Developing the reliability model for availability and ... analysis of a... · analysis of a distillery using Regenerative Point Graphical Technique ... M.M (P.G) College, Fatehabad

Nidhi Chaudhary, Pardeep Goel, Surender Kumar: - Developing the reliability model for availability and behaviour analysis of a distillery using Regenerative Point Graphical Technique

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ISSN (Online):2347-1697 INTERNATIONAL JOURNAL OF INFORMATIVE AND FUTURISTIC RESEARCH ( IJIFR)

Volume -1 Issue -4, December 2013 Research Area: Mathematics, Page No. : 26-40

Developing the reliability model for availability and behaviour

analysis of a distillery using Regenerative Point Graphical Technique

Nidhi Chaudhary

1, Pardeep Goel

2, Surender Kumar

3

*Assistant Professor of Mathematics, KITM College, Karnal

**Department of Mathematics, M.M (P.G) College, Fatehabad

***Research Scholar, JJT University, Jhunjhunu

Abstract

This paper discusses Availability Analysis of Using RPGT Called Regenerative Point

Graphical Technique. The reliability model for availability and behaviour analysis of

distillery plant is developed with a single server which consists of three non-identical units in

which main unit can work in reduced state due to partial failure. The main unit can fail

partially and hence can be in up-state, partially failed state or totally failed state. The system

can work with reduced capacity in a partially failed state. The repairs of the unit, treatment of

the server are considered as perfect. Using RPGT, expressions for the Mean time to system

failure, Total fraction of time for which the system is available, the busy period of the server,

the Number of server’s visits has been evaluated to study the system performance followed

by illustrations, special cases, Tables and Graphs.

Keywords: System Performance, Availability Analysis, Distillery Plant, Base-State, RPGT, Server Busy

Period, MTSF, Primary, Secondary, Tertiary Circuits

Introduction: In this paper the reliability model for availability and behaviour analysis of distillery

industry is developed. The whole industries are divided into three units for a better analysis. Division

of whole industry into three units can be taken up in various combinations. However, study can also

be taken for individual units. One part is considered as main unit and other to be taken as subsidiary

units. In which main unit can work in reduced state after partial failure. The main unit can fail

partially and hence can be in up-state, partially failed state or totally failed state and one of the

subsidiary unit have stand-by and third unit have units in series. The system can work with reduced

capacity in a partially failed state. The repair of the unit, treatment of the server is considered as

perfect. Using RPGT, expressions for the Mean time to system failure, Total fraction of time for

which the system is available, the busy period of the server, the Number of server’s visits has been

evaluated to study the system performance followed by illustrations, special cases, Tables and Graphs.

Gupta, Singh and Vanita [1] defined different types of circuits / cycles like primary, secondary and

tertiary circuits which are located in the transition diagram of the system and introduced the concept

of base state of the system for determining the key parameters of the system using RPGT. Using

RPGT Jindal [2] discussed behaviour and availability analysis of industrial systems. Gupta and Singh

[4 & 5] presented a new approach for availability analysis, behaviour and profit analysis of process

industries. Goel and Singh [3] discussed the availability analysis of the standby complex system

having imperfect switch. Nidhi, Goel & Garg [6] discussed Availability Analysis of a Soap Industry

Using Regenerative Point Graphical Technique (RPGT). In distillery, the first process is malting here.

The barley is made wet and spread on the malting floor to allow the germination process to start.

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Nidhi Chaudhary, Pardeep Goel, Surender Kumar : - Developing the reliability model for availability and behaviour analysis of a distillery using Regenerative Point Graphical Technique

www.ijifr.com PAPER ID: IJIFR/V1/E3/014 IJIFR©2013

27



Second process is grinding in which the dry malt is grinded to make a kind of coarse flour which will

be used in the next operations. Third process is brewing in which the grist will be mixed with the hot

water in the mash turn. Generally the volume of grist is mixed up with 4 volumes of water. In this

operation, 3 successive waters are used, at a temperature between 63 and 95%.Fourth process is

Fermentation of worth in which yeast is added. The action of the yeast on the sugar of the wart will

produce alcohol and carbon-dioxide. Fifth Process is distillation which is used to separate alcohol

from water and other substances contained in the wash. Sixth process is aging in which to have the

right to bear the name of whisky; a grain spirit must be aged at-least for 3 years in a oak cask. Seventh

process is bottling before putting the whisky on the market. Generally, the casks are mixed before

bottling to get a more standardized product like great wines. Last process is Labelling and Packaging

in which containers must be sealed so that any attempt to open the container will leave marks visible

to consumers.

