Performance Analysis Of A System Having One Main Unit And Two Supporting Units

[Volume 1 issue 7 Aug 2013]

Page No.174-188 ISSN :2320-7167

INTERNATIONAL JOURNAL OF MATHEMATICS AND COMPUTER RESEARCH

IJMCR www.ijmcr.in| 1:7 |August|2013|174-188 | 174

Performance Analysis Of A System Having One Main Unit And Two Supporting Units

V K Pathak¹ , Kamal Mehta² and Seema Sahu³ 1. Govt. PG College, Dhamtari (CG)

Sagar Institute of Technology, Bhopal Govt. PG College, Dhamtari (CG)

Abstract

Reliability for functioning units is essential, be it on the parts of memory chips of computers, software or hardware or it can be on the part of heavy machinery in any industrial system. Reliability analysis can give excellent results to improve the maintainability and portability of management design for existing and future product. Extensive reviews of two-component repairable system models have been presented by Lie et al(1977) and Yearout et al.

(1986). In all these models, it is assumed that failure times and repair times of the components are independent.

In this paper, Reliability analysis of a system having one main unit and two supporting units is proposed, assuming that the system fails whenever the main unit fails and system shuts down whenever either both the supporting units fail or main unit and one of the supporting units fail. To improve the reliability of the system, concept of preventive maintenance is also added. Using regenerative point technique various system parameters such as Transition Probabilities, Mean sojourn times, Mean time to system failure, Availability, Busy period of repairman in repairing the failed units etc. are calculated. At last profit analysis is also done. In this paper, failure time distributions are taken to be negative exponential whereas the repair time distributions are arbitrary.

Key words : mean sojourn times, mtsf, availability, busy period.

Introduction : In recent years many reliability models have been studied and evaluated. Engineers and Managers of Industries continuously make some modifications in the configuration and assumptions in the existing model in order to get the estimation of the various parameters such as mean time to system failure, steady state availability, busy period analysis and expected profit etc. which are responsible for making predictions about the production in their industries. In this competitive world of manufacturing there is immense pressure of the manufacturers to improve the quality of their product, to make them foolproof and modify the product frequently, if needed.

Gopalan et [6] al have before this, carried out cost benefit analysis of single server n-unit imperfect switch system with delayed repair. Switching devices play an important role in the cold standby systems. Rander et al [5] have studied the idea of major and minor failures. Singh et al[7] have performed cost benefit analysis of a two unit warm standby system with inspection, repair and post repair. In this particular paper, we have considered one main unit with two supporting units which act like helping partner in the proper functioning of the whole system with the

IJMCR www.ijmcr.in| 1:7 |August|2013|174-188 | 175 assumption that the whole system stops functioning when the main unit fails. The system also fails when either both the supporting units or main unit and one of the supporting unit fails. It is also proposed that after a random period of time the whole system goes for preventive maintenance. This approach can significantly improve the reliability of the working unit with limited overheads. Another way of improving reliability can be adopting an additional main unit in standby mode. But ,the idea here is to propose a system which can reliable and cost less at the same time.

S

₆

S

₃







B(t) a(t)

 g

₃

(t)

S

0

S

1

S

2

 

g

₂

(t)

g

1

(t) ^

 S

₄

  

S

₅

P.M.

^M

0

W

0

S

r

M

₀

W

0

S

0

M

₀

W

r

S

0

M

_g

W

g

S

_g

M

r

W

g

S

_g

S.D

IJMCR www.ijmcr.in| 1:7 |August|2013|174-188 | 176

STATE TRANSITION DIAGRAM

1. Assumptions :

1. The system consists of three units namely, One main unit and two associate units.

2. Any unit can fail when it is put to work.

3. After repair, a unit works as a new.

4. Switching devices are perfect and instantaneous.

5. If main unit fails, the system goes down.

6. If main and any of the associate units fails, the system shuts down.

7. If both the associate units fail, the system again shuts down.

8. There is a single repairman , who repairs the failed unit on the priority basis.

9. The failure time distributions of all the units are taken to be negative exponential whereas the repair time distributions are arbitrary.

