A Parallel System with Priority to Preventive Maintenance over ... - IJEDR

© 2014 IJEDR | Volume 2, Issue 3 | ISSN: 2321-9939

A Parallel System with Priority to Preventive Maintenance over Replacement Subject To Maximum Operation and Repair Times R. Rathee, S.C. Malik Research Scholar, Professor Department of Statistics, M.D. University, Rohtak-124001, India ________________________________________________________________________________________________________ Abstract - The present study deals with profit analysis of a parallel system of two identical units by giving priority to preventive maintenance of one unit over replacement of the other. Each unit has two modes- operative and complete failure. A single server is provided immediately to conduct the repair activities whenever needed. The preventive maintenance of the unit is done after a maximum operation time up to which no failure occurs. The failed unit is replaced by new one in case its repair is not possible by the server in a given maximum repair time. The unit performs with full efficiency as new after repair and preventive maintenance. All random variables are statistically independent. The distributions for failure time, replacement time and the rate by which unit undergoes for preventive maintenance are taken as negative exponential while that of preventive maintenance, repair and replacement rates are assumed as arbitrary with different probability density functions. The semi-Markov process and regenerative point technique are adopted to derive the expressions for some measures of system effectiveness in steady state. The variation of mean time to system failure (MTSF), availability and profit function has been observed graphically for arbitrary values of various parameters and costs. Key Words - Parallel system, preventive maintenance, replacement, priority and profit analysis. I. INTRODUCTION The main intension of the manufacturers is to get maximum profit by selling their products with minimum efforts. And, they have managed it up to a considerable level by using proper operational and repair techniques in their systems. The method of parallel redundancy has been considered as one of the effective strategies for improving performance of the systems and thus profit. Several studies have been conducted on parallel systems under a common assumption that system can work for a long time without requiring any maintenance. But, this assumption seems to be unrealistic as continued operation and ageing of operable systems reduce their performance, reliability and safety. In such a situation preventive maintenance can be conducted after a specific operation time in order to slow the deterioration process. Kishan and Kumar (2009) studied a parallel system with preventive maintenance. The system of parallel units can be made more profitable by making replacement of the failed unit in case server fails to get its repair in a fixed time. Malik and Gitanjali (2012) obtained reliability measures of a parallel system with replacement of the unit subject to maximum repair time. Furthermore, the concept of priority in repair disciplines is one of the best ideas to enhance the profit of the system. Malik and Nandal (2010), Malik and Sureria (2012) and Kumar et al. (2012) have developed reliability models for the standby systems using the concept of priority. However, the idea of priority to preventive maintenance over replacement has not been introduced while analyzing system reliability models of two or more units. Thus, the focus of the present study is to fill up this gap while carrying out profit analysis of a parallel system of two identical units. Each unit has two modes- operative and complete failure. A single server is provided immediately to conduct the repair activities whenever needed. The preventive maintenance of the unit is done after a maximum operation time up to which no failure occurs. The failed unit is replaced by new one in case its repair is not possible by the server in a given maximum repair time. Priority is given to preventive maintenance of one unit over the replacement of the other unit. The unit works as new after repair and preventive maintenance. All random variables are statistically independent. The distributions for failure time, replacement time and the rate by which unit undergoes for preventive maintenance are taken as negative exponential while that of preventive maintenance, repair and replacement rates are assumed as arbitrary with different probability density functions. The semi-Markov process and regenerative point technique are adopted to derive the expressions for some measures of system effectiveness in steady state. The variation of mean time to system failure (MTSF), availability and profit function has been observed graphically for arbitrary values of various parameters and costs. II. NOTATIONS E/ λ α0 β0

: Set of regenerative/ non-regenerative states : Constant failure rate : The rate by which system undergoes for preventive maintenance (called maximum constant rate of operation time) : The rate by which system undergoes for replacement (called maximum constant rate of repair

IJEDR1403063

International Journal of Engineering Development and Research (www.ijedr.org)

