A TWO NON–IDENTICAL UNIT STANDBY COMPLEX SYSTEM

A TWO NON–IDENTICAL UNIT STANDBY COMPLEX SYSTEM

Pradeep Chaudhary

, Ruchi

1,2Department of Statistics, C.C.S.University, Meerut U.P., (India)

ABSTRACT

The present paper deals with a system composed of two non –identical units A and B. Unit-A consists of two non identical components A1 and A2 in series form, whereas unit -B consists of only one component, which is different from the components of unit-A . Initially the unit –A is operative and the unit-B is kept into cold standby. For the successful operation of the system either both the components A₁ and A₂ of unit-A or the unit-B is good. As soon as any component of unit-A fails, the unit-B kept into cold standby becomes operative instantaneously with the help of a perfect and instantaneous switching device. Each component of unit-A as well as unit-B has two modes - Normal (N) and total failure (F). A single repair man is available to repair each component of failed unit-A as well as failed unit-B and the priority in repair is given to unit-A over the unit-B.

The failure rate of each component of unit-A is taken constant with different parameters where as the failure rate of the unit-B is taken general. The distribution of time to repair of each component of unit-A and unit-B are also taken general with different CDF’s. A repaired unit always works as good as new. Using regenerative point technique, various important measures of system effectiveness have been obtained.

Keywords: Availability, Expected Busy Period Of Repair Man, Mean Time To System Failure (MTSF), Net Expected Profit, Reliability.

I. INTRODUCTION

Two unit priority standby system models have been widely studied because they are frequently used in modern business and industrial system. Various authors [1-9] have analyzed the single server two unit priority and ordinary standby system models with different set of assumptions. Multi component system models are analyzed by [9, 10] by using supplementary variable technique.

In present study a system model consisting of two units (A and B) is analyzed in respect of reliability and cost effectiveness measures. Unit-A consists of two non-identical components ( and ) in series configuration so that functioning of both the components of unit-A are necessary for the successful operation of unit-A. The unit- B has single component and it can be used in place of unit-A upon its failure. The unit-B is always used as cold standby i.e. the priority in operation is given to unit-A over the unit-B. A single repairman is always available with the system to repair each component of unit-A as well as the unit-B and the priority in repair is always give to the component of unit-A. The network representation of system model is as shown in Fig. 1.

The S and SD are known as the switching and sensing devices.

The system model is analyzed by using regenerative point technique to obtain the following measures of system effectiveness.

i. Transition probabilities and steady state transition probabilities ii. Mean sojourn time in regenerative states.

In Out

Fig. 1

iv. Point wise and steady state availabilities.

v. The expected busy period of repairman in time interval (0, t).

vi. Net expected profit earned by the system in time interval (0, t) and in steady state.

II. SYSTEM DESCRIPTION AND ASSUMPTIONS

1. The system comprises of two non-identical units A and B. Unit-A consist of two non-identical components and in series form, where as unit-B consists only one component, which is different from the components of unit-A.

2. Initially the unit-A is operative and the unit-B is kept into cold standby.

3. For the successful operation of the system either both the components and of unit-A or the unit-B are good. As soon as any component of unit-A fails, the unit-B kept into cold standby becomes operative instantaneously with the help of a perfect and instantaneous switching device.

4. Each component of unit-A as well as unit-B has two modes- Normal (N) and total failure (F).

5. A single repair man is available to repair each component of the failed unit-A as well as failed unit-B and the priority in repair is given to unit-A over the unit-B.

6. The failure rate of each component of unit-A is taken constant with different parameters where as the failure rate of the unit-B is taken general. The distribution of time to repair of each component of unit-A and unit-B are also taken general with different CDF’s.

7. A repaired unit always works as good as new.

III. NOTATIONS AND STATES OF THE SYSTEM

SD

: Constant failure rate of component . h( )/H( ) : Pdf/cdf of failure time of unit-B.

/ : Pdf/cdf of repair time of component . / : Pdf/cdf of repair time of component .

K( ) : Pdf/cdf of repair time of unit-B.

