Stochastic Modeling of a Computer System with ...

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ISSN (ONLINE): 2250-0758, ISSN (PRINT): 2394-6962 Volume-5, Issue-1, February-2015 International Journal of Engineering and Management Research Page Number: 295-302

Stochastic Modeling of a Computer System with Software Redundancy 1,2

V.J. Munday1, S.C. Malik2 Department of Statistics, M.D. University, Rohtak, Haryana, INDIA

ABSTRACT The semi-Markov process and regenerative point technique are adopted to obtain reliability measures of a computer system by providing software redundancy in cold standby. Initially, hardware and software work together in the system which may fail independently with some probability. There is a single server who repairs the system at hardware failure while software is up-graded as per requirements. The repair and up-gradation activities are performed perfectly and efficiently by the server. The time to hardware and software failures follows negative exponential distribution, whereas the distributions of hardware repair and software up-gradation times are taken as arbitrary with different probability density functions. Graphs are drawn to depict the behaviour of some important performance and economic measures of the system model for arbitrary values of various parameters and costs.

Keywords- Computer System, Software Redundancy, Reliability Measures and Stochastic Model.

I.

INTRODUCTION

The importance of computer systems cannot be denied in the corporate or business world, at the workplace and even in one’s personnel life. They also serve as useful tools for communications and record keeping while saving tons of times of the organizations. But a computer system would not be able to function properly without software that empowers the computer to communicate the results. Therefore, there is a definite need to place emphasis on reliability of computer software. Several techniques have been suggested by the designers and engineers for performance improvement of the systems. The unit wise redundancy technique has

II.

E 𝐸𝐸� O Scs

NOTATIONS

: : : : 295

been considered as one of these in the development of stochastic models for computer systems. Malik and Anand (2010, 12), Malik and Sureria (2012) and Kumar et al. (2013) analyzed computers systems with cold standby redundancy under different failures and repair policies. Also, Malik and Munday (2014) tried to establish a stochastic model for a computer system by providing hardware redundancy in cold standby. The basic interest of the authors in this paper is to evaluate reliability measures of a computer system with software redundancy in cold standby. For this purpose, a stochastic model is developed for a computer system in which hardware and software failures occur independently with some probability. There is a single server who repairs the system at hardware failure while software is up-graded as per requirements. The repair and up-gradation activities are performed perfectly and efficiently by the server. The time to hardware and software failures follows negative exponential distribution, whereas the distributions of hardware repair and software up-gradation times are taken as arbitrary with different probability density functions. The semiMarkov process and regenerative point technique are used to derive the expressions for transition probabilities, mean sojourn times, mean time to system failure (MTSF), availability, busy period of the server due to hardware repair and software up-gradation, expected number of hardware repairs and expected number of software up-gradations. Graphs are drawn to depict the behaviour of some important performance and economic measures of the system model for arbitrary values of various parameters and costs.

Set of regenerative states Set of non-regenerative states Computer system is operative Software is in cold standby Copyright © 2011-15. Vandana Publications. All Rights Reserved.

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a/b 𝜆𝜆1 /𝜆𝜆2 HFUr /HFWr SFUg/SFWUg HFUR/HFWR

: : : : :

SFUG/SFWUG

:

g(t)/G(t) f(t)/F(t) 𝑞𝑞𝑖𝑖𝑖𝑖 (𝑡𝑡)/𝑄𝑄𝑖𝑖𝑖𝑖 (𝑡𝑡)

: : :

𝑞𝑞𝑖𝑖𝑖𝑖 .𝑘𝑘 (𝑡𝑡)/𝑄𝑄𝑖𝑖𝑖𝑖 .𝑘𝑘 (𝑡𝑡)

:

𝑀𝑀𝑖𝑖 (𝑡𝑡)

:

𝑊𝑊𝑖𝑖 (𝑡𝑡)

:

𝜇𝜇𝑖𝑖

:

𝑚𝑚𝑖𝑖𝑖𝑖

:

&  */**

: :

