Cost-Analysis Of A Computer System With Hardware Repair Subject To Inspection And Arbitrary Distributions For Hardware and Software Replacement

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Cost-Analysis Of A Computer System With

Hardware Repair Subject To Inspection

And Arbitrary Distributions For Hardware

and Software Replacement

Jyoti Anand, Ashish Kumar and S.C. Malik

Department of Statistics, M.D. University, Rohtak-124001, Haryana (India)

Email: [email protected]

Abstract

The purpose of the present paper is to carry out Cost-Analysis of a computer system of two identical units. Each unit comprises hardware and software components which may fail completely directly from normal mode. There is a single server who inspects the unit at its hardware failure to see the feasibility of repair. If repair of the hardware components in the unit is not feasible, the components are replaced by similar new one with some replacement time. However, only replacement of the software components in the unit by new one is made at their failure with some replacement time. The failure time of the unit due to hardware and software failure follows negative exponential, while distributions of inspection, repair and replacement time of the components are considered as arbitrary with different probability density functions. There is an independent hardware and software failure in the unit. The expressions for various measures of system effectiveness are derived using semi-Markov process and regenerative point technique. The graphical study of the results obtained for a particular case has been made.

Keywords: Computer System, Hardware Repair, Inspection, Arbitrary Replacement Distribution and Cost- Analysis.

Introduction

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replaced by new one at their failure in negligible time if inspection reveals that repair of the component is not feasible. However, replacement of software components by new one takes some time. But, in practice, replacement of the components at their failure in negligible time is quite impossible. Also, not much work related to the reliability modeling and cost analysis of computer systems with integrated hardware and software components has been reported so far in the literature.

While considering the above facts and to fill up the gap, in the present paper a reliability model for a computer system of two identical units having hardware and software components is developed. The unit may fail completely from normal mode due to the independent failure of hardware and software components. There is a single server who visits the system immediately to do inspection, replacement and repair of the components. The server inspects the unit of its hardware failure to see the feasibility of repair. If repair of the hardware components is not feasible, the components are replaced by new one with some replacement time.

However, only replacement of the software components by new at their failure is made with some replacement time.

The random variables are independent and uncorrelated to each other. The switch devices and repair are perfect. The failure time of the unit due to failure of hardware and software components is distributed exponentially while the distributions of inspection, replacement and repair times are taken as arbitrary. To carry out cost analysis, various reliability characteristics such as mean sojourn times, mean time to system failure (MTSF), availability, busy period of the server due to hardware and software failures, expected number of replacements due to hardware and software failure, expected number of visits by the server and finally the profit considering various costs are determined by adopting semi-Markov process and regenerative point technique. Graphs for the results obtained for a particular case are drawn to depict the behavior of MTSF, availability and profit of the system model.

2. Notations

E : The set of regenerative states

O : The unit is operative and in normal mode Cs : The unit is cold standby

a/b : Probability that the system has hardware / software failure λ1/λ2 : Constant hardware / software failure rate

p/q : Probability that repair of the unit due to hardware failure is not feasible / feasible

FHUr/FHUR : The unit is failed due to hardware and is under repair / under repair continuously from previous state

FHUi/FHUI : The unit is failed due to hardware and is under inspection / under inspection continuously from previous state

FHWi / FHWI : The unit is failed due to hardware and is waiting for inspection/ waiting for inspection continuously from previous state FSURp/FSURP : The unit is failed due to the software and is under replacement/

under replacement continuously from previous state

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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 h(t) / H(t) : pdf / cdf of inspection time of unit due to hardware failure

f(t) / F(t) : pdf / cdf of replacement time of the software f1(t)/ F1(t) : pdf / cdf of replacement time of the hardware

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 transist directly to state Sj so that i ij

j

m

μ

=



and mij =

*

'

( )

(0)

ij ij

tdQ 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, FHUi), S2=(O, FSURp), S3= (O, FHUr), S4=(FSURP,FSWRp), S5= (FHUI, FHWi), S6=(FHUR, FHWi), S7 = (FHUr, FHWI), S8 = (FHUI, FSWRp), S9=(FHUr,FSWRP), S10=(FHUR, FSWRp), S11 = (FSWRP, FHWi), S12 = (O,FHURp), S13=(FHURP,FHWi), S14=(FHURP,FSWRp), S15=(FHURp,FHWI), S16=(FHURp, FSWRP),

