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
Time Aug -99 Sep -99 Oct- 99 Nov- 99 Dec -99 Jan- 00 Feb- Mar- Apr- 00 00 00 May-Jun- 00 00 Jul- 00 Aug- Sep Oct- Nov- Dec 00-00 00 00-00 Jan- 01 Feb- Mar- Apr- 01 01 01 May-Jun- 01 01 Jul- 01 Aug- 01 Demand 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 Lumpiness in spare part demand is the nightmare for inventory managers.
(preventive maintenance) (maintenance plan)
Time Aug -99 Sep -99 Oct- 99 Nov- 99 Dec -99 Jan- 00 Feb- Mar- Apr- 00 00 00 May-Jun- 00 00 Jul- 00 Aug- Sep Oct- Nov- Dec 00-00 00 00-00 Jan- 01 Feb- Mar- Apr- 01 01 01 May-Jun- 01 01 Jul- 01 Aug- 01 Demand 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 Component 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 Lumpiness in spare part demand is triggered by the lumpiness in component repairs. the uncertainty of individual probable defective component generating spare part demand.
(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.
Estimate the demand distribution of spare part pp = tt TT0 dd tt aa tt TT0 tt if aa tt > 0 tt TT 0 DD tt ~BB aa tt, pp tt TT aa tt : amount of a certain kind of component which arrives at the repair organization in time period tt. dd tt : 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. DD tt : the r.v. of spare part demand in time period tt. TT 0 : time set of history. TT: time set of planning.
Determine the order size xx tt in each time period tt ff tt yy tt = min h PP(DD tt = dd)(yy tt dd) + xx tt 0 dd=0 + cc eeee PP(DD tt = dd)(dd yy tt ) + dd=0 ++++++++ +γγ PP(DD tt = dd)ff tt+1 yy tt+1 dd=0 + cc rr xx tt tt TT (1) yy tt+1 = (yy tt dd) + + xx tt tt TT (2) ff TT+1 = ss yy TT+1 (3) DD tt ~BB aa tt, pp tt TT (4) ff tt : the optimal total discounted cost from time tt until the end of time horizon xx tt : order size in time period tt. yy tt : on hand inventory after the arrival of order due in period tt, xx tt 1. h: holding cost per time unit per item. cc rr : regular replenishment cost. cc eeee : emergency replenishment cost. ss: salvage value. γγ: discounted factor.
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). 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 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.
h = 0.2, cc rr = 10, cc eeee = 20, ss = 5, γγ = 1, lead time = 1. Benchmark: spare part demand DD tt ~PPPPPPPPPPPPPP λλ with λλ = tt TT dd 0 tt TT 0 and tt TT. cost With maintenance plan Poisson % cost Red. Total cost Ordering cost Inventory holding cost Penalty cost Salvage value 558147.6 625431.2 10.80% 427300 354080-20.70% 70537.6 82496.2 14.50% 115120 237700 51.60% 54810 48845-12.20% Factors influence the performance of with maintenance plan: 1. pp. Larger failure probability, better performance of the 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.
Problem: Is component foreknowledge(maintenance plan) always available? Experience in Fokker Service. Empirical Guidance.
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 with maintenance plan. The larger the failure probability, the better performance the 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 performance. The availability of maintenance plan is a practical problem in some fields.
Using Maintenance Plan in Spare Part Demand Forecasting and Inventory Control Question / Comments? Thank you