Using Maintenance Plan in Spare Part Demand

Forecasting and Inventory Control

Sha Zhu, Willem van Jaarsveld, Rommert Dekker

Erasmus University Rotterdam

Problem background Model Case study Conclusions

model conclusions case study problem background

Lumpiness in spare part demand is the nightmare for inventory managers.

Time Aug-99

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Dem- and 0 0 0 0 0 3 3 0 0 0 0 0 0 0 0 0 0 4 0 0 0 0 16 4 0

model conclusions case study problem background

(preventive maintenance) (maintenance plan)

model conclusions case study problem background

Lumpiness in spare part demand is triggered by • the lumpiness in component repairs. • the uncertainty of individual probable defective component

generating spare part demand.

Time Aug-99

Sep-99

Oct-99

Nov-99

Dec-99

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Mar-00

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Dem- and 0 0 0 0 0 3 3 0 0 0 0 0 0 0 0 0 0 4 0 0 0 0 16 4 0

Comp-onent repair

1 1 5 0 0 4 7 0 0 7 0 0 5 0 0 0 3 9 0 0 0 3 20 4 1

model conclusions case study problem background

(preventive maintenance) (maintenance plan)

(spare part demand forecasting) (inventory policy)

• Consider a periodic-review, lost sales spare part inventory system in repair organization.

• Parts, installed in the defective components, generate the demand of spare parts if they fail in the initial inspection.

• Demand for a spare part is immediately satisfied by stock if there is one available.

• The stocks are replenished from an external supplier with a constant lead time and regular unit shipment cost.

• If there is a stockout, the demand is satisfied by an emergency shipment with a shorter lead time and unit penalty cost.

• Assume that regular order arrives next time period after that order was made and the supplier has no stock out.

problem background conclusions case study model

problem background conclusions case study model

�̂�𝑝 =∑ 𝑑𝑑𝑡𝑡𝑡𝑡∈𝑇𝑇0∑ 𝑎𝑎𝑡𝑡𝑡𝑡∈𝑇𝑇0

if ∑ 𝑎𝑎𝑡𝑡 > 0𝑡𝑡∈𝑇𝑇0

𝐷𝐷�𝑡𝑡~𝐵𝐵 𝑎𝑎𝑡𝑡, �̂�𝑝 𝑡𝑡 ∈ 𝑇𝑇

Estimate the demand distribution of spare part

𝑎𝑎𝑡𝑡: amount of a certain kind of component which arrives at the repair organization in time period 𝑡𝑡.

𝑑𝑑𝑡𝑡: spare part demand in that component repair in time period 𝑡𝑡.

𝑝𝑝: the part failure probability in the defective component received by the repair organization.

𝐷𝐷𝑡𝑡: the r.v. of spare part demand in time period 𝑡𝑡.

𝑇𝑇0: time set of history.

𝑇𝑇: time set of planning.

problem background conclusions case study model

𝑓𝑓𝑡𝑡 𝑦𝑦𝑡𝑡 = min𝑥𝑥𝑡𝑡≥0

ℎ� 𝑃𝑃(𝐷𝐷�𝑡𝑡 = 𝑑𝑑)(𝑦𝑦𝑡𝑡 − 𝑑𝑑)+∞

𝑑𝑑=0+ 𝑐𝑐𝑒𝑒𝑒𝑒� 𝑃𝑃(𝐷𝐷�𝑡𝑡 = 𝑑𝑑)(𝑑𝑑−𝑦𝑦𝑡𝑡)+

∞

𝑑𝑑=0

++++++++ +𝛾𝛾� 𝑃𝑃(𝐷𝐷�𝑡𝑡 = 𝑑𝑑)𝑓𝑓𝑡𝑡+1 𝑦𝑦𝑡𝑡+1∞

𝑑𝑑=0+ 𝑐𝑐𝑟𝑟 ∙ 𝑥𝑥𝑡𝑡 𝑡𝑡 ∈ 𝑇𝑇 (1)

Determine the order size 𝑥𝑥𝑡𝑡 in each time period 𝑡𝑡

𝑓𝑓𝑡𝑡 ∙ : the optimal total discounted cost from time 𝑡𝑡 until the end of time horizon

𝑥𝑥𝑡𝑡: order size in time period 𝑡𝑡.

