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Cost-Benefit analysis of two-unit cold standby system with two types of failures-open failure due to fog and human error and short-circuit failure due to steep high temperatures

Ashok Kumar Saini

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

Email: [email protected]

Research Article

Abstract               Introduction: A large number of papers on a two unit standby redundant systems have appeared because of the large number of parameters governing system behaviour and the complex environment with uncertainties. In this paper we have taken short-circuit failure mode which is caused due to SH temp. {steep high temperatures) and open failure mode due to fog and human error. When the main unit fails then cold standby system becomes operative. SH temp. cannot occur simultaneously in both the units and after failure the unit undergoes very costly repair facility immediately. The failure time distributions are exponential with p and q as prob. of having SH temp. and Fog and Human error respectively and repair is of two types -Type -I ,Type-II with repair time distributions as arbitrary. Applying the regenerative point technique with renewal process theory we have evaluated (1) MTSF (2) Steady State Availability (3) Busy period analysis (4) Expected number of visits by the repairman and (5) Profit analysis.

Keyword: Cost-benefit, steep high temperature.

INTRODUCTION

In this paper we have taken Short-Circuit failure mode which is caused due to SH temp. (Steep High temperatures) and open failure mode due to Fog and Human errors. When the main operative unit fails then cold standby system becomes operative. SH temp. Cannot occur simultaneously in both the units and after failure the unit undergoes repair facility of low cost in case of short-circuit failure mode immediately. Open failure mode due to Fog and human error may be highly destructive. It may be due to not clear vision and carelessness or he may be unskilled. The repair is very costly in the second failure as there may be accident of major casualty. The repair is done on the basis of first fail first repaired.

ASSUMPTIONS

1. The failure time distributions are general F₁ (t) and F₂ (t) and repair is of two types -Type -I, Type-II with repair time distributions as G ₁(t) and G ₂(t) respectively.
2. The repair starts immediately upon failure of units.
3. The switches are perfect and instantaneous.
4. All random variables are mutually independent.
5. When both the units fail, we give priority to operative unit for repair.
6. The failure of open mode Is due to some human errors.
7. Repairs are perfect and failure of a unit is detected immediately and perfectly.
8. The system is down when both the units are non-operative.
9. After repair, the unit’s works like a new one.

SYMBOLS FOR STATES OF THE SYSTEM

F₁(t)_, F₂ (t): failure time distribution due to SH temp. Mode and open failure mode

G₁(t), G₂(t): repair time distribution due to Type -I, Type-II respectively

Superscripts O, CS, SHTF, OF

Operative, Cold Standby, Steep high temp. failure mode, open failure mode due to human error respectively

Subscripts nsht, sht, he, ur, wr, uR

No steep high temp., Steep high temp., human error, under repair, waiting for repair, under repair continued from previous state respectively

Up states: 0, 1, 2;

Down states: 3,4,5,6

Regeneration point: 0, 1, 2

States of the System

0(O_nsht, CS_nsht)

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

1(SHTF _{sht, ur}, O_nsht)

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

2(OF _{nsht,he, ur}, O_nsht)

The operative unit fails due to human error (open failure mode) and undergoes repair of type II and the standby unit becomes operative with no SH temp..

3(SHTF _sht,uR, OF _{nsht, he,wr})

The first unit fails due to SH temp. and under very costly Type-! Repair is continued from state 1 and the other unit fails due to human errors and is waiting for repair of Type -II.

Figure 1: The State Transition Diagram

4(SHTF _sht,uR, SHTF _sht,wr)

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

5(OF _{nsht, he, uR}, SHTF _{sht, wr}) 

The operating unit fails due to human error (open failure mode) and under repair of Type - II continues from the state 2 and the other unit fails due to SH temp. is waiting for repair of very costly Type- I.

6(OF _nsht,he,uR, OF _nsht,he,wr)

The operative unit fails due to human error and under repair continues from state 2 of Type –II and the other unit is also failed due to human error and is waiting for repair of Type-II and there is no SH temp.

7(O_nsht , OF _{fghe, uR}) 

The operating unit is active and the other unit fails due to open failure mode due to fog and human error and under repair continued of very high costly Type – I.

8(O _nsht, OF _fg,he,ur)

The operative unit is active with no SH temp. and the other unit is failed due to fog and human error and is under repair of very costly Type-II.