The state S0–S3, S12 are regenerative states while the states S4–S11 , S13–S16 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

(

)

(

q

t

dt

Q

p

ij ij ij as

p01= 1

1 2

a

b

λ

+

λ

, p02= 2

1 2

b

a

b

λ

+

λ

,

p13 =

qh

*

(

a

λ

1

+

b

λ

2

)

, p15=

(

)

1

1 2 1 2

1 *

a

h

a

b

a

b

λ

_λ

λ

+

λ





−

+





,

p18=

(

)

2

1 2 1 2

1 *

b

h

a

b

a

b

λ

_λ

λ

+

λ





−

+





, p20 =

f

*

(

a

λ

1

+

b

λ

2

)

,

p24 =

(

)

2

1 2 1 2

1 *

b

f

a

b

a

b

λ

_λ

λ

+

λ





−

+





, p2,11 =

(

)

1

1 2 1 2

1 *

a

f

a

b

a

b

λ

_λ

(4)

p30 =

g

*

(

a

λ

₁

+

b

λ

₂

)

, p36 = 1

(

₁ ₂

)

1 2

1 *

a

g

a

b

a

b

λ

_λ

λ

+

λ





−

+





,

p3,10 = 2

(

₁ ₂

)

1 2

1 *

b

g

a

b

a

b

λ

_λ

λ

+

λ





−

+





p

40

f

(

s

)

∗

=

(1)

For h(t) = 1

1

t

e

θ

−

, f(t) =

θ

e

−θt, g(t) =

α

e

−αt and

f

₁

(

t

)

=

β

e

−βt we have p11.5,15 = 2

1 2 1

pb

a

b

λ

+

λ θ

+

, p11.57 = 2

1 2 1

qb

a

b

λ

+

λ θ

+

, p12.8,16 = 1

1 2 1

pa

a

b

λ

+

λ θ

+

,

p12.89 = 1

1 2 1

qa

a

b

λ

+

λ θ

+

, p22.4 = 2

1 2

b

a

b

λ

+

λ θ

+

, p21.11 = 1

1 2

a

b

θλ

λ

+

λ θ

+

p31.6 = 1

1 2

a

b

λ

+

λ α

+

, p32.10 = 2

1 2

b

a

b

λ

+

λ α

+

(2)

It can be easily verified that p01+p02 = p13+p1,12+p15+p18 = p20+p24+p2,11 = p30+p36+p3,10 = p12,0+p12,1+p12,2 = p13 +p1,12+ p13 +p11.5,15+p11.57+p12.8,16+p12.89 = p20 +p22.4 +p21.11= p30+p31.6+p32.10 = p12,0 +p12,1.13+ p12,2.14= 1

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The mean sojourn times (μi) is the state Si are μ0 =

,

1

2 1

λ

b

a

+

μ1 =

,

1

1 2 1

λ

θ

λ

+

b

+

a

μ2 =

θ

λ

1

+

2

+

1 b

a

, μ3 =

,

1

2 1

λ

α

λ

+

b

+

a

μ

=

λ

1

+

λ

2

+

β

12

1 b

a

(4)

Also

0 02 0

m

+

m

=

μ

,

m

₁₃

+

m

₁_,₁₂

+

m

₁₅

+

m

₁₈

=

μ

₁,

m

₂₀

+

m

₂₄

+

m

_2,11

=

μ

₂

30 36 3,10 3

m

+

m

+

m

=

μ

,

m

₁₂_,₀

+

m

₁₂_,₁₃

+

m

₁₂_,₁₄

=

μ

₁₂ (5) and

1 89 . 12 16 , 8 . 12 57 . 11 15 , 5 . 11 12 , 1

13

+

m

+

m

+

m

+

m

+

m

=

μ

′

m

(say)

2 11 . 21 4 . 22

20

+

m

+

m

=

μ

′

m

(say)

3 10 . 32 6 . 31

30

+

m

+

m

=

μ

′

m

(say),

m

₁₂_,₀

+

m

₁₂_,₁_.₁₃

+

m

₁₂_,₂_.₁₄

=

μ

₁₂

′

(

say

)

(6)

For h(t) = 1

1

t

e

θ

−

,

f

(t) =

θ

e

−θt ,g(t) =

α

e

−αtand

f

1

(

t

)

=

β

e

−βt , we have

)

(

))

