A Two Identical Unit Cold Standby System Subject To Two Types Of Failures Текст научной статьи по специальности «Медицинские технологии»

A Two Identical Unit Cold Standby System Subject To Two

Types Of Failures

Pradeep Chaudhary*, Rashmi Tomar Department of Statistics

Ch. Charan Singh University, Meerut-250004(India) *E-mail ID:pc25jan@gmail.com

Abstract

The paper deals with a system model composed of a two identical unit standby system in which initially one is operative and other is kept as cold standby. Each unit of the system has two possible modes - Normal (N) and Total Failure (F). An operating unit may fail either due to normal or due to chance causes. A single repairman is always available with the system to repair a unit failed due to any of the above causes. The system failure occurs when both the units are in total failure mode. The failure time distributions of a unit failed due to both the causes are taken as exponentials with different parameters whereas the repair time distributions of a failed unit in both types of failure are taken as general with different CDFs. Using regenerative point technique, the various important measures of system effectiveness have been obtained:

Keywords: Reliability, Mean time to system failure, availability, expected busy period of repairman, net expected profit.

1. Introduction

The two unit cold standby systems have been widely studied in the literature of reliability as they are frequently used in modern business and industries. It is obvious that the standby unit is switched to operate when the operating unit fails and the switching device which is used to put the standby unit into operation may be perfect or imperfect at the time of need. In past years various authors including [1,2,4,5,6,9,10,11] analyzed the two identical and non-identical units standby redundant system models with different sets of assumptions such as imperfect switching device, slow switching device, waiting time distribution of repairman, repair machine failure etc. They have analyzed the two identical and nonidentical unit system models by taking the single failure mode of an operating unit i.e. due to normal (ageing effect).

In many realistic situations, the systems are subject to two types of failure .One occurs by a normal cause and the other due to chance cause such as (i) abnormal environmental condition i.e. temperature, pressure, vibration etc. (ii) defective design (iii) misunderstanding the process variables (iv) operator's negligence and mishandling of the system etc. Keeping this fact in view few authors [3,7,8] analyzed the system models assuming two failure modes of each unit.

P. Chaudhary, R. Tomar RT&A, No 1 (52)

A TWO IDENTICAL UNIT COLD STANDBY SYSTEM SUBJECT TO TWO Volume 14, March 2019

TYPES OF FAILURES_____________________________________________________________________

The purpose of the present paper is to deal with a stochastic model of a two identical unit cold standby redundant system subject to two types of failure in each of the operating unit. By using regenerative point technique, the following important measures of system effectiveness are obtained.

i. Transient-state and steady-state transition probabilities.

ii. Mean sojourn time in various regenerative states.

iii. Reliability and mean time to system failure (MTSF).

iv. Point-wise and steady-state availabilities of the system as well as expected up time of the system during time interval (0, t).

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.

2. System Description and Assumptions

1. The system consists of two identical units. Initially, one unit is operative and other is kept as cold standby.

2. Each unit of the system has two possible modes: Normal (N) and Total Failure (F).

3. The switching device used to put the standby unit into operation is always perfect and instantaneous.

4. An operative unit may fail either due to normal cause i.e. due to ageing effect or due to chance cause.

5. The system failure occurs when both the units are in total failure mode.

6. A single repairman is always available at the system to repair a unit failed due to normal cause or due to chance cause.

7. The failure time distributions of the units to reach into the failure mode either due to normal or due to chance cause are taken as exponential whereas the repair time distributions of a unit failed due to both causes are taken as general with different CDF's.

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

3. Notations and States of the System

We define the following symbols for generating the various states of the system-

No,Ns : Unit is in N-mode and operative/standby

Flr, F2r : Unit is in failure mode due to normal cause/due to

chance cause and under repair.

F1w , F2W : Unit is in failure mode due to normal cause/due to

chance cause and waiting for repair.

Considering the above symbols in view of assumptions stated in section-2, the possible states of the system are shown in the transition diagram represented by Fig. 1. It is to be noted that the epochs of transitions into the state S4 from S1, S3 from S1, S5 from S2, S6 from S2 are non-regenerative, whereas all the other entrance epochs into the states of the system are regenerative. The states S0, S1 and S2 are the up-states of the system and the

states S3, S4, S5 and S6 are the failed states of the system.

TRANSITION DIAGRAM

S

5

S2

S

6

s

Up State

□ Failed State

Ф Regenerative Point

Fig. 1

X

Non-Regenerative Point

The other notations used are defined as follows:

E : Set of regenerative states = {S0, S3,S2 }

E : Set of non-regenerative states = {S3 ,S4,S5,S6 }

ttj, ^2 : Constant failure rate of an operative unit due to normal cause/chance cause

Gj (•), G2 (•) : CDF of repair time of failed unit due to normal cause/chance cause.

q;j (•) : p.d.f of transition time from regenerative state S; to Sj.

q(k) (•): p.d.f of transition time from regenerative state S; to Sj via non-regenerative state Sk

Pij (•) : One1 step steady-state transition probability from regenerative state S; to Sj = J qij(u)du.

