Abstract: This note deals with the optimal lot sizing decision at the technology selection stage, and modifies the optimal solution procedure in constant demand case described in Khouja (Omega 2005, 33, 47-53). This note develops an alternative approach to find the optimal lot sizing to improve the study of Khouja (Omega 2005, 33, 47-53). Finally, numerical examples are given to illustrate the result discussed in this study.
INTRODUCTION
Recently, Khouja[1] developed a model to determine the total cost per unit of time and the optimal order quantity at the technology selection stage. The cost of the technology depends on the lot size it can produce. In addition, the model investigated two different types of demand included constant demand and linearly decreasing demand. For convenience, we use notation and assumptions similar to Khouja[1]. Khouja[1] developed the following model for the total cost per unit of time over the life of a mold is:
| (1a) (1b) |
where:
| (2) |
and
| (3) |
Since TC1(Q0)=TC2(Q0) when U0 = uQ0, TC(Q) is continuous and well-defined. All TC1(Q), TC2(Q) and TC(Q) are defined on Q>0. Eq. 2 and 3 yield
| (4) |
| (5) |
| (6) |
and
| (7) |
Equation 5 and 7 imply that TC1(Q)
and TC2(Q) are convex on Q>0. Furthermore, we have
Let
| (8a) (8b) (8c) |
Equation 8a-c imply that TCi(Q) is decreasing on (0, Qi*] and increasing on [Qi*, ∞) for all i = 1, 2. Eq. 4 and 6 yield that:
| (9) |
and
| (10) |
Furthermore, we let
| (11) |
and
| (12) |
Then, we can find Δ1>Δ2 from Eq. 11 and 12. We can obtain optimal lot sizing Q* using following result.
Theorem 1
(A) |
If Δ2>0, then TC(Q*)= TC1(Q1*) and Q* = Q1*. |
(B) |
If Δ1>0 and Δ2≤0, then TC(Q*)= min {TC1(Q1*), TC2(Q2*)}. Hence, Q* is Q1* or Q2* associated with the least cost. |
(C) |
If Δ1≤0, then TC(Q*)= TC2(Q2*) and Q* = Q2*. |
Proof
(A) |
If Δ2>0 then Δ1>0. We have TC1’(Q0)>0 and TC2’(Q0)>0. Eq. 8a-c imply that |
| (i) | TC1(Q) is decreasing on (0, Q1*] and increasing on [Q1*, Q0). |
| (ii) | TC2(Q) is increasing on [Q0, ∞). |
Combining (i), (ii) and Eq. 1a and b, we have that TC(Q) is decreasing on (0, Q1*] and increasing on [Q1*, ∞). Consequently, Q* = Q1*.
(B) |
If Δ1>0 and Δ2≤0. We have TC1’(Q0)>0 and TC2’(Q0)≤0. Eq. 8a-c imply that |
| (i) | TC1(Q) is decreasing on (0, Q1*] and increasing on [Q1*, Q0). |
| (ii) | TC2(Q) is decreasing on [Q0, Q2*] and increasing on [Q2*, ∞). |
Combining (i), (ii) and Eq. 1a and b, we find that
| (iii) | TC(Q) is decreasing on (0, Q1*]. |
| (iv) | TC(Q) is increasing on [Q1*, Q0]. |
| (v) | TC(Q) is decreasing on [Q0 , Q2*]. |
| (vi) | TC(Q) is increasing on [Q2*, ∞). |
Hence TC(Q*)= min {TC1(Q1*), TC2(Q2*)}. Consequently, Q* is Q1* or Q2* associated with the least cost.
(C) |
If Δ1≤0 then Δ2≤0. We have TC1’(Q0)≤0 and TC2’(Q0)≤0. Eq. 8a-c imply that |
(i) |
TC1(Q) is decreasing on (0, Q0). |
(ii) |
TC2(Q) is decreasing on [Q0 , Q2*] and increasing on [Q2*, ∞). |
Combining (i), (ii) and Eq. 1a and b, we have that TC(Q) is decreasing on (0, Q2*] and increasing on [Q2*, ∞). Consequently, Q* = Q2*.
Incorporating the above arguments, we have completed the proof of Theorem 1.
Above Theorem 1 developed in this note is an alternative approach to determine the optimal lot sizing under minimizing the total cost per unit of time. However, Khouja[1] also developed a procedure to find the optimal solution in this situation. Khouja[1] suggested four cases to find the optimal solution. But we find case (d) can be deleted. Since Q1*≥Q0, we can easily obtain the sufficient condition for optimality of TC2(Q) is negative. That is, the Eq. 11 in Khouja[1] does not exist when Q1*≥Q0. It implies that Q2* does not exist. Therefore, case (d) in Khouja’s[1] optimal solution procedure does not exist. Theorem 1 developed in this note explains that after computing the numbers Δ1 and Δ2, we can immediately determine which one of Q1* or Q2* is optimal. Theorem 1 essentially modifies the solution procedure described in Khouja[1].
NUMERICAL EXAMPLES
To illustrate the results, let us apply the proposed method to solve the same numerical examples as Khouja[1]. Let h=$15/unit/year, g(Q)=10+ 8Q + 7Q2 (i.e. C1= 10, C2= 8 and C3= 7) and U= 12 if Q<10; otherwise U= 1.2Q (i.e. U0= 12).
Example 1: When S=$15/setup and D=30 units/year. Then, we have Δ2>0. Using Theorem 1-(A), we get Q* = Q1*= 4. TC(Q*)=TC1(Q1*)= $238.75/year.
Example 2: When S=$30/setup and D=72 units/year. Then, we have Δ1>0, Δ2<0, Q1*= 7 and Q2*= 19. Using Theorem 1-(B), we can find TC1(Q1*) = $711.64/year>TC2(Q2*) = $703.11/year. Therefore, Q* = Q2*= 19 and TC(Q*)=TC2(Q2*)= $703.11/year. These results are different from the numerical example 1 in Khouja[1] under same value of all parameters.
Example 3: When S=$70/setup and D=72 units/year. Then, we have Δ1<0. Using Theorem 1-(C), we get Q* = Q2*= 27. TC(Q*)=TC2(Q2*)= $827.77/year.
ACKNOWLEDGEMENT
We would like to thank the Chaoyang University of Technology to finance this manuscript.