CC BY
34
4.1. Study of the system without intercellular transfer
In this section we simulate our system without the intercellular transfer policy in order to validate our model of cellular system and the developed expressions. The following table summarizes the values of the parameters developed according to the analytical model and that of simulation.
Table 3. Results from the approximate method and simulation
Machines M1a M2a M3a Mb1 Mb2 Mb3 Mb4
Tu*analytic (%) 42.53 62.21 60.91 50.77 41.98 49.63 50.69
Tu * simulation (%) 42.56 62.26 60.96 50.61 41.85 49.47 50.53
Error (%) < 1 %
A(<X>)analytk (%) 96.46 92.54 92.89 96.31 95.8 94.9 97.04
A (œ) simulation (%) 96.53 92.67 92.84 96.01 95.68 94.79 97.37
Error (%) < 0.5 %
The obtained results show that the error between the values given by the analytical formulation and simulation is relatively small. Then the approximation of the cell to a production line working under a continuous and constant load is sufficiently robust to estimate the availability of the autonomous cells.
4.2. Study of the system with intercellular transfer
For the studied system, the strategy consists in applying the intercellular transfer of the cell (a) toward the cell (b) in case of failure of one of machines of the first cell (see Figure 4). We assume that preparation times of intercellular transfer are negligible compared to times between failings and repair machine times. The values of the availability governed by the two types of policy with and without transfer are calculated according to the analytical model. The obtained results show the improvement made by the application of the intercellular transfer policy in term of cell availability (see Figure J).
To evaluate the performance of our policy with simulation tool and to support the results of the analytical model, we select the productivity of the cell, presented in the number of produced pieces, and machine utilization rate as performance criteria (see Table 4).
100 95 90 85 80 75
□ A(«)analytic (%) B A(»)analytic_tr (%)
M1a M 2a M3a Cell (a)
Figure J. Improvement of the availability by the intercellular transfer policy predicted by the analytical model
Table 4. Performance of the cell (a) with and without intercellular transfer policy from simulation
Machine Cell (a)
Mla M2a M3a
Utilization rate (%) Without transfer 42.56 62.26 60.96 44.12 62.95 62.40
With transfer
Number of manufactured products Without transfer 346650 359337
With transfer
The simulation results show that intercellular transfer policy improves the machines utilisation rate and the productivity of the cell. This improvement reflects the augmentation of cell availability.
5. Conclusion
In this paper, an analysis of cellular manufacturing system is presented. The notion of virtual cells and intercellular transfer allowed the development of a solution, which overcomes the effect of failures by continuing the process with the machine of the adjacent cell. Analytical modelling makes possible to determine the expression of the availability of the cell and to explain the improvement obtained by applying the intercellular transfer policy. The simulation results validated our analytical model and proved the effectiveness of the applied policy in the improvement of the system performance.
References
[1] Banerjee, A. & Flynn B. (1987). A Simulation Study of Some Maintenance Policies in a Group Technology Shop. International Journal of Production Research. 25, 1595-1609.
[2] Buzacott, J.A. (1968). Prediction of the efficiency of production systems without internal storage.
International Journal of Production Research. 6, 173-188.
[3] Geonwook, J., Herman, R.L. & Hamid, R.P. (1998). A cellular manufacturing system based on new similarity coefficient which considers alternative routes during machine failure. Computers and Industrial Engineering. 35, 73-76.
[4] Lahoti, A. & Kennedy W.J. (1991). Analyzing the Effect of Installing Diagnostic Equipment in a Manufacturing Cell. Proceeding of the IEEE Annual Reliability and Maintainability Symposium. 10-14.
[5] Savsar, M. (2006). Effects of maintenance policies on the productivity of flexible manufacturing cells.
OMEGA the International Journal of Management Science. 34, 274-282.
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 semi-Markov 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
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 semi-Markov 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.
2. Definition of semi-Markov processes with a discrete state space
Let S be a discrete (finite or countable) state space and let R+ = [0,¥), N0 = {0,1,2,...}. Suppose, that
Xn, Jn, n = 0,1,2,... are the random variables defined on a joint probabilistic space (W , • , P) with values on S and R+ respectively. A two-dimensional random sequence {(Xn, Jn), n = 0,1,2,...} is called a Markov renewal chain if for all /o,....,Vl,i e S, tn e R+, n e ^
1. p{ n+1 = j, Jn+1 £ 11 Xn = i, Jn = tn ,...,X0 = i0, J0 = t0 }
= P{x n+1 = j, J n+1 £ 11 x n = i}= Qij (t), (1)
2. P{X0 = io, J 0= 0} = P{X0 = i'0} = po (2)
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 {Xn : n = 0,1,2,...} is a
homogeneous one-dimensional Markov chain with the transition probabilities
py = P{Xn+! = j IXn = i} = lim Qj(t).