296

Probability that the system has hardware / software failure Hardware/Software failure rate The hardware is failed and under repair/waiting for repair The software is failed and under/waiting for up-gradation The hardware is failed and continuously under repair / waiting for repair from previous state The software is failed and continuously under up-gradation /waiting for up- gradation from previous state pdf/cdf of hardware repair time pdf/cdf of software up-gradation time pdf / cdf of first passage time from regenerative state 𝑆𝑆𝑖𝑖 to a regenerative state 𝑆𝑆𝑗𝑗 or to a failed state 𝑆𝑆𝑗𝑗 without visiting any other regenerative state in (0, t] pdf/cdf of direct transition time from regenerative state 𝑆𝑆𝑖𝑖 to a regenerative state 𝑆𝑆𝑗𝑗 or to a failed state 𝑆𝑆𝑗𝑗 visiting state 𝑆𝑆𝑘𝑘 once in (0, t] Probability that the system up initially in state 𝑆𝑆𝑖𝑖 𝜖𝜖𝜖𝜖 is up at time t without visiting to any regenerative state Probability that the server is busy in the state 𝑆𝑆𝑖𝑖 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 𝑆𝑆𝑖𝑖 which is given by ∞ 𝜇𝜇𝑖𝑖 = 𝐸𝐸(𝑇𝑇) = ∫0 𝑃𝑃(𝑇𝑇 > 𝑡𝑡) 𝑑𝑑𝑑𝑑 = ∑𝑗𝑗 𝑚𝑚𝑖𝑖𝑖𝑖 , where 𝑇𝑇 denotes the time to system failure. Contribution to mean sojourn time (𝜇𝜇𝑖𝑖 ) in state 𝑆𝑆𝑖𝑖 when system transits directly to state 𝑆𝑆𝑗𝑗 so that ′ ∞ 𝜇𝜇𝑖𝑖 = ∑𝑗𝑗 𝑚𝑚𝑖𝑖𝑖𝑖 𝑎𝑎𝑎𝑎𝑎𝑎 𝑚𝑚𝑖𝑖𝑖𝑖 = ∫0 𝑡𝑡𝑡𝑡𝑡𝑡𝑖𝑖𝑖𝑖 (𝑡𝑡) = −𝑞𝑞𝑖𝑖𝑖𝑖∗ (0) Symbol for Laplace-Stieltjes convolution/Laplace convolution Symbol for Laplace Transformation (LT)/Laplace Stieltjes Transformation (LST)

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State Transition Diagram S1

S0 O Scs

g(t)

𝑎𝑎𝑎𝑎1

HFUr Scs

S4 𝑏𝑏𝑏𝑏2

f(t)

SFWUg SFUG

f(t) 𝑏𝑏𝑏𝑏2

S2

f(t) HFWr SFUG

O SFUg

𝑎𝑎𝑎𝑎1

Up-State

S3

Failed State

Regenerative Point

Fig. 1

III.

TRANSITION PROBABILITIES AND MEAN SOJOURN TIMES Simple probabilistic considerations yield the following expressions for the non-zero elements. ∞

𝑝𝑝𝑖𝑖𝑖𝑖 = 𝑄𝑄𝑖𝑖𝑖𝑖 (∞) = ∫0 𝑞𝑞𝑖𝑖𝑖𝑖 (𝑡𝑡)𝑑𝑑𝑑𝑑 𝑝𝑝01 = 𝑎𝑎𝑎𝑎

𝑎𝑎𝑎𝑎 1

𝑝𝑝10 = 𝑔𝑔∗ (0) 𝑝𝑝23 = 𝑎𝑎𝑎𝑎

𝑏𝑏𝑏𝑏 2

1 +𝑏𝑏𝜆𝜆 2

𝑝𝑝02 = 𝑎𝑎𝑎𝑎

,

1 +𝑏𝑏𝜆𝜆 2

𝑝𝑝24 = 𝑎𝑎𝑎𝑎

For 𝑔𝑔(𝑡𝑡) = 𝛼𝛼𝑒𝑒 −𝛼𝛼𝛼𝛼 𝑎𝑎𝑎𝑎𝑎𝑎 𝑓𝑓(𝑡𝑡) = 𝜃𝜃𝑒𝑒 −𝜃𝜃𝜃𝜃 we have 𝑝𝑝21.3 = 𝑎𝑎𝑎𝑎

𝑎𝑎𝑎𝑎 1

1 +𝑏𝑏𝑏𝑏 2

{1 − 𝑓𝑓 ∗ (𝑎𝑎𝜆𝜆1 + 𝑏𝑏𝑏𝑏2 )}

But, 𝑓𝑓 ∗ (0) = 𝑔𝑔∗ (0) = 1 𝑎𝑎𝑎𝑎𝑎𝑎 𝑝𝑝 + 𝑞𝑞 = 1

,

𝑝𝑝20 = 𝑓𝑓 ∗ (𝑎𝑎𝑎𝑎1 + 𝑏𝑏𝜆𝜆2 )

{1 − 𝑓𝑓 ∗ (𝑎𝑎𝑎𝑎1 + 𝑏𝑏𝜆𝜆2 )}

𝑝𝑝31 = 𝑝𝑝42 = 𝑓𝑓 ∗ (0)

