Attitude towards E-learning Technology -- A study in Kerala

(1)

295 Copyright © 2011-15. Vandana Publications. All Rights Reserved.

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

V.J. Munday1, S.C. Malik2

1,2_{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

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 semi-Markov 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.

II. NOTATIONS

E

:

Set of regenerative states

𝐸𝐸�

:

Set of non-regenerative states

O

:

Computer system is operative

(2)

296 Copyright © 2011-15. Vandana Publications. All Rights Reserved.

a/b

:

Probability that the system has hardware / software failure

𝜆𝜆

1

/

𝜆𝜆

2

:

Hardware/Software failure rate

HFUr /HFWr

:

The hardware is failed and under repair/waiting for repair

SFUg/SFWUg

:

The software is failed and under/waiting for up-gradation

HFUR/HFWR

:

The hardware is failed and continuously under repair / waiting

for repair from previous state

SFUG/SFWUG

:

The software is failed and continuously under up-gradation

/waiting for up- gradation from previous state

g(t)/G(t)

:

pdf/cdf of hardware repair time

f(t)/F(t)

:

pdf/cdf of software up-gradation time

𝑞𝑞

𝑖𝑖𝑖𝑖

(

𝑡𝑡

)/

𝑄𝑄

𝑖𝑖𝑖𝑖

(

𝑡𝑡

)

:

pdf / cdf of first passage time from regenerative state

𝑆𝑆

𝑖𝑖

to a

regenerative state

𝑆𝑆

_𝑖𝑖

or to a failed state

𝑆𝑆

_𝑖𝑖

without visiting any

other regenerative state in (0, t]

𝑞𝑞

𝑖𝑖𝑖𝑖.𝑘𝑘

(

𝑡𝑡

)/

𝑄𝑄

𝑖𝑖𝑖𝑖.𝑘𝑘

(

𝑡𝑡

)

:

pdf/cdf of direct transition time from regenerative state

𝑆𝑆

𝑖𝑖

to a

regenerative state

𝑆𝑆

_𝑖𝑖

or to a failed state

𝑆𝑆

_𝑖𝑖

visiting state

𝑆𝑆𝑘𝑘

once in (0, t]

𝑀𝑀

𝑖𝑖

(

𝑡𝑡

)

:

Probability that the system up initially in state

𝑆𝑆

𝑖𝑖

𝜖𝜖𝐸𝐸

is up

at time t without visiting to any regenerative state

𝑊𝑊

𝑖𝑖

(

𝑡𝑡

)

:

Probability that the server is busy in the state

𝑆𝑆

𝑖𝑖

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.

𝜇𝜇

𝑖𝑖

:

The mean sojourn time in state

𝑆𝑆

𝑖𝑖

which is given by

𝜇𝜇𝑖𝑖

=

𝐸𝐸

(

𝑇𝑇

) =

∫ 𝑃𝑃

₀∞

(

𝑇𝑇

>

𝑡𝑡

)

𝑑𝑑𝑡𝑡

=

∑ 𝑚𝑚𝑖𝑖𝑖𝑖

𝑖𝑖

,

where

𝑇𝑇

denotes the time to system failure.

𝑚𝑚

𝑖𝑖𝑖𝑖

:

Contribution to mean sojourn time (

𝜇𝜇

𝑖𝑖

) in state

𝑆𝑆

𝑖𝑖

when system

transits directly to state

𝑆𝑆

_𝑖𝑖

so that

𝜇𝜇𝑖𝑖

=

∑ 𝑚𝑚𝑖𝑖𝑖𝑖

_𝑖𝑖

𝑎𝑎𝑎𝑎𝑑𝑑

𝑚𝑚𝑖𝑖𝑖𝑖

=

∫ 𝑡𝑡𝑑𝑑𝑄𝑄𝑖𝑖𝑖𝑖

₀∞

(

𝑡𝑡

) =

−𝑞𝑞

_{𝑖𝑖𝑖𝑖}∗′

(0)



&

:

Symbol for Laplace-Stieltjes convolution/Laplace convolution

(3)

297 Copyright © 2011-15. Vandana Publications. All Rights Reserved.

State Transition Diagram

Up-State 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.

