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Reliability Of a fc-out-of-n System With A Single Server Extending Non-Preemptive Service To External Customers-
Part II
A. Krishnamoorthy •
Dept. of Mathematics, Cochin University of Science & Technology, Kochi-682022
achyuthacusat@gmail. com
M. K. Sathian •
Dept. of Mathematics, Panampilly Memorial Govt. College, Chalakudy, Thrissur.
sathianmkk@yahoo. com
Narayanan C Viswanath •
Dept. of Mathematics Govt. Engg. College, Thrissur Thrissur 680 009.
narayanan_viswanath@yahoo.com
Abstract
In this paper we study a k-out-of-n system with a single repair facility, which provides service to external customers also. We assume an N-policy for service to failed components(main customers) of the k-out-of-n system starts only on accumulation of N of them. Once started, the repair of external customers is continued until all the components become operational. When not repairing failed components, the server attends external customers(if there is any) who arrive according to a Poisson process. Once selected for service, the external customers receive a service of non-preemptive nature. When there are at least N failed components in the system and/or when the server is busy with failed components, the external customers are not allowed to join the system.Otherwise they join an orbit of infinite capacity. Life time distribution of failed components, service time distribution of main and external customers and the inter retrial time distribution of orbital customers are all assumed to follow independent exponential distributions. Steady state analysis has been carried out and several important system performance measures, based on the steady state distribution, derived. A numerical study comparing the current model with those in which no external customers are considered has been carried out.This study suggests that rendering service to external customers helps to utilize the server idle time profitably, without sacrificing the system reliability.
Keywords: k-out-of-n system; non-preemptive service.
1 Introduction
In this paper, we consider a variant of the model studied in Krishnamoorthy et al. [1]. In part I (see Krishnamoorthy et al. [3]) of this paper we studied the reliability of a k-out-of-n system with a single server rendering non-preemptive service to external customers.In this paper we extend it to
retrial queue of unsatisfied external customers(orbital customers) with linear retrial rate.In effect we replace the infinite queue of external customers in part I by orbital customers and their retrial. However, the stability condition remains the same in both models.
This paper is arranged as follows. In section 2 , we describe the model and in section 3 , its long run behavior is analyzed. The stability condition is derived explicitly in section 3 and computation of the steady state vector using the Neuts-Rao truncation procedure [2] has been discussed. Some important performance measures are derived in section 4 . The effect of rendering service to external customers and N-policy has been studied numerically in section 5 .
2 The retrial model
Here we consider a variant of the model discussed in section 2 of part I by assuming that an arriving external customer either gets immediate service if it finds the server is idle at that time or joins an orbit of infinite capacity, if the server is busy with external customers with < N — 1 failed components of the fc-out-of-n system. As in the model discussed in section 2 of part I, the external customers are not allowed to join the orbit when the server is busy with failed components of the system. An orbital customer retries for service with inter-retrial time following an exponential distribution with parameter 0. All other assumptions and parameters remain the same as in model discussed in section 2 of part I. In this situation the system can be modeled as follows.
Let ^i(t) = the number of external customers in the orbit at time t,
X2(t) = the number of failed components of the fc-out-of-n system, including the one getting service (if any) at time t.
Define
(0, If the server is idle
1, If the server is busy with an external customer
2, If the server is busy with a main customer
Now, X(t) = (X1(t),5(t),X2(t)) forms a continuous time Markov chain on the state space
5 = {(/i, OJ2VA > 0,0 < ;2 < N — 1}U {(/1, IJ2VA > 0,0 < 72 < n — fc + 1} U {(/1,2,7'2)//i > 0,1 < 72 < n — fc + 1}.
Arranging the states lexicographically and partitioning the state space into levels i, where each level i corresponds to the collection of states with number of external customers in the orbit at any time t equal to i, we get an infinitesimal generator of the above chain as
A10 Ao
Q =
A91 Ai
An
A?? Ai
Ao
A
1p
An
The entries of Q are described as below: For i > 0, the transition within level i is represented by the matrix
A,,- =
where
^11 £12 0
£21 £22 £23 0
0 0 £33 A
£41 0 0 a
= A&v — — ¿0/w,D12 =
14
34
14 = *cn(N) 0 rn-k+1(N),D21 = nIN,
- n(°)
J22 = ftIN, D23 = ^Cn(N) 0 rn-k+2-N (1),
D33 = Mn-k+2-N + Ac(n -k + 2-N) 0 r{n-k+2-N)(n -k + 2-N)- ftIn-k+2-
D44 = XEn-k+i + Ac
n-k+1
(n-k + 1)0 rn-k+i(n -k + 1)+vE
D34 = [On-k+2-NX(N-1) ^(n-k + 2-N)],
+ Acn-k+i(r
D41 = VCn-k+i(l)®rN(l).
