spare part modelling – an introduction
DESCRIPTION
Spare part modelling – An introduction. Jørn Vatn. Motivation. For single component optimization models ( wrt ) indicates that there might be beneficial to keep a spare in order to reduce the MDT , and hence the cost of a failure - PowerPoint PPT PresentationTRANSCRIPT
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Spare part modelling – An introduction
Jørn Vatn
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Motivation
For single component optimization models (wrt ) indicates that there might be beneficial to keep a spare in order to reduce the MDT, and hence the cost of a failure
If we only have one component, we can compare the situation with, and without a spare, and find the best solution
In case of many components, there will be a “competition” on achieving the spare in case of simultaneous failures
Thus, we may consider to have more than one spare in stock
What will be the optimal number of spares to keep
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Content Situation 1
A situation with only one maintenance base, where failed components achieve a spare from the stock if available
The failed component is repaired in a workshop, there are infinite number of repair men
Situation 2 Components can not be repaired any longer, and at a critical stock
size, n, we order m new spares There is a lead time before spares arrive Lead time is gamma distributed
What we do not cover More than one maintenance base Several local, and one central stock Many other aspects
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Model assumptions, situation1
Constant rate of failure (total for many components) = Number of spares = s An inventory (stock) holds available spares Failed spares are repaired in the workshop Number of spares in the workshop = X Repair rate for each failed component = Infinite number of repair men
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Modelling
According to Palms theorem X Po( /) Introduce
Probability of shortage
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Modelling, cont
The number of components failed waiting for a spare is denoted the number of backorders = BO
The following recursive regime may thus be used
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Visual basic
Function PrepModel1(lambda As Single, mu As Single, _sMax As Integer, p() As Single, _
R() As Single, EBO() As Single)Dim lm As Singlelm = lambda / mup(0) = Exp(-lm)R(0) = 1 - p(0)EBO(0) = lmFor s = 0 To sMax - 1 p(s + 1) = lm * p(s) / (s + 1) R(s + 1) = R(s) - p(s + 1) EBO(s + 1) = EBO(s) - R(s)Next sEnd Function
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Simple cost model
Cost figures CU = Cost of unavailability of a component per unit time
CS = Capital cost per unit time to keep a spare in stock
Cost equation
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Visual basic
Function OptimizeModel1()Dim p(0 To 10) As SingleDim R(0 To 10) As SingleDim EBO(0 To 10) As SingleDim Cu As SingleDim Cs As SingleDim s As IntegerCu = 10000Cs = 1PrepModel1 0.01, 0.1, 10, p, R, EBOFor s = 0 To 10 Debug.Print s, Cs * s + Cu * EBO(s)Next sEnd Function
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Example result
s Cost 0 999.9999 1 49.37424 2 3.585931 3 3.039493 4 4.001117 5 5.000443 6 6.000523 7 7.000615 8 8.000708 9 9.0008
10 10.00089
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Markov modelling
Since failures and repairs are exponentially distributed an alternative modelling approach will be to use Markov
We may implement different strategies, e.g., an finite number of repair men
We may also introduce semi-Markov models to treat non-exponential repair times, or use virtual states in a phase type modelling approach
Drawbacks It is hard to treat an infinite number of back orders We need manually to specify the transition matrix For huge systems, time and storage capacity is a limitation
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Markov diagram
s 0 -1 -2 -3
l l l l l
smm (s+1)m (s+2)m (s+3)m
….….
Shortage of sparesNumber of spares in stock
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Solutions
Transition matrix State vector Steady state solution Visiting frequencies
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Transition matrix
The indexing generally starts on 0, and moves to r, e.g., there are r +1 system states (we need special indexing)
Each cell in the matrix has two indexes,where the first (row index) represent the ”from” state, whereas the second (column index) represent the “to” state.
The cells represent transition rates from one state to another
aij is thus the transition rate from state i to state j The diagonal elements shall fulfil the condition that all cells
in a row adds up to zero
00 01 0
10 11 1
0 1
r
r
ij
r r rr
a a a
a a a
a
a a a
A
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State probabilities
Let Pi(t) represent the probability that the system is in state i at time t
Now introduce vector notation, i.e. P(t) = [P0(t), P1(t),…,Pr(t)] From the definition of the matrix diagram it might be
shown that the Markov state equations are given by:
P(t) A = d P(t)/d t
These equations may be used to establish both the steady state probabilities, and the time dependent solution
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The steady state solution
In the long run when the system has stabilized we must have that d P(t)/d t = 0, hence
P A = 0 This system of equations is over-determined, hence we
may delete one column, and replace it with the fact thatP0+ P1+…+Pr = 1
Hence, we have
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The steady state solution
P A1 = b
where
and
b = [0,0, …,0,1]
00 01
10 111
0 1
1
1
1r r
a a
a a
a a
A
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Transition matrix
To -3 To -2 To -1 To 0 To 1 To s
From -3 =(s+3)*Mu 0 0 0 0
From -2 =Lambda =(s+2)*Mu 0 0 0
From -1 0 =Lambda =(s+1)*Mu 0 0
From 0 0 0 =Lambda =(s+0)*Mu 0
From 1 0 0 0 =Lambda =(s-1)*Mu
From s 0 0 0 0 =Lambda
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Solution our example
Steady state solution is obtained by P A1 = b
Fraction of time with spare part shortage = 1 is found by: U1 = P-1
Fraction of time with spare part shortage = 2 is found by: U2 = P-2
etc.