J t J
The matrix
Q(t) = [Qj (t): i, j î S][
(3)
(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
10 =Jo = 0,
T n =J1 + ... + Jn , = sup{t n : n e N 0}
A stochastic process {X(t): t > 0} given by the following relation
X(t) =xn for t e [tn,tn+1)
(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 [t n, t n+1) and it is a right-continuous function, from equality X(t n) =X n, it follows that the sequence {X (t n): n = 0,1,2,...} is a Markov chain with the transition probabilities matrix
P = [ pu : i, j e S ]
(6)
The sequence {X(tn): n = 0,1,2,...} is called an embedded Markov chain in a semi-Markov process {X(t): t > 0}. The function
F j (t) = P{t n+1 -t n < t | X(t n ) = i, X(t n+1 ) = j}
Q j (t )
(7)
is a cumulative probability distribution of a random variable Tij that is called holding time of a state i, if the next state will be j . From (11) we have
Qj (t) = PjFj (t).
(8)
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 A = [a j : i, j e S], 0 < -aii = ai < ¥ is the semi-
Markov process with the kernel
Q(t) = [Qj (t): i, j e S] , Where
Qj (t) = pj (1 - e-aiit), t > 0,
a
Pa = — for i ^ j and Pn = 0
a
In the reliability models the parameters and characteristics of a semi-Markov process are interpreted as the reliability characteristics and parameters of the system.
3. Transition probabilities of a semi-Markov process
The transition probabilities of the semi-Markov process are introduced as follows:
Pj (t) = P{X(t) = j | X(0) = i}, i, j e S. (11)
The function
Gt (t) = P{t n+1 -t n < t|X (t n ) = i}= z Qj (t) (9)
jeS
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]
is a cumulative probability distribution of a random variable Tt that is called waiting time of the state i. The waiting time Tt 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
N(t) = n for t e [t n, t n+i)
(10)
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
P{N(t) < ¥} = 1
Pj (t) =5 j [1 - G,. (t)] + ZJ Pkj (t - x)dQlk (x), (12)
keS0
i, j e S .
Using Laplace-Stieltjes transformation we obtain the system of linear equation
Pj(s) = d j[1 - gi(s)] + Z (s)Pij-(s),
keS
i, j e S
where the transforms
Pj (s) = J e-stdPj (t) 0
(13)
are unknown while the transforms
¥
qik (s) = J e-stdQlk (t) , g. (s) = J e(t) 0 0
are given.
Passing to matrices we obtain the following equation
p(s) = [/ - g(s)] + p(s)p(s), where
(14)
P(s) = [Pj (s) : i, j e S] , q(s) = [q (s) : i, j e S],
g(s) = [S j(1 -~i(s)): i, j e S].
In many cases the transitions probabilities Pij(t) and the states probabilities
Pj (t) = P{X(t) = j}, j e S, approach constant values for large t
Pj = lim Pj (t), Pj = lim (t) .
J t J J t J
(15)
(16)
To formulate the appropriate theorem, we have to introduce a random variable
A = min {n e W : X (t „ ) = j},
(17)
That denotes the time of first arrival at state j. A number
f = P{A j <¥|X (t n ) = 1}
(18)
is the probability that the chain that leaves state will sooner or later achieve the state j. As a conclusion of theorems presented by Korolyuk and Turbin [13], we have obtained following theorem
Theorem 1.
Let {X(t) : t > 0} be a semi-Markov process with a discrete state space S and continuous kernel Q(t) = [Qy (t): i, j e SJ. If the embedded Markov
chain {X(tn): n = 0,1,2,...}, contains one positive recurrent class C, such that for each state
i e S, j e C, f = 1 and 0 < E(T) <¥, i e S, then
P = lim Pi (t) = P; = lim p, (t) =
P iE T )
SP jE (Tj )
leS
(19)
where p = [p j, j e S] is the unique stationary
distribution of the embedded Markov chain that satisfies system of equations.
SpP, =p i, j e S Spi =1.