TRANSITION PROBABILITIES

Simple pr p₀₁ = p obabilistic considerations yield the following expressions:

p₀₁ = p _{, ,} p₀₂ = q, p₁₀ = p

p₇₀= G₁^*(λ₁) + qG₁^*(λ₂)

p₂₀= p G₂^*(λ₁) + qG₂^*(λ₂) = p₈₀

p₁₇⁽⁴⁾= p (1- G₁^*(λ₁ )) = p₁₄ = p₇₁⁽⁴⁾

p₂₇⁽⁶⁾= p (1- G₂^*(λ₂ )) = p₂₆ = p₈₇⁽⁶⁾

p₁₈⁽³⁾= q (1- G₁^*(λ₂ )) = p₁₃ = p₇₈⁽³⁾

p₂₈⁽⁵⁾= q (1- G₂^*(λ₂ )) = p₂₅ = p₈₂⁽⁵⁾

clearly

p₀₁ + p₀₂ = 1,                                                                               

p₁₀ + p₁₃ =(p₁₇⁽⁴⁾ ) + p₁₃ = ( p₁₈⁽³⁾) = 1,

p₇₀ + p₇₁ + p₇₈ = 1

p₈₀ + p₈₂⁽⁵⁾ + p₈₇⁽⁶⁾ = 1

And mean sojourn time are                                                                                                                                                        (1)

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

The point wise availability A_i(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) + q₁₁³(t)[c]A₁(t)+ q₁₁⁴(t)[c]A₁(t),

A₂(t) = M₂(t) + q₂₀(t)[c]A₀(t) + [q₂₂⁵(t)[c]+ q₂₂⁶(t)] [c]A₂(t)                                                                                                             

A₇(t) = M₇(t) + q₇₀(t)[c]A₀(t) + q₇₁⁽⁴⁾(t)[c]A₁(t)+ q₇₈⁽³⁾(t)[c]A₈(t) ,

A₈(t) = M₈(t) + q₈₀(t)[c]A₀(t) + q₈₂⁽⁵⁾(t)[c] A₂(t)+ q₈₇⁽⁶⁾(t)] [c]A₇(t)                                                                                                 (7-11)

Taking Laplace Transform of eq. (7-11) and solving for  

 = N₂(s) / D₂(s)                                                                                                                                                          (12)

Where

N₂(s) =  ₀(s)(1 -  ₇₈⁽⁴⁾(s) -  ₈₇⁽⁶⁾(s))-(  ₈₂⁽⁵⁾(s) (  ₂₇⁽⁶⁾(s)  ₇₈⁽³⁾(s) +  ₂₈⁽⁵⁾(s) )

 -  ₇₁⁽⁴⁾(s) (  ₁₇⁽⁴⁾(s) +  ₈₇⁽⁶⁾(s)  ₁₈⁽³⁾(s) ) +  ₇₁⁽⁴⁾(s)  ₈₂⁽⁵⁾(s)

 (  ₁₇⁽⁴⁾(s)  ₂₈⁽⁵⁾(s) -  ₂₇⁽⁶⁾(s)  ₁₈⁽³⁾(s) )]+ (s)₁(s)(1- ₇₈⁽³⁾(s)  ₈₇⁽⁶⁾(s)) + ₁₇⁽⁴⁾(s)(  ₇(s)+ ₇₈⁽³⁾(s)  ₈(s)] + ₁₈⁽³⁾(s)) (  ₇(s) ₈₇⁽⁶⁾(s) –

 ₈(s)) -  ₈₂⁽⁵⁾(s)₁(s)( ₂₇⁽⁶⁾(s)  ₇₈⁽³⁾(s)) + ₁₇⁽⁴⁾(s)(-  ₂(s) ₇₈⁽³⁾(s) +  ₇(s) ₂₈⁽⁵⁾(s)) - ₁₈⁽³⁾(s)(  ₂(s)+  ₇(s) ₂₇⁽⁶⁾(s) +  (s)₂(s)(1- ₇₈⁽³⁾(s)  ₈₇⁽⁶⁾(s)) + ₂₇⁽⁶⁾(s)(  ₇(s)+ ₇₈⁽³⁾(s)  ₈(s)] + ₂₈⁽⁵⁾(s)) (  ₇(s) ₈₇⁽⁶⁾(s) +

 ₈(s)) -  ₇₁⁽⁴⁾(s)₁(s)(- ₂₇⁽⁶⁾(s)-  ₂₈⁽⁵⁾(s) ₈₇⁽⁶⁾(s))+ ₁₇⁽⁴⁾(s) (  ₂(s) ₂₈⁽⁵⁾(s)  ₈(s))- ₁₈⁽³⁾(s)( -  ₂(s) ₈₇⁽⁶⁾(s)+  ₈(s) ₂₇⁽⁶⁾(s)