3199

© 2014 IJEDR | Volume 2, Issue 3 | ISSN: 2321-9939

FUr /FWr FURp UPm WPm FUR/FWR FURP UPM WPM g(t)/G(t) f(t)/F(t) r(t)/R(t) qij (t)/ Qij(t)

: : : : : : : : : : : :

qij.kr (t)/Qij.kr(t)

:

Mi(t)

:

Wi(t)

:

i

:

time) The unit is failed and under repair/waiting for repair The unit is failed and under replacement The unit is under preventive maintenance The unit is waiting for preventive maintenance The unit is failed and under repair / waiting for repair continuously from previous state The unit is failed and under replacement continuously from previous state The unit is under preventive maintenance continuously from previous state The unit is waiting for preventive maintenance continuously from previous state pdf/cdf of repair time of the unit pdf/cdf of preventive maintenance time of the unit pdf/cdf of replacement time of the unit pdf / cdf of passage time from regenerative state Si to a regenerative stateSj or to a failed state Sj without visiting any other regenerative state in (0, t] pdf/cdf of direct transition time from regenerative state Si to regenerative state Sj or to a failed state Sj visiting state Sk, Sr once in (0, t] Probability that the system up initially in state Si  E is up at time t without visiting to any regenerative state Probability that the server is busy in the state Si up to time „t‟ without making any transition to any other regenerative state or returning to the same state via one or more non-regenerative states. The mean sojourn time in state which is given by where

mij

denotes the time to system failure.

: Contribution to mean sojourn time (i) in state Si when system transits directly to state Sj so that

i   mij

and mij =

j

 tdQ (t)   q ij

* ij

'(0)

&  : Symbol for Laplace-Stieltjes convolution/Laplace convolution */** : Symbol for Laplace Transformation /Laplace Stieltjes Transformation The possible transition states of the system model are shown in fig.1 III. TRANSITION PROBABILITIES AND MEAN SOJOURN TIMES Simple probabilistic considerations yield the following expressions for the non-zero elements as 

pij  Qij ()   qij (t )dt 0

p01 

(1)

0  2 (1  g * (   0   0 )), , p02  , p10  g* (  0  0 ), p13  (   0   0 ) 2   0 2   0

0  (1  r * (   0 )), p6,11  p66.11 (1  f * (   0 )), (   0 ) (   0 ) 0  p31  p56  g * (0 ), p49  (1  r * (   0 )), p6,10  p61.10  (1  f * (   0 )), (   0 ) (   0 ) 0  p14  (1  g * (   0   0 )), p11.3  g * ( 0 )(1  g * (   0  0 )), (   0   0 ) (   0   0 ) 0 0 p15  (1  g * (   0   0 )), p16.5  g* (  0 )(1  g * (   0   0 )), (   0   0 ) (   0   0 ) 0 p37  p31.7  p59  1  g * (0 ), p19.5  (1  g * ( 0 ))(1  g * (   0  0 )), (   0   0 )  p60  f * (  0 ), p11.37  (1  g * ( 0 ))(1  g * (   0  0 )), (   0   0 ) p26  p71  p81  p94  p10,1  p11,6  1 (2)

p40  r* (  0 ), p48  p41.8 

It can be easily verified that

p01  p02  p10  p13  p14  p15  p40  p48  p49  p60  p6,10  p6,11  1

p10  p14  p11.3  p11.37  p16.5  p19.5  p40  p41.8  p49  p60  p6,10  p66.11  1

IJEDR1403063

International Journal of Engineering Development and Research (www.ijedr.org)

3200

© 2014 IJEDR | Volume 2, Issue 3 | ISSN: 2321-9939

The mean sojourn times (

0  m01  m02

,

) is in the state Si are

1  m10  m13  m14  m15

6  m60  m6,10  m6,11

,

4'  m40  m41.8  m49 ,

2  m26 , 4  m40  m48  m49 ,

,

9  m94 , 1'  m10  m14  m11.3  m11.37  m16.5  m19.5 , 6'  m60  m6,10  m66.11

(3)

