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Gain-function of two non-identical warm standby system with failure due to non- availability of sunlight 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 Solar System the sunlight plays aa important and vital role. The non-conventional renewable solar energy which is cheap and readily available for use in institutions, hospitals, industries and all sort of equipments and places where energy is required. But for solar energy Sun is the prime source from where solar energy can be generated. During rainy and winter seasons the sun is under the cover of clouds regularly resulting solar penal cells are unable to receive sunlight which causing failure of the system. In the present paper we have taken two non-identical warm standby system with failure due to non- availability of sunlight. When there is non-availability of sunlight the working of unit stops automatically. The failure time distribution is taken as exponential and repair time distribution as general. 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 profit-function. Special case by taking repair as exponential has been derived and graphs are drawn.

Keywords: warm standby, non-availability of sunlight, MTSF, Availability, busy period, Gain-function.

 

INTRODUCTION

In Solar System the sunlight plays the pivotal and vital role. The non-conventional renewable solar energy which is cheap and easily available for use in institutions, hospitals, industries and all sort of places and equipments where energy is required. But for solar energy Sun is the primary source from where solar energy can be generated. The sun is under the cover of clouds almost every day during rainy and winter seasons causing solar penal cells not receiving sunlight and becomes inactive unable to generate solar energy resulting failure of the system.

Assumptions

  • The failure time distribution is exponential whereas the repair time distribution is arbitrary of two non-identical units.
  • The repairs are perfect and starts immediately upon failure of units with repair discipline are FCFS.
  • The operation of the unit stops as soon as there is non-availability of sunlight.
  • The failure of a unit is detected immediately and perfectly.
  • The switches are instantaneous but not perfect.
  • All random variables are mutually independent.

Symbols for states of the System

Superscripts: O, WS, SO, FNASL, SF

Operative, Warm Standby, Stops the operation, Failure due to non-availability of sunlight, Switch failure respectively

Subscripts: nasl, asl, ur, wr, uR

Non-availability of sunlight, availability of sunlight, 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(Oasl, WSnasl ) The first unit is operative due to availability of sunlight and the second unit is warm standby with non-availability of sunlight.

1(SOnasl, Oasl)

The operation of the first unit stops automatically due to non-availability of sunlight and warm standby units starts operating due to availability of sunlight.

2(FNASLur, Oasl)

The first unit fails due to non-availability of sunlight undergoes repair and the second unit continues to be operative due to availability of sunlight.

3(FNASLuR, SOnasl)

The repair of the first unit is continued from state 2 and in the other unit the operation of the unit stops automatically due to non-availability of sunlight.

4(FNASLur, SOnasl)

The one unit fails due to non-availability of sunlight and undergoes repair and the other unit also stops automatically due to non-availability of sunlight.

5(FNASLuR, FNASLwr) 

The repair of the first unit is continued from state 4 and the other unit is failed due to non-availability of sunlight in it and is waiting for repair.

 

Figure 1: The State Transition Diagram

 

 

 

6 (Oasl, FNASLur)

The first unit is operative due to availability of sunlight and the second unit failed due to non-availability of sunlight is under repair.

7(SOnasl, SFur)

The operation of the first unit stops automatically due to non-availability of sunlight and during switchover to the second unit switch fails and undergoes repair.

8(FNASLwr, SFuR)

The repair of failed switch is continued from state 7 and the first unit is failed due to non-availability of sunlight is waiting for repair.

9(Oasl, SOnasl)

The first unit is operative due to availability of sunlight and the operation of warm standby second unit is stopped due to non-availability of sunlight.

10(SOnasl, SFur)

The operation of the first unit stops automatically due to non-availability of sunlight and in the second unit switch fails and undergoes repair.

11(FNASLwr, SFuR)

The repair of the second unit is continued from state 10 and the first unit is failed due to non-availability of sunlight is waiting for repair.

 

TRANSITION PROBABILITIES

Simple probabilistic considerations yield the following expressions :

p01 = , P07 =  

p09 = , p12 = , p14 =  

P20= G1*( λ1), P22(3) = G1*( λ1)=p23, P72 = G2*( λ4),

P72(8) = G2*( λ4)= P78

We can easily verify that

p01 + p07 + p09 = 1, p12 + p14 = 1, p20 + p23 (=p22(3))= 1, p46(6)= 1 p60 = 1,

p72+ P72(5) + p74 = 1, p9,10 =1, p10,2 + p10,2(11) = 1                                                                                                                       (1)

We can easily verify that

p01 + p07 + p09 = 1, p12 + p14 = 1, p20 + p23 (=p22(3))= 1, p46(6)= 1 p60 = 1,

p72+ P72(5) + p74 = 1, p9,10 =1, p10,2 + p10,2(11) = 1 (1)

And mean sojourn time are

µ0 = E(T) =                                                                                                                                                              (2)

 

MEAN TIME TO SYSTEM FAILURE

We can regard the failed state as absorbing

 

,

                                                                                                                                                                               (3-5)

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

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

Where

N1(s) = {  +

D1(s) = 1 -  

Making use of relations (1) and (2) it can be shown that θ0(0) =1, which implies that θ0(t) is a proper distribution.