For i > 0 the transition from level i to i + l is represented by the matrix
0NXN 0 0 0
A0 = 0 ain 0 0
0 0 0 0
For i > l, the transition from level i to i — l is represented by the matrix
A2i = [C
[0 ieiN [o 0
00 0 0]
3 Steady state analysis of the retrial model
3.1 Stability condition
For finding the stability condition for the system study, we apply Neuts-Rao truncation [2] by assuming Aii = A1M and A2i = A2M for all i>M. Then the generator matrix of the truncated system will look like:
Q =
A10 A0 A21 ^11 ^0
A22 A12 Ao
A?
A1
A,
Ao
Ai »,
Ar
Define AM = A0 + A1M + A2M ; then
Am =
D(M) Uu D(M) U12 OD14
D21 D22 D23 0
0 0 D33 D:
D41 0 0 d]
where d(2) = (A + M8)In, D22 = AEN — Let
^m = (nM(0),nM(1),nM(1),nM(2)),where
xM(0) = (nM (0,0), nM (0,1).....nM(0,N — 1)),
nM(1) = (nM(1,0).....nM(1,N — 1)),
ftM(1) = (nM(1,N).....nM(1,n — k + 1)),
nM(2) = (nM(2,1),...,nM(2,n — k + 1)). be the steady state vector of the generator matrix AM. Then the relation nMAM = 0 gives rise to the following equations:
nM(0)D{f + nM(1)D2i + nM(2)D4i = 0,
(1)
*M(0)D1(f+*M (1)022 = 0, (2)
(1)^23 (1)^33 = 0, (3)
(0)^14 + TTM (1)^34 + (2)^44 = 0. (4)
It follows from equation (4) that
^m(2) = "
Substituting for rcM(2) in equation (1), we get It follows from equation (3) that
*M(2) = —^M(0)fl14(fl44)-1 — TTM(1)D34(D44)-1. (5)
ation (1), we get
nM(0)D1(f) + (1)021 — ^M(0)fl14(^44)-1^41 — 7TM(1)D34(D44)-1D41 = 0. (6)
jrM(1) = —^M(1)fl23(^3-31). (7)
Since (13) gives
and hence and
Again from (13),
Since, limM^raMe(D1(?f)) 1e = ^ ew = ew, (17) implies that
(8)
Substituting for 7rM (1) in equation (6), we get
+ (1)021 — ^M(0)014(044)-1^41 +^m(1)023(^33)-1^34(^44)-1^41 = 0.
We notice that the first column of the matrix 041 is — B44e and its all other columns are zero columns. Hence the first column of the matrix (044)-1041 is — e and its all other columns are zero columns. This implies that the first column of the matrix — £>14(044)-1041 is £>14e = Acw(N) and its all other columns are zero columns. In other words — £>14(B44)-1B41 = Acw(N) ® rw(1). Also, the first column of the matrix 034(044)-1D41 is — 034e and its all other columns are zero columns. Since —034e = B33e, the first column of the matrix (033)-1034(044)-1041 is e and its all other columns are zero columns. Hence it follows that 023(033)-1034(044)-1041 is 023e = Acw(N) ® rw(1). Thus equation (8) becomes
^M(0)(01f) + Acw(W) ® rw(1)) + *M(1)(021 + ® rw(1)) = 0. (9)
Adding equations (2) and (9), we get
^M(0)(01f) + Acw(W) ® rw(1) + 0<f) + nM(1)(022 + ^21 + Ac„(N) ® rw(1)) = 0. (10)
Since + = £>22 + 021 = equation (10) reduces to
(*M(0) + nM(1))(A£w + Acw(W) ® rw(1)) = 0. (11)
1
which implies that tcm(0) +rcM(1) is a constant multiple of the steady state vector of the generator matrix + & 7^(1) and hence,
1
"m(0)+^m(1) = v^e;. (12)
where v is a constant. Equation (2) implies that
^M(0) = — rcM(1)022(<))-1. (13)
limrcM(0) = 0. (14)
M^ro
1
lim^^(1) = ^ —(15)
lim ,teM(1)e = vX (16)
M0rcM(0)e = — M0rcM(1)022(£1(f)-1e. (17)
lim M8nM(0)e = - lim nM(l)D22e
1 _ = -v~eN(-AcN(N)- fie)
= ^ + P). (18)
The truncated system is stable if and only if nMAoe < ^пмA2Me,
(19)
nMAoe = XnM(l)e, (20)
nMA2Me = M6nM(0)e. (21)
Making use of equations (16), (18), (20) and (21), the stability condition for the truncated system as M ^ œ is given by
which can be re-arranged as
vX < v(— + u),
I NI,
n a+Nn)
Hence, we conclude that the retrial problem has the same stability condition as the queueing problem, which was obtained in section 3.1 of part I.