𝑝𝑝

𝑖𝑖𝑖𝑖

=

𝑄𝑄

𝑖𝑖𝑖𝑖

(

∞

) =

∫ 𝑞𝑞𝑖𝑖𝑖𝑖

₀∞

(

𝑡𝑡

)

𝑑𝑑𝑡𝑡

𝑝𝑝

01

=

_{𝑎𝑎𝜆𝜆}₁₊𝑎𝑎𝜆𝜆1_{𝑏𝑏𝜆𝜆}₂

,

𝑝𝑝

02

=

_{𝑎𝑎𝜆𝜆}𝑏𝑏𝜆𝜆₁₊2_{𝑏𝑏𝜆𝜆}₂

,

𝑝𝑝

10

=

𝑔𝑔

∗

(0)

𝑝𝑝

20

=

𝑓𝑓

∗

(

𝑎𝑎𝜆𝜆

1

+

𝑏𝑏𝜆𝜆

2

)

𝑝𝑝

23

=

_{𝑎𝑎𝜆𝜆}𝑏𝑏𝜆𝜆2

1+𝑏𝑏𝜆𝜆2

{1

− 𝑓𝑓

∗

₍

_{𝑎𝑎𝜆𝜆}

1

+

𝑏𝑏𝜆𝜆

2

)}

𝑝𝑝

24

=

_{𝑎𝑎𝜆𝜆}𝑎𝑎𝜆𝜆1

1+𝑏𝑏𝜆𝜆2

{1

− 𝑓𝑓

∗

₍

_{𝑎𝑎𝜆𝜆}

1

+

𝑏𝑏𝜆𝜆

2

)}

,

𝑝𝑝

31

=

𝑝𝑝

42

=

𝑓𝑓

∗

(0)

For

𝑔𝑔

(

𝑡𝑡

) =

𝛼𝛼𝑒𝑒

−𝛼𝛼𝑡𝑡

𝑎𝑎𝑎𝑎𝑑𝑑

𝑓𝑓

(

𝑡𝑡

) =

𝜃𝜃𝑒𝑒

−𝜃𝜃𝑡𝑡

we have

𝑝𝑝

21.3

=

_{𝑎𝑎𝜆𝜆}𝑎𝑎𝜆𝜆₁₊1_{𝑏𝑏𝜆𝜆}₂

{1

− 𝑓𝑓

∗

(

𝑎𝑎𝜆𝜆

1

+

𝑏𝑏𝜆𝜆

2

)}

𝑝𝑝

22.4

=

_{𝑎𝑎𝜆𝜆}₁𝑏𝑏𝜆𝜆₊2_{𝑏𝑏𝜆𝜆}₂

{1

− 𝑓𝑓

∗

(

𝑎𝑎𝜆𝜆

1

+

𝑏𝑏𝜆𝜆

2

)}

But,

𝑓𝑓

∗

(0) =

𝑔𝑔

∗

(0) = 1

𝑎𝑎𝑎𝑎𝑑𝑑

𝑝𝑝

+

𝑞𝑞

= 1

It can be easily verified that

𝑝𝑝

01

+

𝑝𝑝

02

=

𝑝𝑝

10

=

𝑝𝑝

20

+

𝑝𝑝

23

+

𝑝𝑝

24

=

𝑝𝑝

31

=

𝑝𝑝

42

=

𝑝𝑝

20

+

𝑝𝑝

21.3

+

𝑝𝑝

22.4

= 1

The mean sojourn times (

𝜇𝜇

_𝑖𝑖

) is the state

𝑆𝑆

_𝑖𝑖

are

𝜇𝜇

0

=

_{𝑎𝑎𝜆𝜆}₁₊1_{𝑏𝑏𝜆𝜆}₂

𝜇𝜇

1

=

1_𝛼𝛼

𝜇𝜇

2

=

�

_{𝑎𝑎𝜆𝜆}₁₊1_{𝑏𝑏𝜆𝜆}₂

�

{1

− 𝑓𝑓

∗

(

𝑎𝑎𝜆𝜆

1

+

𝑏𝑏𝜆𝜆

2

)} =

_{𝑎𝑎𝜆𝜆}₁₊1_{𝑏𝑏𝜆𝜆}₂₊_𝜃𝜃

Also

𝑚𝑚

01

+

𝑚𝑚

02

=

𝜇𝜇

0

,

𝑚𝑚

10

=

𝜇𝜇

1

,

𝑚𝑚

20

+

𝑚𝑚

23

+

𝑚𝑚

24

=

𝜇𝜇

2 S4

S3 f(t)

𝑎𝑎𝜆𝜆1

𝑏𝑏𝜆𝜆2

f(t) S2

𝑏𝑏𝜆𝜆2 _f(t)

g(t)

𝑎𝑎𝜆𝜆1

S0 S1

O SFUg

SFWUg SFUG

HFWr SFUG HFUr

Scs O

(4)

298 Copyright © 2011-15. Vandana Publications. All Rights Reserved.

and

𝑚𝑚

20

+

𝑚𝑚

21.3

+

𝑚𝑚

22.4

=

𝜇𝜇

2′

=

_𝜃𝜃

1

IV. 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

𝜙𝜙

_𝑖𝑖

(

𝑡𝑡

)

