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Cost benefit analysis of two dissimilar warm standby system subject to failure due to atmospheric pressure and gravitational force with switch failure
Ashok Kumar Saini
Associate Professor, Department of Mathematics, B. L. J. S. College, Tosham, Bhiwani, Haryana, INDIA.
Email: [email protected]
Research Article
Abstract Introduction: Development is a multi dimensional process. There are many interrelated and interacting process involved in it – social, economic, political, cultural, educational, technology, etc. Faults in any of the processes affect other processes and hence the overall development. Twounit standby system subject to environmental conditions such as shocks, change of weather conditions etc. have been discussed in reliability literature by several authors due to significant importance in defence, industry etc. In the present paper we have taken twononidentical warm standby system with failure time distribution as exponential and repair time distribution as general. The Role of atmospheric pressure and gravitational force under which the system operates plays significant role on its working. We are considering system under (i) atmospheric pressure and (ii) Gravitational force causing different types of failure requiring different types of repair facilities. Using semi Markov regenerative point technique we have calculated different reliability characteristics such as MTSF, reliability of the system, availability analysis in steady state, busy period analysis of the system under repair, expected number of visits by the repairman in the long run and gainfunction and graphs are drawn.
Keyword: warm standby, atmospheric pressure, gravitational force, switches failure.
INTRODUCTION
The original concept for the reliability was that a chain cannot be stronger then its weakest link. Work based on this idea gave rise to a great improvement in the theory of reliability. The mathematical theory of reliability has grown out of the demands of modern technology and particularly out of experiences in World War II (19391945) with complex military systems although the concept of reliability is as old as man himself.
Assumptions
 The failure time distribution is exponential whereas the repair time distribution is arbitrary of two nonidentical units.
 The repair facility is of four types:
Type I, II repair facility
 when failure due to atmospheric pressure and gravitational force of first unit occurs respectively and
Type III, IV repair facility
 when failure due to atmospheric pressure and gravitational force of the second unit occurs respectively.
 The repair starts immediately upon failure of units and the repair discipline is FCFS.
 The repairs are perfect and start immediately as soon as the atmospheric pressure and gravitational force of the system becomes normal. The atmospheric pressure and gravitational force in both the units do not occur simultaneously.
 The failure of a unit is detected immediately and perfectly.
 The switches are perfect and instantaneous.
 All random variables are mutually independent.
Symbols for states of the System
Superscripts: O, WS, SO, FAP, FGF, SFO
Operative, Warm Standby, Stops the operation, failure due to atmospheric pressure, failure due to gravitational force, Switch failed but operable respectively
Subscripts: nap, ap, gf, ur, wr, uR
No atmospheric pressure, atmospheric pressure, gravitational force, under repair, waiting for repair, under repair continued respectively
Up states: 0, 1, 2, 9;
Down states: 3,4,5,6,7,8,10,11
Regeneration point: 0,1,2,4,7,10
States of the System
0(O_{nap}, WS_{nap})
One unit is operative and the other unit is warm standby and there are no atmospheric pressure and no gravitational force in both the units.
1(SO_{nap}, O_{nap})
The operation of the first unit stops automatically due to atmospheric pressure and warm standby unit’s starts operating with no atmospheric pressure.
2(FAP_{ur}, O_{nap})
The first unit fails and undergoes repair after the atmospheric pressure are over and the other unit continues to be operative with no atmospheric pressure.
3(FAP_{uR}, SO_{uap})
The repair of the first unit is continued from state 2 and the operation of second unit stops automatically due to atmospheric pressure.
4(FAP_{ur}, SO_{gf}) The first unit fails and undergoes repair after the atmospheric pressure are over and the other unit also stops automatically due to gravitational force.
5(FAP_{uR}, FGF_{gf,wr}) The repair of the first unit is continued from state 4 and the other unit is failed due to gravitational force in it and is waiting for repair.
6(O_{nap, }FGFur) The first unit becomes operative with no atmospheric pressure and the second unit is failed due to gravitational force is under repair.
7(SO_{ngf}, SFO_{nap, ur})
The operation of the first unit stops automatically due to gravitational force and during switchover to the second unit switch fails and undergoes repair and there is no atmospheric pressure.
8(FGF_{gf, wr}, SFO_{ngf, uR})
The repair of failed switch is continued from state 7 and the first unit is failed after gravitational force is waiting for repair.
9(O_{nap}, SO_{gf})
The first unit is operative with no atmospheric pressure and the operation of warm standby second unit is stopped automatically due to gravitational force.
10(SO_{gf}, SF_{ur})
The operation of the first unit stops automatically due to gravitational force and the second unit switch fails and undergoes repair after the gravitational force is over.