𝑦𝑦𝑡𝑡: on hand inventory after the arrival of order due in period 𝑡𝑡, 𝑥𝑥𝑡𝑡−1.

ℎ: holding cost per time unit per item.

𝑐𝑐𝑟𝑟: regular replenishment cost. 𝑐𝑐𝑒𝑒𝑒𝑒: emergency replenishment cost.

𝑠𝑠: salvage value. 𝛾𝛾: discounted factor.

𝑦𝑦𝑡𝑡+1 = (𝑦𝑦𝑡𝑡 − 𝑑𝑑)+ + 𝑥𝑥𝑡𝑡 𝑡𝑡 ∈ 𝑇𝑇 (2)

𝑓𝑓𝑇𝑇+1 = −𝑠𝑠 ∙ 𝑦𝑦𝑇𝑇+1 (3)

𝐷𝐷�𝑡𝑡~𝐵𝐵 𝑎𝑎𝑡𝑡, �̂�𝑝 𝑡𝑡 ∈ 𝑇𝑇 (4)

problem background conclusions model case study

• Data set contains information over 100,000 repairs at Fokker Service during the period from 01-01-2000 till 28-02-2010 (122 months, 17012 types of spare parts, 3329 types of components).

• Consider rolling horizon in setting experiment: assume the repair organization could have the component arrival information of the coming 3 months.

• Training set: the first 84 months. Test set: the last 38 months.

• Exclude the cases where more than one spare part of a certain type are installed in the component.

• Spare parts of the same type but installed in different types of components are treated as different types of spare parts.

Component Repaired Piece Part Quantity Used Unit of Measure DOC_DATE

406 1923 1 EA 02-01-2000

1502 2211 6 EA 03-01-2000

problem background conclusions model case study

• ℎ = 0.2, 𝑐𝑐𝑟𝑟 = 10, 𝑐𝑐𝑒𝑒𝑒𝑒 = 20, 𝑠𝑠 = 5, 𝛾𝛾 = 1, lead time = 1.

• Benchmark: spare part demand 𝐷𝐷�𝑡𝑡~𝑃𝑃𝑃𝑃𝑃𝑃𝑠𝑠𝑠𝑠𝑃𝑃𝑃𝑃 𝜆𝜆 with �̂�𝜆 =∑ 𝑑𝑑𝑡𝑡𝑡𝑡∈𝑇𝑇0

𝑇𝑇0 and 𝑡𝑡 ∈ 𝑇𝑇.

• Factors influence the performance of model with maintenance plan:

1. 𝑝𝑝. Larger failure probability, better performance of the model with maintenance plan.

2. Length of time period in which the component arrival information can be obtained by the

repair organization in advance. The longer, the better performance of the model.

cost With maintenance plan Poisson % cost Red.

Total cost 558147.6 625431.2 10.80% Ordering cost 427300 354080 -20.70%

Inventory holding cost 70537.6 82496.2 14.50% Penalty cost 115120 237700 51.60%

Salvage value 54810 48845 -12.20%

problem background case study model conclusions

• Problem: Is component foreknowledge(maintenance plan) always available?

• Experience in Fokker Service.

• Empirical Guidance.

problem background case study model conclusions

• Maintenance plan provides the component arrival information for the repair organization. Using the maintenance plan would improve the spare part demand forecasting and reduce the inventory cost.

• The estimated failure probability is the main factor which influences the performance of the model with maintenance plan. The larger the failure probability, the better performance the model would achieve.

• The length of time period in which the component arrival information can be obtained by the repair organization in advance is another factor which has an impact on the model performance.

• The availability of maintenance plan is a practical problem in some fields.

Using Maintenance Plan in Spare Part Demand Forecasting and Inventory Control

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