𝑏𝑏𝜆𝜆 2

1 +𝑏𝑏𝑏𝑏 2

𝑎𝑎𝜆𝜆 1

1 +𝑏𝑏𝜆𝜆 2

𝑝𝑝22.4 = 𝑎𝑎𝑎𝑎

{1 − 𝑓𝑓 ∗ (𝑎𝑎𝑎𝑎1 + 𝑏𝑏𝑏𝑏2 )} ,

𝑏𝑏𝑏𝑏 2

1 +𝑏𝑏𝑏𝑏 2

{1 − 𝑓𝑓 ∗ (𝑎𝑎𝜆𝜆1 + 𝑏𝑏𝑏𝑏2 )}

It can be easily verified that

𝑝𝑝01 + 𝑝𝑝02 = 𝑝𝑝10 = 𝑝𝑝20 + 𝑝𝑝23 + 𝑝𝑝24 = 𝑝𝑝31 = 𝑝𝑝42 = 𝑝𝑝20 + 𝑝𝑝21.3 + 𝑝𝑝22.4 = 1

The mean sojourn times (𝜇𝜇𝑖𝑖 ) is the state 𝑆𝑆𝑖𝑖 are 𝜇𝜇0 = 𝑎𝑎𝑎𝑎

𝜇𝜇2 = �𝑎𝑎𝑎𝑎

Also

1

1 +𝑏𝑏𝜆𝜆 2

1

1 +𝑏𝑏𝑏𝑏 2

1

� {1 − 𝑓𝑓 ∗ (𝑎𝑎𝑎𝑎1 + 𝑏𝑏𝑏𝑏2 )} = 𝑎𝑎𝑎𝑎

𝑚𝑚01 + 𝑚𝑚02 = 𝜇𝜇0 , 297

𝜇𝜇1 = 𝛼𝛼

1

1 +𝑏𝑏𝑏𝑏 2 +𝜃𝜃

𝑚𝑚10 = 𝜇𝜇1 , 𝑚𝑚20 + 𝑚𝑚23 + 𝑚𝑚24 = 𝜇𝜇2

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and 𝑚𝑚20 + 𝑚𝑚21.3 + 𝑚𝑚22.4 = 𝜇𝜇2′ = IV.

1 𝜃𝜃

RELIABILITY AND MEAN TIME TO SYSTEM FAILURE (MTSF) Let 𝜙𝜙𝑖𝑖 (𝑡𝑡) be the cdf of first passage time from regenerative state 𝑆𝑆𝑖𝑖 to a failed state. Regarding

the failed state as absorbing state, we have the following recursive relations for 𝜙𝜙𝑖𝑖 (𝑡𝑡), 𝜙𝜙0 (𝑡𝑡) = 𝑄𝑄02 (𝑡𝑡) & 𝜙𝜙2 (𝑡𝑡) + 𝑄𝑄01 (𝑡𝑡)

𝜙𝜙2 (𝑡𝑡) = 𝑄𝑄20 (𝑡𝑡) & 𝜙𝜙0 (𝑡𝑡) + 𝑄𝑄23 (𝑡𝑡) + 𝑄𝑄24 (𝑡𝑡)

(1)

Taking LST of above relations (1) and solving for ϕ∗∗ 0 (𝑠𝑠) We have

𝑅𝑅 ∗ (𝑠𝑠) =

1−ϕ ∗∗ 0 (𝑠𝑠) 𝑠𝑠

The reliability of the system model can be obtained by taking Laplace inverse transform of the above equation. The mean time to system failure (MTSF) is given by 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 = lim𝑠𝑠→0

Where

1−ϕ ∗∗ 0 (𝑠𝑠) 𝑠𝑠

𝑁𝑁

= 𝐷𝐷1 1

𝑁𝑁1 = 𝜇𝜇0 + 𝑝𝑝02 𝜇𝜇2 𝑎𝑎𝑎𝑎𝑎𝑎 𝐷𝐷1 = 1 − 𝑝𝑝02 𝑝𝑝20

(2)

(3)

