F.Grabski
Applications of semi-Markov processes in reliability - RTA # 3-4, 2007, December - Special Issue
Grabski Franciszek
Naval University, Gdynia, Poland
Applications of semi-Markov processes in reliability
Keywords
semi-Markov processes, reliability, random failure rate, cold standby system with repair
Abstract
The basic definitions and theorems from the semi-Markov processes theory are discussed in the paper. The semiMarkov processes theory allows us to construct the models of the reliability systems evolution within the time
frame. Applications of semi-Markov processes in reliability are considered. Semi-Markov model of the cold
standby system with repair, semi-Markov process as the reliability model of the operation with perturbations and
semi-Markov process as a failure rate are presented in the paper.
1. Introduction
2. P{ξ 0 = io , ϑ 0 = 0} = P{ξ 0 = i0 } = pio
The semi-Markov processes were introduced
independently and almost simultaneously by P. Levy,
W.L. Smith, and L.Takacs in 1954-55. The essential
developments of semi-Markov processes theory were
proposed by Cinlar [3], Koroluk & Turbin [13],
Limnios & Oprisan [14]. We would apply only semiMarkov processes with a finite or countable state
space. The semi-Markov processes are connected to
the Markov renewal processes.
The semi-Markov processes theory allows us to
construct many
models of the reliability systems
evolution through the time frame.
hold.
From the above definition it follows that a Markov
renewal chain is a homogeneous two-dimensional
Markov chain such that the transition probabilities do
not depend on the second component. It is easy to
notice that a random sequence {ξ n : n = 0,1,2,...} is a
homogeneous one-dimensional Markov chain with the
transition probabilities
p ij = P{ξ n +! = j | ξ n = i} = lim Qij (t ) .
The matrix
[
(4)
Is called a Markov renewal kernel. Both Markov
renewal kernel and the initial distribution define the
Markov renewal chain. This fact allows us to
construct a semi-Markov process.
Let
ξ n , ϑn , n = 0,1,2,... are the random variables defined
on a joint probabilistic space ( Ω , F, P ) with values
on S and R + respectively. A two-dimensional random
sequence {(ξ n , ϑ n ), n = 0,1,2,...} is called a Markov
renewal chain if for all
i0 ,...., i n−1 , i ∈ S , t 0 ,..., t n ∈ R+ , n ∈ N 0 :
= P{ξ n +1 = j , ϑ n +1 ≤ t | ξ n = i} = Qij (t ),
]
Q(t ) = Qij (t ) : i, j ∈ S ,
Let S be a discrete (finite or countable) state space and
let R+ = [0, ∞ ) , N 0 = {0,1,2,...} . Suppose, that
{
(3)
t →∞
2. Definition of semi-Markov processes with a
discrete state space
1. P ξ n+1 = j, ϑn+1 ≤ t | ξ n = i, ϑn = t n ,...,ξ 0 = i0 , ϑ0 = t 0
(2)
τ 0 = ϑ0 = 0,
τ n = ϑ1 + ... + ϑ n , τ ∞ = sup{τ n : n ∈ N 0 }
}
A stochastic process
following relation
(1)
- 60 -
{X (t ) : t ≥ 0}
given by the
F.Grabski
Applications of semi-Markov processes in reliability - RTA # 3-4, 2007, December - Special Issue
X (t ) = ξ n for t ∈ [τ n , τ n+1 )
It means that the process {X (t ) : t ≥ 0} has the finite
number of state changes on a finite period.
Every Markov process {X (t ) : t ≥ 0} with the discrete
space S and the right-continuous trajectories keeping
constant values on the half-intervals, with the
generating matrix of the transition rates
Α = [α ij : i, j ∈ S ] , 0 < −α ii = α i < ∞ is the semi-
(5)
is called a semi-Markov process on S generated by
the Markov renewal chain related to the kernel
Q (t ), t ≥ 0 and the initial distribution p.
Since the trajectory of the semi-Markov process keeps
the constant values on the half-intervals [τ n , τ n+1 ) and
it
is
a
right-continuous
function,
from
equality X (τ n ) = ξ n , it follows that the sequence
{X (τ n ) : n = 0,1,2,...} is a Markov chain with the
transition probabilities matrix
Markov process with the kernel
Q (t ) = [Qij (t ) : i, j ∈ S ] ,
Where
P = [ pij : i, j ∈ S ]
(6)
Qij (t ) = pij (1 − e
The sequence {X (τ n ) : n = 0,1,2,...} is called an
embedded Markov chain in a semi-Markov process
{X (t ) : t ≥ 0} .
The function
p ij =
Fij (t ) = P{ τ n +1 − τ n ≤ t | X (τ n ) = i , X (τ n +1 ) = j}
=
Qij (t )
The transition probabilities of the semi-Markov
process are introduced as follows:
the next state will be j . From (11) we have
Pij (t ) = P{X (t ) = j | X (0) = i}, i, j ∈ S .
(8)
j∈S
(9)
is a cumulative probability distribution of a random
variable Ti that is called waiting time of the state i .
Pij (t ) = δ ij [1 − Gi (t )] + ∑ ∫ Pkj (t − x)dQik ( x),