3.2 Symbols for the States of the System

E : Set of regenerative state

(i=1,2) : Component (i=1,2 ) is good and operative (i=1,2) : Component (i=1,2 ) is good

(i=1,2) : Component (i=1,2 ) is failed and under repair : Unit-B is good and operative /cold standby : Unit-B is failed and under repair

: Unit-B is failed and waiting for repair.

Using the above symbols in view of the assumption stated in section 2, the possible states of the system are as follows :

Up States Failed States

The transition diagram of the system model along with the failure rates & repair time CDF’s, is shown in Fig.2.

It is to be noted that the epochs of entrance from state S1 to S3 and S2 to S4 are non-regenerative.

IV. TRANSITION PROBABILITIES AND MEAN SOJOURN TIMES

Let ( ) , , ……. be the epochs at which the system enters to any state E. By simple probabilistic reasoning, the non-zero elements of Q= (t) may be obtained in the following manner :

4.1 Transition Probabilities

4.1.1One – step Transition Probabilities

To obtain the expressions for (t) (It is also called direct transition probability) we proceed as follows : (t)= Probability that the system transits from state

during time interval (0, t).

= Cdf of transition time from state during time

interval (0,t).

] In particular,

To obtain the expression for (t), we proceed as follows:

For the system to reach into the state to on or before time t, we suppose that the system not transited to up to time u transits to during (u, u+du); u t.

The probability of this event is du. Therefore,

Similarly,

: Up State : Failed State : Regenerative Point : Non-Regenerative Point

Fig. 2:Transition Diagram So

S1

S5 S3

S4 S2

1r 2g O

A , A B

1O 2O s

A , A B

1r 2g wr

A , A B

1O 2O r

A , A B

1g 2r O

A , A B

1g 2r wr

A , A B

1

2

1

2

 

H 

 

K 

 

H

 

G1 

2 

G 

 

G1 

 

G2 

4.1.2Two – step Transition Probabilities

(t) = Probability that the system transits from state to via during time interval (0, t). It is also called indirect transition probability. In particular, To obtain the expression for (t), we proceed as follows :

Suppose that the system transits from state to during (u, u+du); u t, the probability of this event is (u)dH(u). Further suppose that the system passes from to during the interval (v, v+dv) in (u, t).The probability of this event is .

Thus,

= ; on change of limits of integration.

= Similarly,

4.2 Steady state Transition Probabilities

The steady-state transition probabilities can be obtained by using the formulae

(t) and =

Thus,

, . † ;

; ;

)

It can be easily verified that

(1-5)

4.3 Mean Sojourn Time

Mean sojourn time in state is defined as the expected time taken by the system in state before transition to any other state. To calculate the mean sojourn time in state , we observe that so long as the system is in state , there is no transition to state . Hence, if denote the sojourn time in state , then

Similarly

= (mean repair time of component ) = (mean repair time of component )

(6-11)

† The limits of integration are not mentioned whenever they are 0 to ∞.

4.4 Conditional Mean Sojourn Time

To obtain the conditional mean sojourn time in state , given that the system is to transits to state , we define = Conditional mean sojourn time in state / the system transit to state

= =

= (s)⃒s=0

= (s)⃒s=0

In particular,

It can be easily verified that

(12-15)

V. RELIABILITY OF THE SYSTEM AND MEAN TIME TO SYSTEM FAILURE (MTSF)

Let the random variable denotes the time to system failure (TSF) when the system starts functioning from state E. Then the system reliability is given by

= P

To determine the reliability of the system, we regard the failed states of the system as absorbing states. By using simple probabilistic arguments, we observe that is the sum of following mutually exclusive contingencies:

(i).The system remains up in state up to the time t without any transition fro . The probability of this event is

, say

(ii).The system transits from to during time (u, u+du); u t and then starting from state the system should survive during the remaining time (t-u).The probability of this event is

(iii).The system transits from to during time (u, u+du); u t and then starting from state , the system should survive during the remaining time (t-u). The probability of this event is

Therefore,

Similarly,

(16-18) Where,

Taking Laplace Transform of the above set of equations (16-18) and writing the resulting set of algebraic equation in the matrix form as follows (omitting the argument‘s’ for brevity),

=

On solving the above matrix equations for (s), we get

(s)=

* * * * *

0 01 1 02 2

* * * *

01 10 02 20

Z + q Z + q Z 1- q q - q q

(19)

Taking the Inverse Laplace Transform of (19), we can get reliability of the system for known values of the parameters.