pj >(•) : Two step steady-state transition probability from regenerative state S; to Sj via

non-regenerative state Sk = J q(k) (u)du .

nl3 n2 : Mean repair times of operative unit and standby unit

1 The limits of integration are taken to be 0 to ^ whenever they are not mentioned

P. Chaudhary, R. Tomar RT&A, No 1 (52)

A TWO IDENTICAL UNIT COLD STANDBY SYSTEM SUBJECT TO TWO Volume 14, March 2019

TYPES OF FAILURES_________________________________________________________________

= | G1(t)dt and I G2(t)dt

U : Symbol for Laplace Stieltjes Transform, i.e. Qjj (s) = Je_stdQ;j (t)

t

© : Symbol for ordinary convolution i.e. A (t) ©B (t ) = | A (u ) ©B (t-u )du

0

* : Symbol for Laplace Transform. i.e. qi (s) = | e_stq;j (u) du

4. Transition Probabilities and Sojourn Times

(a)

The direct or one step steady-state transition probabilities are as follows

a1

a1 +a2

P01 = I ea 2t a1eaitdt = ——

* rv. 4-f

f -a,t -at j-

P02 =J e 1 a2e 2dt

- ff,-(a1 +a2 )t

a

2

a1 +a2

Pio = je (ai+“2)tdG1 (t) = Gj (щ + a2)

P2o=je (ai+“2)tdG2(t) = G2(a1+a2)

(b) The two step steady-state transition probabilities are given by

p<1l=|a1e-a1uduea2uG1 (u )|

a11 dG1 (t )J e-(a1+“2 )udu

1[1

a1 +a 2 a

i-(a+a)t

dG1 (t)

a1 +a

" [l-^ (aj +a2)]

- rv _ >— -i

Similarly,

p(2) - -

a-

a1 +a2

p252)=■

00

a1 +a2

(l-G^ctj+c^)]

[l-G^cij+c^)]

P21 =—7^[l-G2(<Xi+a2)]

a1 +a2 L J

We observe the following relationship

P01 + P02 =1, P10 + Pu+ P(2) =1,

(1-3)

P20 + P22) + P26)= 1

(a) The mean sojourn times in various states are as follows:

t

0

f -(a, +a2 )tj. 1

V0 =1 e v 1 2 dt =---------

J a, +a2

Similarly,

V = J e-(ai+a >tG1 (t) dt

V 2 =| e-(a' +a2 ),G2 (t) dt

5. Analysis of Characteristics (a) RELIABILITY AND MTSF

Let R (t) be the probability that the system is operative during (0,t) given that at t=0 it starts from state Si e E. By simple probabilistic arguments, we have the following recurrence relations in R (t); i = 0, 1, 2

R0 (t) = Z0 (t) + q01 (t )©R1 (t) + q02 (t )©R2 (t)

Similarly,

R1 (t) = Z1 (t) + q10 (t) ©R 0 (t)

R2 (t) = Z2 (t) + q20 (t)©R0 (t) (4-6)

Where,

Z0 (t) = e (a' +a)t , Zj (t) = e (a' +a)tG1 (t), Z2 (t) = e (a+a2)tG2 (t)

Taking Laplace Transforms of the relation (4-6) and solving the resulting set of algebraic equations for Ro ( s), we get

-(ai +a2 >;

_ -(a' +a2R

(') _ Z0 + q01Z1 + q02Z2 1 - q01q10 - q02q20

We have omitted the argument's' from q* (s) and Z* (s). The expression of mean time to system failure is given by E (T0 ) = limR0 (s)

s^0

(7)

Observing that q* (0) = p- and Z; (0) = Vi, we get

E(Tu) VQ + P01V1 + P02V2 1 - p01p10 - p02p20

(8)

b) AVAILABILITY ANALYSIS

Let Ai (t) be the probability that the system is up at epoch t, when initially it starts operation from state S; e E. Using the regenerative point technique and the tools of Laplace transform, one can obtain the value of A0 (t) in terms of its Laplace transforms i.e. A* (s) given as follows-

A0 (s ) =

NO

D, (s)

(9)

Where,

Ni (s) = Z0 [(i- qn*) (i- q2?*)- q( 2М61*

(6)*

+Z

* * * q02q21 + q01

(1 - q2?)

+Z

[[ 2)*+(1 - q[) q0i

Di (s) = [(1 - q(,)*) (1 - q252)*)- q(4)*q26i)*

- qo2 [г21 +Г - q(1)) q20

The steady-state availability of the system is given by

* , 4 N (s)

A0 = lim sA0 ( s) = lim s----==

0 0 0W s^0 D1 (s)

q01

* (4)* * Л (5)*\~|

q20q12 + q10 (1 - q22 )J

s^0

We observe that

D1 (0) = 0

Therefore, by using L.Hospital's rule the steady state availability is given by

N1 (0)

A0 =

Where,

D1( 0 )

N1 (0)=^0 P10 (1 - p252))- P20P12) P01P12)+P02 (1 - p(1))

+^1

P01

(1 - p252))