The waiting time Ti means the time being spent in
state i when we do not know the successor state.
A stochastic process {N (t ) : t ≥ 0} defined by
i, j ∈ S .
N (t ) = n for t ∈ [τ n , τ n+1 )
(11)
Applying the Markov property of the semi-Markov
process at the jump moments, as a result, we obtain
Markov renewal equation for the transitions
probabilities, [4], [12]
The function
Gi (t ) = P{ τ n +1 − τ n ≤ t | X (τ n ) = i} = ∑ Qij (t )
for i ≠ j and pii = 0
3. Transition probabilities of a semi-Markov
process
is a cumulative probability distribution of a random
variable Tij that is called holding time of a state i , if
Q ij (t ) = p ij Fij (t ) .
αi
), t ≥ 0,
In the reliability models the parameters and
characteristics of a semi-Markov process are
interpreted as the reliability characteristics and
parameters of the system.
(7)
p ij
α ij
−α i it
t
k∈S 0
(12)
Using Laplace-Stieltjes transformation we obtain the
system of linear equation
(10)
~
p ij ( s) = δ ij [1 − g~i ( s)] + ∑ q~ik ( s) ~
p kj ( s),
k ∈S
i, j ∈ S
is called a counting process of the semi-Markov
process {X (t ) : t ≥ 0}.
The semi-Markov process {X (t ) : t ≥ 0} is said to be
regular if for all t ≥ 0
where the transforms
∞
~
pij ( s) = ∫ e − st dPij (t )
P{N (t ) < ∞} = 1
0
are unknown while the transforms
- 61 -
(13)
F.Grabski
Applications of semi-Markov processes in reliability - RTA # 3-4, 2007, December - Special Issue
∞
∞
0
0
where π = [π j ,
q~ik ( s) = ∫ e − st dQik (t ) , g~i ( s) = ∫ e − st dGi (t )
distribution of the embedded Markov chain that
satisfies system of equations.
∑ π i pij = π j , j ∈ S , ∑ π i = 1.
are given.
Passing to matrices we obtain the following equation
~( s) = [ I − ~
~(s ) ~
p
g ( s )] + q
p(s) ,
j ∈ S ] is the unique stationary
i∈S
(20)
i∈S
4. First passage time from the state i to the
states subset A.
(14)
where
The random variable
~( s) = [ ~
~( s) = [q~ ( s ) : i, j ∈ S ] ,
p
pij ( s ) : i, j ∈ S ] , q
ij
Θ A =τ ∆A ,
~
g ( s ) = [δ ij (1 − g~i ( s)) : i, j ∈ S ] .
where
∆ A = min {n ∈ N : X (τ n ) ∈ A},
In many cases the transitions probabilities Pij ( t ) and
the states probabilities
P j (t ) = P{X (t ) = j} , j ∈ S ,
denotes the time of first arrival of semi-Markov
process, at the set of states A.
The function
(15)
Φ iA (t ) = P{Θ A ≤ t | X (0) = i}.
approach constant values for large t
Pij = lim Pij (t ), P j = lim P j (t ) .
t →∞
(16)
t →∞
is the cumulative distribution of the random variable
ΘiA that denotes the first passage time from the state i
to the states subset A.
To formulate the appropriate theorem, we have to
introduce a random variable
∆ j = min {n ∈ N : X (τ n ) = j} ,
Theorem 2. [4], [13]
For the regular semi-Markov processes such that,
(17)
f iA = P{∆ A < ∞ | X (0) = i} = 1,
That denotes the time of first arrival at state j. A
number
f ij = P{∆ j < ∞ | X (τ n ) = i}
t
j∈ A
Applying Laplace-Stieltjes transformation we obtain
the system of linear equations
~
~
φiA (s) = ∑ q~ij ( s) + ∑ φ kA ( s)q~ik ( s), i ∈ A'
j∈A
chain {X (τ n ) : n = 0,1,2,...} , contains one positive
recurrent class C, such that for each state
i ∈ S , j ∈ C , f ij = 1 and 0 < E (Ti ) < ∞, i ∈ S , then
t →∞
t →∞
∑ π j E (T j )
k ∈S 0
i ∈ A'
]
Pij = lim Pij (t ) = Pj = lim Pj (t ) =
(22)
Φ iA (t ) = ∑ Q ij (t ) + ∑ ∫ Φ kA (t − x )dQik ( x),
Theorem 1.
Let {X ( t ) : t ≥ 0} be a semi-Markov process with a
discrete state space S and continuous kernel
Q(t ) = Qij (t ) : i, j ∈ S . If the embedded Markov
π j E (T j )
i ∈ A′ ,
the distributions Φ iA (t ) , i ∈ A′ are proper and they
are the unique solutions of the system of equations
(18)
is the probability that the chain that leaves state i will
sooner or later achieve the state j.
As a conclusion of theorems presented by Korolyuk
and Turbin [13], we have obtained following theorem
[
(21)
k∈A'
with unknown transforms
∞
~
φiA ( s) = ∫ e − st dΦ iA (t ) .
0
(19)
Generating matrix form we get equation
i∈S
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(23)
F.Grabski
Applications of semi-Markov processes in reliability - RTA # 3-4, 2007, December - Special Issue
(I − q~ A' ( s ) ) ~ A' ( s ) = b~ ( s) ,
(24)
5. Semi-Markov model of the cold standby
system with repair
where
The problem is well known in reliability theory
(Barlow & Proschan [1]). The model presented here is
some modification of the model that was considered
by Brodi & Pogosian [2].
~ ( s ) = [ q~ ( s ) : i, j ∈ A]
I = [δ ij : i , j ∈ A' ], q
A'
ij
are the square matrices and
~ ( s) = [~ ( s ) : i ∈ A']T ,
A'
iA

