οΏ½π1βππβ(πβ1)1βπποΏ½
π) οΏ½1βE[πΎπ ]βπ·π₯οΏ½οΏ½+
ππ =1
(ππ£π +ππ )ππ E[ππ ]οΏ½+βππ =1βπ£1π ππ’π (1βE[ππ ])οΏ½1βπ·ποΏ½+ππ1π·πβπποΏ½π(1βπ1βππβ(πβ1)1βπποΏ½
π)
(5.22)
Eq. (5.22) is convex in Q (see Appendix 6 for proof). An iterative procedure will be carried out to determine the level of learning, production quantity and the annual cost for ten cycles of learning. An average of these measures will be used to compare the results with those for the other scenarios studied in the chapter.
Using Eqs. (5.8) and (5.22), the annual cost of the supply chain in a cycle, for the integer multiplier mechanism, would be
E[ππΆπ(π)]=π·οΏ½π΄π£+β
π΄π πΎπ ππ =1 οΏ½
π +βπ£2οΏ½π2βπ1π·π1βππ(1βποΏ½π1βππβ(πβ1)1βπποΏ½
π)(2βππ) οΏ½+ (5.23)
124 π· β οΏ½βπ£1π 2ποΏ½ππ 2EοΏ½ππ 2οΏ½(1+E[πΎπ₯ π ])+πβππππ E[ππ ]π(1βπ1οΏ½π1βππβ(πβ1)1βπποΏ½
π) οΏ½1βE[πΎπ ]βπ·π₯οΏ½οΏ½+
ππ =1
(ππ£π +ππ )ππ E[ππ ]οΏ½+βππ =1βπ£1π πππ (1βE[ππ ])οΏ½1βπ·ποΏ½+
ππ·π1πβπποΏ½π1βππβ(πβ1)1βπποΏ½
(1βππ) +π βππ =1(πΎ2π β1)βπ ππ
For an approximate value of the multipliers in Eq. (5.23), substituting Ksfrom Eq. (5.11) in Eq. (5.23), we get
E[ππΆπ(π)]=π΄π£π·
π +βππ =1οΏ½οΏ½2π΄π π·βπ ππ βπβ2π ππ οΏ½+βπ£2οΏ½π2 βπ1π·π1βππ(1βποΏ½π1βππβ(πβ1)1βπποΏ½
π)(2βππ) οΏ½+ π· β οΏ½βπ£1π 2ποΏ½ππ 2EοΏ½ππ 2οΏ½(1+E[πΎπ₯ π ])+πβππππ E[ππ ]π(1βπ1οΏ½π1βππβ(πβ1)1βπποΏ½
π) οΏ½1βE[πΎπ ]βπ·π₯οΏ½οΏ½+
ππ =1
(ππ£π +ππ )ππ E[ππ ]οΏ½+ βππ =1βπ£1π πππ (1βE[ππ ])οΏ½1βπ·ποΏ½+
ππ·π1πβπποΏ½π1βππβ(πβ1)1βπποΏ½
(1βππ)
(5.24)
The convexity of Eq. (5.24) can be shown in a similar manner to that of Eq. (5.22) (see Appendix 6 for proof). Again, an approximate value of the production quantity and the learning rate would be computed through iterating Eq. (5.25). This will be used to calculate the real- numbered values of multipliers by employing Eq. (5.11). The integer multipliers would be the values βπΎπ β, βπΎπ β respectively. The minimum for the second policy will be found by plugging these integer values in Eq. (5.24). Again, an average of the ten cycles of learning will be calculated here.
5.3.3 Learning in Suppliersβ Quality
In this section, it is assumed that the percentage of defectives per lot, from each supplier decreases following a learning curve. This improved quality may be attributed to the human learning in production and/or inspection at suppliersβ end. Jaber et al. (2008) discovered this behavior in the items of an automotive industry. They showed that the data follows a logistic learning curve of the form
125
πΎ(π) =π+ππππ (5.25)
where a and g are the fit parameters while b is the learning rate and i is the number of shipments.
Substituting this in Eq. (5.9), it becomes
E[ππΆπ(π)] =π·οΏ½π΄π£+βπππ =1π΄π πΎπ οΏ½+ππ·π +π2οΏ½βπ£2οΏ½1βπ·ποΏ½+βπ (πΎπ β1)βπ ππ
π =1 οΏ½
+π· βππ =1οΏ½(ππ£π +ππ )ππ E[ππ ] +βπ£1π 2ποΏ½π’π 2EοΏ½ππ 2οΏ½(1+E[πΎπ₯ π (π)])+οΏ½ππ E[ππ π ]οΏ½ οΏ½1β E[πΎπ (π)]βπ·π₯οΏ½οΏ½οΏ½+βππ =1βπ£1π πππ (1βE[ππ ])οΏ½1βπ·ποΏ½
(5.26)
The second derivative test proves the convexity of above cost function, as in Eq. (5.6). This annual cost function will be iterated through ten shipments. The average level of quality (Ξ³s) will be used to approximate Eq. (5.12) and the procedure to find the multipliers will remain the same as in section 5.2.
5.3.4 Integrated Model
It would be more realistic to consider all the human factors at the same time in our model.
To do this, the inspection errors and learning in supplierβs quality will have to be incorporated in Eq. (5.22), i.e. the case of learning in production. The resulting equation would be
E[ππΆπ(π)]=(π΄π£+βππ =1π΄π )π·
π +βπ£2οΏ½π2βπ1π·π1βππ(1βποΏ½π1βππβ(πβ1)1βπποΏ½
π)(2βππ) οΏ½+ π· β οΏ½βπ£1π 2ποΏ½ππ 2EοΏ½ππ π2 οΏ½(1+E[πΎπ₯ π (π)])+πβππππ E[ππ π]π(1βπ1οΏ½π1βππβ(πβ1)1βπποΏ½
π) οΏ½1βE[πΎπ (π)]β
ππ =1
π·
π₯οΏ½οΏ½+ (ππ£π +ππ )ππ E[ππ π]οΏ½+ βππ =1βπ£1π ππ’π (1βE[ππ π])οΏ½1βπ·ποΏ½+
ππ·π1πβπποΏ½π1βππβ(πβ1)1βπποΏ½
(1βππ)
(5.27)
Similarly, for the integer multiplier mechanism, the annual cost would be
E[ππΆπ(π)]=π·οΏ½π΄π£+β
π΄π πΎπ ππ =1 οΏ½
π +βπ£2οΏ½π2βπ1π·π1βππ(1βποΏ½π1βππβ(πβ1)1βπποΏ½
π)(2βππ) οΏ½+ (5.28)
126 π· β οΏ½βπ£1π 2ποΏ½ππ 2EοΏ½ππ π2 οΏ½(1+E[πΎπ₯ π (π)])+πβππππ E[ππ π]π(1βπ1οΏ½π1βππβ(πβ1)1βπποΏ½
π) οΏ½1βE[πΎπ (π)]β
ππ =1
π·
π₯οΏ½οΏ½+ (ππ£π +ππ )ππ E[ππ π]οΏ½+βππ =1βπ£1π πππ (1βE[ππ π])οΏ½1βπ·ποΏ½+
ππ1π1βπποΏ½π1βππβ(πβ1)1βπποΏ½
(1βππ) +π βππ =1(πΎ2π β1)βπ ππ
Eq. (5.28) is convex in Q. A similar proof to the one in Appendix 6 can be applied here. An average of ten cycles of learning will be computed here for the production size and the annual cost.