11(FGF_{gf, wr}, FGF_{uR})
The repair of the second unit is continued from state 10 and the first unit is failed due to gravitational force is waiting for repair.
Figure 1: The State Transition Diagram
TRANSITION PROBABILITIES
Simple probabilistic considerations yield the following expressions:
p_{01 }= , P_{07} =
p_{09 }= , p_{12 }= , p_{14 }= _{ }
P_{20}= G_{1}^{*}( λ_{1}), P_{22}^{(3)} = G_{1}^{*}( λ_{1})=p_{23}, P_{72} = G_{2}^{*}( λ_{4}),
P_{72}^{(8)} = G_{2}^{*}( λ_{4})= P_{78} _{ }
We can easily verify that
p_{01 }+ p_{07 }+_{ }p_{09 }= 1, p_{12 }+ p_{14 }= 1, p_{20 }+ p_{23 (}=p_{22}^{(3)})= 1, p_{46}^{(6)}= 1_{ }p_{60}^{ }= 1,
p_{72}+ P_{72}^{(5) }+ p_{74 }= 1, p_{9,10} =1, p_{10,2 }+ p_{10,2}^{(11) }= 1 (1)
And mean sojourn time are
µ_{0 }= E(T) = (2)
Mean Time To System Failure
We can regard the failed state as absorbing
,
(35)
Taking LaplaceStiltjes transform of eq. (35) and solving for
= N_{1}(s) / D_{1}(s) (6)
Where
_{ }N_{1}(s) = { +
D_{1}(s) = 1 
Making use of relations (1) and (2) it can be shown that θ_{0}^{*}(0) =1, which implies that θ_{0}(t) is a proper distribution.
MTSF = E[T] = d/ds θ_{0}^{*}(s) = D_{1}^{’}(0)  N_{1}^{’}(0)) / D_{1} (0)
s=0
= ( +p_{01} + p_{01} p_{12} + p_{09} ) / (1  p_{01} p_{12} p_{20} )
where
+ + , + , + ^{(3)},
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), M_{1}(t), M_{2}(t), M_{4}(t) can be found easily.
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_{07}(t)[c]A_{7}(t) + q_{09}(t)[c]A_{9}(t)
A_{1}(t) = M_{1}(t) + q_{12}(t)[c]A_{2}(t) + q_{14}(t)[c]A_{4}(t), A_{2}(t) = M_{2}(t) + q_{20}(t)[c]A_{0}(t) + q_{22}^{(3)}(t)[c]A_{2}(t)
A_{4}(t) = q_{46}^{(3)}(t)[c]A_{6}(t), A_{6}(t) = q_{60}(t)[c]A_{0}(t)
A_{7}(t) = (q_{72}(t)+ q_{72}^{(8)}(t)) [c]A_{2}(t) + q_{74} (t)[c]A_{4}(t)
A_{9}(t) = M_{9}(t) + q_{9,10}(t)[c]A_{10}(t), A_{10}(t) = q_{10,2}(t)[c]A_{2}(t) + q_{10,2}^{(11)}(t)[c]A_{2}(t) (714)
Taking Laplace Transform of eq. (714) and solving for
= N_{2}(s) / D_{2}(s) (15) Where
N_{2}(s) = (1  _{22}^{(3)}(s)) { _{0}(s) + _{01}(s) _{1}(s) + _{09}(s) _{9}(s)}+ _{2}(s){ _{01}(s) _{42}(s) + _{07}(s)_{72}(s) + _{73}^{(8)}(s)) + _{09} (s) _{9,10} (s)( _{10,2} (s) + _{10,2}^{(11)}(s))}
D_{2}(s) = (1  _{22}^{(3)}(s)) { 1  _{46}^{(5)}(s) _{60}(s) ( _{01}(s) _{44} (s) + _{07}(s) _{74}(s))
 _{20}(s)_{01}(s)_{12}(s)_{07}(s)( _{72}(s)) + _{72}^{(8)}(s) + _{09} (s) _{9,10} (s)
( _{10,2} (s) + _{10,2}^{(11)}(s))}
The steady state availability
A_{0} = = =
Using L’ Hospitals rule, we get
A_{0} = = (16)
Where
N_{2}(0)= p_{20}(_{0}(0) + p_{01}_{1}(0) + p_{09} _{9}(0) ) + _{2}(0) (p_{01}p_{12} + p_{07} (p_{72}
+ p_{72}^{(8) }+ p_{09} ))
D_{2}^{’}(0) = p_{20}{ + p_{01} + (p_{01 }p_{14} + p_{07} p_{74} )+ p_{07 } + p_{07} + p_{09}()
+ { 1 ((p_{01}p_{14}^{ }+ p_{07} p_{74 })}
, ,