3.2 Computation of Steady State Vector
We find the steady state vector of {X(t), t > 0}, by approximating it with the steady state vector of the truncated system. Let n = (n0,n1,n2,...) where each ni = (ni(0,0),ni(0,l),... ,nt(0,N -l),nt(l,l),. ,ni(l,n-k + l),ni(2,0), nt(2,l),. ,ni(2,n - k + l)) be the steady state vector of the Markov chain {X(t), t > 0}.
Suppose Au = A1M and A2i = A2M for all i > M. Let nM+r = nM-1Rr+1,r > 0, then from nQ = 0 we get
nM-1A0 + nMA1M + nM+1A2M = 0, nM-1A0 + nM-1RA1M + nM-1R A2M = 0,
nM-1(A0 + RA1M + R2A2M) = 0. Choose R such that A0 + RA1M + R2A2M = 0. We call this R as RM. Also we have
nM-2A0 + nM-1A1M-1 + nMA2M = 0, nM-2A0 + nM-1(A1M-1 + RMA2M) = 0, nM-1 = —nM-2A0(A1M-1 + RMA2M)
= n R
iM-2riM-1
where Next,
RM-1 = A0(A1M-1 + RMA2M)
nM-3A0 + nM-2A1M-2 + nM-1A2M-1 = 0, nM-3A0 + UM-2(A1M-2 + nM-1A2M-1) = 0, UM-2 = -7TM-3A0(A1M-2 + RM-1(A2M-1) = nM-3RM-2.
RM-2 = A0(A1M-2 + RM-1A2M-1)
UoAw + 11^21 = 0
Where
and so on.
Finally
becomes
n0(Aw + R1A21) = 0.
For finding n, first we take n0 as the steady state vector of A10 + R1A21.Then ni for i > l can be found using the recursive formula, ni = ni-1Ri for l < i < M.
Now the steady state probability distribution of the truncated system is obtained by dividing each ni with the normalizing constant
[n0+n1 + -]e = [n0 +n1 + ■■■ + nN-2 + nM-1(I - rm)-1]c.
3.3 Computation of the matrix RM
Consider the matrix quadratic equation which implies
•^0 + + — 0>
The structure of the matrix implies that the matrix has the form:
Rm —
In other words, the non-zero rows of the matrix are those, where the matrix has at least one nonzero entry. Now,
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
2 nM2
0 0 0 0
0 0 0 0
Equation (22) gives rise to the following equations:
^Ml^il ) + ^M2^21 + ^M4^41 = 0,
^M2^23 + ^33 = °>
+ ^M3^34 + ^M4^44 = 0.
From equation (28), we can write From equation(29), we can write
— ^M2^23(^23) .
Substituting for flM3 from (30) in equation (31), we get
Substituting for flM4 from (32) in equation (26), we get
+^M2^21
+^M2^23(^33)-1^34(^44)-1^41 = 0.
Using the same reasoning, that lead us to equation (9), equation (33) becomes
W^f + ^v(W) ® rw(1)) + fiM2(021 + Ac„(N) ® rw(1)) = 0. From (34), it follows that
*M1 = —^M2(^21^cw(W) ® rJV(1))(^1f) + Acw(W) ® rw(1))-1. Substituting for fiM1 in (27), we get
-^m2(021 + ¿cw(W) ® rwL1))LÖ1Lf) + Acw(W) ® rw(1))-1Me/iV
iLM)
-1
That is
^2(^21 +_AcwLN) ® ^(l))^ + Acw(W) ® rwL1))-1D12
+ÄM2Ö22 + — 0.