+ p02p20

+ V2

1(0)= p10 (1 p22 ) + p1^p20 ^0 +(p01 (1 p22) + p02p21)n1

+ (P01P12)+ P02 (1 - PU )) n2

The expected up time of the system in interval (0, t) is given by

Pup (t) = j A0 (u ) du

so that,

P*up (s) =

A0 (s)

(10)

(11)

(12)

(13)

(c)

BUSY PERIOD ANALYSIS

Let Bl (t) and B2 (t) be the probability that the repairman is busy in the repair of a failed

unit due to normal cause and due to chance shock at time t when system initially starts from state Si e E. Using the simple probabilistic arguments in regenerative point technique and

the tools of Laplace transforms, one can obtain the value of Bj (t) and B2 (t) in terms of

*

*

0

s

their Laplace transforms as follows-

B”’(s)=and B»’(s)

where,

N (s)

N3 (s)

* Л (5\ * (6)*

Qoi (!- 422j ) + 402421^

* Л (3 )*\ * (4 )*

qo2 (1 - q(i))+qoiq(2)

N3 (s ) Di (s )

(14-15)

and D1 (s) is already defined in section 5(b). G1

*

and G2* are the L.T. of G1 (t) and G2 (t)

In the long run, the probabilities that the repairman will be busy in repair of normal cause and chance causes are as follows-

B0 =■

where,

'2 (0) >;( 0) and B0 = N3 (0) 0 d 2 ( 0 ) (16-17)

P01 (i - pIS) + p02p21 n1

p02 ( >- p(1) )+p0 1 p ((2) n2

The value of D[ (0) is same as given in expression (12).

The expected busy period of the repairman in repair in repairing during (0, t) are given by

цЬ (t) = } B0 (u) du and цb (t) = j B0 (u) du

so that,

цЬ*( s )=

B0*( s )

and ц b* ( s) =

B0* (s)

(18-19)

2

0

0

s

s

(d) PROFIT FUNCTION ANALYSIS

The net expected total profit incurred by the system in time interval (0, t) is given by P (t) = Expected total revenue in (0, t) - Expected cost of repair in (0, t)

= K0dup (t)- K 1цЬ (t)- К2цЬ (t) (20)

Where, K0 is the revenue per- unit up time by the system during its operation. K 1 and K2 are the amounts paid to the repairman per-unit of time when he is busy in repair of a unit failed due to normal cause and due to chance cause respectively.

The expected total profit incurred per unit time in steady-state is given by

P = K0A0 - K 1 B0 - K2B0 (21)

6. Particular Cases

Case 1: When the repair time of both the units also follow exponential distribution with p.d.fs as follows-

gi (t) = Л1е"^, g2 (t) = Л2е"Л^

The Laplace Transform of above density functions are as given below.

st (s) = G, (s) = Jl, &(s) = G2(s) = ^-

Here Gj (s) are the Laplace-Stieltjes Transforms of the c.d.fs G; (t) corresponding to the

p.d.fs gi (t) .

In view of above, the changed values of transition probabilities and mean sojourn times are given below-

Poi

a1

a1 +a2 ' a1

a1 +a2 + л1'

p02

a2

a1 + a2 a2

a1 +a2 + л1

P10

Л1

a1 +a2 + л1

p20

Л2

a1 +a2 +Л2

p22)=■

a-

a1 +a2 +Л2

a1

a1 +a2 +Л2

1

Vo =-------,

a1 +a2

1

V1 =-----------/

a1 +a2 + Л1

1

V1 =-----------

a1 +a2 +Л2

7. Graphical Study Of Behaviour

The curves for MTSF and profit function are drawn for the two particular cases: case

1 and case 2 in respect of different parameters. In Case 1, when the repair time of unit-1 also follow exponential distribution. We plot curves for MTSF and profit function in Fig.

2 and Fig. 3 w.r.t. for three different values of n and two different values of n while the other parameters are kept fixed as a2 = 0.029. From the curves of Fig. 2 we observe that MTSF increases uniformly as the value of л and л increase and it decreases with the increase in a . Further, we also observed from Fig. 2 that the value of a must be less than 0.012, 0.114 and 0.017 corresponding to % = 0.1,0.2 and 0.3 to achieve at least 300 units of MTSF when n = 0-9 is fixed as. Similarly, we can find the upper bounds of a^ corresponding to the values of n to achieve 300 units of MTSF when л is kept fixed as 0.4.

Similarly, Fig. 3 reveals the variations in profit w.r.t. a for varying values of n and П, when the values of other parameters are kept fixed as a2 = 0.00009, Ko =70, Kj = 40 and K2 = 500. Here also the same trend in respect of aj, л and л are observed as in case of MTSF. From the figure it is clearly observed from the smooth curves, that the system is profitable if the value of parameter aj is less than 0.41, 0.56 and 0.86 respectively for Л = 0.2,0.3 and 0.5 for fixed value of n = 0.9. From dotted curves, we conclude that

system is profitable if the value of parameter at is less than 0.44, 0.60 and 0.90 respectively for ni = 0.2,0.3 and 0.5 for fixed value of n =0.05.

Behaviour of MTSF w.r.t. a for different values of ^ and X2

ai ------->

Fig. 2

Behaviour of PROFIT (P) w.r.t. Щ for different values of ^ and

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