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Profit function of twounit dissimilar warm standby system with failure due to extremely high radiations and 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: In Nuclear Reactor the leakage in form of radiations becomes highly dangerous to the lives of living beings. The radiations from the nuclear reactor are always under serious consideration due to fatal and miserable results to human race. Every precautions and extra care is taken to avoid any misshappening due to radiations. But still due carelessness or due to failure of some equipment in Nuclear Reactors there occur leakage of radiations causing a major casualty. In the present paper we have taken twodissimilar warm standby system with failure due to extremely high radiations. The unit fails due to extremely high radiations. When there are radiations of extremely high magnitude the working of unit stops automatically to avoid excessive damage of the units and when the unit comes in no normal position the repair of the units starts immediately. The failure time distribution is taken as exponential and repair time distribution as genera. Using 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 profitfunction. Special case by taking failure and repair as exponential have been derived and graphs are drawn.
Keyword: warm standby, extremely high radiations, MTSF, Availability, busy period, profit function.
INTRODUCTION
In Nuclear Reactor the leakage in form of radiations becomes highly dangerous to the lives of living beings. The radiations from the nuclear reactor are always under serious consideration due to fatal and miserable results to human race. In the present paper we have taken twodissimilar warm standby system with failure due to extremely high radiations.
Assumptions
 The failure time distribution is exponential whereas the repair time distribution is arbitrary of two nonidentical units.
 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 extremely high radiations of the system become normal. The radiations of both the units do not go extremely high.
 The failure of a unit is detected immediately and perfectly.
 The switches are not perfect and instantaneous.
 All random variables are mutually independent.
Symbols for states of the System
Superscripts: O, WS, SO, FEHR, SFO
Operative, Warm Standby, Stops the operation, Failure due to extremely high radiations, Switch failed but operable respectively
Subscripts: nehr, ehr,ur, wr, uR
No extremely high radiations. Extremely high radiations, 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_{nehr}, WS_{nehr }) One unit is operative and the other unit is warm standby and there is no extremely high temp. in both the units.
1(SO_{nehr}, O_{nehr})
The operation of the first unit stops automatically due to extremely high radiations and warm standby units starts operating and there is no extremely high radiations.
2(FEHR_{ur}, O_{nehr})
The first unit fails and undergoes repair after failure due to extremely high radiations are over and the second unit continues to be operative with no extremely high radiations.
3(FEHR_{uR}, SO_{ehr})
The repair of the first unit is continued from state 2 and in the other unit extremely high radiations occur and stops automatically due to extremely high radiations.
4(FEHR_{ur}, SO_{uehr})
The one unit fails and undergoes repair after the extremely high radiations are over and the other unit also stops automatically due to extremely high radiations.
5(FEHR_{uR}, FEHR_{wr}) The repair of the first unit is continued from state 4 and the other unit is failed due to extremely high radiations in it and is waiting for repair.
Figure 1: The State Transition Diagram
6 (O_{nehr, }FEHR_{ur})
The first unit is operative with no extremely high radiations and the second unit failed due to extremely high radiations is under repair.
7(SO_{nehr}, SFO_{nehr,ur})
The operation of the first unit stops automatically due to extremely high radiations and during switchover to the second unit switch fails and undergoes repair.
8(FEHR_{wr}, SFO_{nehr,uR})
The repair of failed switch is continued from state 7 and the first unit is failed after extremely high radiations and waiting for repair.
9(O_{nehr}, SO_{uehr})
The first unit is operative and the warm standby dissimilar unit is under extremely high radiations
10(SO_{nehr}, FEHR_{ur})
The operation of the first unit stops automatically due to extremely high radiations and the second unit switch fails and undergoes repair after the extremely high radiations is over.
11(FEHR_{wr}, FEHR_{uR})
The repair of the second unit is continued from state 10 and the first unit is failed due to extremely high radiations is waiting for repair.
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)
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}^{*}(0) =(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 extremely high radiations 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 extremely high radiations in (0,t] is
(t) = So that
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_{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 different 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.
COST BENEFIT ANALYSIS
The costbenefit function of the system considering mean uptime, expected busy period of the system under extremely high radiations 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 extremely high radiations 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}: when units automatically stop cost per unit time for which the system is under extremely high radiations
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, extremely high radiations rate increases, the MTSF and steady state availability decreases and the cost function decreased as the failure increases.
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