Chapter 4:Preventive Maintenance

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Chapter 4Preventive Maintenance

• Overview
• Preventive Maintenance
• Replacement Decisions
• Inspection Decisions

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Preventive Maintenance

• Preventive maintenance is a series of preplanned
  tasks performed to counteract known causes of
  potential failures of facility, equipment,
  systems functions
• Objective of preventive maintenance is to
  maximize equipment availability and reliability
• PM can be planned and scheduled based on time,
  use or equipment condition Failure mechanism of
  the equipment ? what task?
• Time based (dominant failure mechanism) ?
  periodic restoration or replacement (simple
  replacement, overhaul)
• Condition based probability of a failure is
  constant regardless of time.Causes design,
  manufacturer, overhaul, installation errors,
  inappropriate initial operating and maintenance
  procedures ? measurement, inspection ?detect
  ?repairs , corrective actions

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Preventive Maintenance
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Elements of Preventive Maintenance
Preventive Maintenance
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Steps for Establishing a PM Program
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Replacement Decisions

• When??at the design stage, break down, evident
  obsolescence
• Repairable items ? repair or replace? ? economic
  analysis Total life cycle cost
  CLCCAcquisitionCInvestment COS Cdisposal
• The failure mechanism ? replacement

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Notations

• Cp cost of preventive maintenance
  Cf cost of breakdown (failure)
  maintenance f(t)
  time-to-failure probability density function
  (p.d.f.) F(t) Equipment or system time to
  failure distribution
  
• r(t) failure rate function N(tp)
  number of failures in the interval (0,tp) N(tp)
  is a random variable
  H(tp) expected number
  of failures in the interval (0,tp) R(t)
  Reliability or survival function
  M(tp) expected value of the truncated
  distribution at tp
• EC(tp) expected cost per cycle
  UEC(tp) expected cost per unit time

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Model 1 Optimal Age-Based Preventive Replacement

• Problem Perform preventive replacement once the
  equipment has reached a specified age tp. When
  failures occur, failure replacements are made.
  Determine the optimal preventive replacement age
  tp to minimize the total expected cost of
  replacing the equipment per unit time.
• Assumptions
• If tp is infinite, no preventive replacement is
  scheduled
• The system is as good as new after preventive
  replacement is performed

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Optimal Age-Based Preventive Replacement

• Applications simple equipment or a single unit
  in which repair/replace at the time of failure
  general overhaul new equipment.
• Model
• Solution golden section method or searching

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Model 2 Optimal Constant-Interval Group
Replacement of Items

• Problem There are a large number of similar
  items which are subject to failure. Whenever an
  item fails it is replaced by a new item. Perform
  group preventive replacement/ maintenance at
  constant intervals of length tp hours. The cost
  of replacing an item under group replacement
  conditions is assumed to be less than that for
  failure replacement. Determine the optimal
  interval tp between group preventive
  replacements to minimize total expected cost per
  unit time
• Assumptions
• N is the total number of items in the system
• Cg is the cost of replacing one item under group
  replacement

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Optimal Constant-Interval Preventive Replacement

• Applications suitable for complex systems such
  as engines and turbines
• Model r(t) failure rate f(t)/R(t)
  EN(tp) expected number of times one item fails
  in interval (0,tp) sum(no. of exp. failures
  which occur in interval (0, tp) when the first
  failure occurs in the i periodPfirst failure
  occuring in interval (i-1,i)

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Examples

• Example 1 Given Cp 5, Cf 10 determine the
  optimal replacement age for the equipment.
  Failures occur according to the normal
  distribution with mean 5 weeks, standard
  deviation 1 week
• Example 2 Given Cg 5, Cf 10. assume that
  there are 100 items in the group. Determine the
  optimal interval between group replacements.
  Failures occur according to the normal
  distribution with mean 5 weeks, standard
  deviation 1 week

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Solutions Example 1

• From model 1 we have

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Solutions Example 2

• From model 2 we have
• C(tp) 1005 100EN(tp) where
• Thus we have the following table

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Inspection Decisions

• Objective obtained useful information (of
  product or indicators) about the state of a
  system ? predict failure, plan further
  maintenance actions.
• Benefits repairs ?, proper planning ?
  disruption ?
• Requirements determine frequency of inspection
  (and/or level of monitoring) ? cost of
  inspectiongtltbenefit of inspection correct
  information, predict failures

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Model 1 Downtime Minimization for a Single
Machine Inspection

• Problem Equipment breaks down from time to time,
  having a loss in production output. to reduce
  the breakdowns, inspections and consequent minor
  repair/replacements can be made. These works cost
  money in terms of materials, wages, and loss of
  production due to scheduled downtime. The problem
  is to determine the inspection policy which
  minimizes the total downtime/unit time incurred
  due to breakdowns and inspections.

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Model 1 Downtime Minimization for a Single
Machine Inspection

• Model total downtime per unit time
• TDT(n) DTr DTi
  Where DTr downtime due to repairs per
  unit time DTi downtime due to inspection per
  unit time n number of inspections per unit
  time ?(n) breakdown rate if n
  inspections is made µ repair rate ?
  inspection rate
• Notes
• For the negative exponential dist. mean time will
  be 1/?,1/µ, 1/?
• If average number of breakdowns per unit time
  k and ?(n) k/n then n (k?/µ)1/2

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Model 2 Profit Maximization for a Single
Machine Inspection

• Problem The machine has a general failure
  distribution. Inspections will reveal the
  condition of the machine and may result in
  reducing the severity of failure. Repairing a
  failed machine incurs a cost of repair Cr. The
  cost of inspection is Ci and p is the profit per
  unit time when the machine is operating. The
  question is how often should this machine be
  inspected to maximize profit
• Model The expected profit per cycle P(T)
  the expected profit without failure
  expected profit with failure

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Profit Maximization for a Single Machine
Inspection

• P(T) P1(T)R(T) P2(T)F(T) (pT
  Ci)R(T)(EpttltTCiCr)F(T) (pT
  Ci)R(T)(EpttltTCiCr)1-R(T)
• We have, the expected profit per unit time
• Solution UP(T) is a function of one variable
  that needs to be maximized

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Examples

• Example 3 if the failure rate is assumed to be
  ?(n) ke-n where k 3 breakdowns/month and mean
  1/µ 0.02month. 1/? 0.005month. Determine the
  optimal inspection frequency to minimize total
  downtime.
• Example 4 suppose we have the following data
• p 10,000 per month
• Ci 75 per month
• Cr 400 per month
• f(t) 0.001/(0.001t1)2 t 0

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Periodic Maintenance Markov Model

• Problem the system can either fail completely
  or undergo periodic PM. The fail system is
  repaired. Determine the system availability,
  probability of the system down for PM and
  probability of system failure, MTTF.
• Assumptions
• System PM, failure and repair rates are constant
• After repair or PM the system is as good as new
• Periodic maintenance is performed on the system
  after every T hours, starting at time 0.

System failed 1
System down for PM p
µp
µ
System working Normally 0
?p
?
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Periodic Maintenance Markov Model

• Model
• Solution Using Laplace transformation
• Availability and MTTF