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 gravitational force in (0,t]
R_{0}(t) = S_{0}(t) + q_{01}(t)[c]R_{1}(t) + q_{07}(t)[c]R_{7}(t) + q_{09}(t)[c]R_{9}(t)
R_{1}(t) = S_{1}(t) + q_{12}(t)[c]R_{2}(t) + q_{14}(t)[c]R_{4}(t),
R_{2}(t) = q_{20}(t)[c]R_{0}(t) + q_{22}^{(3)}(t)[c]R_{2}(t)
R_{4}(t) = q_{46}^{(3)}(t)[c]R_{6}(t), R_{6}(t) = q_{60}(t)[c]R_{0}(t)
R_{7}(t) = (q_{72}(t)+ q_{72}^{(8)}(t)) [c]R_{2}(t) + q_{74} (t)[c]R_{4}(t)
R_{9}(t) = S_{9}(t) + q_{9,10}(t)[c]R_{10}(t), R_{10}(t) = q_{10,2}(t) + q_{10,2}^{(11)}(t)[c]R_{2}(t) (1926)
Taking Laplace Transform of eq. (1926) and solving for
= N_{3}(s) / D_{2}(s) (27)
Where
N_{2}(s) = (1  _{22}^{(3)}(s)) { _{0}(s) + _{01}(s) _{1}(s) + _{09}(s) _{9}(s)} and D_{2}(s) is already defined.
In the long run, R_{0} = (28)
where N_{3}(0)= p_{20}(_{0}(0) + p_{01}_{1}(0) + p_{09} _{9}(0) ) and D_{2}^{’}(0) is already defined.
The expected period of the system under gravitational force in (0,t] is
(t) = So that
The expected Busy period of the server for repairing the failed units under atmospheric pressure by the repairman in (0,t]
B_{0}(t) = q_{01}(t)[c]B_{1}(t) + q_{07}(t)[c]B_{7}(t) + q_{09}(t)[c]B_{9}(t)
B_{1}(t) = q_{12}(t)[c]B_{2}(t) + q_{14}(t)[c]B_{4}(t), B_{2}(t) = q_{20}(t)[c] B_{0}(t) + q_{22}^{(3)}(t)[c]B_{2}(t)
B_{4}(t) = T_{4 }(t)+_{ }q_{46}^{(3)}(t)[c]B_{6}(t), B_{6}(t) = T_{6 }(t)+_{ }q_{60}(t)[c]B_{0}(t)
B_{7}(t) = (q_{72}(t)+ q_{72}^{(8)}(t)) [c]B_{2}(t) + q_{74} (t)[c]B_{4}(t)
B_{9}(t) = q_{9,10}(t)[c]B_{10}(t), B_{10}(t) = T_{10 }(t)+_{ }(q_{10,2}(t) + q_{10,2}^{(11)}(t)[c]B_{2}(t) (29 36)
Taking Laplace Transform of eq. (2936) and solving for
= N_{4}(s) / D_{2}(s) (37)
Where
N_{4}(s) = (1  _{22}^{(3)}(s)) { _{01}(s)_{14}(s) _{4}(s) + _{46 }^{(5)}(s) _{6}(s)) + _{07}^{(3)}(s) _{74}(s)( _{4}(s)
+ _{46}^{(5)}(s) _{6}(s))+ _{09}(s)_{09,10}(s) _{10}(s) )
And D_{2}(s) is already defined.
In steady state, B_{0} = (38)
where N_{4}(0)= p_{20} {( p_{01} p_{14} + p_{07} p_{74}) (_{4}(0) +_{6}(0)) + p_{09} _{10}(0) } and D_{2}^{’}(0) is already defined.
The expected busy period of the server for repair in (0,t] is
(t) = So that (39)
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_{07}(t)[c]P_{7}(t) + q_{09}(t)[c]P_{9}(t)
P_{1}(t) = q_{12}(t)[c]P_{2}(t) + q_{14}(t)[c]P_{4}(t), P_{2}(t) = q_{20}(t)[c]P_{0}(t) + q_{22}^{(3)}(t)[c]P_{2}(t)
P_{4}(t) = q_{46}^{(3)}(t)[c]P_{6}(t), P_{6}(t) = q_{60}(t)[c]P_{0}(t)
P_{7}(t) = L_{7}(t)+ (q_{72}(t)+ q_{72}^{(8)}(t)) [c]P_{2}(t) + q_{74} (t)[c]P_{4}(t)
P_{9}(t) = q_{9,10}(t)[c]P_{10}(t), P_{10}(t) = (q_{10,2}(t) + q_{10,2}^{(11)}(t))[c]P_{2}(t) (4047)
Taking Laplace Transform of eq. (4047) and solving for
= N_{5}(s) / D_{2}(s) (48)
where N_{2}(s) = _{07}(s ) _{7}(s) 1  _{22}^{(3)}(s)) and D_{2}(s) is defined earlier.
