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Applied Mathematical Sciences, Vol. 6, 2012, no. 75, 3723 - 3734 Cost-Benefit Analysis of a Computer System with Priority to S/W Replacement Over H/W Repair J. K. Sureria, S. C. Malik and Jyoti Anand Department of Statistics, M.D. University, Rohtak-124001, Haryana, India [email protected] Abstract. The present paper has been designed with an object to determine reliability and economic measures of a computer system of two identical units-one is operative and the other is kept as spare in cold standby. Each unit has h/w and s/w components which may have independent complete failure from the normal mode. There is a single server who visits the system immediately to do the repair activities whenever needed. The unit is repaired at its h/w failure whereas replacement of the s/w giving some replacement time is made in the unit whenever s/w fails to meet out the requirements. Priority to the replacement of the s/w is given over repair of the h/w components. The failure, repair and replacement time of the h/w and s/w components in the units are independent and uncorrelated random variables. The system model is analysed stochastically in detail using semi-Markov process and regenerative point technique. The numerical results giving some particular values to various costs and other parameters are obtained to carry out cost-benefit analysis. The graphical behaviour of MTSF, availability and profit has been shown with respect to h/w failure for fixed values of other parameters. The results of the present model have also been compared with that of the model Malik and Anand [2011]. Mathematics Subject Classification: 90B25, 60K10 Keywords: Computer System, Independent Failure of H/w and S/w, Priority for Replacement to s/w, Repair of H/w and Cost-Benefit Analysis. 1. Introduction Reliability technology has been playing an important role in the modern age world of industrial growth. The reliability analysis deals with the proper functioning of the components and the system making more effective use of resources at command which in turn ensures the enhanced productivity and better quality of the end product. The computers are long need in most of the industrial sectors; therefore engineers are stressing on the development of reliable computer systems considering various operational and design policies. The technique of cold standby redundancy has been adopted by many researchers as an effective strategy in order to improve the performance and to achieve high reliability of the

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Page 1: Cost-Benefit Analysis of a Computer System with … · Applied Mathematical Sciences ... Replacement to s/w, Repair of H/w and Cost-Benefit Analysis. 1. ... whole system including

Applied Mathematical Sciences, Vol. 6, 2012, no. 75, 3723 - 3734

Cost-Benefit Analysis of a Computer System with

Priority to S/W Replacement Over H/W Repair

J. K. Sureria, S. C. Malik and Jyoti Anand

Department of Statistics, M.D. University, Rohtak-124001, Haryana, India [email protected]

Abstract. The present paper has been designed with an object to determine reliability and economic measures of a computer system of two identical units-one is operative and the other is kept as spare in cold standby. Each unit has h/w and s/w components which may have independent complete failure from the normal mode. There is a single server who visits the system immediately to do the repair activities whenever needed. The unit is repaired at its h/w failure whereas replacement of the s/w giving some replacement time is made in the unit whenever s/w fails to meet out the requirements. Priority to the replacement of the s/w is given over repair of the h/w components. The failure, repair and replacement time of the h/w and s/w components in the units are independent and uncorrelated random variables. The system model is analysed stochastically in detail using semi-Markov process and regenerative point technique. The numerical results giving some particular values to various costs and other parameters are obtained to carry out cost-benefit analysis. The graphical behaviour of MTSF, availability and profit has been shown with respect to h/w failure for fixed values of other parameters. The results of the present model have also been compared with that of the model Malik and Anand [2011]. Mathematics Subject Classification: 90B25, 60K10 Keywords: Computer System, Independent Failure of H/w and S/w, Priority for Replacement to s/w, Repair of H/w and Cost-Benefit Analysis. 1. Introduction Reliability technology has been playing an important role in the modern age world of industrial growth. The reliability analysis deals with the proper functioning of the components and the system making more effective use of resources at command which in turn ensures the enhanced productivity and better quality of the end product. The computers are long need in most of the industrial sectors; therefore engineers are stressing on the development of reliable computer systems considering various operational and design policies. The technique of cold standby redundancy has been adopted by many researchers as an effective strategy in order to improve the performance and to achieve high reliability of the

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3724 J. K. Sureria, S. C. Malik and J. Anand systems. However, this technique has not been used much more in the case of computer systems. Most of the research work has been carried out either considering h/w components or s/w components alone. Friedman and Tran (1992) and Welke et al. (1995) tried to develop a combined reliability model for the whole system including both h/w and s/w. Recently, Malik and Anand [2010] proposed a reliability model of a computer system with cold standby considering independent failure of h/w and s/w. The reliability and availability of multi component repairable systems can be enhanced using the rules of priority to repair activities of components. Malik and Anand [2011] analysed a cold standby computer giving priority to replacement of s/w over repair of h/w subject to inspection.