,

𝜙𝜙

0

(

𝑡𝑡

) =

𝑄𝑄

02

(

𝑡𝑡

)

&

𝜙𝜙

2

(

𝑡𝑡

) +

𝑄𝑄

01

(

𝑡𝑡

)

𝜙𝜙

2

(

𝑡𝑡

) =

𝑄𝑄

20

(

𝑡𝑡

)

&

𝜙𝜙

0

(

𝑡𝑡

) +

𝑄𝑄

23

(

𝑡𝑡

) +

𝑄𝑄

24

(

𝑡𝑡

)

(1)

Taking LST of above relations (1) and solving for

ϕ

₀∗∗

(

𝑠𝑠

)

We have

𝑅𝑅

∗

₍

_𝑠𝑠

_{) =}

1−ϕ0∗∗(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

𝑀𝑀𝑇𝑇𝑆𝑆𝑀𝑀

= lim

𝑠𝑠→01−ϕ0

∗∗₍_𝑠𝑠₎

𝑠𝑠

=

𝑁𝑁1

𝐷𝐷1

(2)

Where

𝑁𝑁

1

=

𝜇𝜇

0

+

𝑝𝑝

02

𝜇𝜇

2

𝑎𝑎𝑎𝑎𝑑𝑑

𝐷𝐷

1

= 1

− 𝑝𝑝

02

𝑝𝑝

20

(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

𝑎𝑎𝑡𝑡

𝑡𝑡

= 0

. The recursive relations for

𝐴𝐴𝑖𝑖

(

𝑡𝑡

)

are given as:

𝐴𝐴

0

(

𝑡𝑡

) =

𝑀𝑀

0

(

𝑡𝑡

) +

𝑞𝑞

01

(

𝑡𝑡

)



𝐴𝐴

1

(

𝑡𝑡

) +

𝑞𝑞

02

(

𝑡𝑡

)



𝐴𝐴

2

(

𝑡𝑡

)

𝐴𝐴

1

(

𝑡𝑡

) =

𝑞𝑞

10

(

𝑡𝑡

)



𝐴𝐴

0

(

𝑡𝑡

)

𝐴𝐴

2

(

𝑡𝑡

) =

𝑀𝑀

2

(

𝑡𝑡

) +

𝑞𝑞

20

(

𝑡𝑡

)



𝐴𝐴

0

(

𝑡𝑡

) +

𝑞𝑞

21.3

(

𝑡𝑡

)



𝐴𝐴

1

(

𝑡𝑡

) +

𝑞𝑞

22.4

(

𝑡𝑡

)



𝐴𝐴

2

(

𝑡𝑡

)

(4)

where

𝑀𝑀

0

(

𝑡𝑡

) =

𝑒𝑒

−(𝑎𝑎𝜆𝜆1+𝑏𝑏𝜆𝜆2)𝑡𝑡

𝑎𝑎𝑎𝑎𝑑𝑑

𝑀𝑀

1

(

𝑡𝑡

) =

𝑒𝑒

𝑀𝑀

��

(

𝑡𝑡

)

Taking LT of relations (4) and solving for

𝐴𝐴

₀∗

(

𝑠𝑠

)

, the steady state availability is given by

𝐴𝐴

0

(

∞

) = lim

𝑠𝑠→0

𝑠𝑠 𝐴𝐴

∗0

(

𝑠𝑠

) =

𝑁𝑁_𝐷𝐷2₂

(5)

Where

𝑁𝑁

2

= (1

− 𝑝𝑝

22.4

)

𝜇𝜇

0

+

𝑝𝑝

02

𝜇𝜇

2

(5)

299 Copyright © 2011-15. Vandana Publications. All Rights Reserved.

VI. BUSY PERIOD OF THE SERVER

(a). Due to Hardware Repair

Let

𝐵𝐵

_𝑖𝑖𝐻𝐻

(

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

𝑆𝑆

_𝑖𝑖

𝑎𝑎𝑡𝑡

𝑡𝑡

= 0

. The recursive relations for

𝐵𝐵

_𝑖𝑖𝐻𝐻

(

𝑡𝑡

)

are as follows:

𝐵𝐵

0𝐻𝐻

(

𝑡𝑡

) =

𝑞𝑞

01

(

𝑡𝑡

)©

𝐵𝐵

1𝐻𝐻

(

𝑡𝑡

) +

𝑞𝑞

02

(

𝑡𝑡

)©

𝐵𝐵

2𝐻𝐻

(

𝑡𝑡

)

𝐵𝐵

1𝐻𝐻

(

𝑡𝑡

) =

𝑊𝑊

1𝐻𝐻

(

𝑡𝑡

) +

𝑞𝑞

10

(

𝑡𝑡

)©

𝐵𝐵

0𝐻𝐻

(

𝑡𝑡

)