(20)
4. First passage time from the state i to the states subset A.
The random variable
Q A =t A ^
where
DA = min{n e N: X(t n) e A},
denotes the time of first arrival of semi-Markov process, at the set of states A. The function
Fia (t) = P{© a < 11X(0) = i}.
(21)
is the cumulative distribution of the random variable QiA that denotes the first passage time from the state i to the states subset A.
Theorem 2. [4], [13]
For the regular semi-Markov processes such that, fM = P{Da <¥|X(0) = i} = 1, i e A', (22)
the distributions FiA (t), i e A are proper and they are the unique solutions of the system of equations
F A (t) = S Q, (t) + S J F kA (t - x)dQlk (x),
jeA keS 0
l e A
Applying Laplace-Stieltjes transformation we obtain the system of linear equations
<~a (s) = Z (s) + Z<~a (s)~it (s), i e A' (23)
jeA AeA'
with unknown transforms
f a (s) = Je-stdFa (t).
Generating matrix form we get equation
30
¥
¥
(l - qA,(s)) ua (s) = b(s), where
I = [ô j : i, j e A'], qa (s) = [pj (s) : i, j e A] are the square matrices and q A'(s) = [2 iA (s): i e A'f ,
(24)
b(s) =
Z bij(s):ieA'
je A
are the one-column matrices of transforms. The formal solution of the equation is
ÏÏ A'(s) = (I - qA'(s))-1b(s).
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 ©iA, i e A'. If the second moments of the waiting times ], i e A' are positive and commonly bounded, and fAA = 1, i e A' , then the expected values of the random variables ©iA, i e A' are the unique solution of equation
(I - Pj H,= Ta,, (25)
where
I = [ôy. : i, j e A'] , PA=[py. : i, j e A']
Ha- = [E(©a ): i e A']T, Ta, = [E(] ) : i e A']T
and the second moments of the time to failure are the unique solution of equation
(I - Pa-) HA'= BA'
where
I = [d ij : j e A'], PA' = [Pj : j e A'.[ HA- = [.E(©2iA ): i e A'], BA, = [¿,. : i e A']
(26)
5. Semi-Markov model of the cold standby system with repair
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].
5.1. Description and assumptions
A system consists of one operating component, an identical stand-by component and a switch, (Figure!).
Figure !. Diagram of the system
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 v 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 g with cumulative distribution function (CDF) G(x) = P(g < x). Let U be a random variable having binary distribution
b(k) = P(U = k) = ak (1 - a)1-k,k = 0,1, 0 < a < 1,
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 h with CDF H(x) = P(h < x).
-F
b, = E(] ) + 2 Z PkE(Tk )E(©kA ).
keA'
T
Figure 2. Reliability evolution of the standby system
Moreover we assume that the all random variables, mentioned above are independent.
5.2. Construction of the semi-Markov model
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
1 - the failed component is repaired, spare is operated
2 - both operating component and spare are "up".
Let 0 =t*,t*,:*,...denote the instants of the states changes, and {Y{t): i > 0} be a random process with the state space S = {0,1,2}, which keeps constant
values on the half-intervals [t*,t *+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 = xn and x, denote the instants of
the system components failures or the instants of whole system renewal. The random process \X{i): i > 0} defined by equation
X(0) = 0,X(0 = r(x„)forie[T„,x„+1)
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
Q(0 =
0 0 O02{t) Qw(t) Quit) o 0
_Q20(t) o2l (t)
For t > 0 we obtain O02(t) = P^<t) = H(t)
Ow(t)=P(q<t,j>(;)
+ P(U = 0,q<t,y <C)
= [1 - G(x)]dF(x) + (1 - a)j' G(x)dF(x)
= F(t)-a^G(x)dF(x) ,
Ou (0 = P(U = 1, q < t, y < C ) = G(x)dF(x) ,
Q20(t)=P(U = 0,q<t) = (l-a)F(t),
Q2l(t)=P(U = l,q<t) = aF(t).
We assume that, the initial state is 2. It means that an initial distribution is
p(0) = [0 o 1].
Hence, the semi-Markov model is constructed.
5.3. The reliability characteristics
The random variable ©,,, that denotes the first passage time from the state i to the states subset A. for / = 2 and A = {0} in our model, represents the time to failure of the system. The function
R(t) = P(e20 >o = i-o20(o, t>o
(27)
is the reliability function of the considered cold standby system with repair.
