Spare part modelling – An
introduction
Jørn Vatn
1
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
2
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
3
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
4
Modelling
 According to Palms theorem X  Po( /)
 Introduce 𝑝 𝑘 = Pr 𝑋 = 𝑘 =
𝜆
𝑘
𝜇
𝑘!
𝑒 −𝜆/𝜇
 𝑝 0 = 𝑒 −𝜆/𝜇
 𝑝 𝑠+1 =
𝜆
𝜇
𝑠+1
𝑝 𝑠
 Probability of shortage 𝑅 𝑠 = 𝑃 𝑋 > 𝑠 =
∞
𝑘=𝑠+1 𝑝(𝑘)
 𝑅 0 = 1 − 𝑝(0)
 𝑅 𝑠+1 =
∞
𝑘=𝑠+2 𝑝
𝑘 =
∞
𝑘=𝑠+1 𝑝(𝑘) −
5
𝑝 𝑠 + 1 = 𝑅 𝑠 − 𝑝(𝑠 + 1)
Modelling, cont
 The number of components failed waiting for a spare is
denoted the number of backorders = BO
 EBO 𝑠 = 𝐸 𝐵𝑂 = 𝐸(max 0, 𝑋 − 𝑠 ) =
 EBO 𝑠 + 1 =
∞
𝑘=𝑠+2(𝑘
∞
𝑘=𝑠+1(𝑘 − 𝑠)𝑝(𝑘)
∞
𝑘=𝑠+1(𝑘 − 𝑠 − 1)𝑝(𝑘)
− 𝑠 − 1)𝑝(𝑘) =
 EBO 𝑠 + 1 = EBO 𝑠 + ∞
𝑘=𝑠+1 −1 𝑝 𝑘 = EBO 𝑠 − 𝑅(𝑠)
 The following recursive regime may thus be used
 𝑝 0 = 𝑒 −𝜆/𝜇
 𝑅 0 = 1 − 𝑝(0)
 EBO 0 = 𝐸 𝑋 = 𝜆/𝜇
 𝑝 𝑠+1 =
𝜆
𝜇
𝑠+1
𝑝 𝑠
 𝑅 𝑠 + 1 = 𝑅 𝑠 − 𝑝(𝑠 + 1)
 EBO 𝑠 + 1 = EBO 𝑠 − 𝑅(𝑠)
6
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 Single
lm = lambda / mu
p(0) = Exp(-lm)
R(0) = 1 - p(0)
EBO(0) = lm
For 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 s
End Function
7
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
 𝐶 𝑠 = 𝐶𝑆 𝑠 + 𝐶𝑈 EBO(𝑠)
8
Visual basic
Function OptimizeModel1()
Dim p(0 To 10) As Single
Dim R(0 To 10) As Single
Dim EBO(0 To 10) As Single
Dim Cu As Single
Dim Cs As Single
Dim s As Integer
Cu = 10000
Cs = 1
PrepModel1 0.01, 0.1, 10, p, R, EBO
For s = 0 To 10
Debug.Print s, Cs * s + Cu * EBO(s)
Next s
End Function
9
Example result
s
0
1
2
3
4
5
6
7
8
9
Cost
999.9999
49.37424
3.585931
3.039493
4.001117
5.000443
6.000523
7.000615
8.000708
9.0008
10
10.00089
10
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 nonexponential 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
11
Markov diagram
Shortage of spares
Number of spares in stock


….
s



0
s
-1
(s+1)
12

-2
(s+2)
-3
(s+3)
….
Solutions




Transition matrix
State vector
Steady state solution
Visiting frequencies
13
 a00
a
Transition matrix
 10
A

 The indexing generally starts on 0,

and moves to r, e.g., there are r +1
 ar 0
system states (we need special indexing)
a01
a11
aij
ar1
a0 r 
a1r 




arr 
 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
14
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
15
The steady state solution
 In the long run when the system has stabilized we must
have that d P(t)/d t = 0, hence
 PA = 0
 This system of equations is over-determined, hence we
may delete one column, and replace it with the fact that
P0+ P1+…+Pr = 1
 Hence, we have
16
The steady state solution
P A1 = b
where
 a00
a
10

A1 


 ar 0
a01
a11
a r1
1
1



1
and
b = [0,0, …,0,1]
17
Transition matrix
To -3
From -3
From -2
From -1
From 0
From 1
From s
=Lambda
0
0
0
0
To -2
=(s+3)*Mu
=Lambda
0
0
0
To -1
0
=(s+2)*Mu
=Lambda
0
0
18
To 0
0
0
=(s+1)*Mu
=Lambda
0
To 1
0
0
0
=(s+0)*Mu
=Lambda
To s
0
0
0
0
=(s-1)*Mu
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
19
Results
Steady state pr.
P-3
7.589E-08
P-2
3.771E-06
P-1
1.508E-04
P0
4.524E-03
P1
9.048E-02
P2
9.048E-01
EBO=
Cost=
BO Contr
3 2.27665E-07
2 7.54159E-06
1 0.000150808
0
0
0
0
0
0
0.000158578
3.585775317
20
Assume only 1 repair man
Shortage of spares
Number of spares in stock


….
s



0


-1

-2

21
-3

….
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
From -3
From -2 =Lambda
From -1
0
From 0
0
From 1
0
From s
0
To -2
=(s+3)*Mu
=Lambda
0
0
0
To -1
0
=(s+2)*Mu
=Lambda
0
22
0
To 0
0
0
=(s+1)*Mu
=Lambda
0
To 1
0
0
0
=(s+0)*Mu
=Lambda
To s
0
0
0
0
=(s-1)*Mu
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
23
Phase type distribution and semi-Markov
 A continuous-time stochastic process is called a semiMarkov 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
24
Diagram for stock
n = Number of components at replenishment
....
Out of stock
....
New order
Out of stock
....
....
....
m = Number of component replenished
25
...
...
Markov diagram, step 1 (failures)

m+n




n

0
26

-1

-2
-3
Markov diagram, step 2 (“repair”)


m+n



n
0

n1


n2


n3
27

-1

-2
-3
Markov diagram, step 2 (“repair”)


m+n



n

0

-1

-2
-3

n1


n2
n3

= 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
28
Markov diagram, step 3 (still failing…)


m+n



n
0

n1



n2



n3

29

-1

-2
-3
Markov diagram, step 4 (complete model)



m+n


n
0



n2







n3
01

30

-2

-3









-11

02

03


-1

n1



-12

-13



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)
31
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()]
32