~

b( s ) =  ∑ q~ij ( s ) : i ∈ A'
 j∈A

5.1. Description and assumptions
A system consists of one operating component, an
identical stand-by component and a switch, (Figure1).
T
1
are the one-column matrices of transforms. The
formal solution of the equation is
−1 ~
~ ( s ) = (I − ~
q A' (s) ) b (s) .
A'
2
Figure 1. Diagram of the system
To solve this equation we use any computer programs,
for example MATHEMATICA. Obtaining the inverse
Laplace transform is much more complicated.
It is essentially simpler to find the expected values
and the second moment of the random variables
Θ iA , i ∈ A′ . If the second moments of the waiting
times Ti , i ∈ A′ are positive and commonly bounded,
and f iA = 1, i ∈ A′ , then the expected values of the
random variables Θ iA , i ∈ A′ are the unique solution
of equation
(I − PA' )
A'
= TA' ,
When the operating component fails, the spare is put
in motion by the switch immediately. The failed unit
(component) is repaired. There is a single repair
facility. The repairs fully restore the components i.e.
the components repairs means their renewals. The
system fails when the operating component fails and
component that was sooner failed in not repaired yet
or when the operating units fail and the switch fails.
We assume that the time to failure of the operating
components are represented by the independent copies
of a non-negative random variable ς with distribution
given by a probability density function (pdf)
f ( x), x ≥ 0 . We suppose that the lengths of the repair
periods of the components are represented by the
identical copies of the non-negative random variables
γ with cumulative distribution function (CDF)
G ( x) = P (γ ≤ x ). Let U be a random variable having
binary distribution
(25)
where
[
]
[
]
I = δ ij : i, j ∈ A' , P A' = p ij : i, j ∈ A' ,
= [E (Θ iA ) : i ∈ A'] , TA' = [E (Ti ) : i ∈ A']
T
A'
T
b( k ) = P(U = k ) = a k (1 − a) 1− k , k = 0,1, 0 < a < 1,
and the second moments of the time to failure are the
unique solution of equation
(I − PA' )
2
A′
= B A'
where U = 0 , when a switch is failed at the moment
of the operating component failure, and U = 1 , when
the switch work at that moment. We suppose that the
whole failed system is replaced by the new identical
system. The replacing time is a non-negative random
variable η with CDF H ( x ) = P (η ≤ x ) .
(26)
where
[
]
[
]
I = δ ij : i, j ∈ A' , P A ' = p ij : i , j ∈ A' ,
2
A′
[
]
= E (Θ 2 iA ) : i ∈ A' , B A ' = [bi : i ∈ A'] ,
T
T
bi = E (Ti ) + 2 ∑ pik E (Tik ) E (Θ kA ) .
k∈ A '
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F.Grabski
Applications of semi-Markov processes in reliability - RTA # 3-4, 2007, December - Special Issue
Q10 (t ) = P (ς ≤ t , γ > ζ )
z
g
1
z3
1
g
z
3
5
+ P (U = 0, ς ≤ t , γ < ζ )
1
z
g
2
z
2
4
g
4
2
= ∫0t [1 − G ( x )]dF ( x) + (1 − a) ∫0t G ( x )dF ( x)
t
1
t
2
t
3
t
t
= F (t ) − a ∫0 G ( x) dF ( x ) ,
t
4
Figure 2. Reliability evolution of the standby system
Q11 (t ) = P(U = 1, ς ≤ t , γ < ζ ) = a ∫0t G ( x )dF ( x) ,
Moreover we assume that the all random variables,
mentioned above are independent.
Q20 (t ) = P (U = 0, ς ≤ t ) = (1 − a) F (t ),
5.2. Construction of the semi-Markov model
Q21 (t ) = P (U = 1, ς ≤ t ) = aF (t ).
To describe reliability evolution of the system, we
have to define the states and the renewal kernel. We
introduce the following states:
0 - the system is failed
We assume that, the initial state is 2. It means that an
initial distribution is
p (0) = [0 0 1] .
1 - the failed component is repaired, spare is operated
Hence, the semi-Markov model is constructed.
2 - both operating component and spare are “up”.
5.3. The reliability characteristics
Let 0 = τ 0∗ , τ 1∗ ,τ 2∗ ,... denote the instants of the states
changes, and {Y (t ) : t ≥ 0} be a random process with
the state space S = {0,1,2} , which keeps constant
The random variable Θ iA , that denotes the first
passage time from the state i to the states subset A,
for i = 2 and A = {0} in our model, represents the
time to failure of the system. The function
values on the half-intervals [τ n∗ ,τ n∗+1 ), 0,1,... and is
right-continuous. The realization of this process is
shown in Figure 1. This process is not semi-Markov,
because the condition (1) of definition (2) is not
satisfied for all instants of the state changes of the
process.
Let us construct a new random process a following
way. Let 0 = τ 0 and τ 1 , τ 2 ,... denote the instants of
the system components failures or the instants of
whole system renewal. The random process
{ X (t ) : t ≥ 0} defined by equation
R(t ) = P (Θ 20 > t ) = 1 − Φ 20 (t ) , t ≥ 0
(27)
is the reliability function of the considered cold
standby system with repair.
System of linear equation (23) for the LaplaceStieltjes transforms of the functions
Φ i 0 (t ) , t ≥ 0, i = 1,2,
in this case is
X (0) = 0, X (t ) = Y (τ n ) for t ∈ [τ n , τ n +1 )
~
~
φ10 ( s) = q~10 ( s ) + φ10 (s)q~11 (s )
is the semi-Markov process.
To have semi-Markov process as a model we must
define its initial distribution and all elements of its
kernel
~
~
φ 20 (s) = q~20 ( s) + φ10 (s)q~21 ( s )
The solution is
0
Q02 (t ) 
 0

Q(t ) = Q10 (t ) Q11 (t )
0 
Q20 (t ) Q21 (t )
0 
q~ ( s)
~
φ10 ( s) = 10~
,
1 − q11 ( s)
q~ (s)q~ (s)
~
φ 20 (s) = q~20 ( s) + 21 ~ 10 .
1 − q11 ( s)
For t ≥ 0 we obtain
Q02 (t ) = P (η ≤ t ) = H (t ),
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(28)
F.Grabski
Applications of semi-Markov processes in reliability - RTA # 3-4, 2007, December - Special Issue
π 1 p12 + π 2 p 21 = π 1 ,
Hence, we obtain the Laplace transform of the
reliability function
~
1 − φ 20 ( s)
~
R (s) =
.
s
π0 =π2,
(29)
π 0 + π 1 + π 2 = 1.
Since, the stationary distribution of the embedded
Markov chain is
The transition probabilities matrix of the embedded
Markov chain in the semi-Markov process
{X (t ) : t ≥ 0} is
 0
P =  p10
 p 20
0
p11
p 21
1
0
0
π0 =
p11
,
2 p11 + p 21
π1 =
p 21
,
2 p11 + p 21
π2 =
p11
.
2 p11 + p 21
(30)
Where
p10 = 1 − p11
p11 = P (U = 1, γ < ζ ) = a ∫0∞ G ( x )dF ( x) ,
p 20 = 1 − a,
Using formula (19) we obtain the limit distribution of
semi-Markov process
p 21 = P(U = 1) = a.
Using formula (9) we obtain the CDF of the waiting
times of Ti , i = 0,1, 2.
G 0 (t ) = H (t ), G1 (t ) = F (t ), G 2 (t ) = F (t ) .
Hence
P0 =
p11 E (η )
p11 E (η ) + p21 E (ς ) + p11 E (ς )
P1 =
p 21 E (ς )
p11 E (η ) + p 21 E (ς ) + p11 E (ς )
P2 =
p11 E (ς )
p11 E (η ) + p 21 E (ς ) + p11 E (ς )
E (T0 ) = E (η ), E (T1 ) = E (ς ), E (T3 ) = E (ς ) .
5.4. Conclusion
The equation (25) in this case has form
1 − p11

 −a
The expectation E (Θ 20 ) denoting the mean time to
failure is
0  E (Θ10 )   E (ς ) 