(

)(

(

1 2 1 1

1 2

1 1

1

θ

λ

β

αθ

β

α

θ

αβ

λ

β

αθ

μ

+

=

′

b

a

q

p

b

a

,

1 2

1 μ

=

_θ

,

μ

1₃

1 α

=

,

β

μ

12

′

=

1

(7)

4. Reliability and Mean Time to System Failure (MTSF)

(5)

φ

0(t) = Q01(t)Ⓢ

φ

1(t) + Q02(t)Ⓢ

φ

2(t)

φ

1(t) = Q13(t)Ⓢ

φ

3(t) + Q1,12(t) Ⓢ

φ

12(t) + Q15(t) + Q18(t)

φ

2(t) = Q20(t)Ⓢ

φ

0(t) + Q24(t) + Q2,11(t),

φ

3(t) = Q30 (t)Ⓢ

φ

0(t) + Q36(t) + Q3,10(t)

φ

12(t) = Q12,0 (t)Ⓢ

φ

0(t) + Q12,13(t) + Q12,14(t) (8) Taking LT of above relation (8) and solving for

φ

~

₀

(

s

)

We have

R*(s) =

s

)

(

~

1 −

φ

₀

(9) The reliability of the system model can be obtained by taking Laplace inverse transform of (9). The mean time to system failure (MTSF) is given by

MTSF =

s

o s

)

(

~

1 lim

−

φ

0

→ =

1

N

D

(10)

where

N1 =

(

μ

0

+

p

02

μ

2

)

+

p

01

(

μ

1

+

p

13

μ

3

+

p

1,12

μ

12

)

D1 =

1 −

p

01

p

13

p

30

−

p

1,12

p

12,0

p

01

−

p

02

p

20

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

A0(t) = M0(t) + q0(t) © A1(t) + q02(t) © A2(t)

A1(t) = M1(t) + A1(t)©[q11.5,15(t) + q11.57(t)] +A2(t)©[q12.8,16(t)+q12.89(t)] + q13(t) © A3(t)+q1,12(t) ©A2(t)

A2(t) = M2(t) + q20(t) © A0(t) + q21.11(t) © A1(t) + q22.4(t) © A2(t) A3(t) = M3(t) + q30(t) © A0(t) + q31.6(t) © A1(t) + q32.10(t) © A2(t)

A12(t) =M12(t) + q12,0(t) © A0(t) + q12,1.13(t) © A1(t) + q12,2.14(t) © A2(t) (11)

where Mi(t) is the probability that the system is up initially in state

S

i

∈

E

is up at time t without visiting to any other regenerative state, we have

( 1 2)

0

( )

a b t

M t

=

e

− λ+ λ , ( 1 2)

1

( )

a b t

M t

=

e

− λ+ λ

H t

, ( 1 2)

2

( )

a b t

M t

=

e

− λ+λ

F t

,

( 1 2)

3

( )

a b t

M t

=

e

− λ+λ

G t

,

_M

₁₂

(

_t

)

=

_e

−(aλ1+bλ2)t

_f

₁

(

_t

)

₍₁₂₎

Taking LT of above relations (11) and solving for

A s

₀*

( )

, the steady state availability is given by

*

0 0

0

( )

lim

( )

s

A

sA s

→

∞ =

2

N

D

=

(13)

where

(6)

+( p13p32.10+ p13 p30 p02+ p1,12(p12,2.14+ p02 p12,0)+ p12.8,16 +p12.89 )μ2+(1-p22.4- p02p20)(μ1+p13μ3 + p1,12μ12)

and

D2 = [p13p20p32.10+(1- p22.4)( p13p30+p1,12 p12,0)+ p1,12 p20 p12,2.14+ p20 (p12.8,16+ p12.89) ]μ0 +( p13p32.10+ p13 p30 p02+ p1,12(p12,2.14+ p02 p12,0) ( p12.8,16 +p12.89 ))

μ

2

′

+(1-p22.4- p02p20)(

μ

1

′

+p13

μ

3

′

+ p1,12

μ

12

′

)

6. Busy Period Analysis for Server (a) Due to Hardware Failure

Let

B

iH

(

t

)

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

B

iH

(

t

)

for are as follows:

H

B

0 (t) = q01(t) © H

B

1 (t)+ q02(t) © H

B

2 (t)