2 Assumptions and Notations:

The following assumptions and notations/symbols are used:

1) The System consists of Three non-identical units ‘A’ ,’B’ and ‘D’ in which ‘A’ can work in

reduced state after partial failure. Unit ‘A’ represents the grinding and flooring, unit ’B’ represents the

processes brewing, mixing, fermentation, distillation and the third unit ‘D’ represents the bottling

labelling and packaging.

2) The unit ‘A’ can fail partially and hence can be in up-state, partially failed state (reduced

state) or totally failed state. The system can work with reduced capacity in a partially failed state and

is repaired when it fails completely or if the server is idle.

3) The unit ‘B’ have cold standby unit ‘C’.

4) Unit ‘D’ which is bottling labelling and packaging have units in series.

5) Priority policy is in order Mainstreams unit, redundant unit.

6) The system is down when any of mainstream units is failed.

7) There is a single repair facility catering to the needs of three units as and when need arises.

8) Nothing can fail further when the system is in failed state. If the mainstream unit fails, the

system is in degraded state and the failed unit is immediately put under repair.

9) If standby unit is on line, and main unit is repaired in the meantime then the main unit is

switched in as on line.

10) Upon failure, if the repairman is busy and if standby unit is also fail, it joins the end of the

queue of failed unit.

11) The distributions of the failure & repair times are exponential and general respectively and

also different for three units. They are also assumed to be independent of each other.

12) Repairs are perfect i.e. the Repair facility never does any damage to the units.

13) A Repaired unit works like a new one.

14) The system is discussed for steady state conditions.

pr/pf:- Probability/transition probability factor.

qi,j(t):- probability density function (p.d.f.) of the first passage time from a regenerative state ‘I’ to a

regenerative state ‘j’ or to a failed state ‘j’ without visiting any other regenerative state in (0,t]

pi,j:- Steady state transition probability from a regenerative state ‘i’ to a regenerative state ‘j’

without visiting any other regenerative state. pi,j= ; where denotes Laplace transformation.

: A circuit formed through un-failed states.



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m- : A circuit (may be formed through regenerative or non-regenerative/failed states) whose terminals

are at the regenerative state m.

m- : A circuit (may be formed through only un-failed regenerative / non- regenerative states) whose

terminals are at the regenerative state m.

: r-th directed simple path from ‘i’-state to ‘j’-state; r takes positive integral values for different paths

from i-state to j-state.

: A directed simple failure free path from ‘ ’-state to i-state.

:pf of the state m reachable from the terminal state m of the m-cycle.

:pf of the state m reachable from the terminal state m of the m- .

Ri(t): Reliability of the system at time t, given that the system entered the un-failed regenerative state

‘i’ at t=0.

Ai(t): Probability that the system is available in up-state at time t, given that the system entered

regenerative state ‘i’ at t=0.

Bi(t): Probability that the server is busy doing a particular job at epoch t, given that the system entered

regenerative state ‘i’ at t=0.

Vi(t): The expected number of visits of the server for a given job in (0,t], given that the system

entered regenerative state ‘i’ at t=0.

Wi(t): Probability that the server is busy doing a particular job at epoch t without transiting to any

other regenerative state ‘i’ through one or more non- regenerative states, given that the system entered

the regenerative state ‘i’ at t=0.

: Mean sojourn time spent in state i, before visiting any other states;

: The total un-conditional time spent before transiting to any other regenerative states, given that the

system entered regenerative state ‘i’ at t=0.

ƞi: Expected waiting time spent while doing a given job, given that the system entered regenerative

state ‘i’ at t=0;

: Fuzziness measure of the ‘j’-state.

λ1/ λ2 / λ3 / λ4/ λ5 : Constant failure rate of the operative units A/A/B/C/D respectively.

g1(t)/g2(t)/ g3(t)/ g4(t) / g5(t) : Probability density function of the Repair rates of units A/A/B/C/D

respectively.

G1(t)/G2(t)/ G3(t)/ G4(t) / G5(t) : Cumulative density function of the Repair rates of failed units

A/A/B/C/D respectively.

A/ Ā/a : Main unit in the operative state/ partial failed state/failed state

B/b : operative state/ failed state.

(C)/c : Standby unit in cold stand-by/ failed state

D/d : operative state/ failed state.