=

1   E (Θ 20 )  E (ς ) 
E (Θ 20 ) = E (ς ) +
The solution is
E (Θ10 ) =
E (ς )
,
1 − p11
E (Θ 20 ) = E (ς ) +
a E (ς )
.
1 − p11
(32)
a E (ς )
,
1 − p11
where
p11 = a ∫0∞ G ( x) dF ( x ) .
(31)
Let us notice, that the cold standby determines
a
increase the meantime to failure 1 +
times.
1 − p11
The limiting availability coefficient of the system is
We will apply theorem 1 to calculate the limit
probability distribution of the state. Now, the system
of linear equation (20) is
A = P1 + P2 =
π 1 p10 + π 2 p 20 = π 0 ,
- 65 -
p 21 E (ς ) + p11 E (ς )
.
p11 E (η ) + p21 E (ς ) + p11 E (ς )
F.Grabski
Applications of semi-Markov processes in reliability - RTA # 3-4, 2007, December - Special Issue
6. Semi-Markov process as the reliability
model of the operation with perturbation
6.2. Semi-Markov model
To construct reliability model of operation, we have to
start from definition of the process states.
Let ei j , i=1,...,n, j=0,1 denotes j-th reliability state
Semi-Markov process as the reliability model of
multi-stage operation was considered by F. Grabski in
[8] and [10]. Many operations consist of some
elementary tasks, which are realized in turn. Duration
of the each task realization is assumed to be positive
random variable. Each elementary operation may be
perturbed or failed. The perturbations increase the
time of operation and the probability of failure as
well.
on i-th step of the operation where, j=0 denotes
perturbation and j=1 denotes success
e2 n+1 - failure (un-success) of the operation
e11 - an initial state.
For convenience we numerate the states
ei 1 ↔ i,
6.1. Description and assumptions
i = 1,..., n
ei 0 ↔ i + n,
Suppose, that the operation consists of n stages which
following in turn. We assume that duration of an i-th
stage, (i = 1, ... , n) is a nonnegative random variable
ξ i ; i = 1,L , n
with a cumulative probability
distribution
i = 1,..., n
e2 n+1 ↔ 2n + 1,
Under the above assumptions, stochastic process
describing of the overall operation in reliability
aspect, is a semi-Markov process { X (t ) : t ≥ 0} with
a space of states S = {1, 2,..., 2n, 2n + 1} and flow
graph shown in Figure 3.
t
Fi (t ) = P(ξ i ≤ t ) = ∫ f i ( x )dx, i = 1,K, n ,
0
where f i (x ) denotes its probability density function
in an extended sense.
Time to failure of the operation on the i-th stage
(component) is the nonnegative random variable η i ,
i = 1, L, n with exponential distribution
P(η i ≤ t ) = 1 − e −λit ; i = 1, K , n .
The operation on each step may be perturbed. We
assume that no more then one event causing
perturbation on each stage of the operation may occur.
Time to event causing of an operation perturbation on
i-th stage is a nonnegative random variable
ζ i ; i = 1, K, n with exponential distribution
1
2
n-1
n
n+
1
n+
2
2n1
2n
2n
+1
Figure 3. Transition graph for n-stage cyclic operation
To obtain a semi-Markov model we have to define all
nonnegative elements of semi- Markov kernel
[
P(ζ i ≤ t ) = 1 − e −α it ; i = 1, K, n .
Q (t ) = Qij (t ) : i, j ∈ S
]
Qij (t ) = P{X (τ n +1 ) = j , τ n +1 − τ n ≤ t | X (τ n ) = i}
The perturbation degreases the probability of the
operation fail. We suppose that time to failure of the
perturbed operation on the i-th stage is the
nonnegative random variable ν i , i = 1,2...., n that
has the exponential distribution with a parameter
β i > λi
First, we define transition probabilities from the state i
to the state j for time not greater than t for i=1,…,n-1.
Qi i +1(t ) = P(ξi ≤ t,ηi > ξi ζi > ξi )
= ∫∫∫ α i e −α i y λi e − λi z f i ( x) dx dy dz
P(ν i ≤ t ) = 1 − e − βi t ; i = 1, K , n.
We assume that the operation is cyclical.
We
assume
that
random
variables
ξ i , , η i ,ν i , ζ i , i = 1,..., n are mutually independent.
D
where
D = {( x, y, z ) : x ≥ 0, y ≥ 0, z ≥ 0,
x ≤ t , z > x, x > y}
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F.Grabski
Applications of semi-Markov processes in reliability - RTA # 3-4, 2007, December - Special Issue
x = u + v,
Since, we have
t
x
∞
0
0
x
This mapping assigns to points from set
Qi i+1(t ) = ∫ fi (x)dx ∫αie−αi ydy ∫ λie−λi zdz
t
∫e
=
− ( λi +α i ) x
y = v, z = w.
∆ = {(u , v, w) : 0 ≤ u ≤ t , v ≥ 0, w > u}
f i ( x ) dx .
the points from plane region D. The Jacobian of this
mapping is
0
For i = n + 1,...,2n − 1 we obtain
J (u , v, w) = 1 .
Qi i + n (t ) = P(ζ i ≤ t ,η i > ζ i , ζ i > ξ i )
Since, we get
∫∫∫ α i e
t
= ∫ α i e −(λi +α i )u [1 − Fi (u )]du , i = 1,..., n − 1.
−α i y
β i e − βi z f i ( x ) dx dy dz
D
0
= ∫∫∫ α i e −α i v β i e − βi w f i (u + v ) J (u, v, w) du dv dw
For i = 1,..., n we get
∆
Qi 2 n+1 (t ) = P (η i ≤ t ,η i < ζ i ,η i < ξ i )
∞
∞
t
0
u
0
t
∞
0
0
= ∫ α i e−α i v dv ∫ βi e− β i wdw∫ fi (u + v )du
t
= ∫ e − βiu du ∫ α i e −α iv f i (u + v) dv .
= ∫ λi e −(λi +α i )u [1 − Fi (u )]du ,
0
i = 1,..., n − 1.
Let us notice that
If on i-th stage a perturbation has happened the
transition probability to next state for time less then or
equal to t is
∫∫ α i e
−α i y
E
Qn +i n + i +1 (t ) = P (ξ i − ζ i ≤ t , ν i > ξ i − ζ i | ξ i > ζ i )
∞
∞
0
y
f i ( x )dx dy = ∫ α i e −α i y ∫ f i ( x) dx
∞
= ∫ α i e −α i y [1 − Fi ( y )]dy
0
=
∫∫∫ α i e
−α i y
βie
− βi z
f i ( x) dx dy dz
D
−α y
∫∫ α i e i f i ( x )dx dy
∞
,
= 1 − ∫ α i e −α i y Fi ( y )dy.
0
E
i = 1,..., n − 1,
Finally, we obtain
t
where
Q n + i n + i +1 (t ) =
D = {( x, y , z ) : x ≥ 0, y ≥ 0, z ≥ 0,
∫e
− βiu
0
∞
[ ∫ α i e −α i v f i (u + v )dv ]du
0
∞
∫αie
0
0 ≤ x − y ≤ t,
z > x − y,
i = 1,..., n − 1.
x > y}
In the same way we get
E = {( x, y ) : x ≥ 0, y ≥ 0, x > y}.
To find the triple integral over the region D, we apply
change of coordinates:
u = x − y , v = y , w = z.
Hence
- 67 -
−α i y
[1 − Fi ( y )]dy
,.
F.Grabski
Applications of semi-Markov processes in reliability - RTA # 3-4, 2007, December - Special Issue
Q n + i 2 n +1 ( t ) = P (ν i ≤ t , ν i < ξ i − ζ i | ξ i > ζ i )
=
∫∫∫ α i e
−α i y
β i e − β i z f i ( x ) dx dy dz
D
∫∫ α i e
−α i y
f i ( x ) dx dy
Q15 (t ) 