H

B

₁ (t)=

W

₁H(t)+(q11.5,15(t)+q11.57(t))© H

B

₁ (t)+(q12.8,16(t)+q12.89(t))© H

B

₂ (t) +q13(t) ©

H

B

3 (t)+ q1,12(t) © H

B

12(t)

H

B

2 (t)= q20(t) © H

B

0 (t) + q21.11(t) © H

B

1 (t) + q22.4(t) © H

B

2 (t)

H

B

3 (t)=

H

W

3 (t) + q30(t) © H

B

0 (t) + q31.6(t) © H

B

1 (t) + q32.10(t) © H

B

2 (t)

H

B

12(t)=

H

W

12 (t) + q12,0(t) © H

B

0 (t) + q12,1.13(t)© H

B

1 (t) + q12,2.14(t) © H

B

2 (t) (14)

where WiH(t) be the probability that the server is busy in state Si due to hardware failure upto 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

)

(

G

1)

qh(t)

(

)

(

F

1)

ph(t)

(

)

(

₁ ( ) ₁ ₁ ( )

) ( 1

2 1 2

1 2

1

_H

_t

_a

_e

_t

_a

_e

_t

e

W

H

=

−aλ+bλ t

+

λ

−aλ+bλ t

©

+

λ

−aλ+bλ t

©

)

(

G

1)

qh(t)

(

)

(

F

1)

ph(t)

(

_b

₂

_e

(a 1 b 2)t

©

₁

_t

+

_b

₂

_e

(a 1 b 2)t

©

_t

+

_λ

− λ+ λ

_λ

− λ+ λ

( 1 2) ( 1 2)

3

( )

1

©1

( )

a b t a b

H

W

t

=

e

− λ+λ

G t

+

_



a e

λ

− λ+λ

_



G t

( 1 2)

2

©1

( )

a b t

b e

λ

− λ+λ

G t





+

_

_

)

(

F

1)

(

)

(

F

1)

(

)

(

)

(

( ) ₁ ₁ ( ) ₁ ₂ ( ) ₁

12

2 1 2

1 2

1

_F

_t

_a

_e

_t

_b

_e

_t

e

t

W

H

=

−aλ+bλ t

+

λ

−aλ+bλ t

©

+

λ

−aλ+bλ t

©

(15)

(b) Due to replacement of the software

Let

B

_iS(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

B

_iS(t):

S

B

₀ (t) = q01(t) © S

B

₁ (t) + q02(t) ©

B

2

(

t

)

S

B

₁ (t)= (q11.5,15(t)+q11.57(t))© S

B

₁ (t)+(q12.8,16(t)+q12.89(t))©

B

2

(

t

)

S

+q13(t) © S

B

₃ (t)+ q1,12(t) © S

B

₁₂(t)

)

(

2

t

B

S =

W

₂S

(

t

)

+q20(t) © S

B

₀ (t) + q21.11(t) © S

B

₁ (t) + q22.4(t)) ©

B

2

(

t

)

(7)

B3S(t) = q30(t) © S

B

₀ (t) + q31.6(t) © S

B

₁ (t) + q32.10(t) ©

B

2

(

t

)

S

B

₁₂(t)= q12,0(t) S

B

₀ (t)+q12,1.13(t)© S

B

₁ (t)+q12,2.14(t)©

B

2

(

t

)

S

(16)

where

W

_iS(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 2)

2

( )

1

©1

( )

2

©1

( )

a b t a b t a b t

S

W

t

=

e

− λ+λ

F t

+

_



a e

λ

− λ+λ



_

F t

+



_

b e

λ

− λ+λ



_

F t

(17) Taking LT of above relations (14) and (16). And, solving for

H

B

0∗ (s) and S

B

0∗ (s), the time for which server is

busy due to repair and replacements respectively is given by ₀ ₀*

0

lim

( )

H H

s

B

sB

s

→

=

= 3

2

H

N

D

(18) And

*

0 0

0

lim

( )

S S

s

B

sB

s

→

=

= 3

2

S

N

D

(19)

Where

)

(

~

))

1 (

(

))

(

~

)

(

~

)

(

~

)(

1 (

2 13 . 1 , 12 0 , 12 01 12

, 1 89 . 12 16 , 8 . 12 3

12 12 , 1 3

13 1

20 02 4 . 22 3

s

W

p

N

s

W

p

s

W

p

s

W

p

N

S S

H H

−

+

=

+

−

=

and D2 is already mentioned.