V. STEADY STATE AVAILABILITY Let 𝐴𝐴𝑖𝑖 (𝑡𝑡) be the probability that the system is in up-state at instant‘t’ given that the system entered regenerative state 𝑆𝑆𝑖𝑖 𝑎𝑎𝑎𝑎 𝑡𝑡 = 0. The recursive relations for 𝐴𝐴𝑖𝑖 (𝑡𝑡) are given as: 𝐴𝐴0 (𝑡𝑡) = 𝑀𝑀0 (𝑡𝑡) + 𝑞𝑞01 (𝑡𝑡)𝐴𝐴1 (𝑡𝑡) + 𝑞𝑞02 (𝑡𝑡) 𝐴𝐴2 (𝑡𝑡) 𝐴𝐴1 (𝑡𝑡) = 𝑞𝑞10 (𝑡𝑡)𝐴𝐴0 (𝑡𝑡) 𝐴𝐴2 (𝑡𝑡) = 𝑀𝑀2 (𝑡𝑡) + 𝑞𝑞20 (𝑡𝑡)𝐴𝐴0 (𝑡𝑡) + 𝑞𝑞21.3 (𝑡𝑡) 𝐴𝐴1 (𝑡𝑡) + 𝑞𝑞22.4 (𝑡𝑡) 𝐴𝐴2 (𝑡𝑡) (4) where ����� (𝑡𝑡) 𝑀𝑀0 (𝑡𝑡) = 𝑒𝑒 −(𝑎𝑎𝜆𝜆 1 +𝑏𝑏𝜆𝜆 2 )𝑡𝑡 𝑎𝑎𝑎𝑎𝑎𝑎 𝑀𝑀1 (𝑡𝑡) = 𝑒𝑒−(𝑎𝑎𝜆𝜆1 +𝑏𝑏𝑏𝑏2 )𝑡𝑡 𝐹𝐹 ∗ Taking LT of relations (4) and solving for 𝐴𝐴0 (𝑠𝑠), the steady state availability is given by 𝑁𝑁 𝐴𝐴0 (∞) = lim𝑠𝑠→0 𝑠𝑠 𝐴𝐴∗0 (𝑠𝑠) = 𝐷𝐷2 (5) 2

Where 𝑁𝑁2 = (1 − 𝑝𝑝22.4 )𝜇𝜇0 + 𝑝𝑝02 𝜇𝜇2 𝐷𝐷2 = (1 − 𝑝𝑝22.4 )𝜇𝜇0 + 𝑝𝑝02 𝜇𝜇2′ + (𝑝𝑝10 𝑝𝑝20 𝜇𝜇2 + 𝑝𝑝21.3 )𝜇𝜇1

298

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(6)

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VI.

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BUSY PERIOD OF THE SERVER

(a). Due to Hardware Repair Let 𝐵𝐵𝑖𝑖𝐻𝐻 (𝑡𝑡) 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 𝑆𝑆𝑖𝑖 𝑎𝑎𝑎𝑎 𝑡𝑡 = 0. The recursive relations for 𝐵𝐵𝑖𝑖𝐻𝐻 (𝑡𝑡) are as follows: 𝐵𝐵0𝐻𝐻 (𝑡𝑡) = 𝑞𝑞01 (𝑡𝑡)©𝐵𝐵1𝐻𝐻 (𝑡𝑡) + 𝑞𝑞02 (𝑡𝑡)©𝐵𝐵2𝐻𝐻 (𝑡𝑡) 𝐵𝐵1𝐻𝐻 (𝑡𝑡) = 𝑊𝑊1𝐻𝐻 (𝑡𝑡) + 𝑞𝑞10 (𝑡𝑡)©𝐵𝐵0𝐻𝐻 (𝑡𝑡) 𝐵𝐵2𝐻𝐻 (𝑡𝑡) = 𝑞𝑞20 (𝑡𝑡)©𝐵𝐵0𝐻𝐻 (𝑡𝑡) + 𝑞𝑞21.3 (𝑡𝑡)©𝐵𝐵1𝐻𝐻 (𝑡𝑡) + 𝑞𝑞22.4 (𝑡𝑡)©𝐵𝐵2𝐻𝐻 (𝑡𝑡) (7) where ������ 𝑑𝑑𝑑𝑑 𝑊𝑊1𝐻𝐻 (𝑡𝑡) = 𝐺𝐺(𝑡𝑡) (b). Due to software Up-gradation Let 𝐵𝐵𝑖𝑖𝑆𝑆 (𝑡𝑡) 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 𝑆𝑆𝑖𝑖 𝑎𝑎𝑎𝑎 𝑡𝑡 = 0. We have the following recursive relations for 𝐵𝐵𝑖𝑖𝑆𝑆 (𝑡𝑡): 𝐵𝐵0𝑆𝑆 (𝑡𝑡) = 𝑞𝑞01 (𝑡𝑡)©𝐵𝐵1𝑆𝑆 (𝑡𝑡) + 𝑞𝑞02 (𝑡𝑡)©𝐵𝐵2𝑆𝑆 (𝑡𝑡) 𝐵𝐵1𝑆𝑆 (𝑡𝑡) = 𝑞𝑞10 (𝑡𝑡)©𝐵𝐵0𝑆𝑆 (𝑡𝑡) 𝐵𝐵2𝑆𝑆 (𝑡𝑡) = 𝑊𝑊2𝑆𝑆 (𝑡𝑡) + 𝑞𝑞20 (𝑡𝑡)©𝐵𝐵0𝑆𝑆 (𝑡𝑡) + 𝑞𝑞21.3 (𝑡𝑡)©𝐵𝐵1𝑆𝑆 (𝑡𝑡) + 𝑞𝑞22.4 (𝑡𝑡)©𝐵𝐵2𝑆𝑆 (𝑡𝑡) (8) where ������ + �𝑏𝑏𝑏𝑏2 𝑒𝑒 −(𝑎𝑎𝑎𝑎 1 +𝑏𝑏𝑏𝑏 2 )𝑡𝑡 ©1�𝐹𝐹(𝑡𝑡) ������ 𝐹𝐹(𝑡𝑡) + �𝑎𝑎𝜆𝜆1 𝑒𝑒 −(𝑎𝑎𝜆𝜆 1 +𝑏𝑏𝑏𝑏 2 )𝑡𝑡 ©1�𝐹𝐹(𝑡𝑡) 𝑊𝑊2𝑆𝑆 (𝑡𝑡) = 𝑒𝑒 −(𝑎𝑎𝑎𝑎 1 +𝑏𝑏𝑏𝑏 2 )𝑡𝑡 ������ ∗ ∗ Taking LT of relations (7) & (8), solving for 𝐵𝐵0𝐻𝐻 (𝑡𝑡) 𝑎𝑎𝑎𝑎𝑎𝑎 𝐵𝐵0𝑆𝑆 (𝑡𝑡). The time for which server is busy due to repair and replacements respectively are given by ∗