D₂(s) = (1 -  ₇₈⁽⁴⁾(s) -  ₈₇⁽⁶⁾(s))-(  ₈₂⁽⁵⁾(s) (  ₂₇⁽⁶⁾(s)  ₇₈⁽³⁾(s) +  ₂₈⁽⁵⁾(s) )

 -  ₇₁⁽⁴⁾(s) ( ₁₇⁽⁴⁾(s) +  ₈₇⁽⁶⁾(s)  ₁₈⁽³⁾(s) ) +  ₇₁⁽⁴⁾(s)  ₈₂⁽⁵⁾(s)

(  ₁₇⁽⁴⁾(s)  ₂₈⁽⁵⁾(s) -  ₂₇⁽⁶⁾(s)  ₁₈⁽³⁾(s) )]+ (s) (s)(1- ₈₇⁽⁶⁾(s) ₇₈⁽³⁾(s)) - ₁₇⁽⁴⁾(s)( ₇₀(s) + ₇₈⁽³⁾(s) ₈₀(s)) –  ₁₈⁽³⁾(s) _{70 87}⁽⁶⁾(s) +  ₈₀(s)) - ₈₂⁽⁵⁾(s) (s)( ₂₇⁽⁶⁾(s) ₇₈⁽³⁾(s)) + ₂₈⁽⁵⁾(s))+( ₁₇⁽⁴⁾(s)( (s) ₇₈⁽³⁾(s)-  ₂₈⁽⁵⁾(s) ₇₀(s))+ ₁₈⁽³⁾(s) ( ₂₀(s)+ ₂₇⁽⁶⁾(s)  ₇₀(s))}] (s)[- ₂₀(s)(1-  ₇₈⁽³⁾(s) ₈₇⁽⁶⁾(s))- ₂₇⁽⁶⁾(s) (  ₇₀(s) + ₇₈⁽³⁾(s) ₈₀(s)) - ₂₈⁽⁵⁾(s)(  ₇₀(s) + ₈₇⁽⁶⁾(s)+ ₈₀(s) - ₇₁⁽⁴⁾(s){  ₁₀(s)( ₂₇⁽⁶⁾(s)+ ₂₈⁽⁵⁾(s)  ₈₇⁽⁶⁾(s)-  ₁₇⁽⁴⁾(s))( ₂₀(s)-  ₂₈⁽⁵⁾ (s)  ₈₀(s))

The steady state availability

A₀ = =  =

 Using L’ Hospitals rule, we get

A₀ =  =                                                                                                                                      (13)

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

(t) =  So that                                                                                                                    (14)

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

(t) = t- (t) So that                                                                                                      (15)

The expected busy period of the server when there is open failure mode failure due to fog and human error 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) +q₁₈⁽³⁾(t) R₈(t) + q₁₇⁽⁴⁾(t)[c]R₇(t) ,

 R₂(t) = S₂(t) + q₂₀(t)[c]R₀(t) +q₂₈⁽⁵⁾(t) R₈(t) +q₂₇⁽⁶⁾(t)][c]R₇(t)

 R₇(t) = S₇(t) + q₇₀(t)[c]R₀ (t) +q₇₁⁽⁴⁾(t) R₁(t) + q₇₈⁽³⁾(t)[c]R₈(t) ,

 R₈(t) = S₈(t) + q₈₀(t)[c]R₀(t) +q₈₂⁽⁵⁾(t) R₂(t) +q₈₇⁽⁶⁾(t)][c]R₇(t)                                                                                          (16-20)

Where

S₁(t) = pG₁(t) e^-(λ₁ + ^λ₂^)t = S₇(t)

S₂(t) = qG₁(t) e^-(λ₁ + ^λ₂^)t = S₈(t)

Taking Laplace Transform of eq. (16-20) and solving for  

 = N₃(s) / D₃(s)                                                                                                                                                            (21)

 Where

N ₃(s) = ₀₁[( ₁(1- ₇₈⁽³⁾ ₈₇⁽⁶⁾ ) + ₁₇⁽⁴⁾( ₇ + ₇₈⁽³⁾ ₈ )+ ₁₈⁽³⁾ ₇ + ₈)]