System of linear equation (23) for the Laplace-Stieltjes transforms of the functions
O,0(0,f>0,/ = 1,2,
in this case is
fco (s) = <?20 (s) + <Ko (s)<72I (S)
The solution is
Ms) =
l"<7ii(X>'
4>20(X> = 9r20(X> +
l"<7ii(X>
(28)
Hence, we obtain the Laplace transform of the reliability function
R (s) =
1 -f20(s)
(29)
The transition probabilities matrix of the embedded Markov chain in the semi-Markov process {X(t): t > 0} is
" 0 0 1"
P = P10 P11 0 (30)
_ P20 P21 0
Where
P10 = 1 - P11
pu = P(U = 1, g < Z ) = aJ¥ G(x)dF(x) ,
P20 = 1 - a, P21 = P(U = 1) = a.
Using formula (9) we obtain the CDF of the waiting times of Ti, i = 0,1,2.
G0 (t) = H (t), Gj(t) = F(t), G2(t) = F(t).
Hence
E(T0) = E(h), E(T) = E(v ), E(T3) = E(V ). The equation (25) in this case has form
"1 - P11 0" " E (©10) " "E (V )"
- a 1 E (© 20 )_ E (V )_
The solution is
E (y )
E (©10) =
1 - P11
E (© 20) = E(y ) +
aE (y )
1 - P11
(31)
We will apply theorem 1 to calculate the limit probability distribution of the state. Now, the system of linear equation (20) is
p 1P10 + p 2 P20 = p 0,
p 1P12 +p 2 P 21 =p l,
p 0 1 2 = 1.
Since, the stationary distribution of the embedded Markov chain is
P11
p =
2 P11 + P21
P21 2 P11 + P 21 !
P11 2 P11 + P21
Using formula (19) we obtain the limit distribution of semi-Markov process
P0 =
P11E (h)
(32)
P11E (h) + P21E (V) + P11E (V)
_P21E (y)_
P11 E(h) + P21E (ç) + P11E (ç)
_P11E (V)_
P11E (h) + P21E (ç) + P11E (ç)
5.4. Conclusion
The expectation E(© 20) denoting the mean time to failure is
P1 =
P2 =
E (© 20) = E(y ) +
aE (y )
1 - P11
where P11 = aJ0° G(x)dF(x) .
Let us notice, that the cold standby determines
a
increase the meantime to failure 1 + -
1 - P11
times.
The limiting availability coefficient of the system is
P21 E(V) + P11E )
A = P1 + P2 =
P11E (h) + P21E (ç) + P11E (ç)
p 0 =p 2
s
6. Semi-Markov process as the reliability model of the operation with perturbation
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.
6.1. Description and assumptions
Suppose, that the operation consists of n stages which following in turn. We assume that duration of an 7-th stage, (/ = 1, ..., n) is a nonnegative random variable 2,,: i = 1.« ,77 with a cumulative probability distribution
0
where fi (x) denotes its probability density function in an extended sense.
Time to failure of the operation on the 7-th stage (component) is the nonnegative random variable r|, , 7 = 1, • ,77 with exponential distribution
P(r1, <t) = 7 = 1,' ,77.
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 7-th stage is a nonnegative random variable i = 1.* ,n with exponential distribution
Pfc, <t)=l-e~a't; 7 = 1,. ,77 .
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 v;, / = 1,2....,// that has the exponential distribution with a parameter
P(v. <i) = l-e Pii; / = 1,' ,77. We assume that the operation is cyclical. We assume that random variables ^, , r), ,v,, Z,, 7 = 1,..., 77 are mutually independent.
6.2. Semi-Markov model
To construct reliability model of operation, we have to
start from definition of the process states.
Let ef i=l.....77, / 0, / denotes j-th reliability state
on i-th step of the operation where, j=0 denotes perturbation and j=l denotes success e^n+l - failure (un-success) of the operation
en - an initial state.
For convenience we numerate the states
2,-j <->/', 7 = 1,...,77 eiQ <H>7+77, 7=1,...,77
Under the above assumptions, stochastic process describing of the overall operation in reliability aspect, is a semi-Markov process {X{t)\ i > 0} with a space of states S = { 1,2,...,2t7,2t7 + 1} and flow graph shown in Figure 3.
To obtain a semi-Markov model we have to define all nonnegative elements of semi- Markov kernel
Qv (t) = P{X( X )I+1 ) = j,z n+1 -x„<t\X(T„) = 7}
First, we define transition probabilities from the state i to the state j for time not greater than t for i=l.....n-1.
