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Profit-function of two-non identical cold standby system under the influence of rainfall and humidity

Ashok Kumar Saini

Associate Professor, Department of Mathematics, BLJS College, Tosham, Bhiwani, Haryana, INDIA.

Email: [email protected]

Research Article

Abstract               Introduction: Reliability is a measure of how well a system performs or meets its design requirements. It is hence the prime concern of all scientists and engineers engaged in developing such a system.. In this paper we have taken FH- failure due to humidity.. When the main unit fails due to Rain Fall then cold standby system becomes operative. Rain Fall cannot occur simultaneously in both the units and after failure the unit undergoes very costly repair facility immediately. Applying the regenerative point technique with renewal process theory the various reliability parameters MTSF, Availability, Busy period, Profit-Function analysis have been evaluated.

Keywords: Cold Standby, FH-failure due to humidity, Rain Fall, first come first serve, MTSF, Availability, Busy period, Profit -Function.

INTRODUCTION

Stochastic behavior of systems operating under changing environments has widely been studied. Jinhua Cao (1989) has studied a man machine system operating under changing environment subject to a Markov process with two states. Dhillon, B.S. and Natesan, J. (1983) studied an outdoor power systems in fluctuating environment. Kan Cheng 1985) has studied reliability analysis of a system in a randomly changing environment. The change in operating conditions viz. fluctuations of voltage, corrosive atmosphere, gravity etc. may make a system completely inoperative. Severe environmental conditions can make the actual mission duration longer than the ideal mission duration. In this paper we have taken FH - Failure due to Humidity and other failure due to Rain Fall. When the main operative unit fails due to Rain Fall then cold standby system becomes operative. Rain Fall cannot occur simultaneously in both the units and after failure the unit undergoes repair facility of very high cost in case of failure due to Rain Fall immediately. Failure due to Humidity may be destructive. The repair is done on the basis of first fail first repaired.

ASSUMPTIONS

1. F1(t) and F2(t) are general failure time distributions due to Rain Fall and Humidity. The repair is of two types -Type -I, Type-II with repair time distributions as G 1(t) and G 2(t) respectively.
2. The Rain Fall are non-instantaneous and it cannot come simultaneously in both the units.
3. Whenever the Rain Fall occur within specified limit of the unit, it works as normal as before. But as soon as there occur Rain Fall of magnitude beyond specified limit of the unit the operation of the unit stops automatically.
4. The repair starts immediately after the Rain Fall of beyond specified limit of the unit are over and works on the principle of first fail first repaired basis.
5. The repair facility does no damage to the units and after repair units are as good as new.
6. The switches are perfect and instantaneous.
7. All random variables are mutually independent.
8. When both the units fail, we give priority to operative unit for repair.
9. FH- Failure due to Humidity when there is humidity in the atmosphere beyond specified limit.
10. Repairs are perfect and failure of a unit is detected immediately and perfectly.
11. The system is down when both the units are non-operative.

SYMBOLS FOR STATES OF THE SYSTEM

F1(t) and F2(t) are the failure rates due to Rain Fall and Humidity respectively

G1(t), G2(t) – repair time distribution Type -I, Type-II due to Rain Fall and Humidity respectively

Superscripts: O, CS, RF, FH

Operative, Cold Standby, Rain Fall, Failure due to Humidity respectively

Subscripts: nrf, rf, hum, ur, wr, uR

No Rain Fall, Rain Fall, Humidity, under repair, waiting for repair, under repair continued from previous state respectively

Up states: 0, 1, 2;

Down states: 3, 4

Regeneration point: 0, 1, 2

Notations

Mi(t) System having started from state I is up at time t without visiting any other regenerative state

Ai (t) state is up state as instant t

Ri (t) System having started from state I is busy for repair at time t without visiting any other regenerative state.

Bi (t) The server is busy for repair at time t.

Hi(t) Expected number of visits by the server for repairing given that the system initially starts from regenerative state i

By Rain Fall we mean Rain Fall beyond the specified limit

States of the System

0(Onrf, CSnrf)

One unit is operative and the other unit is cold standby and there are no Rain Fall in both the units.

1(SORF rf, ur, Onrf)

The operating unit fails due to Rain Fall and is under repair immediately of very costly Type- I and standby unit starts operating with no Rain Fall.