𝐵𝐵0𝐻𝐻 (𝑡𝑡) = lim𝑠𝑠→0 𝑠𝑠 𝐵𝐵0𝐻𝐻 (𝑡𝑡) = ∗

𝑁𝑁𝑆𝑆3

𝐵𝐵0𝑆𝑆 (𝑡𝑡) = lim𝑠𝑠→0 𝑠𝑠 𝐵𝐵0𝑆𝑆 (𝑡𝑡) = 𝐷𝐷

2

𝑁𝑁𝐻𝐻 3 𝐷𝐷2

(9)

(10)

where ∗ ∗ 𝑁𝑁3𝐻𝐻 = 𝑝𝑝02 𝑝𝑝21.3 𝑊𝑊1𝐻𝐻 (0) + 𝑝𝑝01 (1 − 𝑝𝑝22.4 )𝑊𝑊1𝐻𝐻 (0) ∗ 𝑁𝑁3𝑆𝑆 = 𝑝𝑝02 𝑊𝑊2𝑆𝑆 (0)𝑎𝑎𝑎𝑎𝑎𝑎 𝐷𝐷2 𝑖𝑖𝑖𝑖 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚.

VII.

(11)

EXPECTED NUMBER OF HARDWARE REPAIRS

Let 𝑁𝑁𝑁𝑁𝑁𝑁𝑖𝑖 (𝑡𝑡) be the expected number of hardware repairs by the server in (0, t] given that the system entered the regenerative state 𝑆𝑆𝑖𝑖 𝑎𝑎𝑎𝑎 𝑡𝑡 = 0. The recursive relations for 𝑁𝑁𝑁𝑁𝑁𝑁𝑖𝑖 (𝑡𝑡) are given as: 𝑁𝑁𝑁𝑁𝑁𝑁0 (𝑡𝑡) = 𝑄𝑄01 (𝑡𝑡) & [1 + 𝑁𝑁𝑁𝑁𝑁𝑁1 (𝑡𝑡)] + 𝑄𝑄02 (𝑡𝑡) & 𝑁𝑁𝑁𝑁𝑁𝑁2 (𝑡𝑡)

𝑁𝑁𝑁𝑁𝑁𝑁1 (𝑡𝑡) = 𝑄𝑄10 (𝑡𝑡) & 𝑁𝑁𝑁𝑁𝑁𝑁0 (𝑡𝑡)

𝑁𝑁𝑁𝑁𝑁𝑁2 (𝑡𝑡) = 𝑄𝑄20 (𝑡𝑡) & 𝑁𝑁𝑁𝑁𝑁𝑁0 (𝑡𝑡) + 𝑄𝑄21.3 (𝑡𝑡) & 𝑁𝑁𝑁𝑁𝑁𝑁1 (𝑡𝑡) + 𝑄𝑄22.4 (𝑡𝑡) & 𝑁𝑁𝑁𝑁𝑁𝑁2 (𝑡𝑡)