= (x)dxdydz
D
where
D = {(x,y,z): x > 0, v > 0, z > 0,
X < t, Z > X, X > y}
Since, we have
Qi+1\t) = 1f (x)dx \ale_hy dy\ V dz
0 0 x
= \ e_(l'+a)xf (x)dx . 0
For i = n + 1,...,2n _ 1 we obtain
Q„+n(t) = P(Zi £t,hi >Zi,Zi >Xi)
= \a e<A' +a)u [1 _ Ft (u)]du , i = 1,...,n _ 1.
0
For i = 1,...,n we get
Qi2n+1 (t) = P(hi £t,hi <zi,hi <Xi)
= \ V-(X' +a')u [1 _ Ft (u)]du ,
0
i = 1,...,n _ 1.
If on i-th stage a perturbation has happened the transition probability to next state for time less then or equal to t is
Qn+^+1(0 = P(Xi _Zi £t,n>X_Zi IXi >Zi)
\\\ai e ^ p ee _Pi z f (x)dx dy dz
= _D_
" \\a ee -a'yf (x)dxdy '
E
i = 1,...,n _ 1, where
D = {(x,y,z): x> 0,y > 0, z> 0,
0 £ x _ y £ t, z > x _ y, x > y}
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
x = u + V, y = V, z = w. This mapping assigns to points from set
D = {(u, v, w): 0 £ u £ t, v > 0, w > u}
the points from plane region D. The Jacobian of this mapping is
J (u, v, w) = 1. Since, we get \\\a e ^ p e -p'z f (x)dxdy dz
D
= \\\aie~a'Vp ee~p'w f (u + v)J(u, v, w)du dvdw
D
¥ ¥ t = \a ie~a'iVdv \ Pie~biwdw\ f (u + v)du
0 u 0
t ¥ = \ e _bi"du \a e _a'Vft (u + v)dv .
00
Let us notice that
¥ ¥
\\ae~a'yf (x)dxdy = \ate~a'y \ f (x)dx
E 0 y
¥
= \aie^[1 _ Fi (y)]dy
0
¥
= 1 _\aie-a'yFt (y)dy. 0
Finally, we obtain
t ¥
\ e-p'u [\a e-a'Vft (u + v)dv]du
Q (t) = 0_0_
-^n + i n +i+1 v / ¥ '
\a[1 _ F, (y)]dy
0
i = 1,..., n _ 1. In the same way we get
Qn +i 2 n+l(t) = P (ni < t, V, <x,-Z JX, >Zi )
HI a, e ~a y b te ft (x) dx dy dz
_D_
IIate~aiy ft (x)dxdy
E
Q(t) =
0 Q12 (t) Q13 (t) 0 Q15(
Q2i(t) 0 00 Q4i(t) 0 00
0 Q 24 (t) Q 25 (
0 Q34 (t) Q 35 (
0 0 Q45 (
0 0 Q 55 (
i = 1,..., n
where
D = {(x,y,z): x> 0,y > 0,
0 < z < t, z < x - y, x > y}
where
Q12 ()=j e
(l1+a1 )u
dF1 (u),
Q13 (t) = |a 1e+a1)u [1 - F1 (u)]du :
E = {(x,y): x>0,y > 0, x >y}. Hence
t ¥ ¥ Pi 1 e biwdw\aie~aiVdv\ f (u + v)du
Q (t) =_0_w_0_
Sdn +i 2n+1V/ ¥
\aie^ [1 _Ft (y)]dy
0
i = 1,...,n. Similar way we obtain Qn1(t) = P(X n £ t,h n >X n Z n >X n )
= I e +an )u 0
fn (u )du
Q15 (t) = 111 e-(l1i +a1)u [1 - F1 (u)]du
Q21(t) = |e+a2)udF2 (u), 0
Q24 (t ) = j a 2 e -(l2 +a2)u [1 - F2(u)]du:
Q25 (t )=j l e ~(l2 +a2)" [1 - F2{u )]du,
0
Q34 (t) =
t ¥
| e-ß1u [ |a 1e "a1 vf (u + v)dv]du
0 0
ja^ ^ [1 - F1 (y)]dy
Q2n 1(t) = P(X n -Z n < t, V n >X n -Z n I X n >Z n )
t ¥
je-ßnudu jane^fn(u + v)dv
ja ne-anx [1 - Fn (x)] dx
Therefore the semi-Markov reliability model of operation has been constructed.