^M2(-L021 + ACn(N) ® r^Li))^^ + W ® rwLi))-1Me/iV)
+fiM2(-(021 + ® rw(1))(D1(f) + ® rw(1))-1fli2 + D22)
= 0.
We notice that + Acw(W) ® rw(1))e = (£>12 + M0/w)e. and therefore
LÖ21 + Acn(N) ® ^(l))^ + Acw(W) ® rw(1))-1(D12 + M0/w)e
(22) (23)
(24)
(25)
(26)
(27)
(28)
(29)
(30)
(31)
(32)
(33)
(34)
(35)
(36)
(37)
Also,
D22e + (D21 + Acn(N) 0 rN(1))e + Ae = 0.
and hence
(-(D21 + Acn(N) 0 rN(1))(Dl^) + Acn(N) 0 rN(1))-1M9IN)e +
1
(-(D21 + ACn(N) 0 VN(1))(D^M) + ^N(N) 0 TN(1))-1D12 + D22)e (38)
+Ae = 0.
Equation (38) shows that the matrix RM2 is the minimal non-negative solution of the matrix quadratic equation (36). Once obtaining RM2, the matrices RM1,RM3, and RM4 can be found using equations (35), (30) and (31) respectively. Hence the matrix RM can be found. From the form of the matrix D( M), we notice that,
-(D^M)+^N(N)0VN(1))
= M6IN - (AEN - AIN + ACN(N) 0 1)) = Me (1N (AEN - AIN + ACN(N) 0 ^(1))].
and hence
-(D(1M)+AcN(N)0rN(1))-1
1 1 _ +
= (AEN - AIN + ACN(N) 0 tn(1))]-
= (AEn - AIN + ACN(N) 0 1)) + ■■■].
Therefore
lim (-(D1MM) + Acn(N) 0 rN(1))-1MeiN) = In.
and
lim (-(D™ + Acn(N) 0 Tn(1))-1D12) = 0. Hence as M ^ ro equation (36) becomes
R2M2(D21 + ACN(N) 0 TN(1)) + RM2D22 + AIN = 0. (39)
We identify D21 + Acn(N) 0 rN(1) as A2, D22 as A and AIN as A0, which were defined in section 3.2 of part I. Hence equation (39) is the same as equation (24) of section 3.2 of part I. That is the matrix RM tends to the matrix R, the minimal non-negative solution of (24) of section 3.2 of part I, as M ^ ro. This fact can be utilized in determining the truncation level M.
4 System Performance Measures
The following system performance measures were calculated numerically. 1. Fraction of time the system is down,
1.^=0 (nj1(1,n - k + 1) + nji(
Pdown = 1?1=0 (nh(1,n- k + 1)+ nh(2,n -k + 1)).
2. System reliability, PreL = 1 - Pdown
-7i=o inji(i,n- k +1) + nji
= 1 — Zfi=0 (nji (1,n — k + 1) + nji (2,n — k + 1)).
3. Average number of external customers in the orbit,
Norbtt = imi=0 h№;=k1+1 nji(1,h)) + imi=0 h&l-T nji(2J3)).
4. Average number of failed components in the system,
Nfoii = U^1 j3&ri=0 nji(0,h)) + Zj-T frm=0 nji(2J3)).
5. Average number of failed components waiting when server is busy with external
customers
Nfailextb = Ylj3=ko1 jsteT^ nji(0,h)).
6. Expected rate at which external customers joining the system
Eextrate = ^ = 1 (Zj-0+1 nk(0,j3)) + Z^ 0,j3)}.
7. Expected number of external customers on its arrival gets service directly,
^extdirect = 2_/'3=0 ^0(°,73).
8. Fraction of time server is busy with external customers,
^A = 1 (273
^ext&usy — 2/1 = 1 (£73=o+1 n}i(°,73)).
9. Probability that the server is found idle,
£53=0 "o(0,/3) = ""0
TW — ££=0 *o(0,73) — W^o(0,0).
10. Probability that the server is found busy,
— 1 - £r=0 "o(0,A) — 1 - Mro(0,0).
11. Expected loss rate of external customers
04 = (2"3-=fc1+1 "A^)) + 2£=1 (2P*1 *A(0,;'3))}.