In view of the above observations and to strengthen the existing literature on reliability, here we discuss a computer system of two identical units-one is operative and the other is kept as spare in cold standby. Each unit has h/w and s/w components which may have independent complete failure from the normal mode. There is a single server who visits the system immediately whenever needed. The unit is repaired at its h/w failure while replacement of the s/w components giving some replacement time is made in the unit when s/w component fails to meet out the requirements. Priority to the replacement of the software is given over repair of the h/w components. The failure, repair and replacement time of the components in the units are independent and uncorrelated random variables. The switch devices and repairs are perfect. The failure time of the unit due to failure of h/w and s/w components is distributed exponentially while the distributions of repair and replacement times are taken as arbitrary. To carry out the cost-benefit analysis, the expressions for mean sojourn times, mean time to system failure (MTSF), availability, busy period of the server due to h/w repair and replacement of the s/w components, expected number of replacements due to s/w failures and expected number of visits by the server are obtained using semi-Markov process and regenerative point technique. The graphs for a particular case are drawn to show the behaviour of MTSF, availability and profit of the system model. The comparison of the results has also been made with the results obtained for the model Malik and Anand [2011]. 2. Notations E : The set of regenerative states O : The unit is operative and in normal mode Cs : The unit is cold standby λ1/λ2 : Constant hardware / software failure rate FHUr/FHUR : The unit is failed due to hardware and is under repair /under repair continuously from previous state FHWr /FHWR : The unit is failed due to hardware and is waiting for repair/

waiting for repair continuously from previous state

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Cost-benefit analysis of a computer system 3725 FSURp/FSURP : The unit is failed due to the software and is under

replacement/under replacement continuously from previous state

FSWRp/FSWRP: The unit is failed due to the software and is waiting for replacement / waiting for replacement continuously from

previous state f(t) / F(t) : pdf / cdf of replacement time of the software g(t) / G(t) : pdf / cdf of repair time of the unit due to hardware failure qij (t)/ Qij(t) : pdf / cdf of passage time from regenerative state i to a regenerative state j or to a failed state j without visiting any other regenerative state in (0, t] qij.kr (t)/Qij.kr(t) : pdf/cdf of direct transition time from regenerative state i to

a regenerative state j or to a failed state j visiting state k, r once in (0, t]

mij : Contribution to mean sojourn time (µi) in state Si when system transit directly to state Sj so that i ij

j

mµ =∑ and

mij = * '( ) (0)ij ijtdQ t q= −∫

�/ : Symbol for Laplace-Stieltjes convolution/Laplace convolution ~ / * : Symbol for Laplace Steiltjes Transform (LST) / Laplace Transform (LT)

' (desh) : Used to represent alternative result The following are the possible transition states of the system: S0= (O, Cs), S1= (O, FHUr), S2= (O, FSURp), S3= (FHUR, FHWr), S4= (FHWr, FSURp), S5= (FSURP, FSWRp), S6= (FHWr, FSURP) The states S0–S2 and S4 are regenerative states while the states S3, S5 and S6 are non-regenerative as shown in figure 1. 3. Transition Probabilities and Mean Sojourn Times Simple probabilistic considerations yield the following expressions for the non-zero elements

∫∞

=∞=0

)()( dttqQp ijijij as

p01= 1

1 2

aa b

λλ λ+

,p02= 2

1 2

ba b

λλ λ+

,p10= ( )1 2*g a bλ λ+ ,

p13= ( )11 2

1 2

1 *a

g a ba b

λ λ λλ λ

− + +, p14= ( )2

1 21 2

1 *b

g a ba b

λ λ λλ λ

− + +,

p20 = ( )1 2*f a bλ λ+ , p25 = ( )21 2

1 2

1 *b

f a ba b

λ λ λλ λ

− + +,

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3726 J. K. Sureria, S. C. Malik and J. Anand

p26 = ( )11 2

1 2

1 *a

f a ba b

λλ λ

λ λ − + +

,p31 = ( )*g s ,

p40 = p41 = p52 = p61 = ( )*f s (1) It can be easily verified that p01+p02 = p10+p13+p14 = p20+p25+p26 = p31 = p41 = p42 = p52 = p61 = p10 +p11.3 +p14 = p20+p21.6+p22.5 =1 (2) The mean sojourn times (µi) is the state Si are

µ0= ,1

21 λλ ba +µ1= ,

1

21 αλλ ++ baµ2=

θλλ ++ 21

1ba

, 4

1µθ

= (3)