6.3. Two-stage cyclical operation
We will investigate particular case of that model, assuming n = 2 . A transition matrix for the semi-Markov model of the 2-stage cyclic operation in reliability aspect takes the following form
b11 e-b'wdw |a1e ~a1Vdv j f1(u + v)du
Q3 5 (t) = -
ja 1e a1x [1 - F1(x)]dx
t ¥
je-b2"du ja2e~a2Vf2(u + v) ]dv
Q41(t) =-
ja 2 e ~a 2x [1 -F2( x)]dx
b 2 j e-b2wdwja 2 e ~a2Vdv j f2(u + v)du
We will consider 3
Q4 5 (t) = -
Q55 (t) = U(t).
ja 2 e "a2x [1 -F2( x)]dx
¥
¥
0
0
w
¥
0
0
¥
¥
0
0
w
That model allows us to obtain some reliability characteristics of the operation. The random variable Q15 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
~ A' (s) =
Then
(i~5(s) ~15(s)
f25 (s) , ~(s) = ~25 (s)
fi35( s) ~35 (s)
_f~45 (s)_ ~45 (s)_
~A' (s) =
0 ~12(s) ~13 ( s) 0
~2l(s) 0 0 ~24 (s)
0 0 0 ~34( s)
~4i(s) 0 0 0
From the solution of equation (24) we obtain Laplace-Stieltjes transform of the cumulative distribution function of the random variable Q15 denoting time to failure of the operation
f 15 (s) =
~(s)
~(s)
(33)
~(s) = ~15 (s) + ~12 (s)~25 (s) + ~13 (s)~35 (S)
+ ~12 (S)~24 (S)~45 (S) + ~13 (s)~34 (s)~45 (s)
b (s) = 1 - ~12 (s)~21 (s) - ~12 (s)~24 (s)~41 (S)
41 (s).
The Laplace transform of the reliability function is given by the formula
R (s) =
1 f 15 (s)
(34)
6.4. Examples
Example 1 We suppose that
Ft (t) = 1 - e ~Kit, t > 0, i = 1,2
Q12 (t ) =
k,
11 +a1 +k1
Q13 (t ) = Q15 (t ) =
Q21(t ) =
Q2 4 (t) =
a
11 +a1 +k1 I1
11 +a1 +k1 k,
12 +a 2 + k 2
1 - e
1-e
-(^1 +a1 +K1 ) t
-(l1+a1 +K1) t
\
1-e
-(l1+a1 +K1 ) t
. - e
-(l2 +a2 +K 2 )t
a
2_(1 - e -(I2 +a2 +K2 )t
12 +a 2 + K 2
Q25 (t ) = --12-(1 - e-(l2 +a2 +K 2)t ),
12 +a 2 + k 2
Q34 (t ) = bJ+-(1-e-(P1+K1) t ),
P1 +K1
Q35 (t ) = (1-e-(P1+K1) t ),
P1 +K1
Q41(t ) =
K-
P 2 +K 2
Q4 5 (t ) =-
P2
(1 - e-(P2 +K2) t ),
.(1 - e -(P2 +K2) t ),
P 2 + K2
Laplace-Stieltjes transform of these functions are:
~12 (s) =
~13 (s) = ~15 (s) =
~21 (s) =
~24 (s) =
K,
s +11 +a1 + K1 '
a1
s +11 +a1 + k 1 ,
11
s + 1 +a1 + K1 :
K 2
s +12 + a 2 + K 2
a 2
s + 12 +a 2 + k 2
s
~25 (s) =
A
s + A2 +a 2 + k 2
This function is shown on Figure 5.
k,
~34( S) = ~35 (s) =
s + P1 + k 1 P1
s + P1 + k 1
~41 (s) =
~45 (s) =
K,
s + P2 + K2
P 2
s + P 2 + K 2
For K1 = 0,1; k2 = 0,12; A1 = 0,002; A2 = 0,001; a1 = 0,02; a 2 = 0,04; p1 = 0,01; p1 = 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
f 15 (t)
= 0.00712201 e-° 2133571 _ 0.00872613 e180053 1
_ 0.000144434 e _a 1259011 + 0.00374856 e _0'003688691
This function is shown in Figure 4. The reliability function is
R(t) = 3.79823 x 10_16 + 0.0333808 e~0'2133571
200 400 600 800 1000
Figure 5. The reliability function of 2-stage cyclic operation
Mean time to failure we can find solving the matrix equation
(I _ PJ©a,= Ta where
P„
(35)
0 P12 P13 0
P21 0 0 P24
0 0 0 P34 ,
_ P41 0 0 0
E(T)" " E (©15)
E (T2) E (T3) , Q A' = E (© 25) E (© 35)
E (T4)_ E (© 45)
From this equation we obtain the mean time to failure
E(©15) = 275,378 Example 2 Now we assume that
_ 0.0484641 e_a 180053 1 _ 0.0011472 e_a 125901 1
+1.01623 e
_0.00368869 t
0.003 0.0025 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
f0 dla t £ L F (t) = i i , i = 1,2.