Transition diagram of the system

Following the above assumptions and notations, the transition diagram of the system is shown in fig.1

Table 1: Symbolic representation of States

State Symbol Regenerative state/point

Up-state:

Failed state:

Degenerated/Reduced state



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Figure 1: Transition diagram of Distillery Industry

The system can be in any of the following states with respect to the above symbols.

S0 = AB(C)D S1= ĀB(C)D

S2 = aB(C)D S3 = AbCD

S4 =ĀbCD S5 = ĀbcD

S6 = abCD S7 =AbcD

S8 = ĀbCd S9=AbCd

S10=AB(C)d S11=ĀB(C)d

States S0, S1,S2 , S3 ,S4, S5, S6,S7,S8,S9,S10 and S11are regenerative states.

4 Evaluations of parameters of the system:

4.1 Analysis of System:

The key parameters (under steady state conditions) of the system are evaluated by determining a ‘base-

state’ and applying RPGT. The MTSF is determined w. r. t the initial state’0’ and the other parameters

are obtained by using base-state.

4.1.1 Determination of base-state:

From the Transition diagram (fig.1), The Primary, Secondary, Tertiary Circuits at all vertices are shown

in Table-2.



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Table 2: Primary, Secondary, Tertiary Circuits at a Vertex ‘j’

Vertex

‘j’

Primary( CL1) Secondary( CL2) Tertiary( CL3)

0 (0,1,0)

(0,3,0)

(0,1,2,0)

(0,10,0)

(0,1,4,6,3,0)

(1,11,1),(3,4,3),(3,7,3)

(3,9,3)

(1,11,1)

NIL

(1,11,1),

(4,3,4)

,(4,6,3,4),

(3,4,3),(3,7,3),(3,9,3)

NIL

NIL

NIL

NIL

NIL

(3,7,3),(3,9,3)

,(3,7,3),(3,9,3)

NIL

1 (1,0,1)

(1,2,0,1)

(1,4,3,0,1)

(1,4,6,3,0,1)

(1,11,1)

(0,3,0)

(0,10,0)

(0,3,0)

(0,10,0)

(4,3,4)

,(4,6,3,4)

(4,3,4)

,(4,6,3,4)

(3,0,3)

(3,4,3)

(3,7,3),(3,9,3)

(3,4,3),(3,7,3),(3,9,3)

NIL

(3,4,3),(3,7,3),(3,9,3)

NIL

(3,0,3),(3,7,3),(3,9,3)

,(3,0,3),(3,7,3),(3,9,3)

(3,0,3),(3,7,3),(3,9,3),

(3,0,3),(3,7,3),(3,9,3)

(0,10,0)

NIL

NIL

2 (2,0,1,2) (0,1,0), (0,10,0)

(0,3,0)

(0,1,4,6,3,0)

NIL

(3,4,3),(3,7,3),(3,9,3)

(1,11,1),(4,3,4),(4,6,3,4),(3,7,3),(3,9,3)

3 (3,0,1,4,3)

(3,0,3)

(3,4,3),(3,7,3),(3,9,3)

(0,1,0)

(0,1,2,0),(0,10,0)

(0,1,0)

(0,1,2,0),(0,10,0)

NIL

(1,11,1)

NIL

(1,11,1)

NIL

NIL

4 (4,3,4)

(4,6,3,4)

(4,3,0,1,4)

(4,6,3,0,1,4)

(3,0,3)

(3,7,3),(3,9,3)

(3,0,3)

(3,7,3),(3,9,3)

(3,0,3)

(3,7,3)

(3,9,3)

(0,1,0),(0,1,2,0)

(0,10,0)

(3,0,3)

(3,7,3),(3,9,3)

(0,1,0),(0,1,2,0)

(0,10,0)

(0,1,0),(0,1,2,0),(0,10,0)

NIL

(0,1,0),(0,1,2,0),(0,10,0)

NIL

(0,1,0),(0,1,2,0),(0,10,0)

NIL

NIL

(1,11,1),(1,11,1)

NIL

(0,1,0),(0,1,2,0),(0,10,0)

NIL

(1,11,1),(1,11,1)

NIL

5 (5,4,5) (4,3,4),

(4,6,3,4)

(4,3,0,1,4)

(4,6,3,0,1,4)

(3,0,3),(3,7,3),(3,9,3)

,(3,0,3),(3,7,3),(3,9,3)

