CC BY
15
Alka Chaudhary1, Suman Jaiswal2 and Nidhi Sharma PROBABILISTIC ANALYSIS OF A TWO UNIT COLD
PROBABILISTIC ANALYSIS OF A TWO UNIT COLD STANDBY SYSTEM WITH REPAIR AND REPLACEMENT
POLICIES
Alka Chaudhary1, Suman Jaiswal2 and Nidhi Sharma
Department of Statistics Meerut College, Meerut-250001 Ch. Charan Singh University, Meerut-250004 (India) *E-mail ID: alka [email protected], 2E-mail ID: [email protected], E-mail ID: [email protected]
Abstract
The present paper deals with two identical units, one is operative and the other of which other is kept on cold standby. If the operative unit fails, it goes under repair and after repair, it is not considered as good as new. If the unit fails after the first repair, it is replaced with a new unit. A single repairman is always available with the system to repair a failed unit. Failure time, repair time and replacement time distributions are taken as exponential to reduce the complexity of the system model. By using the regenerative point technique, the various important measures of system effectiveness have been obtained and are shown with the help of graph.
Keywords: Cold standby, replacement policy, MTSF, regenerative point, availability, transition probability and mean sojourn time.
1. Introduction
Various researchers have widely used the redundancy technique in systems of identical units to improve system reliability. Since the demand for improving system reliability is increasing, the field of research in reliability theory is becoming more advanced. Many researchers in the field of reliability theory, including [2, 5], have analyzed the model of a two-unit cold standby system. The performance measure of the model of a two-unit system with repair and replacement policies has been studied by various authors [1, 3, 4, 6, 7].
The current paper deals with the study of a probabilistic analysis of a two-unit cold standby system with repair and replacement policies. One unit is operational, while the other is kept on cold standby. Whenever the operative unit fails, it goes into repair. A repaired unit is not considered as good as new. If the repaired unit fails after the first repair, a new unit is installed to replace it. For the purpose of repairing the failed units, a single repairman is always ready with the system. To reduce the complexity of the system, the exponential form of failure time, repair time, and replacement time is taken. After using the regenerative point technique, the following measures of system effectiveness are obtained:
• Transition probabilities and mean sojourn times in various states of the system.
Alka Chaudhary1, Suman Jaiswal2 and Nidhi Sharma RT&A, No 1 (72) PROBABILISTIC ANALYSIS OF A TWO UNIT COLD_V°lume 18, March 2023
• Reliability and mean time to system failure (MTSF).
• Point-wise and steady-state availabilities of the system as well as expected up time of the system during time interval (0, t).
• The expected busy period of repairman during time interval (0, t).
• Net expected profit earned by the system in time interval (0, t) and in steady-state.
2. Model description and assumptions
• The system consists of two identical units, one operational (o) and the other in standby (s).
• A single repairman is always available with the system to repair and replace a failed unit.
• After repair, it is not considered as good as new.
• If the unit fails after the first repair, replace it with a new one.
• The failures of the units are independent and the failure time, repair time and replacement time distributions of the units are as taken exponentials.
3. Notations and states of the system
3.1. Notations
E : Set of regenerative states = {S0 to S3 }.
0 : Constant failure rate of operative unit.
b : Constant failure rate after one time repair of a unit.
a : Repair rate of a unit.
h : Replacement rate of failed unit.
q^j (•) : p.d.f. of transition time from regenerative state S^ toSj.
c k 1
q-j 0 (•): p.d.f. of transition time from regenerative state S^ toSj via non-regenerative state Sk.
~ : Symbol for Laplace-stieltjes transforms i.e. A (s) = J e-stdA (t)
• : Symbol for Laplace-transform i.e.
A* (s) = J e-stA (t) dt
© : Symbol for ordinary convolution i.e. t
A(t)©B(t) = J A(u)B(t-u)du
0
Here, b > 0 , Also the limits of the integration are not mentioned whenever they are 0 to ¥.
3.2. Symbols for the states of the systems
N0,NS : Unit-
Ns, No : Unit-
No, Fr : Unit-
F F w> wr : Unit-
Alka Chaudhary1, Suman Jaiswal2 and Nidhi Sharma RT&A, No 1 (72)
PROBABILISTIC ANALYSIS OF A TWO UNIT COLD_V°lume 18, March 2023
Fr , Fwr : Replaced unit-1 after repair and unit-2 waiting for repair.
Considering the above symbols and keeping in view the assumptions stated in section-2, the possible states of the system model are shown in transition diagram (Figure.1).
