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Benefitbenefit analysis of a twodissimiliar cold standby system subject to rainfall and switch failure with general as well as Erlangk repair facility as first come last serve
Ashok Kumar Saini
Associate Professor, Department of Mathematics, BLJS College, Tosham, Bhiwani, Haryana, INDIA.
Email: [email protected]
Research Article
Abstract Introduction: To transfer a unit from the standby state to the online state, a device known as ‘switching device’ is required. Generally, we assume that 1 the switching device is perfect in the sense that it does not fail and 2 the repair facility is first come first serve basis.. However, there are practical situations where the switching device can also fail and the repair facility is FCLS i.e. First Come Last Serve. We have taken two dissimilar units failure due to rainfall and switch failure distribution as exponential and repair time distribution as General as well as ErlangK phase type. We have find out MTSF, Availability analysis, the expected busy period of the server for repair the failed unit under rainfall in (0,t], expected busy period of the server for repair in(0,t], the expected busy period of the server for repair of switch failure in (0,t], the expected number of visits by the repairman for failure of units in (0,t],the expected number of visits by the repairman for switch failure in (0,t] and cost benefit analysis using regenerative point technique. A special case using failure and repair distributions as exponential is derived and graphs have been drawn.
Keywords: Cold Standby, rainfall, First Come Last Serve, MTSF, Availability, Busy period, costbenefit Function.
INTRODUCTION
To transfer a unit from the standby state to the online state, a device known as ‘switching device’ is required. Generally, we assume that the switching device is perfect in the sense that it does not fail. However, there are practical situations where the switching device can also fail. This has been pointed out by Gnedenko et al (1969).Such system in which the switching device can fail are called systems with imperfect switch. In the study of redundant systems it is generally assumed that when the unit operating online fails, the unit in standby is automatically switched online and the switchover from standby state to online state is instantaneous. In this paper, we have RAINFALL which are noninstantaneous in nature. We assume that the RAINFALL cannot occur simultaneously in both the units and when there occurs RAINFALL of the non –instantaneous nature the operation of the unit stop automatically. Here, we investigate a twounit (nonidentical) cold standby –a system in which offline unit cannot fail with switch failure under the influence of RAINFALL. The RAINFALL cannot occur simultaneously in both the units and when there are less RAINFALL, that is, within specified limit the unit operates as normal as before. But if these are beyond the specified limit the operation of the unit is stopped to avoid excessive damage of the units and as the RAINFALL goes on some characteristics of the stopped unit change which we call failure of the unit. After the RAINFALL are over the failed unit undergoes repair immediately according to first come last served discipline. For example, when a train came on the junction Station it waits for crossing of other train. When the other train came it stops and departs first according to FCLS.
ASSUMPTIONS
The system consists of two dissimilar cold standby units and the rainfall and failure time distribution are exponential with rates λ_{1, }λ_{2 }and λ_{3 }whereas the repairing rates for repairing the failed system due to rainfall and due to switch failure are arbitrary with CDF G_{1 }(t) and G_{2 }(t) respectively.
 The operation of units stops automatically when rainfall occurs so that excessive damage of the unit can be prevented.
 The rainfall actually failed the units. The rainfall are noninstantaneous in nature and it cannot occur simultaneously in both the units.
 The repair facility works on the first come last serve (FCLS) basis.
 The switches are imperfect and instantaneous.
 All random variables are mutually independent.
SYMBOLS FOR STATES OF THE SYSTEM
Superscripts: O, CS, SO, F, SFO
Operative, cold Standby, Stops the operation, Failed, Switch failed but operable respectively
Subscripts: nf, uf,ur, wr, uR
No rainfall, under rainfall, under repair, waiting for repair, under repair continued respectively
Up states: 0,1,2,6; Down states: 3,4,5,7,8
STATES OF THE SYSTEM
0(O_{nf}, CS_{nf})
One unit is operative and the other unit is cold standby and there are no rainfall in both the units.