1. _{01 82}⁽⁵⁾ [( _{1 27}⁽⁶⁾ ₇₈⁽³⁾ + ₂₈⁽⁵⁾) + ₁₇⁽⁴⁾ ₂ + ₂₈⁽⁵⁾ ₇) - ₁₈⁽³⁾( ₂ + ₂₇⁽⁶⁾ ₇ )] + ₀₂[( ₂(1- ₇₈⁽³⁾ ₈₇⁽⁶⁾ ) + ₂₇⁽⁶⁾( ₇ + ₇₈⁽³⁾ ₈ )+ ₂₈⁽⁵⁾( ₇ + ₈₇⁽⁶⁾ ₈)- _{02 71}⁽⁴⁾ [ ₁(- ₂₇⁽⁶⁾ - ₂₈⁽⁵⁾ + ₈₇⁽⁶⁾) ₁₇⁽⁴⁾ ( ₂ + ₂₈⁽⁵⁾ ₈) - ₁₈^{(3 )}(- _{2 87}^{(6 )}+ _{8 27}⁽⁶⁾ )]

 and

D ₃(s) is already defined.

In the long run, R₀ =                                                                                                                                                      (22)

The expected busy period of the system under open failure due to fog and human error in (0,t] is

(t) =  So that

The expected Busy period of the server when there is short-circuit failure due to SH temp. in (0,t]

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

B₁(t) = T₁(t) + q₁₀(t)[c]B₀(t) + q₁₈⁽³⁾(t)B₈(t)+ q₁₇⁽⁴⁾(t)] [c]B₇(t) ,

B₂(t) = T₂(t) + q₂₀(t)[c] B₀(t) + q₂₈⁽⁵⁾(t)B₈(t)+ q₂₇⁽⁶⁾(t)] [c]B₇(t)

B₇(t) = T₇(t) + q₇₀(t)[c]B₀(t) + q₇₁⁽⁴⁾(t)B₈(t)+ q₇₈⁽³⁾(t)] [c]B₈(t) ,

B₈(t) = T₈(t) + q₈₀(t)[c] B₀(t) + q₈₂⁽⁵⁾(t)B₂(t)+ q₈₇⁽⁶⁾(t)] [c]B₇(t)

Where

T₁(t) = pG₂(t) e^-(λ₁ + ^λ₂^)t = T₇(t)

T₂(t) = qG₂(t) e^-(λ₁ + ^λ₁^)t = T₈(t)                                                                                                                                                     (23- 27)

 Taking Laplace Transform of eq. (23-27) and solving for  

   = N₄(s) / D₂(s)                                                                                                                                                         (28)

In steady state, B₀ =                                                                                                                                                      (29)

The expected busy period of the server for repair under short-circuit failure due to SH temp. in (0,t] is

(t) =  So that                                                                                                                                 (30)

The expected number of visits by the repairman for repairing the different 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)] + Q₁₈⁽³⁾(t)[s]H₈(t)+ Q₁₁⁽⁴⁾(t)] [s]H₇(t) ,

 H₂(t) = Q₂₀(t)[s]H₀(t) + Q₂₈⁽⁵⁾(t)[s]H₈(t) +Q₂₇⁽⁶⁾(t)] [c]H₇(t)

 H₇(t) = Q₇₀(t)[s]H₀(t)] + Q₇₁⁽⁴⁾(t)[s]H₁(t)+ Q₇₈⁽³⁾(t)] [s]H₈(t) ,

 H₈(t) = Q₈₀(t)[s]H₀(t) + Q₈₂⁽⁵⁾(t)[s]H₂(t) +Q₈₇⁽⁶⁾(t)] [c]H₇(t)                                                                                                              (31-35)

Taking Laplace Transform of eq. (31-35) and solving for  

= N₅(s) / D₃(s) where                                                                                                                                                              (36)

In the long run, H₀ = N₅(0) / D₃^’(0)                                                                                                                                             (37)

COST BENEFIT ANALYSIS

The cost-benefit function of the system considering mean up-time, expected busy period of the system under SH temp. when the units stops automatically, expected busy period of the server for repair of unit under open failure mode due to fog and human error and expected busy period of the server for repair of unit under short-circuit failure mode due to SH temp. expected number of visits by the repairman for unit failure.

The expected total cost-benefit incurred in (0,t] is

C (t) = Expected total revenue in (0,t] - expected total repair cost in (0,t]

due to open failure mode due to fog and human error for repairing in (0,t].

• expected total repair cost due SCF mode due to SH temp. for repairing the units 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 =  =

 = 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 due to open failure mode due to fog and human error of type- I

K₃: cost per unit time for which the system is under repair due to SCF failure due to SH temp. 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, SH temp. Rate increases, the MTSF and steady state availability decreases and the cost 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. 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.