𝐵𝐵

2𝐻𝐻

(

𝑡𝑡

) =

𝑞𝑞

20

(

𝑡𝑡

)©

𝐵𝐵

0𝐻𝐻

(

𝑡𝑡

) +

𝑞𝑞

21.3

(

𝑡𝑡

)©

𝐵𝐵

1𝐻𝐻

(

𝑡𝑡

) +

𝑞𝑞

22.4

(

𝑡𝑡

)©

𝐵𝐵

2𝐻𝐻

(

𝑡𝑡

)

(7)

where

𝑊𝑊

1𝐻𝐻

(

𝑡𝑡

) =

𝐺𝐺

��

(

𝑡𝑡

)

𝑑𝑑𝑡𝑡

(b). Due to software Up-gradation

Let

𝐵𝐵

_𝑖𝑖𝑆𝑆

(

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

𝑆𝑆

_𝑖𝑖

𝑎𝑎𝑡𝑡

𝑡𝑡

= 0

. We have the following recursive relations for

𝐵𝐵

_𝑖𝑖𝑆𝑆

(

𝑡𝑡

)

:

𝐵𝐵

0𝑆𝑆

(

𝑡𝑡

) =

𝑞𝑞

01

(

𝑡𝑡

)©

𝐵𝐵

1𝑆𝑆

(

𝑡𝑡

) +

𝑞𝑞

02

(

𝑡𝑡

)©

𝐵𝐵

2𝑆𝑆

(

𝑡𝑡

)

𝐵𝐵

1𝑆𝑆

(

𝑡𝑡

) =

𝑞𝑞

10

(

𝑡𝑡

)©

𝐵𝐵

0𝑆𝑆

(

𝑡𝑡

)

𝐵𝐵

2𝑆𝑆

(

𝑡𝑡

) =

𝑊𝑊

2𝑆𝑆

(

𝑡𝑡

) +

𝑞𝑞

20

(

𝑡𝑡

)©

𝐵𝐵

0𝑆𝑆

(

𝑡𝑡

) +

𝑞𝑞

21.3

(

𝑡𝑡

)©

𝐵𝐵

1𝑆𝑆

(

𝑡𝑡

) +

𝑞𝑞

22.4

(

𝑡𝑡

)©

𝐵𝐵

2𝑆𝑆

(

𝑡𝑡

)

(8)

where

𝑊𝑊

2𝑆𝑆

(

𝑡𝑡

) =

𝑒𝑒

𝑀𝑀

��

(

𝑡𝑡

)

+

�𝑎𝑎𝜆𝜆

1

𝑒𝑒

©1

�𝑀𝑀

��

(

𝑡𝑡

)

+

�𝑏𝑏𝜆𝜆

2

𝑒𝑒

©1

�𝑀𝑀

��

(

𝑡𝑡

)

Taking LT of relations (7) & (8), solving for

𝐵𝐵

₀𝐻𝐻∗

(

𝑡𝑡

)

𝑎𝑎𝑎𝑎𝑑𝑑

𝐵𝐵

₀𝑆𝑆∗

(

𝑡𝑡

)

. The time for which server is busy due to

repair and replacements respectively are given by

𝐵𝐵

₀𝐻𝐻

(

𝑡𝑡

) = lim

_𝑠𝑠→₀

𝑠𝑠 𝐵𝐵

₀𝐻𝐻∗

(

𝑡𝑡

) =

𝑁𝑁_𝐷𝐷3𝐻𝐻

2

(9)

𝐵𝐵

0𝑆𝑆

(

𝑡𝑡

) = lim

𝑠𝑠→0

𝑠𝑠 𝐵𝐵

0𝑆𝑆∗

(

𝑡𝑡

) =

𝑁𝑁3

𝑆𝑆

𝐷𝐷2

(10)

where

𝑁𝑁

3𝐻𝐻

=

𝑝𝑝

02

𝑝𝑝

21.3

𝑊𝑊

1𝐻𝐻∗

(0) +

𝑝𝑝

01

(1

− 𝑝𝑝

22.4

)

𝑊𝑊

1𝐻𝐻∗

(0)

𝑁𝑁

3𝑆𝑆

=

𝑝𝑝

02

𝑊𝑊

2𝑆𝑆∗

(0)

𝑎𝑎𝑎𝑎𝑑𝑑

𝐷𝐷

2

𝑖𝑖𝑠𝑠

𝑎𝑎𝑎𝑎𝑎𝑎𝑒𝑒𝑎𝑎𝑑𝑑𝑎𝑎

𝑚𝑚𝑒𝑒𝑎𝑎𝑡𝑡𝑖𝑖𝑚𝑚𝑎𝑎𝑒𝑒𝑑𝑑

.