Total unavailability U = U1 + 2U2 + 3U3
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Results
Steady state pr. BO ContrP-3 7.589E-08 3 2.27665E-07P-2 3.771E-06 2 7.54159E-06P-1 1.508E-04 1 0.000150808P0 4.524E-03 0 0P1 9.048E-02 0 0P2 9.048E-01 0 0EBO= 0.000158578Cost= 3.585775317
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Assume only 1 repair man
s 0 -1 -2 -3
l l l l l
mm m m m
….….
Shortage of sparesNumber of spares in stock
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Structure of transition matrix Define “infinity”, e.g., “- = -3” (as high as feasible) The transition matrix starts with row index “-” There are altogether + 1 + s rows The elements below the diagonal is always If infinite number of repair men, the elements above the
diagonal starts with repair rate ( + s) and decreases with for each cell downwards the diagonal
Each row sums up to 0 in order to find the diagonal elements
To -3 To -2 To -1 To 0 To 1 To sFrom -3 =(s+3)*Mu 0 0 0 0From -2 =Lambda =(s+2)*Mu 0 0 0From -1 0 =Lambda =(s+1)*Mu 0 0From 0 0 0 =Lambda =(s+0)*Mu 0From 1 0 0 0 =Lambda =(s-1)*MuFrom s 0 0 0 0 =Lambda
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Model assumptions, situation 2
Constant rate of failure = Mean lead time when ordering new spares = MLT Lead times are Gamma (Erlang) distributed with
parameters = 4, and = / MLT Note that = 4 may be changed to account for general
value of SD(LT) = ½ / Ordering totally m new spares when stock level equals n
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Phase type distribution and semi-Markov
A continuous-time stochastic process is called a semi-Markov process if the embedded jump chain is a Markov chain, and where the holding times (time between jumps) are random variables with any distribution, whose distribution function may depend on the two states between which the move is made
Semi-Markov processes are hard to work with A phase-type distribution is a probability distribution that
results from a system of one or more inter-related Poisson processes occurring in sequence, or phases. The distribution can be represented by a random variable describing the time until absorption of a Markov process with one absorbing state. Example Erlang distribution
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Diagram for stockn = Number of components at replenishment
. . . .
. . . .
New order
Out of stock
Out of stock
. . . .
. . . .. . .
. . . .
m = Number of component replenished
. . .
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Markov diagram, step 1 (failures)
m + n n 0 -1 -2 -3
l l l l l l l l
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Markov diagram, step 2 (“repair”)
m + n n 0 -1 -2 -3
n1
n2
n3
l l l l l l l l
m
m
mm
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Markov diagram, step 2 (“repair”)
m + n n 0 -1 -2 -3
n1
n2
n3
l l l l l l l l
m
m
mm
m = rate of order, passing 4 steps
The blue states repersent substates in the order
If no components are taken out of stoc, it will be m + n components at the time of replenishment
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Markov diagram, step 3 (still failing…)
m + n n 0 -1 -2 -3
n1
n2
n3
l l l l l l l l
m
m
mm
l
l
l
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Markov diagram, step 4 (complete model)
m + n n 0 -1 -2 -3
n1
n2
n3
l l l l l l l l
01
02
03
-11
-12
-13
m
m
mm
l
m
m
m
m
m
m
m
m
m
m
m
m
m
m
mm
l
l
l
l
l
l
l
l
l
l
l
l
l
l
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Solution for our example
Steady state solution is obtained by P A1 = b
Fraction of time with spare part shortage = 1 is found by: U1 = P-1 + P-11
+ P-12
+ P-13
Fraction of time with spare part shortage = 2 is found by: U2 = P-2 + P-21
+ P-22
+ P-23
etc. Total unavailability formula in Excel:
U = U1 + 2U2 + 3U3
Frequency of running out of spares equals F = (P-0 + P-01
+ P-02
+ P-03
)
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What is ?
Assume that we have a huge number (N) of components with Weibull distributed life times
Consider the situation where t < In this period there is no PM, i.e., we may assume a
corrective strategy The total rate of failures as a function of t is N w(t)
where w(t) = W(t)/ t is the renewal rate Initially N w(t) should form the basis for the total failure
rate, After some time we set = N [1/ + E()]