1(SO_{nf}, O_{nf})
The operation of the first unit stops automatically due to rainfall and cold standby units starts operating.
2(SO_{nf}, SFO_{nf,ur})
The operation of the first unit stops automatically due to rainfall and during switchover to the second unit switch fails and undergoes repair.
3(F_{ur}, O_{uf})
The first one unit fails and undergoes repair after the rainfall are over and the other unit continues to be operative with no rainfall.
4(F_{u}_{r}, SO_{uf})
The one unit fails and undergoes repair after the rainfall are over and the other unit also stops automatically due to rainfall.
5(F_{uR}, F_{wr})
The repair of the first unit is continued from state 4 and the other unit is failed due to rainfall in it and is waiting for repair.
6(F_{uR}, SO_{uf})
The repair of the first unit is continued from state 3 and in the other unit rainfall occur and stops automatically due to rainfall.
7(F_{wr}, SFO_{uR})
The repair of failed switch is continued from state 2 and the first unit is failed after rainfall and waiting for repair.
TRANSITION PROBABILITIES
Simple probabilistic considerations yield the following expressions:
p_{01}= , p_{02} =
p_{13}= , p_{14}=
p_{23}= λ_{1}G_{2}^{*}( λ_{2}), p_{23}^{7}= λ_{2}G_{2}^{*}( λ_{2}), p_{24}= _{2}^{*}( λ_{2}),
p_{30}= G_{1}^{*}( λ_{1}), p_{33}^{6}= _{1}^{*}( λ_{1})
p_{43} = G_{1}^{*}( λ_{2}), p_{43}^{5} = G_{1}^{*}( λ_{2}) (1)
Figure 1: The State Transition Diagram
We can easily verify that
p_{01 }+ p_{02}= 1, p_{13} + p_{14}= 1, p_{23}+ p_{23}^{1 }+ p_{24}= 1, p_{30 }+p_{33}^{6 }= 1,
p_{43}+ P_{43}^{5}= 1 (2)
And mean sojourn time are
µ_{0 }= E(T) = = 1/ λ_{1}
Similarly
µ_{1 = }1/ λ_{2}, µ_{2 }= ,
µ_{4 }= (3)
MEAN TIME TO SYSTEM FAILURE
We can regard the failed state as absorbing
, (46)
Taking LaplaceStieltjes transforms of eq. (46) and solving for
= N_{1}(s) / D_{1}(s) (7)
Where
N_{1}(s) = { +
D_{1}(s) = 1 
_{ }
Making use of relations (1) and (2) it can be shown that =1, which implies
that is a proper distribution.
MTSF = E [T] = = (D_{1}^{’}(0)  N_{1}^{’}(0)) / D_{1} (0)
s=0
= ( +p_{01} + p_{01} p_{13} ) / (1  p_{01} p_{13} p_{30} ) (8)
where
+ , + , + ^{(1)} + ,, + ^{(6)}
+ ^{(5)}
AVAILABILITY ANALYSIS