2(FH nrf,hum, ur, Onrf)

The operative unit fails due to FH resulting from Humidity and undergoes repair of type II and the standby unit becomes operative with no Rain Fall.

3(RFrf,uR, FH nrf, hum,wr)

The first unit fails due to Rain Fall and under very costly Type-! repair is continued from state 1 and the other unit fails due to FM resulting from Humidity and is waiting for repair of Type -II.

4(RF rf,uR, RFrf,wr)

The one unit fails due to Rain Fall is continues under repair of very costly Type - I from state 1 and the other unit also fails due to Rain Fall. is waiting for repair of very costly Type- I.

5(FH nrf, hum, uR, RF rf, wr)

The operating unit fails due to Humidity (FH mode) and under repair of Type - II continues from the state 2 and the other unit fails due to Rain Fall is waiting for repair of very costly Type- I.

6(FH nrf,hum,uR, FH nrf,hum,wr)

The operative unit fails due to FH resulting from Humidity and under repair continues from state 2 of Type –II and the other unit is also failed due to FH resulting from Humidity and is waiting for repair of Type-II and there is no Rain Fall.

Figure 1: The State Transition Diagram

Transition Probabilities

Simple probabilistic considerations yield the following expressions:

p01 = , p02 =

p10 = , p13 =p11(3)= p11(4)=

p25 = p22(5)= p22(6) =

clearly

p01 + p02 = 1,

p10 + p13 =(p11(3) ) + p14 = ( p11(4)) = 1,

p20 + p25 = (p22(5)) + p26 =(p22(6)) = 1                                                                                                                                             (1)

And mean sojourn time are

µ0 = E(T) =                                                                                                                                                            (2)

Mean Time To System Failure

Ø0(t) = Q01(t)[s] Ø1(t) + Q02(t)[s] Ø2(t)

Ø1(t) = Q10 (t)[s] Ø0(t) + Q13(t) + Q14(t)

Ø2(t) = Q20 (t)[s] Ø0(t) + Q25(t) + Q26(t)                                                                                                                                    (3-5)

We can regard the failed state as absorbing

Taking Laplace-Stiljes transform of eq. (3-5) and solving for

ø0*(s)  = N1(s) / D1(s)                                                                                                                                                                       (6)

Where

N1(s) = Q01*[ Q13 * (s) + Q14 * (s) ] + Q02*[ Q25 * (s) + Q26 * (s) ]

D1(s) = 1 - Q01* Q10* - Q02* Q20*

Making use of relations (1) and (2) it can be shown that ø0*(0)  =1, which implies that ø0*(t)  is a proper distribution.

MTSF = E[T] =  (s)      = (D1(0) - N1(0)) / D1 (0)

s=0

= ( +p01  + p02 ) / (1 - p01 p10 - p02 p20 )

where

= 1 + 2, 1= 0 + 3 + 4

++

AVAILABILITY ANALYSIS

Let Mi(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 M0(t), M1(t), M2(t) can be found easily.

The point wise availability Ai(t) have the following recursive relations

A0(t) = M0(t) + q01(t)[c]A1(t) + q02(t)[c]A2(t)

A1(t) = M1(t) + q10(t)[c]A0(t) + q11(3)(t)[c]A1(t)+ q11(4)(t)[c]A1(t),

A2(t) = M2(t) + q20(t)[c]A0(t) + [q22(5)(t)[c]+ q22(6)(t)] [c]A2(t)                                                                                          (7-9)

Taking Laplace Transform of eq. (7-9) and solving for

= N2(s) / D2(s)                                                                                                                                                           (10)

Where

N2(s) =  0(s)(1 -  11(3)(s) -  11(4)(s)) (1-  22(5)(s)-  22(6)(s)) +  (s)1(s)

[ 1-  22(5)(s)-  22(6)(s)] + 02(s)  2(s)(1- 11(3)(s) - 11(4)(s))

D2(s) = (1 -  11(3)(s)-  11(4)(s)) { 1 -  22(5)(s) - 22(6)(s) )[1-( 01(s)  10 (s))(1-

11(3)(s)-  11(4)(s))]

A0 =  =  =

Using L’ Hospitals rule, we get

A0 =  =                                                                                                                                      (11)

The expected up time of the system in (0,t] is

(t) =  So that                                                                                                    (12)

The expected down time of the system in (0,t] is

(t) = t- (t) So that                                                                                                      (13)

The expected busy period of the server when there is FH-failure resultlng from Humidity in (0,t]