IV. RELIABILITY AND MEAN TIME TO SYSTEM FAILURE (MTSF)

 i(t) be the cdf of first passage time from regenerative state Si to a failed state. Regarding the failed state as absorbing state, we have the following recursive relations for  i(t): Let

0 (t)  Q01 (t)& 1  Q02 (t) 1 (t)  Q10 (t)& 0  Q14 (t)& 4 (t )  Q13 (t )  Q15 (t) 4 (t)  Q40 (t)& 0  Q48 (t)  Q49 (t ) Taking LST of above relation (4) and solving for

R* (s) 

(4)

(s), we have

1   ( s) s **

(5) The reliability of the system model can be obtained by taking Inverse Laplace transform of (5). The mean time to system failure (MTSF) is given by

1   ** ( s) N  s 0 s D

MTSF  lim

(6)

Where

N  0  p011  p01 p14 4 and D  1  p01 p10  p01 p14 p40

(7)

V. STEADY STATE AVAILABILITY Let Ai(t) be the probability that the system is in up-state at instant „t‟ given that the system entered regenerative state Si at t = 0.The recursive relations for

Ai (t )

are given as:

A0 (t )  M 0 (t)  q01 (t )A1 (t )  q02 (t )A2 (t ) A1 (t )  M1(t)  q10 (t )A0 (t )  q14 (t )A4 (t )  (q11.3 (t )  q11.37 (t ))A1(t )  q16.5 (t )A6 (t)  q19.5 (t )A9 (t) A2 (t )  q26 (t )A6 (t ) A4 (t )  M 4 (t )  q40 (t )A0 (t )  q41.8 (t )A1 (t )  q49 (t )A9 (t ) A6 (t )  M 6 (t )  q60 (t )A0 (t )  q61.10 (t )A1 (t )  q66.11 (t )A6 (t ) A9 (t )  q94 (t )A4 (t ) (8)

Where

M 0 (t )  e(2 0 )t , M1 (t )  e( 0  0 )t G(t ) , M 4 (t )  e( 0 )t R(t ) , M 6 (t )  e( 0 )t F(t ) (9)

Taking LT of above relations (8) and solving for A 0*(s). The steady state availability is given by

A0 ()  lim sA0* ( s)  s 0

N1 D1

(10)

Where

N1  0 X  {1(1  p01)  4 ( p14  p19.5 )}Y  6Z and

D1  (0  2 p02 ) X  {1' (1  p49 )  4' (p14  p19.5 )  9 (p14p49  p19.5 )}Y 6' Z (11) VI. BUSY PERIOD ANALYSIS FOR SERVER (a) Due to Repair Let BiR (t ) be the probability that the server is busy in repair the unit at an instant„t‟ given that the system entered regenerative state Si at t=0.The recursive relations for BiR (t ) are as follows:

B0R (t )  q 01 (t )B1R (t )  q02 (t )B2R (t )

IJEDR1403063

International Journal of Engineering Development and Research (www.ijedr.org)

3201

© 2014 IJEDR | Volume 2, Issue 3 | ISSN: 2321-9939

B1R (t )  W1 (t)  q10 (t )B0R (t )  q14 (t )B4R (t )  (q11.3 (t )  q11.37 (t ))B1R (t )  q16.5 (t )B6R (t)  q19.5 (t )B9R (t) B2R (t )  q26 (t )B6R (t ) B4R (t )  q40 (t )B0R (t )  q 41.8 (t )B1R (t )  q 49 (t )B9R (t )

B6R (t )  q60 (t )B0R (t )  q 61.10 (t )B1R (t )  q 66.11 (t )B6R (t ) B9R (t )  q 94 (t )B4R (t )

(12)

Where

W1 (t )  e( 0  0 )t G(t )  (e( 0  0 )t 1)G(t )  (0e( 0  0 )t 1)G(t ) Taking LT of above relations (12) and solving for

B0R* (s) .The

(13)

time for which server is busy due to repair is given

by

B0R ()  lim sB0R* (s)  s 0

N2 D1

(14)

Where

N2  W1* (0)(1  P49 ) Y and D1 is already mentioned.