Let M_{i}(t) be the probability of the system having started from state i is up at time t without making any other regenerative state belonging to E. By probabilistic arguments, we have
The value of M_{0 }(t) = λ_{1} λ_{3}t M_{1 }(t)= λ_{1} λ_{2}t
M_{3 }(t) λ_{1} _{1} ). (9)
The point wise availability A_{i}(t) have the following recursive relations
A_{0}(t) = M_{0}(t) + q_{01}(t)[c]A_{1}(t) + q_{02}(t)[c]A_{2}(t)
A_{1}(t) = M_{1}(t) + q_{13}(t)[c]A_{3}(t) + q_{14}(t)[c]A_{4}(t),
A_{2}(t) = {q_{23}(t) + q_{23}^{(7)}(t)}[c]A_{3}(t) +q_{33}^{(6)}(t)[c]A_{3}(t)
A_{4}(t) = {q_{43}(t) + q_{43}^{(5)}(t)[c]A_{3}(t) (10 14)
Taking Laplace Transform of eq. (1014) and solving for
= N_{2}(s) / D_{2}(s) (15)
Where
N_{2}(s) = (1  _{33}^{(6)}(s)) _{0}(s) +[ _{01}(s) _{1}(s) + _{13}(s) _{14}(s) _{43}(s) + _{43}^{(5)}(s) _{02}(s){ _{23}(s)) + _{23}^{(1)} (s) _{24} (s)( _{43} (s) + _{43}^{(5)}(s))}] _{3}(s)
D_{2}(s) = (1  _{33}^{(6)}(s))  _{30}(s) _{01}(s){ _{13}(s) _{14} (s) ( _{43}(s) _{43}^{(5)}(s)) }
+ _{20}(s) _{23}(s) _{23}^{(7)}(s) _{24}(s)( _{43}(s)) + _{43}^{(5)}(s) )}]
The steady state availability
A_{0} = = =
Using L’ Hospitals rule, we get
A_{0} = = (16)
Where
N_{2 }(0) = p_{30} _{0}(0) + p_{01} _{1}(0) _{3}(0) )
D_{2}^{’ }(0) = + p_{01} + p_{14} + p_{02}( + p_{24} )] p_{30}
The expected up time of the system in (0,t] is
(t) = So that (17)
The expected down time of the system in (0,t] is
(t) = t (t) So that (18)
The expected busy period of the server for repairing the failed unit under rainfallin (0,t]
R_{0}(t) = q_{01}(t)[c]R_{1}(t) + q_{02}(t)[c]R_{2}(t)
R_{1}(t) = S_{1}(t) + q_{13}(t)[c]R_{3}(t) + q_{14}(t)[c]R_{4}(t),
R_{2}(t) = S_{2}(t) + q_{23}(t)[c]R_{3}(t) + q_{23}^{(7)}(t)[c]R_{3}(t) + q_{24}(t)[c]R_{4}(t)
R_{3}(t) = q_{30}(t)[c]R_{0}(t) +q_{33}^{(6)}(t)[c]R_{3}(t),
R_{4}(t) = S_{4}(t) + (q_{43}(t)+ q_{43}^{(5)}(t)) [c]R_{3}(t) (1923)
Where
S_{1}(t)= λ_{1} λ_{2}t, S_{2}(t) = λ_{1}t _{2}(t), S_{4}(t) = λ_{1}t _{1}(t) (24)
Taking Laplace Transform of eq. (1923) and solving for
= N_{3}(s) / D_{2}(s) (25)
Where
N_{3}(s) = (1  _{33}^{(6)}(s)) [ _{01}(s)( _{1}(s) + _{14}(s) _{4}(s)+ _{02}(s)( _{2}(s) + _{24}(s) _{4}(s))] andD_{2}(s) is already defined.
In the long run, R_{0} = (26)
whereN_{3}(0)= p_{30}[ p_{01}( _{1}(0) + p_{14} _{4}(0) ) + p_{02 }( _{2}(0) + p_{24} _{4}(0) ) and D_{2}^{’}(0) is already defined.