Also

01 02 0m m µ+ = , 10 13 14 1m m m µ+ + = , 20 25 26 2m m m µ+ + = (4) and

10 11.3 12.4 1m m m µ′+ + = (Say), 20 21.6 22.5 2m m m µ′+ + = (Say) (5) 4. Reliability and Mean Time to System Failure (MTSF) Let φ i (t) be the c.d.f. of first passage time from regenerative state i to a failed state. Regarding the failed state as absorbing state, we have the following recursive relations for φ i(t): ( ) ( ) ( ) ( )∑∑ +φ=φ

kk,i

jjj,ii tQttQt (6)

where j is an un-failed regenerative state to which the given regenerative state i can transit and k is a failed state to which the state i can transit directly. Taking LST of above relation (6) and solving for )(

~0 sφ

We have

R*(s) =ss)(

~1 0φ− (7)

The reliability of the system model can be obtained by taking Laplace inverse transform of (7). The mean time to system failure (MTSF) is given by

MTSF =ss

os

)(~

1lim 0φ−→

= 1

1

ND

,where (8)

N1 = 0 01 1 02 2p pµ µ µ+ + , D1 = 01 10 02 201 p p p p− − 5. 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 i at t = 0. The recursive relations for Ai (t) are given as

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Cost-benefit analysis of a computer system 3727

( ) ( ) ( ) ( )( ),n

i i ji jj

A t M t q t A t= + ∑ (9)

where j is any successive regenerative state to which the regenerative state i can transit through n≥1(natural number) transitions. Where Mi (t) is the probability that the system is up initially in state iS E∈ is up at time t without visiting to any other regenerative state, we have

( )1 20 ( ) a b tM t e λ λ− += , ( )1 2

1( ) ( )a b tM t e G tλ λ− += , ( )1 22 ( ) ( )a b tM t e F tλ λ− += (10)

Taking LT of above relations (9) and solving for *0 ( )A s , the steady state

availability is given by

*0 00( ) lim ( )

sA sA s

→∞ = 2

2

ND

= (11)

where N2 = (p10p20+ p10p21.6)µ0 +( p01p20+ p21.6)µ1 +( p10p02)µ2 And D2 = (p10p20 + p10 p21.6) µ0 + (p01p20+p21.6) 1µ′+ (p10 p02) 2µ′ + (p14 p20 + p14 p21.6) 4µ 6. Busy Period Analysis for Server (a) Due to Hardware Failure Let )(tBHi be the probability that the server is busy in repairing the unit due to hardware failure at an instant ‘t’ given that the system entered state i at t = 0. The recursive relations )(tBHi for are as follows:

( ) ( ) ( ) ( )( )Hi ,W t nH H

i ji jj

B t q t B t= + ∑ (12)

where j is any successive regenerative state to which the regenerative state i can transit through n≥1(natural number) transitions. Where Wi

H (t) be the probability that the server is busy in state Si due to hardware failure up to time t without making any transition to any other regenerative state or returning to the same via one or more non-regenerative states and so

1 2 1 2( ) ( )1 1G( ) ( 1)G( )a b t a b tHW e t a e tλ λ λ λλ− + − += + (13)

(b) Due to replacement of the software Let S

iB (t)be the probability that the server is busy due to replacement of the software at an instant ‘t’ given that the system entered the regenerative state i at t = 0. We have the following recursive relations for S

iB (t):

( ) ( ) ( ) ( )( )i ,W t nS S S

i ji jj

B t q t B t= + ∑ (14)

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3728 J. K. Sureria, S. C. Malik and J. Anand where j is any successive regenerative state to which the regenerative state i can transit through n≥1(natural number) transitions. Where S

iW (t) be the probability that the server is busy in state Si due to replacement of the software up to time t without making any transition to any other regenerative state or returning to the same via one or more non-regenerative states and so

( ) ( ) ( )1 2 1 2 1 22 1 2( ) ( ) ( 1) ( ) ( 1) ( )a b t a b t a b tSW t e F t a e F t b e F tλ λ λ λ λ λλ λ− + − + − += + + ,

4 ( ) ( )SW t F t= (15)

Taking L.T. of above relations (13) and (14). And, solving for H

B ∗0 (s)

andS

B ∗0 (s), the time for which server is busy due to repair and replacements

respectively is given by

*0 00

lim ( )H H

sB sB s

→= = 3

2

HND

(16)

and

*0 00

lim ( )S S

sB sB s

→= = 3

2

SND

(17)

where

3 1 01 20 21.6 3 01 02 2 14 10 20 21.6 4( )[ ], ( ) ( ) ( ) ( )H H S S SN W s p p p N p p W s p p p p W s= + = + +% % % And D2 is already mentioned. 7. Expected Number of Replacements of the Units Due to Software Failure Let )(tR Si be the expected number of replacements of the failed software by the server in (0, t] given that the system entered the regenerative state i at t = 0. The recursive relations for )(tR Si are given as

( ) ( ) ( )( ),nH H

i j ji jj

R t q t R tδ = +∑ (18)

Where j is any regenerative state to which the given regenerative state i transits and jδ =1, if j is the regenerative state where the server does job afresh, otherwise

jδ = 0.