7. Expected Number of Replacements of the Units (a) Due to Hardware Failure

Let

R

_iH

(

t

)

be the expected number of replacements of the failed hardware components by the server in (0, t] given that the system entered the regenerative state i at t = 0. The recursive relations for

R

_iH

(

t

)

are given as

)

(

0

t

R

H = Q01(t)Ⓢ

R

1

(

t

)

H

+ Q02(t)Ⓢ

R

2

(

t

)

H

)

(

1

t

R

H =Q11.57(t)Ⓢ

R

1

(

t

)

H

+Q11.5,15(t)Ⓢ[1+

R

1

(

t

)

H

]+Q12.8,16(t)Ⓢ[1+

R

2

(

t

)

H ] +Q12.89(t)ⓈR2H(t)+ Q13(t)ⓈR3H(t) + Q1,12(t)Ⓢ[1+

R

12

(

t

)

H ]

)

(

2

t

R

H = Q20(t)Ⓢ

R

0

(

t

)

H

+ Q21.11(t)Ⓢ

R

1

(

t

)

H

+ Q22.4(t)Ⓢ

R

2

(

t

)

H R3H(t) = Q30(t)Ⓢ

R

0

(

t

)

H

+ Q31.6(t)Ⓢ

R

1

(

t

)

H

+ Q32.10(t)Ⓢ

R

2

(

t

)

H

)

(

12

t

R

H =Q12,0(t)Ⓢ[1+

R

0

(

t

)

H

]+Q12,1.13(t)Ⓢ[1+

R

1

(

t

)

H

]+ Q12,2.14(t)Ⓢ

R

2

(

t

)

H

(8)

(b) Due to Software Failure

Let

R

_iS

(

t

)

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

R

_iS

(

t

)

are given as

)

(

0

t

R

S = Q01(t) Ⓢ

R

1

(

t

)

S

+ Q02(t) Ⓢ

R

2

(

t

)

S

)

(

1

t

R

S = (Q11.5,15(t)+Q11.57(t))Ⓢ

R

1

(

t

)

S

+(Q12.8,16(t) +Q12.89(t))

R

2

(

t

)

S

+ Q13(t)Ⓢ

R

3

(

t

)

S + Q1,12(t)Ⓢ[1+

R

12

(

t

)

S ]

)

(

2

t

R

S = Q20(t)Ⓢ[1+

R

0

(

t

)

S

] + Q21.11(t)Ⓢ

R

1

(

t

)

S

+ Q22.4(t)Ⓢ

R

2

(

t

)

S

)

(

3

t

R

S = Q30(t)Ⓢ

R

0

(

t

)

S

+ Q31.6(t)Ⓢ

R

1

(

t

)

S

+ Q32.10(t)Ⓢ

R

2

(

t

)

S

)

(

12

t

R

S = Q12,0(t)Ⓢ

R

0

(

t

)

S

+Q12,1.13(t)Ⓢ

R

1

(

t

)

S

+ Q12,2.14(t)Ⓢ

R

2

(

t

)

S

(21)

Taking LT of relations (20) and (21). And, solving for

R

~

₀H

(

s

)

and

R

~

₀S

(

s

)

. The expected number of replacements per unit time to the hardware and software failures are respective of given by

0 0

0

( )

lim

( )

H H

s

R

sR

s

→

∞ =



= 4

2

H

N

D

(22)

0

( )

lim

₀ 0

( )

S S

s

R

sR s

→

∞ =



= 4

2

S

N

D

(23)

Where

)

1 (

))

(

)(

1 (

20 02 4 . 22 12

, 1 4

13 . 1 , 12 0 , 12 12 , 1 12 , 1 16 , 8 . 12 15 , 5 . 11 20 02 4 . 22 4

p

N

p

N

S H

−

=

+

−

=

+

p

₂₀

(

p

₁₂_.₈_,₁₆

+

p

₁₂_.₈₉

+

p

₁₃

(

1 −

p

₀₁

p

₃₀

−

p

₃₁_.₆

)

+

p

₁_,₁₂

(

p

₁₂_,₂_.₁₄

+

p

₀₂

p

₁₂_,₀

))

and D2 is already mentioned.