2. Symbols and Notations: M is main unit and W and S are Associate units

E0

=

State of the system at epoch t=0

E=

set of regenerative states S0 S6

 =

Job arrival Rate

 ) (t

q_ij Probability density function of transition time from S_itoS_j

 ) (t

Q_ij Cumulative distribution function of time to transition time from S_itoS_j

 )

i(t



Cumulative distribution function of time to system failure when starting from E₀ S_iE state

 )

i(t



Mean Sojourn time in the state E₀ S_i E

 ) (t

B_i Repairman is busy in the repair at time t /E₀ S_iE

3 

2 1/r /r

r Constant repair rate of Main Unit / Associate units respectively.

 / / /

 

 =

Failure rate of Main Unit / Associate units respectively.

IJMCR www.ijmcr.in| 1:7 |August|2013|174-188 | 177 )

( / ) ( / )

( _{2 3}

1 t g t g t

g

=

Probability density function of repair time of Main Unit / Associate units respectively.

) ( / ) ( / )

( _{2 3}

1 t G t G t

G

=

Cumulative distribution function of repair time of Main Unit / Associate units respectively.

a(t) =

Probability density function of preventive maintenance .

b(t) =

Probability density function of preventive maintenance completion time.

A(t) =

Cumulative distribution function of preventive maintenance.

B(t) =

Cumulative distribution function of preventive maintenance completion time.

=

Symbol for Laplace -stieltjes transform

=

Symbol for Laplace-convolution

M

_o

/M

_g

/M

_r

=

Main unit under operation / good and non –operative mode / repair state

W

_o

/W

_g

/W

_r

=

Associative unit under operation / good and non –operative mode / repair state

S

_o

/S

_g

/S

_r

=

Associative unit under operation / good and non –operative mode / repair state

P.M =

System under preventive maintenance

S.D =

System under shutdown

Up states -

S

0

= (M

g

, W

g

, S

g

) ; S

1

= (M

o

, W

o

, S

o

) ; S

2

= (M

o

, W

r

, S

o

); S

0

= (M

o

, W

o

, S

r

) Down states -

S

₄

= (M

_r

, W

_g

, S

_g

); S

₅

= (S.D.) ; S

₆

= (P.M.)

3.

Transition Probability : -

Using markovian regenerative process , Simple probabilistic considerations yields the following non zero transition probabilities -

1 ^

^



^

^

0

01

e dt

p 

^t

; 2. ( ) [ 1

^*

( )]

0

12

a x

dt X t A e

p  

^

^

^Xt

 ^ 

s

c

IJMCR www.ijmcr.in| 1:7 |August|2013|174-188 | 178

3. ( ) [ 1

^*

( )]

0

13

a x

dt X t A e

p  

^

^

^Xt

 ^  ^{; 4. ( ) [ 1}

^*

^{( )]}

0

14

a x

dt X t A e

p  

^

^

^t

 ^ 

5. ( )

^*

( )

0

16

e a t dt a x

p  

^{ Xt}

 ^{; 6. ( )}

^*2

^{( )}

0 2

21

e g t dt g Z

p  

^{ Zt}



7. ( ) [ 1

*2

( )]

0

2

25

Z e G t dt g Z

p 

^



^Zt

  ^{; 8. ( )}

^*3

^{( )}

0 3

31

e g t dt g Y

p 

^



^Yt



9. ( ) [ 1

*3

( )]

0

3

35

Y e G t dt g Y

p  

^{ Yt}

  ^{; 10. p } 

^

^{g t dt}

0 1

40

( )

11. p ^ 

^

e

^{ t}

dt

0 50





; 12. p ^ 

^

b t dt

0

60

( ) Eq. 1- 12

where

X 











;Y 







;Z 







5. Mean Sojourn Time & Mean Time to System Failure:

Let 𝜇

_𝑖

in the state 𝑆

_𝑖

be defined as time that system continuous to be in state 𝑆

_𝑖

before transiting to any other states. If T denotes the Sojourn time in state 𝑆

_𝑖

, then

𝜇

_𝑖

= 𝐸 𝑡 = ∅ 𝑇 < 𝑡 𝑑𝑡

∞

0

Using above relation we can obtain the following equation-

 



^

¹

0

0

 

^

^e

^{ t}

^dt 

;

1 [ 1 ( )]

)

(

₂^*

0 2

*

2

g Z

dt Z t G

e

^Zt

 

 

^



)]

( 1 1 [ )

(

₃^*

0 3

*

3

g Y

dt Y t G

e

^Yt

 

 

^



;

( ) 1

0 1 4

 

^

^{G t dt} 



;

 



^



^

¹

0

5

e

^t

dt



( ) 1

0

4

 