Гч
h S F F
5
F F R' wr a
P S,
No,Fr
0 : Up State О : Failed State • : Regenerative
Figure.1: Transition diagram
4. Transition probabilities The direct or one-step steady state transition probabilities can be obtained as follows:
P., = lim Q--(t) У У
Similarly,
P01 = I9e-etdt, p15 = j6e-(a+e)tdt =
p32 = jae-(a+P)t dt =
a+e
a
a + p'
P41 = Jhe htdt =1, It can be easily verified that,
P01 = P23 = P41 = P53 =1
Pj2 = jae (a+e)t dt = a
a+e
P01 = jee"etdt = 1
P34 = ibe-(a+p)tdt = -&
a + p
p53 = jae at dt = 1
P12 + P15 =1, 58
P32 + P34 =1
(1-3)
S
a
a
e
S
4
Alka Chaudhary1, Suman Jaiswal2 and Nidhi Sharma RT&A, No 1 (72) PROBABILISTIC ANALYSIS OF A TWO UNIT COLD_V°lume 18, March 2023
5. Mean sojourn times
The mean sojourn time y in state S; is defined as the expected time taken by the system in state S; before making the transition into any other state. If T denotes the sojourn time in state S;, then mean sojourn time in state S; is;
y = IP (T > t) dt (4)
Therefore,
yo=j e-etdt=e=y 2 -ia+0)^t -_L_
y = J =
a + 9
f -(a+R)t 1 y = Wdt =--
3 a + R
y4 = j e-htdt =1 h
= j e-atdt = - (5-9)
a
6. Analysis of characteristics 6.1 Reliability of the system and MTSF
Let R. (t) be the probability that the system is operative during (0, t) given that at t = 0, it starts from state S; î E. .To obtain it, we assume the failed states S4 and S5 as absorbing. Now using simple probabilistic arguments we have the following recurrence relations in R. (t) ; i =0, 1,2,3.
R0(t) = Z0(t) + q01(t)© Rj(t) Rj(t) = Z1(t) + q12(t)© R2(t) R2(t) = Z2(t) + q23(t)© R3(t)
R3(t) = Z3(t) + q32(t)© R2(t) (10-13)
Where, Z0 = e-0t, ^ = e-(a+0)t, Z2 = e-0t, Z3 = e-(a+b}t
Taking Laplace transforms of relations (10-13) and simplifying the resulting set of algebraic equations for R0 (s) , we get;
R* (sN1(s) Ro (s )_ D1(S)
Where,
Nl(s) =(1 - q*23q32)Z0 + q0i(1 - q23q32)Zl + q01q12Z2 + q01q12q23Z3 Dj(s) = 1 - q^3q32 (14)
Alka Chaudhary1, Suman Jaiswal2 and Nidhi Sharma RT&A, No 1 (72) PROBABILISTIC ANALYSIS OF A TWO UNIT COLD_V°lume 18, March 2023
Here, also for brevity we have omitted the argument's' from q* (s),Z0 (s) . The expression of mean time to system failure is given by,
E (To )=SmK (s)
Observing q* (0) = pj, Z0 (0) = y0, We have,
E [t ] = P34(yo + yi) + P12(V2 + ¥3) (15)
p34
6.2. Availability analysis
Let Aj (t) be the probability that the system is up (operative) at epoch t, when system initially starts from state S; e E. Using the basic probabilistic concepts in regenerative point technique as in case of reliability, one can obtain the recurrence relations for Aj (t); i=0, 1, —, 5. Taking the Laplace
Transformations and solving the resulting set of algebraic equations for AQ (s), we get,
A^Nf (16)
Where,
N2© =(1-q23q32 -q4lq12q23q34 -q41q15q53q34)Z0 + (q01 -q01q32q23)Zl + (q01q12 + q01q15q53q32)Z2 + (q01q12q23 + q01q15q53)Z3
and
D2(s) =1 -q23q32 -q12q23q34q41 -q15q53q34q41 The steady state availability of the system is given by,
An = lim sA* (s) = lim
_ s _ lim sN2 (s)
s®0 s A° (S) s®) S D2 (s )
As D2(0) = 0, so by using L. Hospitals rule, we get
N (0)
Ac (17)
0 D2 (0)
Where,
N2(0) = p34yj + (p12 + p15p32)y2 + y3
Alka Chaudhary1, Suman Jaiswal2 and Nidhi Sharma RT&A, No 1 (72) PROBABILISTIC ANALYSIS OF A TWO UNIT COLD_V°lume 18, March 2023
and
d2(0)=p34(vi + y 4)+(Pi2 + pi5p32)y 2 + y3 + PI5P34 y 5 (18)
The expected up time of the system in interval (0, t) are given by; t
Hup (t) = 1A0 (u)du 0
So that
¿>P (s) = ^ (19)
6.3. Busy period analysis
Let B1 (t) and b2 (t)be the respective probabilities that the unit is under repair and under replacement at time t due to the failure unit, when the system initially starts from regenerative states S; e E. Using the probabilistic arguments as in case of availability, we develop the recurrence relations for B| (t) and b2 (t); i=0,1,2,3,4,5. Then, taking the Laplace Transforms of these recurrence relations and solving the resulting algebraic equations for B0* (s) and b2* (s)we get;
B0* W-and B2* W-Ü (20-21>
Where,
N3(s) -(qoi-qoiq32q23)Zi +^«23 +qoiq53q^)Z3 (22)
N4(s) - (q^ii+¿q^i^i (23)
In the long run, the probabilities that the repairman will be busy are respectively given by:-
1 N (o) 2 nA (0) b0 - and b2 - (24)
0 D (0) 0 d2 (0) ( )
Where,
N3(0) = p34yj + y3 N4(0) = p34yj
The value of D2 (0)is same as given by expression (18).