(15)

(b) Due to Replacement Let BiRp (t ) be the probability that the server is busy in replacement the unit at an instant „t‟ given that the system entered regenerative state Si at t=0.The recursive relations for BiRp (t ) are as follows:

B0Rp (t )  q 01 (t )B1Rp (t )  q02 (t )B2Rp (t )

B1Rp (t )  q10 (t )B0Rp (t )  q14 (t )B4Rp (t )  (q11.3 (t )  q11.37 (t ))B1Rp (t )  q16.5 (t )B6Rp (t)  q19.5 (t )B9Rp (t) B2Rp (t )  q26 (t )B6Rp (t )

B4Rp (t )  W4 (t )  q40 (t )B0Rp (t )  q 41.8 (t )B1Rp (t )  q 49 (t )B9Rp (t )

B6Rp (t )  q60 (t )B0Rp (t )  q 61.10 (t )B1Rp (t )  q 66.11 (t )B6Rp (t ) B9Rp (t )  q 94 (t )B4Rp (t )

(16)

Where

W4 (t )  e( 0 )t R(t )  (e( 0 )t 1) R(t )

Taking LT of above relations (16) and solving for

(17)

B0Rp* ( s) .The

time for which server is busy due to replacement is

given by

N3 D1

(18)

and D1 is already mentioned.

(19)

B0Rp ( s )  lim sB0Rp* ( s )  s 0

Where

N3  W4* (0)( p14  p19.5 )Y

(c) Due to Preventive Maintenance Let BiP (t ) be the probability that the server is busy in preventive maintenance the unit at an instant „t‟ given that the system entered regenerative state Si at t=0.The recursive relations for BiP (t ) are as follows:

B0P (t )  q 01 (t )B1P (t )  q02 (t )B2P (t )

B1P (t )  q10 (t )B0P (t )  q14 (t )B4P (t )  (q11.3 (t )  q11.37 (t ))B1P (t )  q16.5 (t )B6P (t)  q19.5 (t )B9P (t) B2P (t )  W2 (t )  q26 (t )B6P (t ) B4P (t )  q40 (t )B0P (t )  q 41.8 (t )B1P (t )  q 49 (t )B9P (t )

B6P (t )  W6 (t )  q60 (t )B0P (t )  q 61.10 (t )B1P (t )  q 66.11 (t )B6P (t ) B9P (t )  W9 (t)  q 94 (t )B4P (t ) IJEDR1403063

(20)

International Journal of Engineering Development and Research (www.ijedr.org)

3202

© 2014 IJEDR | Volume 2, Issue 3 | ISSN: 2321-9939 Where

W2 (t )  W9 (t )  F (t ) and W6 (t )  e( 0 ) t F (t )  (0e( 0 )t 1) F (t )  (e( 0 )t 1) F (t ) Taking LT of above relations (20) and solving for

B0P* (s) .The

(21)

time for which server is busy due to preventive maintenance is

given by

B0P ()  lim sB0P* ( s )  s 0

N4 D1

(22)

Where

N4  W2* (0) p02 X  W6* (0) Z W9* (0)(p14 p49  p19.5 ) Y and D1 is already mentioned.

