Optimal screening procedures for items with a random number of defects

Ji Hwan & Maxim Finkelstein

IMA Journal of Management Mathematics2026https://doi.org/10.1093/imaman/dpag012article
AJG 2
Weight
0.50

What the paper says

In this paper, we study optimal screening procedures for items with random number of defects. Each defect that causes an item’s failure during this screening procedure is repaired/removed. Upon observing the numbers of removed defects, a decision is made whether to discard an item or to justify its future field operation. It is shown that these decisions depend on the distribution of the number of defects in an item. Three discrete distributions are considered: negative binomial, Poisson and binomial. It is shown, e.g. for the negative binomial case, that screening out of items with any number of removed defects improves the quality of remaining items. On the other hand, for the Poisson distribution of defects, there is no need to screen out items, as the distribution of the number of remaining defects does not depend on the number of the removed defects. The optimal screening policies to minimize the corresponding expected cost functions for each case are analyzed. The numerical illustrations of the obtained results are provided. Through the numerical examples, it is shown that the optimal screening policy significantly differs depending on the distribution of the number of defects in an item as well as the involved costs.

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https://doi.org/https://doi.org/10.1093/imaman/dpag012

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@article{ji2026,
  title        = {{Optimal screening procedures for items with a random number of defects}},
  author       = {Ji Hwan & Maxim Finkelstein},
  journal      = {IMA Journal of Management Mathematics},
  year         = {2026},
  doi          = {https://doi.org/https://doi.org/10.1093/imaman/dpag012},
}

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Evidence weight

0.50

Balanced mode · F 0.40 / M 0.15 / V 0.05 / R 0.40

F · citation impact0.50 × 0.4 = 0.20
M · momentum0.50 × 0.15 = 0.07
V · venue signal0.50 × 0.05 = 0.03
R · text relevance †0.50 × 0.4 = 0.20

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