Taking L.S.T. of relations (18). And, solving for )(~

0 sR S . The expected number of replacements per unit time to the software failures is given by

0 00( ) lim ( )S S

sR sR s

→∞ = % = 4

2

SND

(19)

Where 4 14 01 20 02 10SN p p p p p= + and D2 is already mentioned.

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Cost-benefit analysis of a computer system 3729 8. Expected Number of Visits by the Server Let Ni(t) be the expected number of visits by the server in (0, t] given that the system entered the regenerative state i at t = 0. The recursive relations for Ni(t) are given as

( ) ( ) ( )( ),n

i j ji jj

N t q t N tδ = +∑ (20)

where j is any regenerative state to which the given regenerative state i transits and jδ =1, if j is the regenerative state where the server does job afresh, otherwise jδ = 0. Taking LST of relation (20) and solving for 0 ( )N s% . The expected number of visits per unit time by the server is given by

0 00( ) lim ( )

sN sN s

→∞ = % = 5

2

ND

(21)

where N5 = p10p20 + p10 p21.6 and D2 is already specified. 9. Cost-Benefit Analysis The profit incurred to the system model in steady state can be obtained as P = 0 0 1 0 2 0 3 0 4 0

H S SK A K B K B K R K N− − − − (22) where K0 = Revenue per unit up-time of the system K1 = Cost per unit time for which server is busy due to hardware failure K2 = Cost per unit time for which server is busy due to software failure K3 = Cost per unit replacement of the failed software K4 = Cost per unit visit by the server and 0 0 0 0 0, , , ,H S SA B B R N are already defined.

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3730 J. K. Sureria, S. C. Malik and J. Anand

State Transition Diagram

Fig. 1

Up-state Failed state Regenerative point

FHUR FHWr

O CS

O FHUr

FHWr FSURP

O FSURp

FHWr FSURp

FSURP FSWRp

aλ1

aλ1

bλ2

aλ1

g(t)

g(t)

f(t) bλ2

f(t)

S5

S6

S4 S2

S3

S1 S0

f(t)

bλ2

f(t)

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Cost-benefit analysis of a computer system 3731

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3732 J. K. Sureria, S. C. Malik and J. Anand

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Cost-benefit analysis of a computer system 3733 10. Conclusion The numerical results considering particular values to the parameters are obtained to carry out the cost-benefit analysis of a computer system giving priority to replacement of s/w over h/w repair. The graphs for mean time to system failure (MTSF), availability and profit are drawn with respect to h/w failure rate (λ1) for the fixed values of other parameters as shown in figures 2, 3 and 4 respectively. These figures indicate that MTSF, availability and profit increase with the increase of h/w repair rate (α) and s/w replacement rate (θ) for a=0.7 and b=0.3. And, their values become more by interchanging the values of a and b. However, values of these measures increase further if h/w and s/w failure rates increase. It can also be seen that when values of a and b are interchanged, the system becomes more profitable for λ1 >0.01. 11. Comparative Study Considering particular values of the parameters, graphs for the difference of MTSF and profit of the present model and that of the model Malik and Anand [2011] are drawn under the same set of assumptions as shown in figures 5 and 6. Figure 5 shows that MTSF of the present model is less than that of the model Malik and Anand [2011]. However, if we increase repair rate (α) of the h/w from 2.5 to 3.5, the present model will have more value of MTSF. Figure 6 indicates that present model has more value of profit.

Hence, study reveals that a computer system in which repair of h/w is done immediately without getting inspection will be more profitable as compared to the computer system where inspection of h/w component is carried out to see the feasibility of repair. Acknowledgement. The authors are grateful to the University Grants Commission (UGC) New Delhi,, India for providing financial Assistance to carry out this research work under Major Research Project.

References

[1] Friedman, M.A. and Tran, P.: Reliability techniques for combined hardware / software systems. Proceedings of Annual Reliability and Maintainability Symposium(1992), pp. 209-293.

[2] Welke, S.R.; Johnson, B.W. and Aylar, J.H.: Reliability modeling of hardware / software systems, IEEE Transactions on Reliability(1995), Vol. 44, No. 3, pp. 413-418.

[3] Mailk, S.C and Anand, Jyoti: Reliability and Economic Analysis of a Computer System with Independent Hardware and Software Failures. Bulletin of Pure and Applied Sciences (2010).Vol.29E (No.1), pp.141-153.

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3734 J. K. Sureria, S. C. Malik and J. Anand

[4] Malik, S.C and Anand, Jyoti: Reliability Modeling of a Computer System with Priority for Replacement at Software Failure over Repair Activities at H/W Failure. International Journal of Statistics and System, ISSN 0973-2675, Vol. 6,Number 3(2011),pp.315-325.

Received: February, 2012