Chapter 5: Preventive Maintenance: Mathematical Models

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Chapter 5 Preventive Maintenance Mathematical
Models

• Overview
• Replacement analysis
• Inspection decisions
• Imperfect Preventive Maintenance
• Delay-time modeling

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Replacement Analysis

• 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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Replacement Analysis 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, ( f(t)
  truncated 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
(Type I Policy)

• 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 suitable for simple equipment or a
  single unit in which repair at the time of
  failure (or replacement) could nearly correspond
  to general overhaul.
• Model

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Optimal Age-Based Preventive Replacement

• SolutionGolden section method Minimize
  g(t) Subject to a ? t ? b
• Step1.
• Choose an allowable final tolerance level ?.
• Let a1,b1a,b and calculate ?1 a1
  (1-?)(b1-a1) ?1 a1 ?(b1-a1) 0 ? ? ?
  1
• Evaluate g(?1) and g(?1)
• Let k 1 go to step 2
• Step 2. If bk ak lt ?, stop as the optimal
  solution is t (ak bk)/2. Else, if g(?k) gt
  g(?k) go to step 3 if g(?k) ? g(?k) go to step 4

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Optimal Age-Based Preventive Replacement

• Step 3.
• Let ak1 ?k and bk1 bk. Furthermore, let
  ?k1 ?k and ?k1 ak1 ?(bk1-ak1).
• Evaluate g(?k1) and go to step 5
• Step 4.
• Let ak1 ak and bk1 ?k. Furthermore, let ?
  k1?k and ? k1 ak1 (1-?)(bk1 - ak1).
• Evaluate g(? k1) and go to step 5
• Step 5. kk1 go to step 2

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Model 2 Optimal Constant-Interval Preventive
Replacement (Type II Policy- Group Replacement)

• Problem perform group preventive replacement/
  maintenance at constant intervals of length tp
  hours, regardless of the number of intervening
  failures prior to tp. Determine the optimal
  interval tp between group preventive
  replacements to minimize total expected cost per
  unit time
• Assumptions
• When failure happens, minimal repair is
  performed. Minimal repair does not change the
  failure rate of the system
• Preventive replacement/maintenance renews the
  system.

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Optimal Constant-Interval Preventive Replacement

• Applications suitable for complex systems such
  as engines and turbines
• Model
• Solution golden section method

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Extension of Policy I

• Optimal Type I General Policy replace the system
  after (k-1) repairs. For a system subjected to
  (i-1) repairs, it is repaired (or replaced if
  ik) at time of failure or at maintenance age Ti
  (Ti is the number of hours from the last repair
  or replacement) whichever occurs first.

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Extension of Policy II

• Optimal Type II General Policy replace the
  system after (k-1) repairs. For a system
  subjected to (i-1) repairs, it is always repaired
  (or replace if i k) at age Ti. In case of
  failure, a minimal repair is carried out

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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 Optimal Inspection Schedule Minimizing
Expected Cost for a Single Machine

• Problem perform inspections at times x1,x2, ,
  until a failed equipment is detected. Determine
  an optimal inspection schedule (x1,x2,, , xn) to
  minimize total costs per unit time associated
  with inspection, repair and non-detection of
  failed equipment.

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Optimal Inspection Schedule Minimizing Expected
Cost for a Single Machine

• Model
• Solution take the partial derivative respect to
  xi and set to 0

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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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Model 3 Minimize Expected Cost with Minimal
Repair

• Problem Each production cycle begins with a new
  machine (or one overhauled to the condition good
  as new). After a production period, the process
  may shift to an out-of-control state, which is
  equivalent to machine failure. Then, a
  maintenance action is performed (minimal repair)
  on the machine to bring the process under
  control. Inspection is performed at times x1,x2,
  , xn to observe the machine (process0 in order
  to take appropriate maintenance action. Three
  types of costs are consideredreplacement or
  overhaul cost Cr, the cost of inspection Ci, the
  cost of repair Cf and the increased cost per unit
  time of running in an out-of-control state s. the
  objective is to determine the optimal value of n
  and xi, i 1, 2, , n such that the expected
  total operating cost per unit time is minimized

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Minimize Expected Cost with Minimal Repair

• Assumptions
• The life of the machine is a random variable with
  probability density function f(t)
• The repair times are negligible and the repair
  brings the machine back to an in-control state.
  The repairs do not change the failure
  distribution (i.e., the repairs are minimal)

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Minimize Expected Cost with Minimal Repair

• Model
• Notes
• The model can be simplified if xn T
• If the failure rate follows an exponential
  distribution, then inspection are equally
  spaced. That is xi1 - xi xi xi-1, for all i