(12)

Taking LST of relations (12) and solving for 𝑁𝑁𝑁𝑁𝑁𝑁0∗∗ (𝑠𝑠). The expected number of hardware repair is given by 𝑁𝑁 𝑁𝑁𝑁𝑁𝑁𝑁0 = lim𝑠𝑠→0 𝑠𝑠𝑁𝑁𝑁𝑁𝑁𝑁0∗∗ (𝑠𝑠) = 4 (13) 𝐷𝐷 2

Where 𝑁𝑁4 = 𝑝𝑝01 (1 − 𝑝𝑝22.4 )𝑎𝑎𝑎𝑎𝑎𝑎 𝐷𝐷2 𝑖𝑖𝑖𝑖 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚.

VIII.

(14)

EXPECTED NUMBER OF SOFTWARE UP-GRADATIONS

Let 𝑁𝑁𝑁𝑁𝑁𝑁𝑖𝑖 (𝑡𝑡) be the expected number of software up-gradations in (0, t] given that the system entered the regenerative state 𝑆𝑆𝑖𝑖 𝑎𝑎𝑎𝑎 𝑡𝑡 = 0. The recursive relations for 𝑁𝑁𝑁𝑁𝑁𝑁𝑖𝑖 (𝑡𝑡) are given as follows: 𝑁𝑁𝑁𝑁𝑁𝑁0 (𝑡𝑡) = 𝑄𝑄01 (𝑡𝑡) & 𝑁𝑁𝑁𝑁𝑁𝑁1 (𝑡𝑡) + 𝑄𝑄02 (𝑡𝑡) & [1 + 𝑁𝑁𝑁𝑁𝑁𝑁2 (𝑡𝑡)] 𝑁𝑁𝑁𝑁𝑁𝑁1 (𝑡𝑡) = 𝑄𝑄10 (𝑡𝑡) & 𝑁𝑁𝑁𝑁𝑁𝑁0 (𝑡𝑡)

𝑁𝑁𝑁𝑁𝑁𝑁2 (𝑡𝑡) = 𝑄𝑄20 (𝑡𝑡) & 𝑁𝑁𝑁𝑁𝑁𝑁0 (𝑡𝑡) + 𝑄𝑄21.3 (𝑡𝑡) & 𝑁𝑁𝑁𝑁𝑁𝑁1 (𝑡𝑡) + 𝑄𝑄22.4 (𝑡𝑡) & 𝑁𝑁𝑁𝑁𝑁𝑁2 (𝑡𝑡) 299

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(15)

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Taking LST of relations (15) and solving for 𝑁𝑁0∗∗ (𝑠𝑠). The expected numbers of software up-gradation are given by 𝑁𝑁 𝑁𝑁𝑁𝑁𝑁𝑁0 (∞) = lim𝑠𝑠→0 𝑠𝑠𝑁𝑁𝑁𝑁𝑁𝑁0∗∗ (𝑠𝑠) = 5 (16) 𝐷𝐷 2

Where 𝑁𝑁5 = 𝑝𝑝02 (1 − 𝑝𝑝22.4 ) 𝑎𝑎𝑎𝑎𝑎𝑎 𝐷𝐷2 𝑖𝑖𝑖𝑖 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚

IX.

(17)

COST-BENEFIT ANALYSIS

The profit incurred to the system model in steady state can be obtained as: 𝑃𝑃 = 𝐾𝐾0 𝐴𝐴0 − 𝐾𝐾1 𝐵𝐵0𝐻𝐻 − 𝐾𝐾2 𝐵𝐵0𝑆𝑆 − 𝐾𝐾3 𝑁𝑁𝑁𝑁𝑁𝑁0 − 𝐾𝐾4 𝑁𝑁𝑁𝑁𝑁𝑁0

(18)

Where

𝐾𝐾0 = 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝑝𝑝𝑝𝑝𝑝𝑝 𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢 𝑢𝑢𝑢𝑢 − 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑜𝑜𝑜𝑜 𝑡𝑡ℎ𝑒𝑒 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠

𝐾𝐾1 = 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 𝑝𝑝𝑝𝑝𝑝𝑝 𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑓𝑓𝑓𝑓𝑓𝑓 𝑤𝑤ℎ𝑖𝑖𝑖𝑖ℎ 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑖𝑖𝑖𝑖 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 𝑑𝑑𝑑𝑑𝑑𝑑 𝑡𝑡𝑡𝑡 ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟

𝐾𝐾2 = 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 𝑝𝑝𝑝𝑝𝑝𝑝 𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑓𝑓𝑓𝑓𝑓𝑓 𝑤𝑤ℎ𝑖𝑖𝑖𝑖ℎ 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑖𝑖𝑖𝑖 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 𝑑𝑑𝑑𝑑𝑑𝑑 𝑡𝑡𝑡𝑡 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑢𝑢𝑢𝑢 − 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝐾𝐾3 = 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 𝑝𝑝𝑝𝑝𝑝𝑝 𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑜𝑜𝑜𝑜 𝑡𝑡ℎ𝑒𝑒 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎

𝐾𝐾4 = 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 𝑝𝑝𝑝𝑝𝑝𝑝 𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢 𝑢𝑢𝑢𝑢 − 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑜𝑜𝑜𝑜 𝑡𝑡ℎ𝑒𝑒 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑎𝑎𝑎𝑎𝑎𝑎 𝐴𝐴0 , 𝐵𝐵0𝐻𝐻 , 𝐵𝐵0𝑆𝑆 , 𝑁𝑁𝑁𝑁𝑁𝑁0 , 𝑁𝑁𝑁𝑁𝑁𝑁0 𝑎𝑎𝑎𝑎𝑎𝑎 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑.

X.

PARTICULAR CASES

Suppose 𝑔𝑔(𝑡𝑡) = 𝛼𝛼𝑒𝑒 −𝛼𝛼𝛼𝛼 𝑎𝑎𝑎𝑎𝑎𝑎 𝑓𝑓(𝑡𝑡) = 𝜃𝜃𝑒𝑒 −𝜃𝜃𝜃𝜃 We can obtain the following results: 𝑁𝑁 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀(𝑇𝑇0 ) = 1 𝐷𝐷1

𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴(𝐴𝐴0 ) =

𝑁𝑁2 𝐷𝐷2

𝑁𝑁𝐻𝐻 3 𝐷𝐷2 𝑁𝑁𝑆𝑆3

𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 𝑑𝑑𝑑𝑑𝑑𝑑 𝑡𝑡𝑡𝑡 ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 (𝐵𝐵0𝐻𝐻 ) = 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 𝑑𝑑𝑑𝑑𝑑𝑑 𝑡𝑡𝑡𝑡 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 �𝐵𝐵0𝑆𝑆 � =

𝐷𝐷2

𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛 𝑜𝑜𝑜𝑜 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑎𝑎𝑎𝑎 ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 (𝑁𝑁𝑁𝑁𝑁𝑁0 ) =

𝑁𝑁4 𝐷𝐷2

𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛 𝑜𝑜𝑜𝑜 𝑢𝑢𝑢𝑢 − 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑎𝑎𝑎𝑎 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑟𝑟𝑟𝑟 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 (𝑁𝑁𝑁𝑁𝑁𝑁0 ) =

Where 𝑁𝑁1 = (𝑎𝑎𝜆𝜆 𝑁𝑁2 = 𝑎𝑎𝜆𝜆

𝑎𝑎𝜆𝜆1 +2𝑏𝑏𝜆𝜆2 +𝜃𝜃 +𝑏𝑏𝜆𝜆 1 2 )(𝑎𝑎𝜆𝜆1 +𝑏𝑏𝜆𝜆2 +𝜃𝜃)

𝐷𝐷1 =

1

𝐷𝐷2 =

1 +𝑏𝑏𝜆𝜆2

𝑁𝑁3𝐻𝐻 = (𝑎𝑎𝜆𝜆 𝑁𝑁4 = (𝑎𝑎𝜆𝜆 300

𝑎𝑎𝜆𝜆1

1 +𝑏𝑏𝜆𝜆2 )𝛼𝛼

𝑎𝑎𝜆𝜆1 (𝑎𝑎𝜆𝜆1 +𝜃𝜃) +𝑏𝑏𝜆𝜆 1 2 )(𝑎𝑎𝜆𝜆1 +𝑏𝑏𝜆𝜆2 +𝜃𝜃)

(𝑎𝑎𝜆𝜆1 +𝑏𝑏𝜆𝜆2 )(𝑎𝑎𝜆𝜆1 +𝑏𝑏𝜆𝜆2 +𝜃𝜃)−𝜃𝜃𝜃𝜃𝜆𝜆2 (𝑎𝑎𝜆𝜆1 +𝑏𝑏𝜆𝜆2 )(𝑎𝑎𝜆𝜆1 +𝑏𝑏𝜆𝜆2 +𝜃𝜃)