The mean time to system failure (MTSF) is given by

Using the result

and

We get

^{0 01 1 02 2}

01 10 02 20

ψ + p ψ + p ψ 1- p p - p p



VI. AVAILABILITY ANALYSIS 6.1 For unit A

Let (t) be the probability that the system is up at epoch t due to the operation of unit-A, when initially the system starts operation from state ЄE. By using simple probabilistic arguments, we observe that is the sum of following contingencies :

(i).The system remains up in state up to time t without making any transition from .The probability of this event

is , say

(ii).The system transits from state to during time (u, u+du); u t and then starting from state ,it is observed to be up at epoch t, the probability of this event is

(iii) The system transits from state to during time (u, u+du); u t and then starting from state he probability of this event is,

Therefore,

Similarly,

(21-26) Where,

Taking Laplace Transform of the above equations (21-26) and writing the solution of resulting set of algebraic equations in the matrix form as follows (omitting the argument‘s’ for brevity) :

On solving for ) by Cramers rule, we have

(27) Where,

And

Taking the Inverse Laplace Transform of (27), we can get availability of the system due to the working of unit- A for known values of the parameters.

In the long run, the steady state probability that the system will be operative due to operation of unit-A is given by

= ) = ) =

Now,

Therefore, by L Hospital rule, the steady state availability of the system due to operation of unit-A is given by

(28)

Where,

And

The expected up time of the system, due to unit-A is given by

(29)

So that

(30)

6.2 For unit B

Let (t) be the probability that the system is up at epoch t due to the operation of unit-B, when initially the system starts operation from state ЄE. By using simple probabilistic arguments, we observe that is the sum of following contingencies :

(i).The system transition from during time (u, u+du); u t and then starting from state it is observed to be up at epach t due to unit-B, The probability of this event is

(ii).The system transits from state to during time (u, u+du); u t and then starting from state , system is observed to be up at epoch t due to unit-B, the probability of this event is

Therefore,

Similarly,

(31-36) Where,

,

Taking Laplace Transform of the above relations (31-36) and writing the solution of resulting set of algebraic equation in the matrix form as follows (omitting the argument‘s’ for brevity) :

=

On solving for ), we have

(37)

and D₁(s) is same as defined above in section 6.1

Taking the Inverse Laplace Transform of (37), we can get availability of the system due to the operation of unit- B for known values of the parameters.

In the long-run, the steady state probability that the system will be operative due to operation of unit- B is given by

= ) = ) =

As

= 0

Therefore, by L Hospital rule, the steady state availability due to operation of unit-B is given by

(38)

Now to obtain and , we observe

And (0) = Thus,

And is same as defined in section 6.1.

The expected up time of the system, due to unit-B is given by

(39) So that,

(40)

VII. BUSY PERIOD ANALYSIS 7.1 For Repair of Component

Let (t) be the probability that the repair facility is busy in the repair of a failed component at epoch t, when initially the system starts operation from state ЄE. Using elementary probabilistic arguments, we observe that is the sum of the following contingencies:

(i).The system transits from during time (u, u+du); u t and then starting from state , it is observed that the component will be under repair at epoch t. The probability of this event is,

(ii).The system transits from state to during time (u, u+du); u t and then starting from state ,it is observed that the component will be under repair at epoch t, the probability of this event is

Therefore,

Similarly,

(41-46) Taking Laplace Transform of the above relations (41-46) and solving the resulting set of algebraic equation for

we get

(47) Where,

and is defined in previous section.

Taking the Inverse Laplace Transform of (47), we can get the probability that the repair man is busy in the repair of component A₁ , the steady-state probability that the repairman will be busy in the repair of component A₁ is given by

As, =0, therefore by L Hospital rule, the steady state probability that the repairman will be busy in the repair of component A₁ is given by

(48) Where,

1

is same as defined in section 6.1.

