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

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**MODULE 2: WARRANTY COST ANALYSIS**Professor D.N.P. Murthy The University of Queensland Brisbane, Australia**WARRANTY COST ANALYSIS**• Modeling Warranty Costs • Perspectives and Cost Bases • Probabilistic Elements • Cost Models for the FRW and PRW • Other Cost Models • Information Needs**WARRANTY MODELING**• Reasons for modeling • Marketing (Sales) • Economic (Cost) • Engineering (Design) • Operational (Servicing) • Many disciplines involved**COST MODELS**Costs depend on: • Type of warranty • Failure pattern of items • Repairability • Failure pattern of replaced or repaired items • Incidental costs**FAILURES**• Occur randomly • Depend on • Product characteristics • Usage rate • Age • Maintenance • Environmental factors**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)**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)**MODEL ASSUMPTIONS**• All claims made • All claims are legitimate • Claims are made instantaneously • Instantaneous repair or replacement • Fixed cost per claim**MODEL ASSUMPTIONS**• Identical items • Independence (statistical) • No brand switching • All parameters known**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-lx x > 0.**EXPONENTIAL DISTRIBUTION**• Mean Time to Failure (MTTF): m = 1/l • Median Time to Failure: .6931/l • Constant failure rate, l Example: If the failure rate is known to be l = 1.25 per year, then MTTF = .8 yr.**WEIBULL DISTRIBUTION**More General Distribution F(x) = 1 - exp{-(lx)b} x > 0 Failure rate may be • decreasing (b < 1) • constant (b = 1) • increasing (b > 1)**WEIBULL DISTRIBUTION**MTTF and standard deviation depend of b and l. Example: l = 1.25, as before. If b = 1 (exponential), m = 0.8 and s = 0.8. If b = .5 (DFR), m = 1.6 and s = 4. If b = 2 (IFR), m = 0.71 and s = 0.37.**SOME NOTATION**mT = 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.) cs = seller’s average cost per item cr = average repair cost**NONRENEWING FRW**NONREPAIRABLE ITEMS Basis: Seller’s total cost per unit sold Ave. cost = cs[1 + M(W)] (W = length of warranty period)**NONRENEWING FRW**NONREPAIRABLE ITEMS Example: TV picture tube, exponential time to failure with MTTF = 6.5 years l = 1/6.5 = .1538. Take W = 6 months and 1 year, and cs = $67.20.**NONRENEWING FRW**NONREPAIRABLE ITEMS [Cont.] For the exponential, M(t) = lt. Thus: Six-month warranty: Ave. cost = $67.20[1+.1538(.5)] = $72.37 One-year warranty: Ave. cost = $67.20[1+.1538] = $77.54**NONRENEWING FRW**REPAIRABLE ITEMS Additional assumption: repaired items are “good-as-new” (i.e., have same failure distribution as new items). Ave. cost = cs + crM(W)**NONRENEWING PRW**• Equivalent to rebate form • Non-repairable items • cb = buyer’s cost of item (selling price) Ave. cost to seller = cs + cb[F(W) - mW/W] Formulas for mW are needed. (See: Book on Warranty Cost Analysis)**NONRENEWING PRW**Example: Same as previously, with cb = $105.00. One-year warranty. From formula in text, for W = 1, mW = .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.**INFORMATION NEEDS**• Form of distribution • Parameter values • Type of warranty • Rectification policy • Other cost information**DATA MAY INCLUDE**• Test data • Claims data • Data on similar products • Part and component data • Vendor data • Subjective information**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**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**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),Case Studies in Reliability and Maintenance, Wiley, New York[Contains two case studies]**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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