In the long run, P_{0} = (49 )
where N_{5}(0)= p_{20} p_{07} _{4}(0) and D_{2}^{’}(0) is already defined.
The expected busy period of the server for repair of the switch in (0,t] is
(t) = So that (50)
The expected number of visits by the repairman for repairing the nonidentical units in (0,t]
H_{0}(t) = Q_{01}(t)[c]H_{1}(t) + Q_{07}(t)[c]H_{7}(t) + Q_{09}(t)[c]H_{9}(t)
H_{1}(t) = Q_{12}(t)[c][1+H_{2}(t)] + Q_{14}(t)[c][1+H_{4}(t)], H_{2}(t) = Q_{20}(t)[c]H_{0}(t) + Q_{22}^{(3)}(t)[c]H_{2}(t)
H_{4}(t) = Q_{46}^{(3)}(t)[c]H_{6}(t), H_{6}(t) = Q_{60}(t)[c]H_{0}(t)
H_{7}(t) = (Q_{72}(t)+ Q_{72}^{(8)}(t)) [c]H_{2}(t) + Q_{74} (t)[c]H_{4}(t)
H_{9}(t) = Q_{9,10}(t)[c][1+H_{10}(t)], H_{10}(t) = (Q_{10,2}(t)[c] + Q_{10,2}^{(11)}(t))[c]H_{2}(t) (5158)
Taking Laplace Transform of eq. (5158) and solving for
= N_{6}(s) / D_{3}(s) (59)
Where
N_{6}(s) = (1 – _{22}^{(3)*}(s)) { _{01}(s)_{12}(s)_{14}(s)) _{09} (s) _{9,10} (s)}
D_{3}(s) = (1  _{22}^{(3)*}(s)) { 1  (_{01}(s) _{14} (s) + _{07}(s) _{74}(s))_{46}^{(5)*}(s) _{60}(s)}
 _{20}(s)_{01}(s)_{12}(s)_{07}(s)( _{72}(s)) + _{72}^{(8)}(s) +
_{09} (s)_{9,10} (s) ( _{10,2} (s) +Q _{10,2}^{(11)*}(s))}
In the long run, H_{0} = (60)
where N_{6}(0)= p_{20} (p_{01 }+ p_{09}) 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)[c]V_{1}(t) + Q_{07}(t)[c]V_{7}(t) + Q_{09}(t)[c]V_{9}(t)
V_{1}(t) = Q_{12}(t)[c]V_{2}(t) + Q_{14}(t)[c]V_{4}(t), V_{2}(t) = Q_{20}(t)[c]V_{0}(t) + Q_{22}^{(3)}(t)[c]V_{2}(t)
V_{4}(t) = Q_{46}^{(3)}(t)[c]V_{6}(t), V_{6}(t) = Q_{60}(t)[c]V_{0}(t)
V_{7}(t) = (Q_{72}(t)[1+V_{2}(t)]+ Q_{72}^{(8)}(t)) [c]V_{2}(t) + Q_{74} (t)[c]V_{4}(t)
V_{9}(t) = Q_{9,10}(t)[c]V_{10}(t), V_{10}(t) = (Q_{10,2}(t) + Q_{10,2}^{(11)}(t))[c]V_{2}(t) (6168)
Taking LaplaceStieltjes transform of eq. (6168) and solving for
= N_{7}(s) / D_{4}(s) (69)
where N_{7}(s) = _{07} (s) _{72} (s) (1 – _{22}^{(3)*}(s)) and D_{4}(s) is the same as D_{3}(s)
In the long run, V_{0} = (70)
where N_{7}(0)= p_{20} p_{07 }p_{72} and D’_{3}(0) is already defined.
COSTBENEFIT ANALYSIS
The gain function of the system considering mean uptime, expected busy period of the system under atmospheric pressure when the units stops automatically, expected busy period of the server for repair of unit due to gravitational force 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 due to atmospheric pressure in (0, t ] when the units automatically stop in (0,t]
 expected busy period of the system under gravitational force
 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 nonidentical 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 repair due to atmospheric pressure when units automatically stop,
K_{4}: Cost per unit time for which the system is under repair due to gravitational force,
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 analyzed graphically that when the failure rate due to voltage fluctuations, failure rate due to steep high acoustics increases, the MTSF and steady state availability decreases and the cost function decreased as the failure increases.
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