The expected busy period for the repair facility in repair of component A_1, during (0,t) is given by

du (49)

So that

(50)

7.2 For Repair of Component

Let (t) be the probability that the repair facility is busy in the repair of a failed component at epoch t,

(i).The system transits from during time (u, u+du); u t and then starting from state , it is observed that the component will be under repair at epoch t, The probability of this event is

(ii).The system transits from state to during time (u, u+du); u t and then starting from state ,it is observed that the component will be under repair at epoch t, the probability of this event is

Therefore,

Similarly,

(56) Taking Laplace Transform of the above (51-56) equations and solving the resulting set of algebraic equation for

, we get

(57) Where,

Taking Inverse Laplace Transform of (57), we can get the probability that the repairman will be busy at epoch t for known values of the parameters. The steady state probability that the repairman will be busy in repair of component given by

As, =0, therefore by L Hospital rule, the steady state probability that the repairman will be busy in the repair of component is given by

(58) Where,

The expected busy period for the repair facility in the repair of component , during (0,t) is given by

(59) So that

(60)

7.3 For Repair of Unit B

Let (t) be the probability that the repair facility is busy in the repair of a failed unit B at epoch t, when initially the system starts operation from state ЄE. Using elementary probabilistic arguments, we observe that

is the sum of the following contingencies:

(i).The system transits from during time (u, u+du); u t and then starting from state , it is observed that the unit B will be under at epoch t, the probability of this event is,

(ii).The system transits from state to during time (u, u+du); u t and then starting from state ,it is observed

that the unit-B will be under repair at epoch t, the probability of this event is

Therefore,

Similarly,

(61-66) Where,

Taking Laplace Transform of the above equations (61-66) and solving the resulting set of algebraic equation for we get

(67) Where,

Taking the Inverse Laplace Transform of (67), we can get the probability that the repairman is busy at epoch t for known values of the parameters. The steady state probability that the repair man is busy in repair of unit-B is given by

As, , therefore by L Hospital rule, we have

The expected busy period for the repair facility in repair of unit-B during (0,t) is given by

(69) So that

(70)

VII. COST BENEFIT ANALYSES

Let denote per unit up time revenues by the system due to the operation of unit-A and unit-B respectively. are the repair costs per unit of time when component component and unit- B respectively are under repair.

Then the net expected profit incurred by the system during time interval (0,t) is defined as

The net expected profit incurred by the system per unit of time in steady state is given by

REFERENCES

[1] L. R. Goel, Rakesh Gupta, and S. K. Singh, Cost analysis of a two unit priority standby system with imperfect switch and arbitrary distributions, Microelectron Reliab., 25(1), 1985, 65-69.

[2] L. R. Goel, Rakesh Gupta and S. K. Singh, Cost analysis of a two unit cold standby system with two types of operation and repair, Microelectron Reliab., 25(1), 1985, 71-75.

[3] L. R. Goel, Rakesh Gupta and S. K. Singh, Cost analysis of a two unit standby system with delayed replacement and better utilization of units, Microelectron Reliab., 25(1), 1985, 81-86.

[4] L. R. Goel, Rakesh Gupta and S. K. Singh, Analysis of two units cold standby system with three model, Microelectron Reliab., 23(6), 1983, 1041-1044.

[5] L. R.Goel, Rakesh Gupta and S. K. Singh, Profit analysis of a cold standby system with two repair distributions, Microelectron Reliab., 25(3), 1985, 467-472

[6] L. R. Goel, Rakesh Gupta, Analysis of a two unit standby system with three modes and imperfect

[7] L. R. Goel, Rakesh Gupta, Availability analysis of two unit cold standby system with two switching failure modes, Microelectron Reliab., 24(3), 1984, 419-423.

[8] L. R. Goel, S. G. Sharma, and Rakesh Gupta, Cost analysis of a two unit cold standby system under different weather conditions, Microelectron Reliab., 25(4),1985, 655-659.

[9] Rakesh Gupta and Pawan Kumar,Analysis of two unit series subsystem with one standby system model, Gujarat Statistical Review, 29(1), 2002,87-100.

[10] Rakesh Gupta, C. P. Bajaj, and S. M. Sinha, A single server multi component two unit cold standby system with inspection and imperfect switching device, Microelectron Reliab., 26(5), 1986, 873-877.