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

N0(t) = Q01(t) Ⓢ[1+N1(t)] + Q02(t) Ⓢ[1+N2(t)]

N1(t) = [Q11.57(t)+Q11.5,15(t)]ⓈN1(t )[Q12.8,16(t)+Q12.89(t)]ⓈN2(t) +Q13(t) ⓈN3(t) +Q1,12(t) N12(t)

N2(t) = Q20(t) N0(t) + Q21.11(t) N1(t) + Q22.4(t) N2(t)

N3(t) = Q30(t) N0(t) + Q32.10(t) N2(t) + Q31.6(t) N1(t)

N12(t)=Q12,0(t) N0(t)+ Q12,1.13(t) N1(t)+Q12,2.14(t) N2(t) (24)

(9)

₀ ₀

0

( )

lim

( )

s

N

sN s

→

∞ =



= 5

2

N

D

(25)

where

N5 = (1-p22.4) (p13p30+p1,12p12,0 )+ p20 (p13p32.10+ p1,12 p12,2.14 +p12.8+p12.89) and D2 is already specified.

9. Economic 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 5 0

H S H S

K A

−

K B

−

K B

−

K R

−

K R

−

K N

(26)

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 hardware component K4 = Cost per unit replacement of the failed software

K5 = Cost per unit visit by the server and 0

,

0

,

0

,

0

,

0

,

0

H S H S

A B

B

R

N

are already defined.

10. Particular Case

Suppose g(t) =

a

e

-at,

h

t

e

1t

1

)

(

=

θ

−θ ,

f

(

t

)

=

θ

e

−θtand

f

1

(

t

)

=

β

e

−βt We can obtain the following results

MTSF (T0) = 1

1

N

D

, Availability (A0) =

2

N

D

Busy period due to hardware failure

( )

₀ 3

2

H

N

B

D

=

Busy period due to software failure

( )

0 3 2

S S

N

B

D

=

Expected number of replacements at hardware failure

( )

₀ 4

2

H

N

R

D

=

Expected number of replacements at software failure

( )

₀ 4

2

S S

N

R

D

=

Expected number of visits by the server

( )

₀ 5

2

N

D

=

(27)

Where

1 2

1 1 2 1 1

2 1 2 2

1 1 1 2

1 2

1 1 2 1

1

)

)(

)((

(

))

(

)

(

)(

(

R

q

p

b

a

b

a

b

a

b

a

b

a

b

a

b

a

N

λ

θ

λ

θ

λ

β

α

αβ

λ

θ

λ

β

λ

θ

λ

θ

λ

+

−

+

(10)

)

)(

(

₁ ₂ ₁ ₂ ₁ ₂ ₁ ₁ ₂ ₁ ₂

1

=

a

λ

+

b

λ

a

λ

+

b

λ

+

θ

a

λ

+

b

λ

+

θ

a

λ

+

b

λ

+

α

a

λ

+

b

λ

+

β

R

1 2 1 2 1 2 1 2 1 1 1 2 1 2 1 1 2 1 2 1 1

))

(

)

)(

((

)

(

)

)(

2 (

R

p

q

b

a

b

a

b

a

b

a

b

a

b

a

b

a

b

a

D

λ

α

λ

β

λ

β

α

θ

λ

θ

β

λ

α

λ

θ

λ

θ

λ

+

=

1 1 2 1 2 1 1 2 1 1 2 1 2 1 2 1 1 1 2 1 2 1 2 1 2 2 1 2 1 1 2 1 2 1 2 2 1 1 2 1 2 1 2 2 2 1 2

)

))(

(

))

(

)(

(

)

)(

(

)

)(

(

)

)(

(

)

)(

)((

(

)

(

R

b

a

q

p

q

p

b

a

b

a

b

a

b

a

b

a

b

a

b

a

b

a

p

b

a

b

q

p

b

a

b

a

b

q

D

αβθθ

λ

β

α

θ

β

α

θ

αβ

λ

β

αθ

β

λ

α

λ

θ

λ

θ

λ

θ

λ

θ

λ

α

λ

αβθ

β

λ

θ

α

λ

θ

αβθ

αβ

α

β

λ

θ

λ

θαβθ

β

λ

αβ

λ

θ

+

=

1 2 1 2 1 2 1 1 2 1 2 1 1 1 2 1 1 2 1 2 1 2 1 2 1 1 2 1 2 1 2 2 1 1 1 2

)

)(

(

)

)(

(

)

)(

(

)

(

)

)(

(

)

