# Non-identical k-out-of-n:G redundant system with Weibull Law.

NOTATION

n : number of components in the system

S : number of spare units.

c : cost of each component / unit.

d : down time cost of the system.

p : unit module reliability (q = 1 -- p)

MTTF : mean time to failure of the system

MTTR : mean time to repair of the system

I(S) : MTTF in the case of k-out-of-n:G system

T(S) : average cost

[R.sub.o] : system reliability without spares.

INTRODUCTION

Weibull Law was extensively used in life-testing and reliability problems. In large number of cases, or the components that experience sudden failures of a random nature, the exponential failure model is generally applicable. However Weibull distribution as a failure law for many devices, is widely used for reliability calculations. Through appropriate choice of parameter variety of failure rate behavior can be modeled using Weibull failure law. As a special case, constant failure / random failure rate(exponential failure) is covered by this law and in addition to this the early failure(decreasing failure rate) wear-out failure (increasing failure rate) phenomena is explained by Weibull law.

k-out-of-n:G REDUNDANT SYSTEM Assumptions:

1. The system is composed of (S+n) units.

2. The units of the system are in two states: good / bad.

3. The system works effectively if at least k of its (S+n) units function correctly or equivalently that the system functions correctly if there are almost (S+k-1) of its (S+n) units failed.

4. The fault detection and switch over mechanism is perfect , never fails, always gives the correct answer and always switches properly.

5. Repair / replacement of the system follows exponential distribution with a failure rate of [mu] and the system will be in perfect condition after repair / replacement is completed.

6. Components in the system are non-identical with the failure rate

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

The reliability of the system is given by:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (1)

The average cost of the system is given by

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (2)

Therefore system average cost is a function of spares and optimal value of S is S * satisfies the condition

T(S) [less than or equal to] T(S + 1) and T(S -1 [??] T(S)) .............. (3)

An equivalent condition satisfying (3) is given by

[I(S)/A(S)][I(S) + 1/[mu] + A(S)] + (S + 4/[mu] [greater than or equal to] d/[mu]c .................... (4)

The cost of the component c is taken as Rs.20,000 /- and for the selected values of d, [mu] and [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], the optimal spares S* is obtained using equation (4) and the results are shown in the tables (1) &(2).

The results in table(1) which pertains to b=1 (belongs to exponential law and represents constant failure rate [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (t)) agrees with the results already discussed as a particular case of the modal with exponential law. The results in total agrees with all the comments in the case of the exponential model discussed by Ravi & Chari which are applicable as a particular case of Weibull law.

For the value of shape parameter b=2 ,Weibull distribution indicates increased failure rate as a function of time t and table (2) shows the results. Therefore it is observed that as 't' increases the reliability of the system decreases rapidly. A fewer number of spares are suggested when the down time cost of the system is very high compared to the cost of individual component. In addition to that at a allow value of the ratio c/d when t is large ,number of spares suggested are larger to increase the reliability level of the system. Overall the number of optimal spares suggested in this case is smaller which reasonably explains the fact that the components are under the influence of wear-out failure(increased failure rate)stage. Hence the modal does not encourage large number of optimum spares since they obviously increase cost and becomes cost ineffective.

CONCLUSION

This paper discusses the cost-effective optimum design of k-out-of-n:G redundant system with non-identical components. The components are assumed to have monotonically increasing failure rate. Thus paper considers the optimal design of system with components following weibull hazard rate which is more appropriate in real situations. This paper clearly indicates that if the system is under the influence of increased failure rate there is no improvement in reliability and hence the optimal spares are not suggested by the method.

REFERENCES:

[1] H Pham. 'Optimal Cost--effective Design of Triple--Modular_Redundancy with Spares Systems, IEEE Transactions on Reliability, Vol 42 no3, September 1993.

[2] E Balaguruswamy. 'Reliability Engineering ,' Tata McGraw-Hill 1984, p134.

[3] M.Ravi Kumar and A.A.Chari "Cost-Effective Optimum Design of n-order non-Identical Redundant systems" The Inatitute of Engineers(India). Vol. 81, Sept. 2000

[4] M.Ravi Kumar et all "Cost -Effective Optimum Design of n-order non-Identical Redundant systems with Weibull Law" Advances in Theoretical and Applied Mathematics, Vol 7, no 4, November 2012 .

M. Ravi Kumar , K. Mallikarjunudu, P.Vijajasekhara Reddy, and Y. Ramesh Babu

Dept of Basic Sciences, G .Pulla Reddy Engineering College (Autonomous), Kurnool--518002, Andhra Pradesh, India. ravikumar.gprec@gmail.com
```Table 1: Optimal spares S * for selected values of [theta]=5,
[[gamma].sub.1] = 0.01 , [[gamma].sub.2] = 0.02,
[[gamma].sub.3] =0.03, [[gamma].sub.4] =0.03, [mu] = 0.02
and b=1 (k = 3, n = 4)

c/d                 S *                t=1                   t=2
RN (S *)   [R.sub.o]   RN (S *)   [R.sub.o]

0.02                4        1        0.9967        1        0.9875
0.04                2        1                    0.9994
0.06                1      0.9967                 0.9987
0.08                0      0.9967                 0.9875
0.1                 0      0.9967                 0.9875

c/d                             t=3
RN (S *)   [R.sub.o]

0.02                   1        0.9734
0.04                 0.9988
0.06                 09960
0.08                 0.9734
0.1                  0.9734

Table 2: Optimal spares S * for selected values of
[theta]=5, [gamma] [mu]= 0.02 and b=2 (k = 3, n = 4)

c/d    t=1, [[gamma].sub.1] =0.0002, t=10, [[gamma].sub.1] =0.0002,
[[gamma].sub.2] =0.0008,      [[gamma].sub.2] =0.0008,
[[gamma].sub.3] =0.0018,      [[gamma].sub.3] =0.0018,
[[gamma].sub.4] = 0.0016      [[gamma].sub.4] = 0.0016

S *   [R.sub.N]   [R.sub.o]   S *   [R.sub.N]   [R.sub.o]
(S *)                         (S *)

0.02   0        1           1        0      0.7537      0.7537
0.04   0        1                    0      0.7537
0.06   0        1                    0      0.7537
0.08   0        1                    0      0.7537
0.1    0        1                    0      0.7537

c/d    t=100, [[gamma].sub.1] =0.02,
[[gamma].sub.2] =0.08,
[[gamma].sub.3] =0.18,
[[gamma].sub.4] = 0.04

S *   [R.sub.N]   [R.sub.o]
(S *)

0.02   1        0            0
0.04   0        0
0.06   0        0
0.08   0        0
0.1    0        0
```