(11)

VII. EXPECTED NUMBER OF HARDWARE REPAIRS

Let

𝑁𝑁𝐻𝐻𝑅𝑅𝑖𝑖

(

𝑡𝑡

)

be the expected number of hardware repairs by the server in (0, t] given that the system

entered the regenerative state

𝑆𝑆𝑖𝑖

𝑎𝑎𝑡𝑡

𝑡𝑡

= 0

. The recursive relations for

𝑁𝑁𝐻𝐻𝑅𝑅

_𝑖𝑖

(

𝑡𝑡

)

are given as:

𝑁𝑁𝐻𝐻𝑅𝑅

0

(

𝑡𝑡

) =

𝑄𝑄

01

(

𝑡𝑡

)

&

[1 +

𝑁𝑁𝐻𝐻𝑅𝑅

1

(

𝑡𝑡

)] +

𝑄𝑄

02

(

𝑡𝑡

)

&

𝑁𝑁𝐻𝐻𝑅𝑅

2

(

𝑡𝑡

)

𝑁𝑁𝐻𝐻𝑅𝑅

1

(

𝑡𝑡

) =

𝑄𝑄

10

(

𝑡𝑡

)

&

𝑁𝑁𝐻𝐻𝑅𝑅

0

(

𝑡𝑡

)

𝑁𝑁𝐻𝐻𝑅𝑅

2

(

𝑡𝑡

) =

𝑄𝑄

20

(

𝑡𝑡

)

&

𝑁𝑁𝐻𝐻𝑅𝑅

0

(

𝑡𝑡

) +

𝑄𝑄

21.3

(

𝑡𝑡

)

&

𝑁𝑁𝐻𝐻𝑅𝑅

1

(

𝑡𝑡

) +

𝑄𝑄

22.4

(

𝑡𝑡

)

&

𝑁𝑁𝐻𝐻𝑅𝑅

2

(

𝑡𝑡

)

(12)

Taking LST of relations (12) and solving for

𝑁𝑁𝐻𝐻𝑅𝑅

₀∗∗

(

𝑠𝑠

)

. The expected number of hardware repair is given by

𝑁𝑁𝐻𝐻𝑅𝑅

0

= lim

𝑠𝑠→0

𝑠𝑠𝑁𝑁𝐻𝐻𝑅𝑅

0∗∗

(

𝑠𝑠

) =

(13)

Where

𝑁𝑁

4

=

𝑝𝑝

01

(1

− 𝑝𝑝

22.4

)

𝑎𝑎𝑎𝑎𝑑𝑑

𝐷𝐷

2

𝑖𝑖𝑠𝑠

𝑎𝑎𝑎𝑎𝑎𝑎𝑒𝑒𝑎𝑎𝑑𝑑𝑎𝑎

𝑚𝑚𝑒𝑒𝑎𝑎𝑡𝑡𝑖𝑖𝑚𝑚𝑎𝑎𝑒𝑒𝑑𝑑

.

(14)

VIII. EXPECTED NUMBER OF SOFTWARE UP-GRADATIONS

Let

𝑁𝑁𝑆𝑆𝑁𝑁𝑖𝑖

(

𝑡𝑡

)

be the expected number of software up-gradations in (0, t] given that the system entered the

regenerative state

𝑆𝑆

_𝑖𝑖

𝑎𝑎𝑡𝑡

𝑡𝑡

= 0

. The recursive relations for

𝑁𝑁𝑆𝑆𝑁𝑁𝑖𝑖

(

𝑡𝑡

)

are given as follows:

𝑁𝑁𝑆𝑆𝑁𝑁

0

(

𝑡𝑡

) =

𝑄𝑄

01

(

𝑡𝑡

)

&

𝑁𝑁𝑆𝑆𝑁𝑁

1

(

𝑡𝑡

) +

𝑄𝑄

02

(

𝑡𝑡

)

&

[1 +

𝑁𝑁𝑆𝑆𝑁𝑁

2

(

𝑡𝑡

)]

𝑁𝑁𝑆𝑆𝑁𝑁

1

(

𝑡𝑡

) =

𝑄𝑄

10

(

𝑡𝑡

)

&

𝑁𝑁𝑆𝑆𝑁𝑁

0

(

𝑡𝑡

)

(6)

Taking LST of relations (15) and solving for

𝑁𝑁

₀∗∗

(

𝑠𝑠

)

. The expected numbers of software up-gradation are given

by

𝑁𝑁𝑆𝑆𝑁𝑁

0

(

∞

) = lim

𝑠𝑠→0

𝑠𝑠𝑁𝑁𝑆𝑆𝑁𝑁

0∗∗

(

𝑠𝑠

) =

(16)