)(

)((

(

R

b

a

b

a

b

a

b

a

q

p

b

a

b

a

b

a

b

a

b

q

p

b

a

b

a

b

q

p

b

a

N

λ

θ

λ

α

β

λ

θ

λ

α

λ

β

λ

θ

α

λ

θ

λ

θ

β

λ

θ

β

θ

λ

α

λ

θ

βα

α

β

λ

θ

λ

θ

+

=

1 1 2 1 2 1 2 1 2 1 1 2 1 1 1 2 1 1 2 1 2 1 1 2 1 1 2 1 2 1 1 2 1 2 1 1 2 1 1 3

)

(

)

))(

)(

(

))

)(

(

)

)(

(

)

)(

(

)

)(

(

R

b

a

b

a

b

a

b

a

b

a

q

p

a

b

a

b

a

q

b

a

b

a

p

b

a

b

a

b

a

b

a

b

a

N

H

θ

λ

αβ

θ

λ

β

λ

α

λ

θ

λ

β

α

λ

θ

β

λ

θ

α

λ

β

α

λ

θ

β

λ

α

θ

λ

β

λ

α

λ

θ

λ

αβ

λ

−

+

−

+

−

+

−

+

−

+

=

)

)(

(

)

(

1 2 1 2 1 1 2 1 2 3

θ

λ

θ

λ

+

=

b

a

b

a

p

b

a

b

N

S

)

)(

(

))

(

)

)(

((

2 1 1 2 1 2 1 1 1 1 2 1 2 1 1 4

β

λ

θ

λ

β

λ

θ

λ

β

λ

+

=

b

a

b

a

b

a

b

a

b

a

p

a

N

H

)

)(

(

)

(

)

(

1 2 1 2 1 2 1 1 2 1 2 2 1 1 1 4

θ

λ

θ

λ

θ

λ

θ

λ

θ

+

=

b

a

b

a

b

a

b

a

b

a

p

N

S

(11)

State Transition Diagram

Fig. 1

Up-state

Failed state

Regenerative point

FHUI

FHWi

O

Cs

O

FHUi

FHUR

FHWi

O

FSURp

O

FHUr

FHWI

FSWRP

FHWi

FSURP

FSWRp

FHUR

FSWRp

∩

a

λ

1

a

λ

1

b

λ

2

b

λ

2

b

λ

2

a

λ

1

a

λ

1

ph(t)

f(t)

qh(t)

f(t)

f

1

(t)

g (t)

qh(t)

f(t)

S

4

S

10

S

6

S

3

S

2

S

11

S

7

S

5

S

1

S

0

FHUr

FSWRP

FHURp

FSWRP

FHUI

FSWRp

O

FHURp

FHURP

FSWRp

FHURP

FHWi

g(t)

f

1

(t)

ph(t)

FHURp

FHWI

S

15

g (t)

qh(t)

b

λ

2

b

λ

2

f

1

(t)

a

λ

1

f

1

(t)

f

1

(t)

ph(t)

g(t)

S

9

S

8

S

16

S

13

S

12

(12)

GRAPH BETWEEN FAILURE RATE AND AVAILABILITY

0.999 0.9992 0.9994 0.9996 0.9998 1 1.0002

0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09

FAILURE RATE(λ1)

A V

A

I

L

A B I

L

I

T

Y

a=0.7,b=0.3,λ2=0.005,θ=20,

θ1=10,α=2.5,β=30,p=0.3,q=0.7 a=0.7,b=0.3,λ2=0.005,θ=20,

θ1=10,α=2.5,β=30,p=0.3,q=0.7 a=0.7,b=0.3,λ2=0.005,θ=30,

θ1=10,α=2.5,β=30,p=0.3,q=0.7

a=0.3,b=0.7,λ2=0.005,θ=20,

θ1=20,α=2.5,β=30,p=0.3,q=0.7 a=0.7,b=0.3,λ2=0.005,θ=20,

θ1=10,α=3.5,β=30,p=0.3,q=0.7 a=0.7,b=0.3,λ2=0.02,θ=20,

θ1=10,α=2.5,β=30,p=0.3,q=0.7 a=0.3,b=0.7,λ2=0.005,θ=20,

θ1=10,α=2.5,β=30,p=0.3,q=0.7 a=0.7,b=0.3,λ2=0.005,θ=20,

θ1=10,α=2.5,β=30,p=0.7,q=0.3

Fig.3

GRAPH BETWEEN MTSF AND FAILURE RATE

0 20000 40000 60000 80000 100000 120000 140000

0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09

FAILURE RAT E(λ1)

