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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

MODULE 2: WARRANTY COST ANALYSIS

Professor D.N.P. Murthy

The University of Queensland

Brisbane, Australia

warranty cost analysis
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
WARRANTY MODELING
  • Reasons for modeling
      • Marketing (Sales)
      • Economic (Cost)
      • Engineering (Design)
      • Operational (Servicing)
  • Many disciplines involved
cost models
COST MODELS

Costs depend on:

  • Type of warranty
  • Failure pattern of items
  • Repairability
  • Failure pattern of replaced or repaired items
  • Incidental costs
failures
FAILURES
  • Occur randomly
  • Depend on
      • Product characteristics
      • Usage rate
      • Age
      • Maintenance
      • Environmental factors
cost calculation basis
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
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
MODEL ASSUMPTIONS
  • All claims made
  • All claims are legitimate
  • Claims are made instantaneously
  • Instantaneous repair or replacement
  • Fixed cost per claim
model assumptions14
MODEL ASSUMPTIONS
  • Identical items
  • Independence (statistical)
  • No brand switching
  • All parameters known
probability
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
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
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 distribution18
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
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
NONRENEWING FRW

NONREPAIRABLE ITEMS

Basis: Seller’s total cost per unit sold

Ave. cost = cs[1 + M(W)]

(W = length of warranty period)

nonrenewing frw21
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 frw22
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 frw23
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
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 prw25
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
INFORMATION NEEDS
  • Form of distribution
  • Parameter values
  • Type of warranty
  • Rectification policy
  • Other cost information
data may include
DATA MAY INCLUDE
  • Test data
  • Claims data
  • Data on similar products
  • Part and component data
  • Vendor data
  • Subjective information
other cost models
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
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
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 reading31
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.