^ |1 dla t > L/
It means that the duration of the stages are determined and they are equal Xi = Li for i = 1,2. In this case the elements of Q(t) are:
Q12 (t ) =
0
_(A1+a1 ) L1
for t £ L for t > L,
e
Qj3 (t ) =
a,
A1 +a1
1 - e
-(^1+aj ) t
Q15 (t ) =
A +a1
A +a1 A +a1
a1 (1 - e-(^i+a, ) Lj
1-e
-(A1+a1 ) t
1-e
-(A1+a1) L|
for t < L1
for t > L1
for t < L1
for t > L1
O J 0 for t < L2
Q21 (t) = |e_(A2 +a2) L2 for t > L„
Q24 (t )=
a
A2 +a 2 a1 A2 +a 2
1-e
-(A2 +a2) t
1-e
-(A2 +a2) L2
for t < L
for t > L
~13 (s) =
~15 (s) =
a1 a e ~(A1+a1+s )lJ
s + A1 + a1 s + A1 + a1
Aj A e ~(A1+a1+s )lJ
s + A1 + a1 s + A1 + a1
~21(s) = e ~(A2 +a 2+s )L2
~24(s) =
a.
a 2e
-(A2 +a 2+s) L2
s + A2 +a2 s + A2 +a2
~25 (s) =
A2e
-(A2 +a 2 +s )L2j
s + A2 +a 2 s + A2 +a 2
~ axe
q34(s) = —
a1L1 ( 1 - e-bj -aj+s)Lj ^
1 - e ~ajLj
s + P1 -a1
Q25 (t ) =
A2 (j - -(A2 +a2) t
Q34 (t)=
A2 +a2
A2 + a2
„-ah
1-e
-(A2 +a2) L2
for t < L
for t > L
(Pi -aj)(1-e-*)
a1e aL
[(Pi -aj)(1-e )
(-e-(Pl-a1)')
(-e-(p1-ai) L )
0 < t < l
t > L
~35(s) =
Pi
e-№1+s)L1 e~a1L1(1- e-(p1-a1+s)L1) ^
1-e'
e-^-e"
1)
s + P1
s + P, -a1
~41 (s) =
a 2 e
a2L2 ( 1 - e-(P2 -a2 +s)L2 ^
1-e
-a2L2
s + P2-a2
~45(s) = P2
(1-e-(P2+s)L2 „-a2L^! ^-(P -a2+s)L2^ ^
1 - e-a2L2
e-a2L2(1- e
2)
s+P2
s + P2 -a2
Q35(t)=
1
(1 -e-11)
1
(1 -e-^1)
1-e-* Ll^ (j-e-:P1-c'1)t
' (ft-a)
1-e_PlLl l-P-a. 1-e-(P1-"1)L1)
(P1 -a1)
0<t < L
t >L1
Q4j(t ) =
-a2L2
(P2 -a2)(1-e-a2L2)
(1 - e -(P2 -a2)t), 0 <t< L2
a 2e
-a2L2
St5^) =
1
(1 - e~a2L2)
1
(1 - e^2)
(P2 -a2)(1 -e-a2i2) (j-e"P* )-
(i-e-(P2-a2)L2 )
P2e~a2L2
(P2 -a2)
(j- e-(P2 -a2)t)
(j - e-P2L1
(P2 -a2)
1 - e
(P2 -C2)L1
t > L2
0 < t < l2 , t > l
The Laplace'a-Stieltjes transform of these functions are:
~12(s) = e-(Aj+aj+s )Lj
The mean time to failure we can find solving the matrix equation (35), where
p12 = e-(Aj+aj)Lj
a1 (1 - e-(Aj+aj)Lj) A1 (1 - e-(Aj+aj)Lj) P13 =-:- P15 ="
A1 + a1
p 2j = e-(A2 +a 2) L2
A1 + a1
P 24 =-
a 2(1 - e '(A2 +a 2) L2) A2 + a 2
P 25 =■
A2(1 - e+a 2) L21) A2 + a 2
-a1L1
P34 =
1-e
-a1L1
( j - e-(P1-a1)Lj ^
Pj -a1
A
2
A
e
2
p35 = . _a,r, ('
_Pi£i
1 _ e _aiil
bi
P 41 =
P 45 =
e ~a1L1 (1 _ e-(Pl_«l)L1
Pi _a 1
e~a2L2 ^ 1 _ e"(P2_a2)¿2 ^
i -aoLo
1 _ e 22
P 2
i -aoLo
1 _ e 22
P 2 _a 2
œ 1 _ e-P2L2 e -a 2^2(1 _ e_ P _a2 )L2 ) ^ P2 P2_a2
P51 = 1
E (T1) =
1 e
-(a1+l1)L1
a 1 +11 a 1 +11
E (T2) =
1 e
-(a 2+I2) ¿2
E (T3) = _"v ^34
a 2 +12 a 2 +12
d(~34 (s) + ~35 (s))
ds
I s=0
E (T4) = _~v ^41
d(~41 (s) + ~45 (s))
ds
For the same parameters K1 = 0,1; k 2 = 0,12 ;
= 0,002; 12 = 0,001; a1 = 0,02; a2 = 0,04; P1 = 0,01; P1 = 0,01,
1 .