^

^{B t dt} 

 Eq. 13- 18

Time to system failure can be regarded as the first passage time to the failed state. To obtain it we regarded to down state as absorbing states. Using argument as for the regenerative process, we obtain the following recursive relation for 𝜋

_𝑖

(𝑡)

as follows:

( ) ( ) ( ) ( ) ( )

~

0

t t

Q u dQ u t

t

_{ij i}

t

ij i

i

 

    

) ( )

( )

( _{01 1}

~

0 t Q t



t





S

S

IJMCR www.ijmcr.in| 1:7 |August|2013|174-188 | 179 )

( ) ( ) ( )

( ) ( )

( )

( ₁₆

~ 14

~ 13 3

~ 12 2

~

1 t Q t



t Q t



t Q t Q t



) ( ) ( )

( )

( ₂₅

~ 1 21

~

2 t Q t



t Q t



) ( ) ( )

( )

( ₃₅

~ 1 31

~

3 t Q t



t Q t

 Eq. 19- 22

In matrix Form





















 



























































35

~ 25

~

16

~ 14

~

3

~ 2

~ 1

~ 0

~

31

~ 21

~

13

~ 12

~ 01

~ 0

1 0

0

0 1

0

1 0

0 0

1

Q Q

Q Q

Q Q

Q Q

Q









Then, we calculate the value of

D

₁

( s )

,

N

₁

( s )

,

D

₁

( 0 )

and

N

₁

( 0 )

as follows :

31

~ 13

~ 21

~ 12

~

1(s) 1 Q Q Q Q

D   

31 13 21 12

1(0) 1 p p p p

D   

Eq. 23

] [

)

( ₃₅

~ 13

~ 25

~ 12

~ 16

~ 14

~ 01

~

1 s Q Q Q Q Q Q Q

N    

35 13 25 12 16 14

1(0) p p p p p p

N    

Eq. 24

To Calculate the MTSF , we use the following formula :

31 13 21 12

13 3 12 2 1 35 13 25 12 16 14 0

1 ' 1 '

1

1

) (

) 0 (

) 0 ( ) 0 (

p p p p

P P

P P P P P MTSF P

D N MTSF D















 

 







 Eq. 25

6. Availability Analysis:

Let 𝑀

_𝑖

𝑡 denote the probability that the system is up initially in regenerative state 𝑆

_𝑖

at epoch t without passing through any other regenerative state. It might return to itself through one or more non regenerative states so that either it continues to remain in regenerative state without visiting any regenerative state including itself by probability arguments.

We observe that the entry to any of the state 𝑆

₀

, 𝑆

₁

, 𝑆

₂

𝑎𝑛𝑑 𝑆

₃

is a regenerative point. 𝐴

_𝑖

𝑡 is defined as the probability that the system is up in state 𝑆

₀

, 𝑆

₁

, 𝑆

₂

𝑎𝑛𝑑 𝑆

₃

at epoch it.

To obtain it consider all possible consequences.

S S

S S

IJMCR www.ijmcr.in| 1:7 |August|2013|174-188 | 180

1. Probability that the system initially up is 𝑆

₀

is up at epoch t without transiting to any other

regenerative state in E which is 𝑀

₀

𝑡 .

2. Probability that the system transits to 𝑆

_𝑖

in E during (u,u+du) and then starting from 𝑆

₀

it is up at epoch t which is

) ( )

( ) (

0

0 0

0

q t A t u du q A t

A

_i

t

i i



i

^{ }

 Thus we have

) ( )

( ) ( )

( _{0 01 1}

0 t M t q t A t

A  

) ( )

( ) ( )

( ) ( )

( ) ( )

( ) ( )

( _{1 12 2 13 3 14 4 16 5}

1 t M t q t A t q t A t q t A t q t A t

A     

) ( )

( ) ( )

( ) ( )

( _{2 21 1 25 5}

2 t M t q t A t q t A t

A   

) ( )

( ) ( )

( ) ( )

( _{3 31 1 35 5}

3 t M t q t A t q t A t

A   

) ( )

( )

( _{40 0}

4 t q t A t

A 

) ( )

( )

( _{50 0}

5 t q t A t

A 

) ( )

( )

