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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()]