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MODULE 2: WARRANTY COST ANALYSIS Professor D.N.P. Murthy The University of Queensland Brisbane, Australia

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WARRANTY COST ANALYSIS Modeling Warranty Costs Perspectives and Cost Bases Probabilistic Elements Cost Models for the FRW and PRW Other Cost Models Information Needs

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WARRANTY MODELING Reasons for modeling Marketing (Sales) Economic (Cost) Engineering (Design) Operational (Servicing) Many disciplines involved

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COST MODELS Costs depend on: Type of warranty Failure pattern of items Repairability Failure pattern of replaced or repaired items Incidental costs

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FAILURES Occur randomly Depend on Product characteristics Usage rate Age Maintenance Environmental factors

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FAILURES UNDER WARRANTY

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COST CALCULATION BASIS Per item sold Item plus replacements under warranty Life cycle of item over fixed time horizon Per unit of time (when sales occur over time)

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ANALYSIS OF SELLER’S COSTS Costs also depend on proportion of legitimate claims made proportion of bogus claims servicing policy administrative costs incidental costs (e.g., shipping)

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MODELLING PROCESS

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SIMPLE MODEL

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DETAILED MODEL

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MODEL ASSUMPTIONS All claims made All claims are legitimate Claims are made instantaneously Instantaneous repair or replacement Fixed cost per claim

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MODEL ASSUMPTIONS Identical items Independence (statistical) No brand switching All parameters known

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PROBABILITY Involves life distribution of the items X = time to failure (lifetime) of the item F(x) = P(X < x) Most common distribution: exponential F(x) = 1 - e - x x > 0.

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EXPONENTIAL DISTRIBUTION Mean Time to Failure (MTTF): Median Time to Failure:.6931/.6931/ Constant failure rate, Constant failure rate, Example: If the failure rate is known to be = 1.25 per year, then MTTF =.8 yr.

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WEIBULL DISTRIBUTION More General Distribution F(x) = 1 - exp{-( x) } x > 0 Failure rate may be decreasing ( < 1) constant ( = 1) increasing ( > 1)

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WEIBULL DISTRIBUTION MTTF and standard deviation depend of and. Example: = 1.25, as before. If = 1 (exponential), = 0.8 and = 0.8. If =.5 (DFR), = 1.6 and = 4. If = 2 (IFR), = 0.71 and = 0.37.

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SOME NOTATION T = Average time to failure of all items that fail with lifetimes less than T M(T) = Expected number of failures in the interval from 0 to T (“Renewal function;” gotten from tables.) c s = seller’s average cost per item c r = average repair cost

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NONRENEWING FRW NONREPAIRABLE ITEMS Basis: Seller’s total cost per unit sold Ave. cost = c s [1 + M(W)] (W = length of warranty period)

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NONRENEWING FRW NONREPAIRABLE ITEMS Example: TV picture tube, exponential time to failure with MTTF = 6.5 years = 1/6.5 =.1538. Take W = 6 months and 1 year, and c s = $67.20. = 1/6.5 =.1538. Take W = 6 months and 1 year, and c s = $67.20.

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NONRENEWING FRW NONREPAIRABLE ITEMS [Cont.] For the exponential, M(t) = t. Thus: Six-month warranty: Ave. cost = $67.20[1+.1538(.5)] = $72.37 Ave. cost = $67.20[1+.1538(.5)] = $72.37 One-year warranty: Ave. cost = $67.20[1+.1538] = $77.54 Ave. cost = $67.20[1+.1538] = $77.54

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NONRENEWING FRW REPAIRABLE ITEMS Additional assumption: repaired items are “good-as-new” (i.e., have same failure distribution as new items). Ave. cost = c s + c r M(W)

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NONRENEWING PRW Equivalent to rebate form Non-repairable items c b = buyer’s cost of item (selling price) Ave. cost to seller = c s + c b [F(W) - W /W] Formulas for W are needed. (See: Book on Warranty Cost Analysis)

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NONRENEWING PRW Example: Same as previously, with c b = $105.00. One-year warranty. From formula in text, for W = 1, W =.0694. Ave. cost = $67.20 + 105[1 - e -.1538 -.0674] = $74.88. Comments: (1) This is less than for FRW (2) Cost is less for renewing PRW. (2) Cost is less for renewing PRW.

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INFORMATION NEEDS Form of distribution Parameter values Type of warranty Rectification policy Other cost information

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DATA MAY INCLUDE Test data Claims data Data on similar products Part and component data Vendor data Subjective information

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OTHER COST MODELS Many other warranties Distributions other than exponential Two- and higher-dimensional versions Life-cycle cost models Other cost bases Discounting to present value Indifference pricing

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REDUCING WARRANTY COSTS Improve reliability Better quality control Optimal servicing strategy (e.g., repair versus replace) Monitoring warranty claims – prevention of fraud Effective management of warranty logistics

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FURTHER READING Blischke, W. R., and D. N. P. Murthy (1994), Warranty Cost Analysis, Marcel Dekker, Inc., New York Blischke, W.R. and D.N.P. Murthy (eds). (2002), [Contains two Blischke, W.R. and D.N.P. Murthy (eds). (2002), Case Studies in Reliability and Maintenance, Wiley, New York [Contains two case studies]

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FURTHER READING For more recent literature, see: DNP Murthy and I Djamaludin, Product warranty – A review, International Journal of Production Economics, 79 (2002), 231-260.

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