( _{60 0}

6 t q t A t

A 

Eq. 25- 32

Taking in the matrix form of above equation



















































































































































6

* 5

* 4

* 3

* 2

* 1

* 0

*

6

* 5

* 4

* 3

* 2

* 1

* 0

*

1 0

0 0

0 60 0

*

0 1

0 0

0 50 0

*

0 0

1 0

0 40 0

*

35 0 0 *

1 31 0

0 *

25 0 0 *

0 21 1

0 *

16 0 *

14

* 13

* 12

1 * 0

0 0

0 0

01 0 1 *

M M M M M M M

A A A A A A A

q q q

q q

q q

q q

q q

q

C

C

C

C C C C

C C

C

C

C

C

IJMCR www.ijmcr.in| 1:7 |August|2013|174-188 | 181 Then we calculate the values of

D

₂

( s )

,

N

₂

( s )

,

D

₂

( 0 )

and

N

₂

( 0 )

34 )

1 ( ) 0 (

* 3 *

* * 2 *

* * 1 ) *

*

*

* 1 *

0 ( ) * (

33 ]

) (

[ ) 1

( ) 0 (

*] ) *

*

*

* ( *

*

* [ *

) *

*

*

* 1 *

( ) 2(

13 3 12 2 1 31 13 21 12 0

2

13 01 12

01 01

31 13 21 12 2

16 35 13 25 12 14 35

13 25 12 2

60 16 35

13 25 12 50 40 14 01 35

13 25 12

Eq p

p p

p p p N

q M q

q M q

M q q

q q M q

s N

Eq p

p p p p p p

p p p D

q q q

q q

q q q q q q

q q

s q D









    











































To obtain the value of

35 Eq.

) 1

)(

( ) 0 ' (

13 3 12 2 1 31 13 21 12

2 5

p p p p p p

D  

o

         

Thus the steady state Availability of the system

36 ) Eq.

1 )(

(

) 1

) ( (

) 0 '(

) 0 ) (

*( lim 0

) (

13 3 12 2 1 31 13 21 12 5

13 3 12 2 1 31 13 21 12 0

0

2 2 0 0

p p

p p p p

p p

p p p A p

D s N s A

A

o

   































 



 





7. Busy Period Analysis :

(a) Busy period repairman for performing Normal repair :

Let 𝑊_𝑖 𝑡 denote the probability that the repairman is busy initially with repair in regenerative state S4

and remains busy at epoch t without transiting to any other state or returning to itself through one or more regenerative state. By probabilistic argument, we have

) ( ) (t G t W_i  _i

Developing Similarly relationship as in availability for normal repair, we have

43 - 37 Eq.

) ( )

( ) (

) ( )

( ) (

) ( )

( ) ( ) (

) ( )

( ) ( )

( ) ( ) (

) ( )

( ) ( )

( ) ( ) (

) ( )

( ) ( )

( ) ( )

( ) ( )

( ) (

) ( )

( ) (

1 0 60

' 6

1 0 50

' 5

1 0 40

4 1

4

1 5 35

1 1 31

3 1

3

1 5 25

1 1 21

2 1

2

1 6 16

1 4 14

1 3 13

1 2 12

1 1

1 1 01

' 0

t B t

q t B

t B t

q t B

t B t

q t W t B

t B t

q t B t

q t W t B

t B t

q t B t

q t W t B

t B t

q t B t

q t B t

q t B t

q t B

t B t

q t B





























 _C

C C

C C

C

C

C

C

C

C C

IJMCR www.ijmcr.in| 1:7 |August|2013|174-188 | 182 In Matrix form, we have :











































































































































0 0

4

* 3

* 2

* 0 0

6 1

* 5

1

* 4

1

* 3

1

* 2

1

* 1

1

* 0

1

*

1 0

0 0

0 60 0

*

0 1

0 0

0 50 0

*

0 0

1 0

0 40 0

*

35 0 0 *

1 31 0

0 *

25 0 0 *

0 21 1

0 *

16 0 *

14

* 13

* 12

1 * 0

0 0

0 0

01 0 1 *

W W W

B B B B B B B

q q q

q q

q q

q q

q q

q

44 Eq.

) 0 (

*) ( *

4 ) *

* ( *

3 ) *

* ( *

2 ) * (

14 4 13 3 12 2 3

01 14 13

12 01 01 3

p p

p N

q W q

q W q

q W q

s N







 









To find the steady state the fraction of time for which the repairman of busy with repair , we first calculate

4

* 4 3

* 3 2

*

2

  ; W   ; W  

W Eq. 44-47

Therefore in long run the fraction of fine for the repairman in busy with the normal repair is given by-