The expected period of the system under rainfall in (0,t] is
(t) = So that (27)
The expected Busy period of the server for repair of dissimilar units by the repairman in (0,t]
B_{0}(t) = q_{01}(t)[c]B_{1}(t) + q_{02}(t)[c]B_{2}(t)
B_{1}(t) = q_{13}(t)[c]B_{3}(t) + q_{14}(t)[c]B_{4}(t),
B_{2}(t) = q_{23}(t)[c] B_{3}(t) + q_{23}^{(7)}(t)[c]B_{3}(t) + q_{24}(t)[c] B_{4}(t)
B_{3}(t) = T_{3}(t)+q_{30}(t)[c] B_{0}(t) + q_{33}^{(6)}(t)[c]B_{3}(t)
B_{4}(t) = T_{4 }(t)+{ q_{43}(t) + q_{43}^{(5)}(t)} [c]B_{3}(t) (2832)
Where
T_{3}(t) = λ_{2}t _{1}(t) T_{4}(t) = λ_{1}t _{1}(t) (33)
Taking Laplace Transform of eq. (2832) and solving for
= N_{4}(s) / D_{2}(s) (34)
Where
N_{4}(s) = _{3}(s) [ _{01}(s) _{13}(s) _{14}(s) _{43}(s) + _{43 }^{(5)}(s))}+ _{02}(s) _{23}(s)
_{23}^{(7)}(s) _{24}(s) ( _{43}(s) + _{43 }^{(5)}(s))}]+ _{4}(s) [ _{01}(s) _{44}(s) _{33}^{(6)}(s)
_{02}(s) _{24 }(s)(1 _{33}^{(6)}(s)
And D_{2}(s) is already defined.
In steady state, B_{0} = (35)
whereN_{4}(0)= _{3}(0)+ _{4}(0) { p_{30}(p_{01}p_{14} + p_{02} p_{24}) } and D_{2}^{’}(0) is already defined.
The expected busy period of the server for repair in (0,t] is
(t) = So that (36)
The expected Busy period of the server for repair of switch in (o,t]
P_{0}(t) = q_{01}(t)[c]P_{1}(t) + q_{02}(t)[c]P_{2}(t)
P_{1}(t) = q_{13}(t)[c]P_{3}(t) + q_{14}(t)[c]P_{4}(t),
P_{2}(t) = L_{2 }(t)+ q_{23}(t)[c]P_{3}(t) + q_{23}^{(7)}(t)[c]P_{3}(t)+q_{24}(t)[c]P_{4}(t)
P_{3}(t) = q_{30}(t)[c]P_{0}(t) + q_{33}^{(6)}(t)[c]P_{3}(t),
P_{4}(t) = (q_{43}(t)+ q_{43}^{(5)}(t)) [c]P_{3}(t) (3741)
Where L_{2 }(t) = λ_{1}t _{2}(t)
Taking Laplace Transform of eq. (3741) and solving for
= N_{5}(s) / D_{2}(s) (43)
Where N_{5}(s) = _{02}(s ) _{2}(s) 1  _{33}^{(6)}(s)) andD_{2}(s) is defined earlier.
In the long run, P_{0} = (44 )
whereN_{5}(0)= p_{30} p_{02} _{2}(0) and D_{2}^{’}(0) is already defined.
The expected busy period of the server for repairof the switch in (0,t] is
(t) = So that (45)
The expected number of visits by the repairman for repairing the different units in (0, t]
H_{0}(t) = Q_{01}(t)[s]H_{1}(t) + Q_{02}(t)[s]H_{2}(t)
H_{1}(t) = Q_{13}(t)[s][1+H_{3}(t)] + Q_{14}(t)[s][1+H_{4}(t)],
H_{2}(t) = [Q_{23}(t) +Q_{23}^{(7)}(t)] [s][1+H_{3}(t)] + Q_{24}(t)[s][1+ H_{4}(t)]
H_{3}(t) = Q_{30}(t)[s]H_{0}(t) + Q_{33}^{(6)}(t)[s]H_{3}(t),
H_{4}(t) = (Q_{43}(t)+ Q_{43}^{(5)}(t)) [s]H_{3}(t) (4650)
Taking Laplace Transform of eq. (4650) and solving for
= N_{6}(s) / D_{3}(s) (51)
Where
N_{6}(s) = (1 – _{33}^{(6)*}(s) ){ _{01}(s) _{13}(s) _{14} (s) _{02}(s) _{24}(s)
_{23} (s) _{23}^{(7)}(s)
D_{3}(s) = (1  _{33}^{(6)*}(s)) – _{30}(s) _{01} (s) { _{13}(s) _{14}(s) ( _{43}^{*}(s) _{43}^{(5)}(s))}
+ _{02}(s) _{23}(s) _{23}^{(7)}(s) _{24}(s)( _{43}(s)) + _{43}^{(5)}(s)) }]
In the long run, H_{0} = (52)
Where N_{6}(0)= p_{30}and D’_{3}(0) is already defined.