R0(t) = q01(t)[c]R1(t) + q02(t)[c]R 2(t)

R1(t) = S1(t) + q01(t)[c]R1 (t) + [q11(3)(t) + q11(4)(t)[c]R1(t),

R2(t) = q20(t)[c]R0(t) + [q22(6)(t)+q22(5)(t)][c]R2(t)                                                                                                                  (14-16)

Taking Laplace Transform of eq. (14-16) and solving for

= N3(s) / D3(s)                                                                                                                                                            (17)

Where

N 3(s) = 01(s)  1(s) and

D 3(s)= (1 -  11(3)(s)-  11(4)(s)) – 01(s) is already defined.

In the long run, R0 =                                                                                                                                                      (18)

The expected period of the system under FH-failure resulting from Humidity in (0,t] is

(t) =  So that

The expected Busy period of the server when there is Rain Fall in (0,t]

B0(t) = q01(t)[c]B1(t) + q02(t)[c]B2(t)

B1(t) = q01(t)[c]B1(t) + [q11(3)(t)+ q11(4)(t)] [c]B1(t),

B2(t) = T2(t) + q02(t)[c] B2(t) + [q22(5)(t)+ q22(6)(t)] [c]B2(t)

T2(t) = e- λ1t G2(t)                                                                                                                                                                               (19- 21)

Taking Laplace Transform of eq. (19-21) and solving for

= N4(s) / D2(s)                                                                                                                                                           (22)

Where

N4(s) = 02(s)  2(s))

In steady state, B0 =                                                                                                                                                      (23)

The expected busy period of the server for repair in (0,t] is

(t) =  So that                                                                                                                                (24)

The expected number of visits by the repairman for repairing the non-identical units in (0,t]

H0(t) = Q01(t)[s][1+ H1(t)] + Q02(t)[s][1+ H2(t)]

H1(t) = Q10(t)[s]H0(t)] + [Q11(3)(t)+ Q11(4)(t)] [s]H1(t),

H2(t) = Q20(t)[s]H0(t) + [Q22(5)(t) +Q22(6)(t)] [c]H2(t)                                                                                                             (25-27)

Taking Laplace Transform of eq. (25-27) and solving for

= N6(s) / D3(s)                                                                                                                                                           (28)

In the long run, H0 =                                                                                                                                                      (29)

PROFIT-FUNCTION ANALYSIS

The Profit-Function analysis of the system considering mean up-time, expected busy period of the system under Rain Fall when the units stops automatically, expected busy period of the server for repair of unit under Humidity, expected number of visits by the repairman for unit failure.

The expected total Profit-Function incurred in (0,t] is

C(t) = Expected total revenue in (0,t] - expected total repair cost in (0,t] due to Rain Fall failure

• expected total repair cost due to FH- failure resulting from Humidity for repairing the units in (0,t ]
• expected busy period of the system under Rain Fall when the units automatically stop in (0,t]
• expected number of visits by the repairman for repairing of non-identical the units in (0,t]

The expected total cost per unit time in steady state is

C =  =

= K1A0 - K 2R0 - K 3B0 - K 4H0

Where

K1: revenue per unit up-time,

K2: cost per unit time for which the system is under repair of type- I

K3: cost per unit time for which the system is under repair of type-II

K4: cost per visit by the repairman for units repair.

CONCLUSION

After studying the system, we have analysed graphically that when the failure rate due to Humidity and failure rate due to Rain Fall increases, the MTSF and steady state availability decreases and the Profit-function decreased as the failure increases.

REFERENCES

1. Barlow, R.E. and Proschan, F., Mathematical theory of Reliability, 1965; John Wiley, New York.
2. Gnedanke, B.V., Belyayar, Yu.K. and Soloyer, A.D., Mathematical Methods of Relability Theory, 1969 ; Academic Press, New York.
3. Dhillon, B.S. and Natesen, J, Stochastic Anaysis of outdoor Power Systems in fluctuating environment, Microelectron. Reliab.. 1983; 23, 867-881.
4. Cao, Jinhua, Stochatic Behaviour of a Man Machine System operating under changing environment subject to a Markov Process with two states, Microelectron. Reliab., 1989; 28, pp. 373-378.
5. Kan, Cheng, Reliability analysis of a system in a randomly changing environment, Acta Math. Appl. Sin. 1985, 2, pp.219-228.