) 0 '(

) 0 ) (

1( 0 * lim ) (

2 3 0

1 0

D t N s B

B    

13 3 12 2 1 31 13 21 12 5

14 4 13 3 12 1 2

0 ( ) ( )(1 )

p p

p p p p

p p

B p

o

   























 



Eq. 48

(b) Busy period repairman performing for Shutdown repair

Developing similar relationships as in availability for shutdown repair, we have

)

( )

( )

(

_{01 1}²

2

0

t q t B t

B 

) ( )

( ) ( )

( ) ( )

( ) ( )

( )

(

_{12 2}² _{13 3}² _{14 4}² _{16 6}²

2

1

t q t B t q t B t q t B t q t B t

B    

) ( )

( ) ( )

( )

(

_{21 1}² _{25 5}²

2

2

t q t B t q t B t

B  

55 - 49 Eq.

) ( )

( ) (

) ( )

( )

(

) ( )

( ) (

) ( )

( ) ( )

( ) (

2 0 60

2 6

2 0 50

5 2

5

2 0 40

2 4

2 5 35

2 1 31

2 3

t B t

q t B

t B t

q W t B

t B t

q t B

t B t

q t B t

q t B













C C C

C C

C

C C

C C

C C

IJMCR www.ijmcr.in| 1:7 |August|2013|174-188 | 183 Taking Matrix form











































































































































0 5

* 0 0 0 0 0

6 2

* 5

2

* 4

2

* 3

2

* 2

2

* 1

2

* 0

2

*

1 0

0 0

0 60 0

*

0 1

0 0

0 50 0

*

0 0

1 0

0 40 0

*

35 0 0 *

1 31 0

0 *

25 0 0 *

0 21 1

0 *

16 0 *

14

* 13

* 12

1 * 0

0 0

0 0

01 0 1 *

W

B B B B B B B

q q q

q q

q q

q q

q q

q

*)]

*

* ( *

[ * 5 ) * ( , that

So N₄ s W q₀₁ q₁₂ q₂₅ q₁₃ q₃₅

To find the steady state the fraction of time for which the repairman of busy with repair , we first calculate 56 Eq.

57 Eq.

) (

) 0

( _{5 12 25 13 35}

4 p p p p

N 





Therefore in long run the fraction of fine for the repairman in busy with the shutdown repair is given by

58 ) Eq.

1 )(

(

) ) (

(

) 0 '(

) 0 ) (

2(

* lim 0

) 2(

* lim 0

) (

13 3 12 2 1 31 13 21 12 5

35 13 25 12 2 5

0

2 4 0

0 2

0

p p

p p p p

p p p B p

D s N sB

t s s B

B

o

   

















 



 

 





(c) Similarly to calculate the Busy Period of repairman performing the Preventive maintenance

16 01 5

* 6 *

) *

(s W q q

N 

To find the steady state the fraction of time for which the repairman of busy with repair , we first calculate

6

* 6

  W

5

* 5

 

W

IJMCR www.ijmcr.in| 1:7 |August|2013|174-188 | 184 N5(0)



6p16

Therefore, in long run ,for the fraction of time the repairman in busy with the preventive maintentance is given by

13 3 12 2 1 31 13 21 12 5

16 3 6

0

2 5 0

0 3

0

) 1

)(

) ( (

) 0 ' (

) 0 ) (

3 (

* lim 0

) 3 (

* lim 0

) (

p p

p p p p B p

D s N s B

t s s B

B

o

   















 



 

 





7. Particular Cases:

When all repair time distributation are n-phases Erlangian distributation . i.e.,

Density Function And Survial Function





^









 



¹

0 1

! ) ) (

(

! ; 1 ) ) (

(

n

j

t nr j i i

t nr n i i

i

j

e t t nr

n G e t nr t nr

g

i i

And other distributation are negative exponential

t t

t

t b t e A t e B t e

e t

a( )



^ , ( )



^ , ( ) ^ , ( ) ^

For n=1

2

2 21 16

14 13

12

60 50

40 01

;

;

;

;

1

; 1

; 1

; 1

r Z p r p X

p X p X

p X

p p

p p

 

 

 

 

 

























3 35

3 3 31 2

25 ; ;

r Y p Y r Y p r r Z p Z

 

 

 

We can see that

1

; 1

;

1 _{21 25 31 35}

16 14 13

12 p  p  p  p  p  p  p 

p and

IJMCR www.ijmcr.in| 1:7 |August|2013|174-188 | 185

 