Q 25 (t ) 
Q 35 (t ) ,

Q 45 (t ) 
Q 55 (t ) 
Q12 (t ) Q13 (t )
0
 0

0
0
Q 24 (t )
Q 21 (t )
0
0
Q 34 (t )
Q (t ) =  0

0
0
0
Q 41 (t )
 0
0
0
0

,
E
where
i = 1,..., n
Q12 (t ) = ∫ e −( λ1+α1 )u dF1 (u ) ,
t
where
0
D = {( x, y , z ) : x ≥ 0, y ≥ 0,
t
0 ≤ z ≤ t,
Q13 (t ) = ∫ α 1e −(λ1i +α1 )u [1 − F1 (u )]du ,
z < x − y , x > y}
0
E = {( x, y ) : x ≥ 0, y ≥ 0, x > y}.
Q15 (t ) = ∫ λ1e −( λ1i +α1 )u [1 − F1 (u )]du,
t
0
Hence
t
t
Qn +i 2 n+1 (t ) =
βi ∫ e
− βi w
∞
dw ∫ α i e
−α i v
dv ∫ f i (u + v)du
w
0
∞
∫αie
−αi y
Q21 (t ) = ∫ e
∞
−( λ2 +α 2 ) u
dF2 (u ) ,
0
0
Q2 4 (t ) = ∫ α 2 e −(λ2 +α 2 )u [1 − F2 (u )]du ,
t
[1 − Fi ( y )]dy
0
0
i = 1,..., n.
Q25 (t ) = ∫ λ 2 e −( λ2 +α 2 )u [1 − F2 (u )]du,
t
Similar way we obtain
0
Qn 1 (t ) = P(ξ n ≤ t ,η n > ξ n ζ n > ξ n )
t
t
= ∫e
− ( λn +α n ) u
Q34 (t ) =
f n (u ) du
∫e
− β1u
0
∞
[ ∫ α 1 e −α1v f 1 (u + v )dv ]du
0
∞
∫ α 1e
0
−α i y
,
[1 − F1 ( y )]dy
0
Q2 n 1 (t ) = P(ξ n − ζ n ≤ t , ν n > ξ n − ζ n | ξ n > ζ n )
t
t
=
∫e
− β nu
0
∞
0
∞
Q3 5 (t ) =
du ∫ α n e −α i v f n (u + v )dv
∫αne
−α n x
∞
∞
w
0
β1 ∫ e − βi w dw ∫ α 1e −α1v dv ∫ f 1 (u + v )du
0
∞
∫ α1
.
e −α1x [1 −
,
F1 ( x )]dx
0
[1 − Fn ( x)] dx
0
t
Therefore the semi-Markov reliability model of
operation has been constructed.
Q4 1 (t ) =
∫e
− β 2u
∞
du ∫ α 2 e −α 2v f 2 (u + v) ]dv
0
0
∞
∫α 2e
−α 2 x
,
[1 −F2 ( x)]dx
0
6.3. Two-stage cyclical operation
t
∞
∞
We will consider 3
β 2 ∫ e − β 2w dw ∫ α 2 e −α 2v dv ∫ f 2 (u + v)du
w
0
,
Q4 5 (t ) = 0 ∞
−α 2 x
[1 −F2 ( x)]dx
∫α 2e
We will investigate particular case of that model,
assuming n = 2 . A transition matrix for the semiMarkov model of the 2-stage cyclic operation in
reliability aspect takes the following form
0
Q55 (t ) = U (t ).
- 68 -
F.Grabski
Applications of semi-Markov processes in reliability - RTA # 3-4, 2007, December - Special Issue
That model allows us to obtain some reliability
characteristics of the operation. The random
variable Θ15 denoting the first passage time from state
1 to state 5 in our model, means time to failure of the
operation. The Laplace-Stieltjes transform for the
cumulative distribution function of that random
variable we will obtain from a matrix equation (25).
In this case we have A′ = {1,2,3,4}, A = {5} and
Then
~
φ15 ( s ) 
 q~15 ( s ) 
~

 q~ ( s) 
~
25
~ ( s ) = φ 25 ( s ) , b
,
(s) =  ~
A'
φ~ ( s) 


q
s
(
)
35
 ~35 
~

φ 45 ( s )
 q 45 ( s) 
q~12 ( s ) q~13 ( s)
 0
~
0
0
~ ( s ) =  q 21 ( s )
q
A'
 0
0
0
~
0
0
 q 41 ( s )

~
q 24 ( s) 
.
q~34 ( s) 

0 
0
From the solution of equation (24) we obtain LaplaceStieltjes transform of the cumulative distribution
function of the random variable Θ15 denoting time to
failure of the operation
a~ ( s)
~
φ15 ( s) = ~
,
b ( s)
(33)
a~ ( s ) = q~15 ( s) + q~12 ( s )q~25 ( s ) + q~13 ( s) q~35 ( s )
+ q~12 ( s) q~24 ( s )q~45 ( s ) + q~13 ( s) q~34 ( s ) q~45 ( s)
~
b ( s ) = 1 − q~12 ( s )q~21 ( s) − q~12 ( s )q~24 ( s) q~41 ( s)
κ1