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Model 4 Coordinating the inspection of a Group
of Machines

• Problem develop a schedule of inspections for a
  group of machines. There are two types of costs
  a setup cost incurred whenever a review to
  determine which machines to inspect takes place,
  regardless of the number number of machines
  inspected, and a setup cost associated with each
  machine to be inspected. The second cost is the
  failure cost, which consists of the cost of
  repair and the incurred cost for running in an
  out-of-control state. The time between two
  consecutive setups is the basic cycle. The
  objective is to determine the inspection time Ti
  for machine i as a multiple of the basic cycle so
  as to minimize the expected cost per unit time

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Coordinating the inspection of a Group of Machines

• Assumptions
• A cycle schedule is repeated every T time units
• A periodic review at equal intervals of length
  T0 is made to determine which machine should be
  inspected at cost A, called the setup cost
• The failure distribution function of a machine is
  exponential distribution, then the inspection
  interval is assume to be constant for each
  machine. That means ti,j1- ti,j ti,j ti,j-1
  Ti for all j

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Coordinating the inspection of a Group of Machines

• Model UEC( T0, T1, T2, , TN)
  total expected cost/T (setup cost
  inspection cost failure cost)
  /T
  
  where Ti kT0, k is an integer

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Imperfect Preventive Maintenance

• Assumptions maintenance will improve the
  equipment condition at a certain degree but will
  not restore it to a new state unless the
  equipment is completely replaced
• Single unit system, is considered , operate for
  an infinite horizon
• The unit start to operate at time 0. It has a
  failure distribution F(t) and a density function
  f(t)
• The failure rate r(t) f(t)/1-F(t) is
  monotonely increasing
• The unit is maintained preventively at times kT,
  k 1, 2, , Tgt0

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Imperfect Preventive Maintenance

• The unit undergoes only minimal repairs at
  failures between preventive maintenance. Minimal
  repair does not change the failure rate (type II
  PM policy)
• The repair and PM times are negligible
• The cost of PM is Cp and the repair cost is Cf
• EC denotes the expected cost per cycle, ECD
  denotes expected cycle duration, and UE the
  expected cost per unit time
• Models the effect of preventive maintenance on
  the state of the equipment

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Model 1

• Equipment after PM has the same failure rate as
  before PM with probability P or is as good as new
  with probability q 1-P. If the equipment
  survived up to time jT, where j preventive
  maintenance. Then
• Where ECD is
  the expected cycle duration up to perfect PM.
  Simplify UEC,
  
• Differentiating UEC(P,T) and set to 0 we get
  

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Model 2 The Equipment Age Reduces by x Units of
Time after Each PM

• 0 ? x ? T, if x T equipment is as good as new
  (perfect PM), x 0 equipment is as bad as old
• If the equipment is replaced after N intervals
  with a replacement cost Cr

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Model 3 The Age and the Failure Rate of a Piece
of Equipment Are Reduced by an Amount
Proportional to the PM Cost Cp

• In the case of the age reduction, the
  relationship between new age and old age
  y1Cp/C0(yT) where C0 is initial cost of the
  equipment and yT is the age just before PM, and
  y is the age just after PM

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Model 3

• In the case of the failure rate reduction to
  1-Cp/C0r(yT) by each PM when it was r(yT)
  before PM , in the steady state we have
  1-Cp/C0r(yT) r(y)

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Delay-time Modeling

• Two-stages failure process
• A defect becomes detectable
• The detectable defect gives rise eventually to
  failure of the equipment
• h period between the defect is first
  detectable to failure delay-time ? p.d.f f(h) ?
  model (inspection period T, down time, UEC)

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Delay-time Modeling

• Assumptions
• An inspection is performed every T units of time
  at cost CI and requires d time units dltltT
• Inspections are perfect in that any defects
  present within the plant are identified
• Defects that are identified at inspection will be
  repaired within the inspection period
• The initial instant at which a defect may be
  assumed to first arise within the plant (known as
  the time of origin of the fault) is uniformly
  distributed over time since the last inspection
  and independent of h. Faults arise at the rate of
  k per unit time.
• The probability density function of delay time
  f(h) is known

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Delay-time Modeling

• If a fault arises during (0, T h) then it is
  repaired as a breakdown repair with probability
  (T h/T) given that a fault will arise.
  Otherwise, it is considered as an inspection
  repair. The probability of a fault arising as a
  breakdown P(T) is given by

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Delay-time Modeling

• Model 1 Minimize expected downtime per unit time
  to be incurred in period T
  where db average downtime of breakdown
  repair k the arrival rate of defects per
  unit time
• Model 2 Minimize expected total cost per unit
  time