Where

𝑁𝑁

5

=

𝑝𝑝

02

(1

− 𝑝𝑝

22.4)

𝑎𝑎𝑎𝑎𝑑𝑑

𝐷𝐷

2

𝑖𝑖𝑠𝑠

𝑎𝑎𝑎𝑎𝑎𝑎𝑒𝑒𝑎𝑎𝑑𝑑𝑎𝑎

𝑚𝑚𝑒𝑒𝑎𝑎𝑡𝑡𝑖𝑖𝑚𝑚𝑎𝑎𝑒𝑒𝑑𝑑

(17)

IX. COST-BENEFIT ANALYSIS

The profit incurred to the system model in steady state can be obtained as:

𝑃𝑃

=

𝐾𝐾

0

𝐴𝐴

0

− 𝐾𝐾

1

𝐵𝐵

0𝐻𝐻

− 𝐾𝐾

2

𝐵𝐵

0𝑆𝑆

− 𝐾𝐾

3

𝑁𝑁𝐻𝐻𝑅𝑅

0

− 𝐾𝐾

4

𝑁𝑁𝑆𝑆𝑁𝑁

0

(18)

Where

𝐾𝐾

0

=

𝑅𝑅𝑒𝑒𝑅𝑅𝑒𝑒𝑎𝑎𝑅𝑅𝑒𝑒

𝑝𝑝𝑒𝑒𝑎𝑎

𝑅𝑅𝑎𝑎𝑖𝑖𝑡𝑡

𝑅𝑅𝑝𝑝 − 𝑡𝑡𝑖𝑖𝑚𝑚𝑒𝑒

𝑚𝑚𝑓𝑓

𝑡𝑡ℎ𝑒𝑒

𝑠𝑠𝑎𝑎𝑠𝑠𝑡𝑡𝑒𝑒𝑚𝑚

𝐾𝐾

1

=

𝐶𝐶𝑚𝑚𝑠𝑠𝑡𝑡

𝑝𝑝𝑒𝑒𝑎𝑎

𝑅𝑅𝑎𝑎𝑖𝑖𝑡𝑡

𝑡𝑡𝑖𝑖𝑚𝑚𝑒𝑒

𝑓𝑓𝑚𝑚𝑎𝑎

𝑤𝑤ℎ𝑖𝑖𝑖𝑖ℎ

𝑠𝑠𝑒𝑒𝑎𝑎𝑅𝑅𝑒𝑒𝑎𝑎

𝑖𝑖𝑠𝑠

𝑏𝑏𝑅𝑅𝑠𝑠𝑎𝑎

𝑑𝑑𝑅𝑅𝑒𝑒

𝑡𝑡𝑚𝑚

ℎ𝑎𝑎𝑎𝑎𝑑𝑑𝑤𝑤𝑎𝑎𝑎𝑎𝑒𝑒

𝑎𝑎𝑒𝑒𝑝𝑝𝑎𝑎𝑖𝑖𝑎𝑎

𝐾𝐾

2

=

𝐶𝐶𝑚𝑚𝑠𝑠𝑡𝑡

𝑝𝑝𝑒𝑒𝑎𝑎

𝑅𝑅𝑎𝑎𝑖𝑖𝑡𝑡

𝑡𝑡𝑖𝑖𝑚𝑚𝑒𝑒

𝑓𝑓𝑚𝑚𝑎𝑎

𝑤𝑤ℎ𝑖𝑖𝑖𝑖ℎ

𝑠𝑠𝑒𝑒𝑎𝑎𝑅𝑅𝑒𝑒𝑎𝑎

𝑖𝑖𝑠𝑠

𝑏𝑏𝑅𝑅𝑠𝑠𝑎𝑎

𝑑𝑑𝑅𝑅𝑒𝑒

𝑡𝑡𝑚𝑚

𝑠𝑠𝑚𝑚𝑓𝑓𝑡𝑡𝑤𝑤𝑎𝑎𝑎𝑎𝑒𝑒

𝑅𝑅𝑝𝑝 − 𝑔𝑔𝑎𝑎𝑎𝑎𝑑𝑑𝑎𝑎𝑡𝑡𝑖𝑖𝑚𝑚𝑎𝑎

𝐾𝐾

3

=

𝐶𝐶𝑚𝑚𝑠𝑠𝑡𝑡

𝑝𝑝𝑒𝑒𝑎𝑎

𝑅𝑅𝑎𝑎𝑖𝑖𝑡𝑡

𝑎𝑎𝑒𝑒𝑝𝑝𝑎𝑎𝑖𝑖𝑎𝑎

𝑚𝑚𝑓𝑓

𝑡𝑡ℎ𝑒𝑒

𝑓𝑓𝑎𝑎𝑖𝑖𝑎𝑎𝑒𝑒𝑑𝑑

ℎ𝑎𝑎𝑎𝑎𝑑𝑑𝑤𝑤𝑎𝑎𝑎𝑎𝑒𝑒

𝐾𝐾

4

=

𝐶𝐶𝑚𝑚𝑠𝑠𝑡𝑡

𝑝𝑝𝑒𝑒𝑎𝑎

𝑅𝑅𝑎𝑎𝑖𝑖𝑡𝑡

𝑅𝑅𝑝𝑝 − 𝑔𝑔𝑎𝑎𝑎𝑎𝑑𝑑𝑎𝑎𝑡𝑡𝑖𝑖𝑚𝑚𝑎𝑎

𝑚𝑚𝑓𝑓

𝑡𝑡ℎ𝑒𝑒

𝑓𝑓𝑎𝑎𝑖𝑖𝑎𝑎𝑒𝑒𝑑𝑑

𝑠𝑠𝑚𝑚𝑓𝑓𝑡𝑡𝑤𝑤𝑎𝑎𝑎𝑎𝑒𝑒

𝑎𝑎𝑎𝑎𝑑𝑑

𝐴𝐴

0,

𝐵𝐵

0𝐻𝐻

,

𝐵𝐵

0𝑆𝑆

,

𝑁𝑁𝐻𝐻𝑅𝑅

0,

𝑁𝑁𝑆𝑆𝑁𝑁

0

𝑎𝑎𝑎𝑎𝑒𝑒

𝑎𝑎𝑎𝑎𝑎𝑎𝑒𝑒𝑎𝑎𝑑𝑑𝑎𝑎

𝑑𝑑𝑒𝑒𝑓𝑓𝑖𝑖𝑎𝑎𝑒𝑒𝑑𝑑

.

X. PARTICULAR CASES

Suppose

𝑔𝑔

(

𝑡𝑡

) =

𝛼𝛼𝑒𝑒

−𝛼𝛼𝑡𝑡

𝑎𝑎𝑎𝑎𝑑𝑑

𝑓𝑓

(

𝑡𝑡

) =

𝜃𝜃𝑒𝑒

−𝜃𝜃𝑡𝑡

We can obtain the following results:

𝑀𝑀𝑇𝑇𝑆𝑆𝑀𝑀

(

𝑇𝑇

0) =𝑁𝑁1 𝐷𝐷1

𝐴𝐴𝑅𝑅𝑎𝑎𝑖𝑖𝑎𝑎𝑎𝑎𝑏𝑏𝑖𝑖𝑎𝑎𝑖𝑖𝑡𝑡𝑎𝑎

(

𝐴𝐴

0

) =

𝐵𝐵𝑅𝑅𝑠𝑠𝑎𝑎

𝑃𝑃𝑒𝑒𝑎𝑎𝑖𝑖𝑚𝑚𝑑𝑑

𝑑𝑑𝑅𝑅𝑒𝑒

𝑡𝑡𝑚𝑚

ℎ𝑎𝑎𝑎𝑎𝑑𝑑𝑤𝑤𝑎𝑎𝑎𝑎𝑒𝑒

𝑓𝑓𝑎𝑎𝑖𝑖𝑎𝑎𝑅𝑅𝑎𝑎𝑒𝑒

(

𝐵𝐵

0𝐻𝐻

) =

𝑁𝑁3

𝐻𝐻

𝐷𝐷2

𝐵𝐵𝑅𝑅𝑠𝑠𝑎𝑎

𝑃𝑃𝑒𝑒𝑎𝑎𝑖𝑖𝑚𝑚𝑑𝑑

𝑑𝑑𝑅𝑅𝑒𝑒

𝑡𝑡𝑚𝑚

𝑠𝑠𝑚𝑚𝑓𝑓𝑡𝑡𝑤𝑤𝑎𝑎𝑎𝑎𝑒𝑒

𝑓𝑓𝑎𝑎𝑖𝑖𝑎𝑎𝑅𝑅𝑎𝑎𝑒𝑒

�𝐵𝐵

0𝑆𝑆

�

=

𝑁𝑁

3 𝑆𝑆

𝐷𝐷

2

𝐸𝐸𝐸𝐸𝑝𝑝𝑒𝑒𝑖𝑖𝑡𝑡𝑒𝑒𝑑𝑑

𝑎𝑎𝑅𝑅𝑚𝑚𝑏𝑏𝑒𝑒𝑎𝑎

𝑚𝑚𝑓𝑓

𝑎𝑎𝑒𝑒𝑝𝑝𝑎𝑎𝑖𝑖𝑎𝑎

𝑎𝑎𝑡𝑡

ℎ𝑎𝑎𝑎𝑎𝑑𝑑𝑤𝑤𝑎𝑎𝑎𝑎𝑒𝑒

𝑓𝑓𝑎𝑎𝑖𝑖𝑎𝑎𝑅𝑅𝑎𝑎𝑒𝑒

(

𝑁𝑁𝐻𝐻𝑅𝑅

0) =

𝑁𝑁

_𝐷𝐷

4 2

𝐸𝐸𝐸𝐸𝑝𝑝𝑒𝑒𝑖𝑖𝑡𝑡𝑒𝑒𝑑𝑑

𝑎𝑎𝑅𝑅𝑚𝑚𝑏𝑏𝑒𝑒𝑎𝑎

𝑚𝑚𝑓𝑓

𝑅𝑅𝑝𝑝 − 𝑔𝑔𝑎𝑎𝑎𝑎𝑑𝑑𝑎𝑎𝑡𝑡𝑖𝑖𝑚𝑚𝑎𝑎

𝑎𝑎𝑡𝑡

𝑠𝑠𝑚𝑚𝑓𝑓𝑡𝑡𝑤𝑤𝑎𝑎𝑎𝑎𝑒𝑒

𝑓𝑓𝑎𝑎𝑖𝑖𝑎𝑎𝑅𝑅𝑎𝑎𝑒𝑒

(

𝑁𝑁𝑆𝑆𝑁𝑁

0) =

𝑁𝑁

_𝐷𝐷

5 2

Where

𝑁𝑁

1

=

₍_{𝑎𝑎𝜆𝜆}₁₊𝑎𝑎𝜆𝜆_{𝑏𝑏𝜆𝜆}1₂+2₎₍_{𝑎𝑎𝜆𝜆}𝑏𝑏𝜆𝜆₁2₊+_{𝑏𝑏𝜆𝜆}𝜃𝜃₂₊_𝜃𝜃₎

𝐷𝐷

1

=

(𝑎𝑎𝜆𝜆₍1_{𝑎𝑎𝜆𝜆}+𝑏𝑏𝜆𝜆₁₊2_{𝑏𝑏𝜆𝜆})(𝑎𝑎𝜆𝜆₂₎₍1_{𝑎𝑎𝜆𝜆}+𝑏𝑏𝜆𝜆₁₊2_{𝑏𝑏𝜆𝜆}+𝜃𝜃₂)₊−𝜃𝜃𝑏𝑏𝜆𝜆_𝜃𝜃₎ 2

𝑁𝑁

2

=

_{𝑎𝑎𝜆𝜆}₁1₊_{𝑏𝑏𝜆𝜆}₂

𝐷𝐷

2

=

𝛼𝛼𝜃𝜃(𝑎𝑎𝜆𝜆1_{𝛼𝛼𝜃𝜃}+𝜃𝜃₍)+(_{𝑎𝑎𝜆𝜆}₁𝜃𝜃𝑎𝑎𝜆𝜆₊_{𝑏𝑏𝜆𝜆}1₂+₎₍𝛼𝛼𝑏𝑏𝜆𝜆_{𝑎𝑎𝜆𝜆}₁2₊)(_{𝑏𝑏𝜆𝜆}𝜃𝜃𝑎𝑎𝜆𝜆₂₊1+_𝜃𝜃₎𝑏𝑏𝜆𝜆2+𝜃𝜃)

𝑁𝑁

3𝐻𝐻

=

₍_{𝑎𝑎𝜆𝜆}₁𝑎𝑎𝜆𝜆₊_{𝑏𝑏𝜆𝜆}1₂₎_𝛼𝛼

𝑁𝑁

3𝑆𝑆

=

₍_{𝑎𝑎𝜆𝜆}₁𝑏𝑏𝜆𝜆₊_{𝑏𝑏𝜆𝜆}1₂₎_𝜃𝜃

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XI. 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

K0= 15000, K11000, K2= 700, K3= 1500, K4=

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.

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.

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0 50 100 150 200 250 300

0.010.020.030.040.050.060.070.080.09 0.1

MT

S

F

Hardware Failure Rate (

λ

1)

MTSF Vs H/w Failure Rate (

λ

1)

λ2=0.001,α=2,θ=5,a=0.6,b=0.4 λ2=0.002

θ=7 a=0.4,b=0.6

0.95 0.96 0.97 0.98 0.99 1 1.01

0.010.020.030.040.050.060.070.080.09 0.1

A

vai

labilit

y

Hardware Failure Rate (

λ

1)

Availability Vs Hardware Failure rate

(

λ

1)

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

Fig. 3

14100 14200 14300 14400 14500 14600 14700 14800 14900 15000 15100

0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1

P

rof

it

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

Fig. 4

Fig. 2

K0= 15000, K1= 1000,

K2= 700, K3= 1500,