The expected number of visits by the repairman for repairing the switch in (0, t]
V_{0 }(t) = Q_{01}(t)[s]V_{1}(t) + Q_{02}(t)[s][1+V_{2}(t)]
V_{1}(t) = Q_{13}(t)[s]V_{3}(t) + Q_{14}(t)[s]V_{4}(t),
V_{2}(t) = Q_{24}(t)[s][1+V_{4}(t)] +[ Q_{23}(t) + Q_{23}^{(7)}(t)[s][1+V_{3}(t)]
V_{3}(t) = Q_{30}(t)[s]V_{0}(t) + Q_{33}^{(6)}(t)[s]V_{3}(t) (5357)
Taking LaplaceStieltjes transform of eq. (5357) and solving for
= N_{7}(s) / D_{4}(s) (58)
Where N_{7 }(s) = (1 – _{33}^{(6)*}(s) ){ _{01}(s) _{14}(s) _{43} (s) _{02}(s) _{24}(s) +Q^{*}_{ 02}(s)( _{23} (s) _{23}^{(7)}(s)
and D_{4}(s) is the same as D_{3}(s)
In the long run, V_{0} = (59)
Where N_{7 }(0)= p_{30}[p_{01} p_{14} p_{43 + }p_{02} ] and D’_{3}(0) is already defined.
COST BENEFIT ANALYSIS
The costbenefit function of the system considering mean uptime, expected busy period of the system under rainfall when the units stops automatically, expected busy period of the server for repair of unit and switch, expected number of visits by the repairman for unit failure, expected number of visits by the repairman for switch failure. The expected total costbenefit incurred in (0, t] is, C (t) = Expected total revenue in (0,t]  expected total repair cost for switch in (0,t]
 expected total repair cost for repairing the units in (0,t ]
 expected busy period of the system under rainfall when the units automatically stop in (0,t]
 expected number of visits by the repairman for repairing the switch in (0,t] expected number of visits by the repairman for repairing of the units in (0,t]
The expected total cost per unit time in steady state is
C = =
= K_{1}A_{0}  K_{2}P_{0 } K_{3}B_{0}  K_{4}R_{0 } K_{5}V_{0}  K_{6}H_{0 }
Where
K_{1}: revenue per unit uptime,
K_{2}: cost per unit time for which the system is under switch repair
K_{3}: cost per unit time for which the system is under unit repair
K_{4}: cost per unit time for which the system is under rainfall when units automatically stop.
K_{5}: cost per visit by the repairman for which switch repair,
K_{6}: cost per visit by the repairman for units repair.
CONCLUSION
After studying the system, we have analysed graphically that when the failure rate, rainfall rate increases, the MTSF and steady state availability decreases and the cost function decreased as the failure increases.
REFERENCES
 Gnedanke, B.V., Belyayar, Yu.K. and Soloyer, A.D., Mathematical Methods of Relability Theory,1969 ; Academic Press, New York.
 Barlow, R.E. and Proschan, F., Mathematical theory of Reliability, 1965; John Wiley, New York.
 Dhillon, B.S. and Natesen, J, Stochastic Anaysis of outdoor Power Systems in fluctuating environment, Microelectron. Reliab. 1983; 23,867881.
 Goel, L.R., Sharma, G.C. and Gupta, Rakesh Cost Analysis of a TwoUnit standby system with different weather conditions, Microelectron. Reliab., 1985; 25, 665659.
 Goel, L.R., Sharma G.C. and Gupta Parveen, Stochastic Behaviour and Profit Anaysis of a redundant system with slow switching device, Microelectron Reliab., 1986; 26, 215219.