 





 

 



; 1

; 1 1

; 1

; 1

; 1 1

6 5

1 4

3 3

2 2

1 0







 

 

 



r

r Y r

Z X

So, we Calculate ,

2 1 3 1 4

1 2 3 3

1 2 1

2 1 3 2 1 2 3

2 3 2

2 1

3 3

3 2

1

2 2 1 3 1

* 4 2 0 2

2 1 3 1

1

* 3 1 0

2 2 1 3 1

2 1 3

3 2 1 1

; ) (

; ) (

; 1) (1

1

; 1

;

; 1

; 1 1

)]

( ) [

( ) ;

) ( (

)

; ( ] [

L YL L

ZL r M

L L r M

M M

L L r L L r K

L L K

ZL K

r L Z

r L Y

L X Where

M K L K M B M

M K L K M

L B M

period Busy

M K L K M

K k L

ty Availabili K

K K MTSF L



 



 



























 





































 

 

 



 

 

 











 

 

8. Profit Analysis:

The profit analysis of the system can be carried out by considering the expected busy period of repairman in repair of the unit in [0,t] . Therefore,

G(t)= total revenue earned by the system in [0,t]- Expected repair cost in [0,t]







^{ }









t b

t b t

up

b b

up

dt t B t dt

t B t dt

t A t Where

C C

t C

0 0 21 2

0 0 1 1

0 0

2 3 1 2 1

) ( )

(

; ) ( )

(

; )

( )

( ) (













IJMCR www.ijmcr.in| 1:7 |August|2013|174-188 | 186 Figure 1.1

Figure 1.2 MTSF vs. failure rate

0 10 20 30 40 50 60

1 2 3 4 5 6 7

Failure Rare

MTSF

Availability vs. failure rate

0 20 40 60 80 100 120 140

1 2 3 4 5 6 7

Failure rate

Availability

IJMCR www.ijmcr.in| 1:7 |August|2013|174-188 | 187 Figure 1.3

Discussion :

A pioneering work in this direction involving component of two-unit system was initiated by Harris(1968). We have considered a two unit redundant system in which failure times of the component are taken to be exponential to derive mean time to system failure by using regenerative point technique for arbitrary repair time distribution.

9. Conclusion:

The mean time to system failure (MTSF) and availability of the system decreases rapidly with the increase of failure rates



&



for fixed values of other parameters. However, it is noted that values of profit decreases with the increase in the failure rate but increases as and when repair rates

r

₁

& r

₂ increase. With preventive maintenance the reliability of the system increases considerably.

10. References :

1. Chander S. & Bhardwaj R.K.( 2009). “Reliability and economic amalysis of 2-out of-3 redundant system with priority to repair.” African Journal of Mathematics and Computer Science Research, vol 2(11), pp 230-236.

2. Wang K.H. , Hsieh C.H. and Liou(2006). “Cost benefit analysis of series systems with cold standby components and repairable service station.” Quality Technology Quantitative Management , vol. 3, pp 77-92.

3. Yadavally V.S.S. & Bekker A. (2005). “Bayesian study of two component system with common cause shock failure.” Asia-Pacific Journal of Operation Research, vol 22(1), pp 105-119.

4. Rajamanickam Paul S. & Chandrasekar B.( 1997). “Reliability Measures for two unit systems with a dependent structure for failure and repair times.”, Microelectron Reliab., vol 37, no. 5, pp 829-833.

5. Rander M.C., Kumar Ashok & Tuteja R.K.( 1992). “ A two unit cold standby system with major and minor failures and preparation time in the case of major failure”, Microelectron Reliab., vol 32, no. 9, pp 1199- 1204.

Profit vs. failure rate

0 10 20 30 40 50 60 70

1 2 3 4 5 6 7

Failure Rate

Profit

IJMCR www.ijmcr.in| 1:7 |August|2013|174-188 | 188 6. Gopalan M.N. & Waghmare S.S.( 1992). “Cost benefit analysis of single server n-unit imperfect switch system

with delayed repair”, Microelectron Reliab., vol 32, no. 2, pp 59-63.

7. Singh S.K. , Goel S.K. & Gupta Rakesh(1986). “Cost bebefit analysis of a two unit warm standby system with inspection, repair and post repair”, IEEE transactions on Reliability, vol. 35, no.1, pp 49-53.