MTSF = E[T] = d/ds θ0*(0)   = (D1(0) - N1(0)) / D1 (0)

 

 s=0

 = ( +p01  + p01 p12  + p09 ) / (1 - p01 p12 p20 )

 where

 +  + ,  + , + (3),

 

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), M4(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) + q07(t)[c]A7(t) + q09(t)[c]A9(t)

A1(t) = M1(t) + q12(t)[c]A2(t) + q14(t)[c]A4(t), A2(t) = M2(t) + q20(t)[c]A0(t) + q22(3)(t)[c]A2(t)

A4(t) = q46(3)(t)[c]A6(t), A6(t) = q60(t)[c]A0(t)

A7(t) = (q72(t)+ q72(8)(t)) [c]A2(t) + q74 (t)[c]A4(t)

A9(t) = M9(t) + q9,10(t)[c]A10(t), A10(t) = q10,2(t)[c]A2(t) + q10,2(11)(t)[c]A2(t)                                                                               (7-14)

Taking Laplace Transform of eq. (7-14) and solving for  

  = N2(s) / D2(s)                                                                                                                                                                                 (15)

Where

N2(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))}

D2(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

A0 =  =  =

Using L’ Hospitals rule, we get

A0 =  =                                                                                                                                           (16)

Where

N2(0)= p20(0(0) + p011(0) + p09 9(0) ) + 2(0) (p01p12 + p07 (p72

 + p72(8) + p09 ))

D2(0) = p20{  + p01 + (p01 p14 + p07 p74 )+ p07  + p07  + p09()

                + { 1- ((p01p14 + p07 p74 )}

, ,  

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 non-availability of sunlight in (0,t]

R0(t) = S0(t) + q01(t)[c]R1(t) + q07(t)[c]R7(t) + q09(t)[c]R9(t)

R1(t) = S1(t) + q12(t)[c]R2(t) + q14(t)[c]R4(t),

R2(t) = q20(t)[c]R0(t) + q22(3)(t)[c]R2(t)

R4(t) = q46(3)(t)[c]R6(t), R6(t) = q60(t)[c]R0(t)

R7(t) = (q72(t)+ q72(8)(t)) [c]R2(t) + q74 (t)[c]R4(t)

R9(t) = S9(t) + q9,10(t)[c]R10(t), R10(t) = q10,2(t) + q10,2(11)(t)[c]R2(t)                                                                                                (19-26)

Taking Laplace Transform of eq. (19-26) and solving for  

 = N3(s) / D2(s)                                                                                                                                                                   (27)

Where

N2(s) = (1 -  22(3)(s)) {  0(s) + 01(s) 1(s) + 09(s) 9(s)} and D2(s) is already defined.

In the long run, R0 =                                                                                                                                                        (28)

where N3(0)= p20(0(0) + p011(0) + p09 9(0) ) and D2(0) is already defined.

The expected period of the system under non-availability of sunlight 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]

B0(t) = q01(t)[c]B1(t) + q07(t)[c]B7(t) + q09(t)[c]B9(t)

B1(t) = q12(t)[c]B2(t) + q14(t)[c]B4(t), B2(t) = q20(t)[c] B0(t) + q22(3)(t)[c]B2(t)

B4(t) = T4 (t)+ q46(3)(t)[c]B6(t), B6(t) = T6 (t)+ q60(t)[c]B0(t)

B7(t) = (q72(t)+ q72(8)(t)) [c]B2(t) + q74 (t)[c]B4(t)

B9(t) = q9,10(t)[c]B10(t), B10(t) = T10 (t)+ (q10,2(t) + q10,2(11)(t)[c]B2(t)                                                                                              (29- 36)

Taking Laplace Transform of eq. (29-36) and solving for 

 = N4(s) / D2(s) (37)

Where

N4(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 D2(s) is already defined.

In steady state, B0 =                                                                                                                                                       (38)

where N4(0)= p20 {( p01 p14 + p07 p74) (4(0) +6(0)) + p09 10(0) } and D2(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]

P0(t) = q01(t)[c]P1(t) + q07(t)[c]P7(t) + q09(t)[c]P9(t)

P1(t) = q12(t)[c]P2(t) + q14(t)[c]P4(t), P2(t) = q20(t)[c]P0(t) + q22(3)(t)[c]P2(t)

P4(t) = q46(3)(t)[c]P6(t), P6(t) = q60(t)[c]P0(t)

P7(t) = L7(t)+ (q72(t)+ q72(8)(t)) [c]P2(t) + q74 (t)[c]P4(t)

P9(t) = q9,10(t)[c]P10(t), P10(t) = (q10,2(t) + q10,2(11)(t))[c]P2(t)                                                                                                           (40-47)

Taking Laplace Transform of eq. (40-47) and solving for

  = N5(s) / D2(s)                                                                                                                                                                  (48)

where N2(s) = 07(s )  7(s) 1 -  22(3)(s)) and D2(s) is defined earlier.