−( λ +α +κ ) t 
1 − e 1 1 1 ,
λ1 + α1 + κ1 

Q13 (t ) =
α1

−(λ +α +κ ) t 
1 − e 1 1 1 ,
λ1 + α1 + κ1 

Q15 (t ) =
λ1

−( λ +α +κ ) t 
1 − e 1 1 1 ,
λ1 + α 1 + κ 1 

Q21 (t ) =
κ2
1 − e − (λ2 +α 2 +κ 2 )t ,
λ2 + α 2 + κ 2
Q2 4 (t ) =
α2
1 − e −(λ2 +α 2 +κ 2 )t ,
λ2 + α 2 + κ 2
Q2 5 (t ) =
λ2
1 − e − (λ2 +α 2 +κ 2 )t ,
λ2 + α 2 + κ 2
(
)
(
)
(
)
(
)
(
)
Q34 (t ) =
κ1
1 − e −(β1+κ1) t ,
β1 + κ1
Q35 (t ) =
β1
1 − e −(β1 +κ1) t ,
β1 + κ1
(
)
(
)
Q4 1 (t ) =
κ2
1 − e − ( β 2 +κ 2 ) t ,
β2 + κ2
Q4 5 (t ) =
β2
1 − e −( β 2 +κ 2 ) t ,
β2 + κ2
Laplace-Stieltjes transform of these functions are:
− q~13 ( s )q~34 ( s )q~41 ( s).
The Laplace transform of the reliability function is
given by the formula
~
1 − φ15 (s)
~
R (s) =
.
s
Q12 (t ) =
(34)
q~12 ( s ) =
κ1
,
s + λ1 + α 1 + κ 1
q~13 ( s ) =
α1
,
s + λ1 + α 1 + κ 1
q~15 ( s) =
λ1
,
s + λ1 + α 1 + κ 1
q~21 ( s ) =
κ2
,
s + λ2 + α 2 + κ 2
q~24 ( s ) =
α2
,
s + λ2 + α 2 + κ 2
6.4. Examples
Example 1
We suppose that
Fi (t ) = 1 − e −κ it , t ≥ 0, i = 1,2 .
- 69 -
F.Grabski
q~25 ( s) =
q~34 ( s) =
Applications of semi-Markov processes in reliability - RTA # 3-4, 2007, December - Special Issue
λ2
,
s + λ2 + α 2 + κ 2
This function is shown on Figure 5.
1
κ1
,
s + β1 + κ 1
0.8
0.6
0.4
β1
q~35 ( s ) =
,
s + β1 + κ 1
0.2
200
400
600
800
1000
q~41 ( s ) =
κ2
,
s + β2 + κ2
Figure 5. The reliability function of 2-stage cyclic
operation
q~45 ( s) =
β2
.
s + β2 + κ2
Mean time to failure we can find solving the matrix
equation
For κ1 = 0,1 ; κ 2 = 0,12 ; λ1 = 0,002 ;
(I − PA' ) Θ A' = TA'
λ2 = 0,001;
α1 = 0,02 ;
α 2 = 0,04 ; β1 = 0,01 ; β1 = 0,01 ,
applying
(33) and (34), with help of
MATHEMATICA computer program, we obtain the
density function and the reliability function as inverse
Laplace transforms .
The density function is given by the formula
(35)
where
 0

p
PA' =  21
 0

 p 41
φ15 (t )
p12
p13
0
0
0
0
0
0
 E (T1 ) 


E (T2 ) 

,
T=
 E (T3 ) 


 E (T4 ) 
= 0.00712201 e −0.213357 t − 0.00872613 e −0.180053 t
− 0.000144434 e −0.125901 t + 0.00374856 e −0.00368869 t
0 

p 24 
,
p34 

0 
 E (Θ15 ) 
 E (Θ )
25 
.
Θ A' = 
 E (Θ 35 ) 


 E (Θ 45 )
From this equation we obtain the mean time to failure
This function is shown in Figure 4.
The reliability function is
E(Θ15 ) = 275, 378
Example 2
Now we assume that
R(t ) = 3.79823 × 10 −16 + 0.0333808 e −0.213357 t
0 dla t ≤ Li
Fi (t ) = 
, i = 1,2 .
1 dla t > Li
− 0.0484641 e −0.180053 t − 0.0011472 e −0.125901 t
+ 1.01623 e −0.00368869 t
It means that the duration of the stages are determined
and they are equal ξ i = Li for i = 1,2.
In this case the elements of Q(t) are:
0.003
0.0025
for t ≤ L1
 0
Q12 (t ) =  −(λ1+α ) L1
1
e
for t > L1
0.002
0.0015
0.001
0.0005
200
400
600
800
1000
Figure 4. The density function the time to failure of
2-stage cyclic operation
- 70 -
F.Grabski
Applications of semi-Markov processes in reliability - RTA # 3-4, 2007, December - Special Issue
 α1 
−( λ +α ) t 
1 − e 1 1  for t ≤ L1

 λ + α1 

Q13 (t ) =  1
α
−( λ +α ) L 

1

1 − e 1 1 1  for t > L1
 λ1 + α1 

 λ1 
−( λ1+α ) t 
1
 for t ≤ L1
 λ + α 1 − e

1
1
Q15 (t ) = 
λ
−
+
)
(
λ
α
L


1

1 − e 1 1 1  for t > L1
 λ1 + α1 

 α2 
−( λ +α ) t 
1 − e 2 2  for t ≤ L 2

λ + α 2 

Q24 (t ) =  2
α
−(λ2 +α2 ) L2 

1

 for t > L2
1 − e

 λ2 + α 2 
(
0 ≤ t ≤ L1
,
t >L
p13 =
)
(
)
(
− ( λ +α + s ) L
 1 − e −( β1 −α1+ s ) L1

 s + β −α
1
1





β1  1− e−(β1+s)L1 e−α1L1 (1− e−(β1−α1+s)L1 ) 


−

s + β1 − α1
1− e−α1L1  s + β1

 1 − e − ( β 2 −α 2 + s ) L2

 s + β −α
2
2





−( β −α2 +s) L2
β2 1− e−(β2 +s)L2 e−α 2L2 (1− e
) 
−

s + β2 −α2
1− e−α2L2  s + β2

α 1 (1 − e − (λ1 +α1 ) L1 )
λ1 + α 1
p15 =
p 21 = e − (λ2 +α 2 ) L2



β e−a2L2
1
1 − e−β 2t − 2
1 − e−( β2 −α 2 )t  , 0 ≤ t ≤ L2

−a2L2 
(β2 − α2 )
)
 (1 − e

Q45(t ) = 

β2e−a2L2 
1
−( β2 −α 2 ) L 
− β2 L1

1 , t>L
1− e
−
1 − e
2

(1 − e−a2L2 ) 
(β2 − α2 ) 



(
λ2
λ e 2 2 21
− 2
s + λ2 + α 2
s + λ2 + α 2
p12 = e −( λ1 +α1 ) L1
)
(
q~25 ( s ) =
The mean time to failure we can find solving the
matrix equation (35), where
)
(
α2
α e − (λ2 +α 2 + s ) L2
− 2
s + λ2 + α 2
s + λ2 + α 2
q~45(s) =