𝑁𝑁5 𝐷𝐷2

𝛼𝛼𝛼𝛼(𝑎𝑎𝜆𝜆1 +𝜃𝜃)+(𝜃𝜃𝜃𝜃𝜆𝜆1 +𝛼𝛼𝛼𝛼𝜆𝜆2 )(𝜃𝜃𝜃𝜃𝜆𝜆1 +𝑏𝑏𝜆𝜆2 +𝜃𝜃) 𝛼𝛼𝛼𝛼(𝑎𝑎𝜆𝜆1 +𝑏𝑏𝜆𝜆2 )(𝑎𝑎𝜆𝜆1 +𝑏𝑏𝜆𝜆2 +𝜃𝜃)

𝑁𝑁3𝑆𝑆 = (𝑎𝑎𝜆𝜆

𝑁𝑁5 = (𝑎𝑎𝜆𝜆

𝑏𝑏𝜆𝜆1

1 +𝑏𝑏𝜆𝜆2 )𝜃𝜃

𝑏𝑏𝜆𝜆2 (𝑎𝑎𝜆𝜆1 +𝜃𝜃) +𝑏𝑏𝜆𝜆 1 2 )(𝑎𝑎𝜆𝜆1 +𝑏𝑏𝜆𝜆2 +𝜃𝜃)

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XI.

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CONCLUSION

The behaviour of some important performance measures such as MTSF, availability and profit with respect to hardware failure rate (𝜆𝜆1 ) has been observed for arbitrary values of various parameters including K 0 = 15000, K1 1000, K 2 = 700, K 3 = 1500, K 4 = 1200 with a=0.6 and b=0.4 as shown respectively in figures 2, 3 and 4. It is revealed that these measures go on decreasing with the increase of hardware and software failure rates. But, their values increase with the increase of hardware repair rate (α) and up-gradation rate (θ). On the other hand, if the values of a and b are interchanged i.e. a=0.4 and b=0.6, than MTSF and availability of the system increase while profit declines. Hence the study reveals that a computer system in which software redundancy is provided in cold standby be more profitable if it has more chances of hardware failure may because of the less hardware repairable cost.

301

REFERENCES [1] Anand, Jyoti and Malik, S.C. (2012): Analysis of a Computer System with Arbitrary Distributions for H/W and S/W Replacement Time and Priority to Repair Activities of H/W over Replacement of the S/W, International Journal of Systems Assurance Engineering and Management, Vol.3 (3), pp. 230-236. [2] Kumar, Ashish; Anand, Jyoti and Malik, S.C. (2013): Stochastic Modeling of a Computer System with Priority to Up-gradation of Software over Hardware Repair Activities. International Journal of Agricultural and Statistical Sciences, Vol. 9(1), pp. 117-126. [3] Malik, S.C. and Anand, Jyoti (2010): Reliability and economic analysis of a computer system with independent hardware and software failures. Bulletin of Pure and Applied Sciences, Vol. 29E(01), pp.141-153. [4] Malik, S.C. and Munday, V.J. (2014): Stochastic Modeling of a Computer System with Hardware Redundancy. International Journal of Computer Applications, Vol. 89(7), pp. 26-30. [5] Malik, S.C. and Sureria, J.K. (2012): Profit Analysis of a Computer System with H/W Repair and S/W Replacement. International Journal of Computer Applications, Vol.47(1), pp. 19-26.

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MTSF Vs H/w Failure Rate (λ1) 300 250

λ2=0.001,α=2,θ=5,a=0.6,b=0.4

200

λ2=0.002 θ=7

150

a=0.4,b=0.6

MTSF

100 50 0

0.010.020.030.040.050.060.070.080.09 0.1

Fig. 2 Hardware Failure Rate (λ1)

Availability Vs Hardware Failure rate (λ1)

λ2=0.001,α=2,θ=5,a=0.6,b=0.4 λ2=0.002 α=3 θ=7 a=0.4,b=0.6

1.01 1 0.99

Availability

0.98 0.97 0.96 0.95

0.010.020.030.040.050.060.070.080.09 0.1

Fig. 3

Hardware Failure Rate (λ1)

Profit Vs Hardware Failure Rate (λ1) λ2=0.001, α=2, θ=5, a=0.6, b=0.4 λ2=0.002 α=3 θ=7 a=0.4, b=0.6

15100 15000 14900 14800 14700 14600 14500 14400 14300 14200 14100

Profit

K 0 = 15000, K1 = 1000, K 2 = 700, K 3 = 1500, K 4 = 1200 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1

Hardware Failure Rate (λ1)

302

Fig. 4

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