(23)

VII. EXPECTED NUMBER OF REPAIRS Let Ri (t ) be the expected number of repairs by the server in (0, t] given that the system entered the regenerative state Si at t = 0. The recursive relations for Ri (t ) are given as:

R0 (t )  Q01 (t )& R1 (t )  Q02 (t )& R2 (t ) R1 (t )  Q10 (t ) & (1  R0 (t ))  Q14 (t ) & R4 (t )  Q11.3 (t ) & (1  R1 (t ))  Q11.37 (t ) & R1 (t )  Q16.5 (t ) & (1  R6 (t))  Q19.5 (t ) & R9 (t)

R2 (t )  Q26 (t )& R6 (t ) R4 (t )  Q40 (t )& R0 (t )  Q41.8 (t )& R1 (t )  Q49 (t )& R9 (t ) R6 (t )  Q60 (t )& R0 (t )  Q61.10 (t )& R1 (t )  Q66.11 (t )& R6 (t ) R9 (t )  Q94 (t )& R4 (t )

(24)

** Taking LST of above relations (24) and solving for R0 ( s) .The expected no. of repairs per unit time by the server are giving by

R0 ()  lim sR0** (s)  s 0

N5 D1

(25)

Where

N5  ( P10  P11.3  P16.5 )(1  P49 )Y and D1 is already mentioned. VIII.

(26)

EXPECTED NUMBER OF REPLACEMENTS

Let Rpi (t ) be the expected number of replacements by the server in (0, t] given that the system entered the regenerative state Si at t = 0. The recursive relations for Rpi (t ) are given as:

Rp0 (t )  Q01 (t )& Rp1 (t )  Q02 (t )& Rp2 (t ) Rp1 (t )  Q10 (t ) & Rp0 (t )  Q14 (t ) & Rp4 (t )  Q11.3 (t ) & Rp1 (t )  Q11.37 (t ) & (1  Rp1 (t ))  Q16.5 (t ) & Rp6 (t)  Q19.5 (t ) & Rp9 (t)

Rp2 (t )  Q26 (t )& Rp6 (t ) Rp4 (t )  Q40 (t )& (1  Rp0 (t ))  Q41.8 (t )& (1  Rp1(t ))  Q49 (t )& Rp9 (t ) Rp6 (t )  Q60 (t )& Rp0 (t )  Q61.10 (t )& Rp1 (t )  Q66.11 (t )& Rp6 (t ) Rp9 (t )  Q94 (t )& Rp4 (t )

(27)

Taking LST of above relations (27) and solving for Rp0** ( s ) .The expected number of replacements per unit time by the server is giving by

Rp0 ()  lim sRp0** ( s)  s 0

N6 D1

(28)

Where

N6  Y (1  p49 )( p14  p11.3  p19.5 ) and D1 is already mentioned.

(29)

IX. EXPECTED NUMBER OF PREVENTIVE MAINTENANCES Let Pi (t ) be the expected number of preventive maintenance by the server in (0, t] given that the system entered the regenerative state Si at t = 0. The recursive relations for Pi (t ) are given as:

IJEDR1403063

International Journal of Engineering Development and Research (www.ijedr.org)

3203

© 2014 IJEDR | Volume 2, Issue 3 | ISSN: 2321-9939

P0 (t )  Q01 (t )& P1 (t )  Q02 (t )& P2 (t ) P1 (t )  Q10 (t ) & P0 (t )  Q14 (t ) & P4 (t )  (Q11.3 (t )  Q11.37 (t )) & P1 (t )  Q16.5 (t ) & P6 (t)  Q19.5 (t ) & P9 (t)

P2 (t )  Q26 (t )& (1  P6 (t )) P4 (t )  Q40 (t )& P0 (t )  Q41.8 (t )& P1(t )  Q49 (t )& P9 (t ) P6 (t )  Q60 (t )& (1  P0 (t ))  Q61.10 (t )& (1  P1 (t ))  Q66.11 (t )& (1  P6 (t )) P9 (t )  Q94 (t )& (1  P4 (t )) Taking LST of above relations (30) and solving for

P0** ( s) .The

(30)

expected number of preventive maintenances per unit time by the

server is giving by

P0 ()  lim sP0** ( s)  s 0

N7 D1

(31)

Where

N7  p02 X  ( p14 p49  p19.5 )Y  Z and D1 is already mentioned.