α 2e
1 − e −( β2 −α2 )t , 0 ≤ t ≤ L2

 (β 2 − α 2 )(1 − e −a2 L2 )
Q41 (t ) = 
α 2 e −a2 L2

1 − e −(β2 −α 2 ) L2 , t > L2
 (β − α )(1 − e −a2 L2 )
2
2

− a2L2
q~24 ( s ) =
α e −α 2 L2
q~41 ( s ) = 2 −α L
1− e 2 2
)
(
λ1
λ e − (λ1 +α1 + s ) L1
− 1
s + λ1 + α 1
s + λ1 + α 1
q~35(s) =
 1  −β t β1e−a1L1

1−e 1 −
1−e−(β1−α1)t  , 0 ≤t ≤ L1

−a1L1 
(β1 −α1)
(1− e ) 

Q35(t) = 
−a1L1


e
β
1
L
β
−


β
α
(
−
−
1

1−e 1 1)L1  , t > L1
1−e 1  − 1
(1−e−a1L1 ) 
β
α
(
)
−
 1 1



)
q~15 ( s ) =
α 1e −α1L1
~
q34 ( s ) =
1 − e −α1L1
 λ2 
−( λ +α ) t 
1 − e 2 2  for t ≤ L 2

λ + α 2 

Q25 (t ) =  2
λ
−
+
)


(
λ
α
L
1

1 − e 2 2 2  for t > L2
 λ2 + α 2 

(
α1
α e − (λ1 +α1 + s ) L1
− 1
s + λ1 + α 1
s + λ1 + α 1
q~21 ( s) = e −( λ2 +α 2 + s ) L2 ,
for t ≤ L2
 0
Q21 (t ) =  −(λ2 +α 2 ) L2
for t > L 2
e

α1e−a1L1
−( β1−α1 ) t
),
 β −α − −a1L1 (1− e
( 1 1 )(1 e )
Q34 ()
t =
α1e−a1L1

(1− e−(β1−α1) L1 ) ,
(β1 −α1 )(1 − e−a1L1 )
q~13 ( s) =
)
)
p 24 =
α 2 (1 − e − (λ2 +α 2 ) L2 )
λ2 + α 2
p 25 =
λ 2 (1 − e 2 2
λ2 + α 2
p 34 =
α 1e −α1L1  1 − e − ( β1−α1 ) L1

1 − e −α1L1  β 1 − α 1
The Laplace’a-Stieltjes transform of these functions
are:
q~12 ( s) = e − (λ1 +α1+ s ) L1
- 71 -
− ( λ +α ) L21
)




λ1 (1 − e − (λ1 +α1 ) L1 )
λ1 + α1
F.Grabski
p 35 =
β1
1−
−
p 41 =
p 45 =
e −α1L1
Applications of semi-Markov processes in reliability - RTA # 3-4, 2007, December - Special Issue
To assess reliability of the many stage operation we
can apply a semi-Markov process. Construction of the
semi-Markov model consist in defining a kernel of
that process. A way of building the kernel for the
semi-Markov model of the many stage operation is
presented in this paper. From Semi-Markov model we
can obtain many interesting parameters and
characteristics for analysing reliability of the
operation.
From presented examples we get conclusion that for
the determined duration of the stages, mean time to
failure of the operation is essentially greater than for
exponentially distributed duration of the stages with
the same expectations.
− β1L1
1− e
(
β1
e −α1L1 (1 − e − ( β1−α1) L1
)
β1 − α 1
α 2 e −α 2 L2  1 − e − ( β 2 −α 2 ) L2

1 − e −α 2L2  β 2 − α 2




β2
1 − e −α 2 L2
7. Semi-Markov process as a failure rate
 1 − e − β 2 L2 e −α 2L2 (1 − e −( β −α 2 ) L2 ) 


−


β
β
α
−
2
2
2


The reliability function with semi-Markov failure rate
was considered by Kopoci ski & Kopoci ska [11],
Kopoci ska [12] and by Grabski [4], [6], [9]. Suppose
that the failure rate {λ (t ) : t ≥ 0} is the semi-Markov
process with the discrete state space S = {λ j : j ∈ J },
p51 = 1
J = {0,1,..., m} or J = {0,1,2,...} ,
with the kernel
e − (α1 +λ1 ) L1
1
E (T1 ) =
−
α 1 + λ1
α 1 + λ1
E (T2 ) =
0 ≤ λ0 < λ1 < ...
Q (t ) = [Qij (t ) : i, j ∈ J ]
− (α 2 + λ2 ) L2
e
1
−
α 2 + λ2
α 2 + λ2
E (T3 ) = −
d (q~34 ( s ) + q~35 ( s))
ds
E (T4 ) = −
d (q~41 ( s) + q~45 ( s))
ds
and the initial distribution p = [ pi : i ∈ J ] .
We define a conditional reliability function as
  t


Ri (t) = Eexp − ∫ (u)du λ(0) = λi  , t ≥ 0,

  0

s =0
s =0
i ∈J
In [6] it is proved, that for the regular semi-Markov
process {λ (t ) : t ≥ 0} the conditional reliability
functions Ri (t ) , t ≥ 0, i ∈ J defined by (17), satisfy
the system of equations
For the same parameters κ1 = 0,1 ; κ 2 = 0,12 ;
λ1 = 0,002 ; λ2 = 0,001; α 1 = 0,02 ; α 2 = 0,04 ;
β1 = 0,01 ; β1 = 0,01 ,
1
1
and L1 =
i L2 =
, the mean time to failure of
κ1
κ1
the operation is
t
Ri (t) = e−λi t [1− Gi (t)] + ∑∫ e−λi xRj (t − x)dQij (x), i ∈ J
j 0
Applying the Laplace transformation we obtain the
system of linear equations
E (Θ15 ) = 366.284 .
1
~
~
~
Ri (s) =
− Gi (s + λi ) + ∑Rj (s)q~ij (s + λi ), i ∈ J ,
j
s + λi
In previous case the mean time to failure is
E (Θ15 ) = 275, 378 .
where
6.5. Conclusion
∞
∞
~
~
Ri ( s) = ∫ e − s t Ri (t )dt , Gi ( s ) = ∫ e − s t Gi (t ) dt ,
It means that for the determined duration of the stages
mean time to failure of the operation is essentially
greater than for exponentially distributed duration of
the stages with the same expectations.
0
- 72 -
0
F.Grabski
Applications of semi-Markov processes in reliability - RTA # 3-4, 2007, December - Special Issue
1
− g~0 (s + λ0 ) ,
~ ( s)]= 
[I − q
λ