12. Expected service completion rate of external customers,
= £2£=o (2"3-=0+1 "7i(0,73)).
13. Expected number of external customers when server is busy with external customers
«6 = Z™1=oA(Z73='))+1 "71(0,73)).
14. Expected successful retrial rate
07 = 0-271=1 (253=1) "/1(0,73)).
5 Numerical study of the performance of the system
5.1 The effect of N policy on the server busy probability
A comparison of Table 1 of part I, which report the behaviour of server busy probability with variation in the N-policy level, with that of part II shows that the models described in section 2 of part I and its variant where external customers are sent to the orbit, which was described in section 2 of part II have similar behaviour as far as the server busy probability is considered. Comparison of Table 3 of part I, which report the variation in the fraction of time the server remains busy with external customers with increase in N, with table 2 of part II also points to similar behaviour for both models. Table 4 of part I and table 3 of part II indicate that the two models have similar reliability.
5.2 Cost Analysis
As in the case of the queueing model discussed in section 2 of part I, we analyzed a cost function for the retrial model for finding an optimal value for the N-policy level. For defining the cost function, let C1 be the cost per unit time incurred if the system is down, C2 be the holding cost per unit time per external customer in the orbit, C3 is the cost incurred for starting failed components service after accumulation of N of them, C4 be the cost due to loss of 1 external customer, C5 be the holding cost per unit time of one failed component, C6 be the cost per unit time if the server is idle. We define the cost function as:
Expected cost per unit time
c3
= C1 • Pdown + C2 • N0rMt + C4 • 04 + C5 • N/aa + — + C6 • PWfe.
"S
where is found exactly in the same lines as in section 4.1 of part I.
Our numerical study, as presented in Table 4, show that an optimal value for N can be found for different parameter choices and also that this optimal value happens to be a much smaller value like N = 6. This shows the care needed in selecting the N-policy level.
6 Conclusion
We analyzed a k-out-of-n system where the server renders service to external customers also. In the case of a system where a minimum number of working components is necessary for its operation, the service to external customer should be carefully managed so that it does not affect the system reliability much. Krishnamoorthy et al. [1] managed to do that by introducing an N-policy , in which the ongoing service of an external customer is preempted at the moment when N failed components have accumulated for repair. Differing from Krishnamoorthy et al.[1], here we considered a non-preemptive service for external customers thereby making their service more attractive. We analyzed two models: one in which the external customers joins a queue and another in which they moving to an orbit of infinite capacity. Our numerical study showed that rendering non-preemptive service to external customers has not affected the system reliability much, thereby re-asserted that the same could be an effective idea for utilizing the server idle time and there by earning more profit to the system. Analysis of a cost function has helped us in finding an optimal value for the N-policy level.
Table 1: Variation in the server busy probability when external customers are allowed k = 20, A =
4,A = 3.2,= 5.5,= 8, 9 = 5
N n=45 n=50 n=55 n=60
1 0.82349 0.82352 0.82353 0.82353
3 0.82995 0.82999 0.83 0.83
5 0.83222 0.83228 0.83229 0.83229
7 0.83328 0.83336 0.83338 0.83338
9 0.83385 0.83398 0.83401 0.83401
11 0.83417 0.83437 0.83442 0.83442
13 0.8343 0.83463 0.8347 0.83471
15 0.83424 0.83479 0.8349 0.83493
