2012
DOI: 10.1016/j.orl.2012.06.009
|View full text |Cite
|
Sign up to set email alerts
|

A primal–dual algorithm for computing a cost allocation in the core of economic lot-sizing games

Abstract: 5We consider the economic lot-sizing game with general concave ordering cost functions. It is 6 well-known that the core of this game is nonempty when the inventory holding costs are linear. 7The main contribution of this work is a combinatorial, primal-dual algorithm that computes a cost 8 allocation in the core of these games in polynomial time. We also show that this algorithm can be 9 used to compute a cost allocation in the core of economic lot-sizing games with remanufacturing 10 under certain assumption… Show more

Help me understand this report

Search citation statements

Order By: Relevance

Paper Sections

Select...
2
1
1
1

Citation Types

0
7
0

Year Published

2013
2013
2016
2016

Publication Types

Select...
5

Relationship

1
4

Authors

Journals

citations
Cited by 10 publications
(7 citation statements)
references
References 11 publications
0
7
0
Order By: Relevance
“…Lemma 1 (Chen andZhang 2006, Gopaladesikan et al 2012). Let x * and α * respectively be optimal solutions to f r (R, s r−1 ) and its dual, obtained using the algorithm by Gopaladesikan et al (2012). Then, x * and α * satisfy the following properties:…”
Section: The Economic Lot Sizing Gamementioning
confidence: 99%
See 3 more Smart Citations
“…Lemma 1 (Chen andZhang 2006, Gopaladesikan et al 2012). Let x * and α * respectively be optimal solutions to f r (R, s r−1 ) and its dual, obtained using the algorithm by Gopaladesikan et al (2012). Then, x * and α * satisfy the following properties:…”
Section: The Economic Lot Sizing Gamementioning
confidence: 99%
“…. , (a m , b m )} with a 1 = 1 and b m = T , obtained by the algorithm of Gopaladesikan et al (2012).…”
Section: Dynamic Cost Allocationmentioning
confidence: 99%
See 2 more Smart Citations
“…If the production is made at one period, then the capacity is 3 + η and the total cost is κ(3 + η) + 1; if the production is made at two periods, then the capacity is (3 + η)/2 and the total cost is κ(3 + η)/2 + 2; if the production is made at three periods, then the capacity is (3 + η)/3 and the total cost is κ(3 + η)/3 + 3. For κ = 1/2, it is optimal to set the capacity level at 3 + η when η ∈ [0, 1], the capacity level at (3 + η)/2 when η ∈ [1,9] and the capacity level at (3 + η)/3 when η ≥ 9. Thus, the optimal capacity level may decrease with demand.…”
Section: Inventory Carry Overmentioning
confidence: 99%