− g~ ( s + λ )
1
1

 1
∞
q~ij ( s ) = ∫ e − s t dQij (t ) .
0
In matrix notation we have
where
~
~
~ ( s)] R
[I − q
( s) = H( s ) ,
λ
∞
g~0 ( s) = q~01 ( s) = ∫ e − s t dG0 (t ) ,
0
where
∞
g~1 ( s) = q~10 ( s) = ∫ e − s t dG1 (t ) ,
[I − ~
q λ ( s )]
0
1 − q~00 (s + λ0 ) − q~01 (s + λ0 ) − q~02 ( s + λ0 ) L
 − q~ (s + λ ) 1 − q~ (s + λ ) − q~ ( s + λ ) L
1
11
1
12
1

=  ~10
 − q 20 (s + λ2 ) − q~21 (s + λ 2 ) 1 − q~22 ( s + λ2 ) L


M
M
M
O

~
 R0 ( s ) 

~
R1 ( s) 
~

R(s) = ~
 R2 ( s) 


 M 
~
 R ( s )
~
R ( s ) =  ~0  ,
 R1 ( s) 
~
 s+1λ − G0 ( s + λ0 )
~
0
H( s) =  1
.
~
 s +λ1 − G1 ( s + λ1 ) 
~

 1
 s + λ − G0 ( s + λ0 )
0


1
~

− G1 ( s + λ1 ) 
~
H (s ) =  s + λ1


 1
~

− G1 ( s + λ2 ) 

 s + λ2


M
The solution of (20) takes the form
~
R0 (s)
=
The conditional mean times to failure we obtain from
the formula
~
µ i = lim Ri ( p), p ∈ (0, ∞ ), i ∈ J
p→0+
1
s +λ0
[
]
~
~
− G0 (s + λ0 ) + g~0 (s + λ0 ) s +1λ − G1(s + λ1)
1
,
1− g~ (s + λ )g~ (s + λ )
0
0
1
1
~
R1 (s)
(21)
The unconditional mean time to failure has a form
=
1
s + λ0
[
0
N
µ = ∑ P(λ (0) = λi ) µ i .
i =1
]
~
~
− G1 (s + λ1 ) + g~1 (s + λ1 ) s +1λ − G0 (s + λ0 )
0
.
1 − g~ (s + λ ) g~ (s + λ )
0
1
1
The Laplace transform of the unconditional reliability
function is
7.1. Alternating random process as a failure
rate
~
~
~
R ( s ) = p 0 R0 ( s ) + p1 R1 ( s ).
Assume that the failure rate is a semi-Markov process
with the state space S = {λ0 , λ1} and the kernel
Example 3.
Assume that
G0 (t ) 
 0
Q(t ) = 
,
0 
G1 (t )
t
t
0
0
G0 (t ) = ∫ g 0 ( x) dx , G1 (t ) = ∫ g1 ( x) dx ,
where G0 (t ), G1 (t ), are the cumulative probability
distribution functions with nonnegative support.
Suppose that at least one of the functions is absolutely
continuous with respect to the Lebesgue measure. Let
p = [ p0 , p1 ] be an initial probability distribution of
the process. That stochastic process is called the
alternating random process. In that case the matrices
from the equation (20) are
where
g 0 ( x) =
g1 (t ) =
- 73 -
β 0α 0 α 0 −1 − β 0 x
x
e
,
Γ(α 0 )
β 1α1 α1 −1 − β1 x
,
x e
Γ(α 2 )
x ≥ 0,
x ≥ 0.
F.Grabski
Applications of semi-Markov processes in reliability - RTA # 3-4, 2007, December - Special Issue
Suppose that an initial state is λ0 . Hence the initial
1
distribution is p(0) = [1 0] and the Laplace
transform of the unconditional reliability function
0.8
is R ( s ) = R0 ( s ) .
form of
0.4
~
~
0.6
Now the equation (20) takes the
0.2
20


1



α
β1 1

−
α1
 ( s + β1 + λ1 )

  R~ ( s )
0
−
α 

( s + β 0 + λ0 ) 




  R~ ( s ) 
1
 1 

60
80
100
Figure 6. The reliability function from example 2
α
β0 0

α
β1 0
 1 −
 s + λ 0 (s + λ ) ( s + β + λ )α 0
0
0
0
=
α1
 1
β1
−

α1
 s + λ1 (s + λ1 ) (s + β 1 + λ1 )
40
The corresponding density function
f (t ) = − R' (t )
is shown in Figure 7







0.04
0.03
0.02
0.01
For
20
40
60
80
100
Figure 7. The density function from example 2
α 0 = 2 , α 1 = 3 , β 0 = 0.2 , β 1 = 0.5 , λ 0 = 0, λ1 = 0.2 ,
we have
8. Conclusion
The semi-Markov processes theory is convenient for
description of the reliability systems evolution
through the time. The probabilistic characteristics of
semi-Markov processes are interpreted as the
reliability coefficients of the systems. If A represents
the subset of failing states and i is an initial state, the
random variable ΘiA designating the first passage
time from the state i to the states subset A, denotes
the time to failure of the system. Theorems of semiMarkov processes theory allows us to find the
reliability characteristic, like the distribution of the
time to failure, the reliability function, the mean time
to failure, the availability coefficient of the system
and many others. We should remember that semiMarkov process might be applied as a model of the
real system reliability evolution, only if the basic
properties of the semi-Markov process definition are
satisfied by the real system.

1
0.04
0.04  1
0.125
−
−
+
2
2
3
s s(s + 0.2) (s + 0.2) s + 0.2 (s + 0.2)(s + 0.7) 
~
R0(s) =
0.04 0.125
1−
(s + 0.2)2 (s + 0.7)3
Using the MATHEMATICA computer program we
obtain the reliability function as the inverse Laplace
transform.
R(t ) = 1.33023 exp( −0.0614293 t )
+ exp( −0.02 t )(1.34007 ⋅ 10 −14 + 9.9198 ⋅ 10 −15 t )
− 2 exp( −0.843935 t )[0.0189459 cos( 0.171789 t )
+ 0.00695828sin( 0.171789 t )]
References
− 2 exp( −0.37535 t )[0.146168 cos(0.224699 t )
+ 0.128174 sin( 0.224699 t )]
Figure 6 shows the reliability function.
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F.Grabski
Applications of semi-Markov processes in reliability - RTA # 3-4, 2007, December - Special Issue
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