In the long run, P0 =                                                                                                                                                        (49 )

where

N5(0)= p20 p07 4(0)

and D2(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 non-identical units in (0,t]

H0(t) = Q01(t)[c]H1(t) + Q07(t)[c]H7(t) + Q09(t)[c]H9(t)

H1(t) = Q12(t)[c][1+H2(t)] + Q14(t)[c][1+H4(t)], H2(t) = Q20(t)[c]H0(t) + Q22(3)(t)[c]H2(t)

H4(t) = Q46(3)(t)[c]H6(t), H6(t) = Q60(t)[c]H0(t)

H7(t) = (Q72(t)+ Q72(8)(t)) [c]H2(t) + Q74 (t)[c]H4(t)

H9(t) = Q9,10(t)[c][1+H10(t)], H10(t) = (Q10,2(t)[c] + Q10,2(11)(t))[c]H2(t)                                                                          (51-58)

Taking Laplace Transform of eq. (51-58) and solving for  

 = N6(s) / D3(s)                                                                                                                                                                   (59)

Where

N6(s) = (1 –  22(3)*(s)) { 01(s)12(s)14(s))  09 (s) 9,10 (s)}

D3(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, H0 =  (60 )

where N6(0)= p20 (p01 + p09) and D’3(0) is already defined.

The expected number of visits by the repairman for repairing the switch in (0,t]

V0(t) = Q01(t)[c]V1(t) + Q07(t)[c]V7(t) + Q09(t)[c]V9(t)

V1(t) = Q12(t)[c]V2(t) + Q14(t)[c]V4(t), V2(t) = Q20(t)[c]V0(t) + Q22(3)(t)[c]V2(t)

V4(t) = Q46(3)(t)[c]V6(t), V6(t) = Q60(t)[c]V0(t)

V7(t) = (Q72(t)[1+V2(t)]+ Q72(8)(t)) [c]V2(t) + Q74 (t)[c]V4(t)

V9(t) = Q9,10(t)[c]V10(t), V10(t) = (Q10,2(t) + Q10,2(11)(t))[c]V2(t) (61-68)

Taking Laplace-Stieltjes transform of eq. (61-68) and solving for  

 = N7(s) / D4(s) (69)

where N7(s) =  07 (s) 72 (s) (1 –  22(3)*(s)) and D4(s) is the same as D3(s)

In the long run, V0 =  (70)

where N7(0)= p20 p07 p72 and D’3(0) is already defined.

 

GAIN-FUNCTION ANALYSIS

The Gain- function of the system considering mean up-time, expected busy period of the system under non-availability of sunlight 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 non-identical units failure, expected number of visits by the repairman for switch failure.

The expected total Gain-function 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 non-availability of sunlight 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 non-identical units in (0,t]

The expected total cost per unit time in steady state is

C = =

 = K1A0 - K2P0 - K3B0 - K4R0 - K5V0 - K6H0

Where

K1: revenue per unit up-time,

K2: cost per unit time for which the system is under switch repair

K3: cost per unit time for which the system is under unit repair

K4: when units automatically stop cost per unit time for which the system is under non-availability of sunlight

K5: cost per visit by the repairman for which switch repair,

K6: cost per visit by the repairman for non-identical units repair.

 

CONCLUSION

After studying the system, we have analyzed graphically that when the failure rate due to non-availability of sunlight and failure rate due to switch failure increases, the MTSF and steady state availability decreases and the Gain-Function also decreased as the failure increases.

 

REFERENCES

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  3. Gnedanke, B.V., Belyayar, Yu.K. and Soloyer, A.D., Mathematical Methods of Relability Theory, 1969 ; Academic Press, New York.
  4. Goel, L.R., Sharma,G.C. and Gupta, Rakesh Cost Analysis of a Two-Unit standby system with different weather conditions, Microelectron. Reliab., 1985; 25, 665-659.
  5. 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, 215-219.
  6. Gupta, Rakesh and Goel, Rakesh, Profit Analysis of a Two-Unit Cold Standby system under Different Weather Conditions, Microelectron. Reliab., 1991;31, pp.18-27.
  7. Nakagawa, T., A 2-Unit Repairable Redundant System with Switching Failure, IEEE, Trans. Reliab., 1977; R-26, 2, pp.128-130.

 

 

 
 
 
 
 
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