M

T

S

F

a=0.3,b=0.7,λ2=0.005,θ=20,

θ1=10,α=2.5,β=30,p=0.3,q=0.7 a=0.7,b=0.3,λ2=0.005,θ=20,

θ1=10,α=3.5,β=30,p=0.7,q=0.3

a=0.7,b=0.3,λ2=0.005,θ=20,

θ1=10,α=2.5,β=20,p=0.3,q=0.7

a=0.7,b=0.3,λ2=0.005,θ=20,

θ1=20,α=2.5,β=30,p=0.3,q=0.7 a=0.7,b=0.3,λ2=0.02,θ=20,

θ1=10,α=2.5,β=30,p=0.3,q=0.7 a=0.7,b=0.3,λ2=0.005,θ=20,

θ1=20,α=2.5,β=20,p=0.3,q=0.7

a=0.7,b=0.3,λ2=0.005,θ=30,

θ1=10,α=2.5,β=30,p=0.3,q=0.7 a=0.7,b=0.3,λ2=0.005,θ=20,

θ1=10,α=2.5,β=30,p=0.7,q=0..3

(13)

11. Conclusion

The mean time to system failure (MTSF) decline rapidly with the increase of hardware and software failure rates λ1 and λ2 respectively keeping fixed values of other parameters as shown in Figure 2. However, MTSF increases with the increase of the repair rate (α), replacement rate (θ) of the unit at software failure, inspection rate (θ1) of the unit at hardware failure and replacement rate (β) of the unit at hardware failure. Further, it is interesting to note that a system with a>b has more MTSF as compared to the system a<b.

Figures 3 and 4 respectively indicate that availability and profit of the system model decrease with the increase of failure rates λ1 and λ2. But their values increase when repair rate (α), replacement rate (θ and

β), inspection rate (θ1) increase. On the basis of the results obtained for a particular case, it is concluded that a computer system with replacement at independent hardware and software failures can be made more profitable:

(i) by giving preference to the replacement of the hardware components by new one instead of their repair.

(ii) by controlling the failure of hardware components.

References

[1] Friedman, M.A. and Tran, P. (1992), Reliability techniques for combined hardware / software systems, Proceedings of Annual

Reliability and Maintainability Symposium, pp. 209-293.

[2] Welke, S.R.; Johnson, B.W. and Aylar, J.H. (1995), Reliability modeling of hardware / software systems, IEEE Transactions on

Reliability, Vol. 44, No. 3, pp. 413-418.

[3] Lai, C.D.; Xie, M.; Poh, K.L.; Dai, Y.S. and Yang, P. (2002), A model for availability analysis of distributed software / hardware

systems, Information and Software Technology, Vol. 44, pp. 343-350.

GRAPH BETWEEN PROFIT(P) AND FAILURE RATE(λ1)

14930

14940

14950

14960

14970

14980

14990

15000

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

FAILURE RATE (λ1)

PR

O

F

IT

a=0.7,b=0.3,λ2=0.005,θ=30,

θ1=10,α=3.5,β=30,p=0.3,q=0. a=0.7,b=0.3,λ2=0.02,θ=20,

θ1=10,α=2.5,β=30,p=0.3,q=0.7

a=0.7,b=0.3,λ2=0.005,θ=20,

θ1=10,α=2.5,β=20,p=0.3,q=0.7

a=0.7,b=0.3,λ2=0.005,θ=20,

θ1=10,α=2.5,β=30,p=0.3,q=0.7 a=0.3,b=0.7,λ2=0.005,θ=20,

θ1=10,α=2.5,β=30,p=0.3,q=0.7

a=0.7,b=0.3,λ2=0.005,θ=20,

θ1=10,α=2.5,β=30,p=0.7,q=0. a=0.7,b=0.3,λ2=0.005,θ=20,

θ1=20,α=2.5,β=30,p=0.3,q=0.7

a=0.7,b=0.3,λ2=0.005,θ=20,

θ1=10,α=3.5,β=30,p=0.3,q=0.

(14)

[4] Mailk,S.C and Anand,Jyoti (2010),Reliability And Economic Analysis Of A Computer System With Independent Hardware And