1
and L1 = — i L2 = —, the mean time to failure of
k
K,
the operation is
E (Q15) = 366.284. In previous case the mean time to failure is
E (Q15) = 275,378. 6.5. Conclusion
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.
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.
7. Semi-Markov process as a failure rate
The reliability function with semi-Markov failure rate was considered by Kopocinski & Kopocinska [11], Kopocinska [12] and by Grabski [4], [6], [9]. Suppose that the failure rate {A(t) : t > 0} is the semi-Markov process with the discrete state space S = {A} : j e J}, J = {0,1,...,m} or J = {0,1,2,...}, 0 £ A0 < A1 < ... with the kernel
Q(t) = [Qj (t): i, j e J]
and the initial distribution p = [pi : i e J]. We define a conditional reliability function as
R (t) = E
exp - J n(u)du 111(0) = 1
t > 0, i e J
In [6] it is proved, that for the regular semi-Markov process {A(t) : t > 0} the conditional reliability
functions Ri (t) , t > 0, i e J defined by (17), satisfy
the system of equations
R(t) = e-At[1_G(t)]+Z\e-XixR](t_x)dQ(x),i e J
j 0
Applying the Laplace transformation we obtain the system of linear equations
R (s) _ Gi (s+1 )+ZR. (s)~;. (s+1i ), i e J,
s+1 . j j
where
Ri (s) = Je-stRi (t)dt, Gi (s) = Je'stGt (t)dt,
s=0
¥
¥
q,]{s) = \e-stdO,]{t).
o
In matrix notation we have [l-qx(s)]R(s) = H(s), where
1 -g0(s + X0) -gl(s + \) 1
where
l(s) = qm(s)= ¡e-stdG0(t),
[I-qJ*)]
g1(i) = ?10(i) = je-stdG.it).
l-q00(s + X0) -qm(s + X0) -q02(s + X0) -qw(s + \) l-qu(s + \) -^(s + Xj) -q2n(s + X2) -q2l(s + X2) 1 -q22(s + X2)
R (s) =
Rois) R^s) R2(s)
H(5) =
5 + Xf
--G0(s + X0)
1
S + Xj
_l_
s + X„
-G^ + XJ
- Gj (s + Xo )
The conditional mean times to failure we obtain from the formula
R (s) =
H (s) =
Rois) R1(s)_
-^--G0(s + X0)
The solution of (20) takes the form
■ï+V)
l-g0(5+A<:i)g1(5+A1)
H,■= hm RtiplpeiO,*), i^J (21)
The unconditional mean time to failure has a form
N ;=1
7.1. Alternating random process as a failure rate
Assume that the failure rate is a semi-Markov process with the state space S = {An, X,} and the kernel
Rxis)
Q (0 =
o G0 (t ) G,(0 0
l-gn^ + Xo^^ + Xj)
The Laplace transform of the unconditional reliability function is
R(s) = p0R0(s) + p1R1(s).
Example 3. Assume that
t t G0 (0 = J g0 ix)dx , Gi if) = J gi ix)dx,
o o
where Gn(/). Gx (I), 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 = \-PoP\ ] 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
P
a0
= x > 0,
r(a0)
«i
IV
g,(0 = -^xai-Vßl \ x > 0. r(a2)