The expected busy period of skilled repairman and regular repairman during (0,t) are given by,
^b(t) = }B0(u)du and ^2(t) = JB2(u)du
0 0
PROBABILISTIC ANALYSIS OF A TWO UNIT COLD
Alka Chaudhary1, Suman Jaiswal2 and Nidhi Sharma
So that
(25-26)
7. Profit function analysis
We are now in the position to obtain the net expected profit incurred during time (0, t) by considering the characteristics obtained in earlier sections. Let us consider,
Ko = revenue per unit time by the system when it is operative. Ki = cost per unit time when repairman is busy in repair of failed unit. K2 = cost per unit time when repairman is busy in replacement of failed unit. Then, the net expected profit incurred during time (0, t),
The Behavioral characteristics of MTSF and Profit function for different values of Failure rate 0 shown in figures 2 and 3 respectively.
In figure 2, the graphical analysis of MTSF in relation to failure rate 0 for three different values of repair rate a i.e. 0.09, 0.35 and 0.85 and for two fixed values of b at 0.05 and 0.09 has been plotted.
The graphical analysis of Profit function in relation to failure rate 0 for different values of repair rate a i.e. 0.35, 0.55 and 0.95 and for fixed values of b at 0.32 and 0.45 has been shown in figure 3.
In the analysis at figure 2 and 3, the MTSF and Profit function shows a decline as the value of failure rate increases and when the value of repair rate increases, the graph shows the increase in the value of MTSF and Profit function.
P(t) = Expected total revenue in (0,t) - Expected amount spent on repair in (0,t)
= K0 p up(t) - K^ b(t) - K2 p 2(t)
1
The expected profit per-unit time in steady state is given by,
8. Graphical study of system behaviour
Alka Chaudhary1, Suman Jaiswal2 and Nidhi Sharma PROBABILISTIC ANALYSIS OF A TWO UNIT COLD
Figure.2: Behavior ofMTSF with respect toCL , P and 0
According to the dotted curves in the MTSF graph in relation to failure rate, in order to achieve MTSF for at least 300 units, the failure rate 0 must be less than 0.01, 0.02, and 0.04 for values of a 0.09, 0.35, and 0.85 whenp is fixed at 0.09.
Alka Chaudhary1, Suman Jaiswal2 and Nidhi Sharma PROBABILISTIC ANALYSIS OF A TWO UNIT COLD
Similarly, for smooth curves the upper bounds for failure rate (0) 0.012, 0.031 and 0.081 corresponding to a i.e., 0.09, 0.35and 0.85 when b is fixed at 0.05.
Figure 3 shows that the system is profitable only when the failure rate 0 is less than 0.045, 0.058, and 0.078, respectively, for values of a 0.35, 0.55, and 0.95 when b fixed at 0.45.
Similarly, we conclude from smooth curves that the system is profitable only when failure rate (0 ) is less than 0.05, 0.067 and 0.098 respectively for values of repair rate a i.e., 0.35, 0.55and 0.95 when b is fixed at 0.032.
References
[1] Chander, S. (2007). MTSF and profit evaluation of an electric transformer with inspection for online repair and replacement. Journal of Indian Soc. Stat. Operation Research, 28(1-4):33-43.
[2] Gupta, R. and Tyagi, A. (2014). A two identical unit cold standby system with switching device and geometric failure and repair time distributions. Aligarh Journal of Statistics, 34: 55-66.
[3] Gupta, U. and Kumar, P. (2019). Performance measures of a two non-identical unit system model with repair and replacement policies. RT & A, 14(1):53-62.
[4] Kumar, P., Bharti, A. and Gupta, A. (2012). Reliability analysis of a two non-identical unit system with repair and replacement having correlated failures and repairs. Journal of Informatics and Mathematical Sciences, 4(3):339-350.
[5] Mahmoud, M.A.W and Moshref, M.E. (2010). On a two-unit cold standby system considering hardware, human error failures and preventive maintenance. Mathematical and Computer Modeling, 51(5-6):736-745.
[6] Sharma, N., joorel, J. P. S. (2019). Study of stochastic model of a two unit system with inspection and replacement under multi failure. RT & A, 14(3):31-38.
[7] Singh, M. and Chander, S. (2005). Stochastic analysis of reliability models of an electric transformer and generator with priority and replacement. Jr. Decision and Mathematical Sciences, 10(1-3):79-100.