17 0.83394 0.83486 0.83505 0.83509
19 0.83325 0.83483 0.83515 0.83521
21 0.83192 0.83465 0.8352 0.83531
23 0.82945 0.83424 0.83518 0.83538
0.838 0.836 0.834 0.832 0.83 0.828 0.826 0.824 0.822
0 10 20 30
Table 2: Effect of the N-policy level on the fraction of time server is busy with external customers
k = 20,A = 4, A = 3.2,p = 3.2,p = 8, e = 5
N n=40 n=45 n=50 n=55 n=60
1 0.09635 0.09628 0.09626 0.09626 0.09626
3 0.10287 0.10276 0.10273 0.10273 0.10273
5 0.10523 0.10506 0.10503 0.10502 0.10502
7 0.10644 0.10618 0.10612 0.10611 0.10611
9 0.10725 0.10685 0.10676 0.10675 0.10674
11 0.10798 0.10732 0.10719 0.10716 0.10716
13 0.10879 0.10772 0.1075 0.10746 0.10745
15 0.10991 0.10811 0.10775 0.10768 0.10766
17 0.11461 0.10858 0.10798 0.10786 0.10783
19 0.11983 0.10925 0.10822 0.10801 0.10797
21 0.1103 0.10851 0.10815 0.10808
23 0.11208 0.10893 0.10831 0.10818
25 0.11522 0.10959 0.1085 0.10828
27 0.1107 0.10877 0.10839
29 0.11265 0.1092 0.10852
31 0.11615 0.10991 0.1087
33 0.11116 0.10898
35 0.1134 0.10945
37 0.11026
39 0.11172
41 0.11435
Table 3: Variation in the system reliability with increase in N k = 20, A = 4, A = 3.2,^ = 5.5,^ = 8,
9 = 5
N n=40 n=45 n=50 n=55 n=60
1 0.99963 0.99993 0.99998 1 1
3 0.99948 0.99989 0.99998 1 1
5 0.99924 0.99985 0.99997 0.99999 1
7 0.99885 0.99977 0.99995 0.99999 1
9 0.9982 0.99964 0.99993 0.99998 1
11 0.99712 0.99942 0.99988 0.99998 1
13 0.9953 0.99905 0.99981 0.99996 0.99999
15 0.99217 0.99843 0.99968 0.99994 0.99999
17 0.9769 0.99736 0.99947 0.99989 0.99998
19 0.9955 0.99909 0.99982 0.99996
21 0.99223 0.99844 0.99968 0.99994
23 0.98638 0.9973 0.99945 0.99989
25 0.97578 0.99528 0.99905 0.99981
27 0.99165 0.99833 0.99966
29 0.98509 0.99705 0.9994
31 0.97315 0.99475 0.99894
33 0.99058 0.99812
35 0.98297 0.99663
37 0.99393
39 0.989
Table 4: Analysis of a cost function n = 50, X = 3.2,^ = 5.5,£ = 8, Cx = 2000, C2 = 1000, C3 =
800, C4 = 1000, C5 = 10, C6 = 200, 0 = 5
N X = 4 X = 4.5 X = 5
1 6235.23047 6440.20947 6671.65918
2 6137.3877 6343.84668 6576.75928
3 6109.98389 6317.7207 6551.88965
4 6102.75391 6311.82178 6547.30566
5 6102.27734 6312.30322 6548.71436
6 6104.71094 6315.28613 6552.17676
7 6108.70947 6319.521 6556.51709
8 6113.67188 6324.50439 6561.33057
9 6119.2749 6329.98047 6566.44873
10 6125.32666 6335.80176 6571.76465
11 6131.69824 6341.87891 6577.22021
12 6138.31006 6348.14307 6582.78711
13 6145.10449 6354.55762 6588.43018
14 6152.04492 6361.09961 6594.13086
15 6159.104 6367.74854 6599.88428
17 6173.53564 6381.33594 6611.51611
19 6188.38672 6395.33936 6623.31689
21 6203.78809 6409.88037 6635.37354
23 6220.13477 6417.44531 6647.98535
25 6238.73828 6443.09375 6662.8042
27 6266.49854 6471.54688 6690.0752
29 6356.05566 6571.71631 6799.88672
31 7073.24658 7340.11523 7618.78223
Acknowlegment: Krishnamoorthy acknowledges financial support from Kerala State Council for Science, Technology and Environment, Grant
No.001/KESS/2013/CSTE, under the Emeritus Scientist scheme.
Authors would like to thank Prof.V.Rykov for some suggestion which improved the presentation of the paper.
References
[1] A. Krishnamoorthy, M. K. Sathian and C. N. Viswanath; Reliability of a k-out-of n system with repair by a single server extending service to external customers with pre-emption; Electronic Journal "Reliability:Theory and Applications" (Gnedenko forum,volume 11, June 2016,pp.61-93)
[2] M.F. Neuts and B.M. Rao; Numerical investigation of a multiserver retrial model; Queueing Systems; Volume 7, Number 2, 169-189, June 1990.
[3] A. Krishnamoorthy, M. K. Sathian and C. N. Viswanath; Reliability of a k-out-of n system with a single server extending non-preemptive service to external customers- Part I; To appear in Electronic Journal "Reliability:Theory and Applications" (Gnedenko forum,September 2016).