(32)

Where

X  (1  p49 ){(1  p11.3  p11.37 )(1  p66.11)  p61.10 p16.5}  p41.8 (1  p66.11 )( p14  p19.5 ) Y  p01(1  p66.11)  p61.10 p02 Z  (1  p49 ){ p01 p16.5  (1  p11.3  p11.37 ) p02}  p02 p41.8 ( p14  p19.5 )

(33)

X. PROFIT ANALYSIS The profit incurred to the system model in steady state can be obtained as

P  K0 A0  K1B0R  K2 B0Rp  K3 B0P  K4 R0  K5 Rp0  K6 P0

(34)

Where P = Profit of the system model K0 = Revenue per unit up-time of the system K1 = Cost per unit time for which server is busy due to repair K2 = Cost per unit time for which server is busy due to replacement K3 = Cost per unit time for which server is busy due to preventive maintenance K4 = Cost per unit time repair K5 = Cost per unit time replacement K6 = Cost per unit time preventive maintenance XI. CONCLUSION The results for some important reliability measures have been evaluated for the particular case g (t ) =  e t , r (t ) =  e   t ,

f (t ) =  e t . Graphs are drawn to show the behavior of MTSF, availability and profit with respect to failure rate (λ) as shown in figures 2, 3 and 4 respectively. It is observed that MTSF, availability and profit go on decreasing with the increase of failure rate (λ) and the rate (α0) by which unit undergoes for preventive maintenance while their values increase with the increase of repair rate (θ) and replacement rate (β). MTSF and availability keep on increasing as the rate (β0) by which unit undergoes for replacement increases while system becomes less profitable. Also, there is no effect of preventive maintenance rate (α) on MTSF whereas system becomes more profitable with the increase of preventive maintenance rate (α). Thus, the study reveals that a parallel system of two identical units in which priority to preventive maintenance is given over replacement can be made more reliable and profitable to use by increasing repair, preventive maintenance and replacement rates. REFERENCES [1] R. Kishan and M. Kumar, “Stochastic analysis of a two-unit parallel system with preventive maintenance”, Journal of Reliability and Statistical Studies, vol. 22, pp. 31- 38, 2009. [2] S.C. Malik and P. Nandal, “Cost-analysis of stochastic models with priority to repair over preventive maintenance subject to maximum operation time”, Learning Manual on Modeling, Optimization and Their Applications, Edited Book, Excel India Publishers, pp 165-178, 2010. [3] S.C. Malik and Kumar Ashish, “Stochastic modeling of a computer system with priority to PM over S/W replacement subject to maximum operation and repair times”, International Journal of Computer Applications, vol.43 (3), pp. 27-34, 2012. [4] S.C. Malik and J.K. Sureria, “Probabilistic analysis of a computer system with priority to h/w repair over s/w replacement”, International Journal of Statistics and Analysis, vol.2 (4), pp. 379-389, 2012.

IJEDR1403063

International Journal of Engineering Development and Research (www.ijedr.org)

3204

© 2014 IJEDR | Volume 2, Issue 3 | ISSN: 2321-9939 [5] S.C. Malik and Gitanjali, “Cost-benefit analysis of a parallel system with arrival time of the server and maximum repair time”, International Journal of Computer Applications, vol.46 (5), pp. 39-44, 2012.

State Transition Diagram UPm WPm

UPM WPm

FWR FURp r(t)

f(t)

f(t) O UPm

f(t)

2λ

O O

O FUr

g(t)

g(t)

FUR FWr

g(t)

r(t)

λ

λ

r(t)

f(t) UPM FWr

O FURp

FUR WPm

λ

FURP FWr

f(t) UPm FWRp

Up-State Failed State Regenerative State

Fig. 1

IJEDR1403063

International Journal of Engineering Development and Research (www.ijedr.org)

3205

© 2014 IJEDR | Volume 2, Issue 3 | ISSN: 2321-9939

IJEDR1403063

International Journal of Engineering Development and Research (www.ijedr.org)

3206