(3,0,3),(3,7,3),(3,9,3),(0,1,0),(0,1,2,0),(0,3,0),(0,10,0)

(3,0,3),(3,7,3),(3,9,3),(0,1,0),(0,1,2,0),(0,3,0),(0,10,0)



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6 (6,3,4,6)

(6,3,0,1,4,6)

(3,4,3),

(3,7,3),(3,9,3),(3,4,3)

(3,0,1,4,3)

(3,0,3)

(3,0,1,4,3)

,(3,0,3)

(3,7,3),(3,9,3)

(0,1,0)

(0,1,2,0)

(0,10,0),(1,11,1)

NIL

(0,1,0),(0,1,2,0),(0,10,0),(1,11,1)

(0,1,0),(0,1,2,0),(0,10,0)

(0,1,0),(0,1,2,0),(0,10,0),(1,11,1)

,(0,1,0),(0,1,2,0),(0,10,0)

NIL

(1,11,1)

(1,11,1)

NIL

7 (7,3,7) (3,0,1,4,3)

(3,4,3), (3,7,3),(3,9,3)

(3,0,3)

(0,1,0),(0,1,2,0),(0,10,0),(1,11,1)

NIL

(0,1,0),(0,1,2,0),(0,10,0)

8 (8,4,8) (4,3,4)

,(4,6,3,4)

(4,3,0,1,4)

(4,6,3,0,1,4)

(3,0,3),(3,7,3),(3,9,3)

,(3,0,3),(3,7,3),(3,9,3)

(3,0,3),(3,7,3),(3,9,3),(0,1,0),(0,1,2,0),(0,3,0),(0,10,0)

(3,0,3),(3,7,3),(3,9,3),(0,1,0),(0,1,2,0),(0,3,0),(0,10,0)

9 (9,3,9) (3,4,3), (3,7,3)

(3,0,1,4,3)

(3,0,3)

NIL

(0,1,0),(0,1,2,0),(0,10,0),(1,11,1)

(0,1,0),(0,1,2,0),(0,10,0)

10 (10,0,10) (0,1,0)

,(0,1,2,0)

,(0,3,0),

(0,1,4,6,3,0)

(1,11,1),

(1,11,1),

(3,4,3),(3,7,3),(3,9,3)

,(3,4,3),(3,7,3),(3,9,3),(4,6,3,4)

11 (11,1,11) (1,0,1),

(1,2,0,1),

(1,4,3,0,1)

(1,4,6,3,0,1)

(0,10,0),(0,3,0)

,(0,10,0) ,(0,3,0)

(0,10,0),(0,3,0),(4,3,4),(4,6,3,4),(3,7,3),(3,9,3)

(0,10,0),(0,3,0),(4,3,4),(4,6,3,4),(3,7,3),(3,9,3)

In the Transition diagram of fig.1,there are five, five, one, five, four, one, two, one, one, one, one &

one primary circuits are at vertices 0,1,2,3,4,5,6 ,7,8,9,10&11 respectively and ten, twelve, four, six,

eighteen, four, fourteen, five, four, four, four& four secondary circuits are at vertices

0,1,2,3,4,5,6,7,8,9,10 &11 respectively and seven, twenty three, eight, two, sixteen, twenty, sixteen,

seven, twenty, seven, nine& sixteen tertiary circuits are at vertices 0,1,2,3,4,5,6,7,8,9,10 & 11

respectively. Since there are largest no. of primary circuits at the vertex ‘3’with lesser number of

secondary & tertiary circuits. Therefore ‘3’ is a base-state.

Table 3: Primary, Secondary, Tertiary Circuits w.r.t Simple Paths (Base-State’3’)



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Vertex (3 j0):P0

(P1) (P2)

1 (3 0):{3,0}

{0,1,0}

{0,1,2,0}

{0,10,0}

{1,11,1}

{1,11,1}

NIL

2 (3 1):{3,0,1}

{0,1,0}

{0,1,2,0}

{0,10,0},{1,11,1}

{1,11,1}

{1,11,1}

NIL

3

(3 2):{3,0,1,2} {0,1,0}

{0,1,2,0}

{0,10,0},{1,11,1}

{1,11,1}

{1,11,1}

NIL

4 (3 4):{3,4}

(3 4):{3,0,1,4}

NIL

{0,1,0}

{0,1,2,0}

{0,10,0},{1,11,1}

NIL

{1,11,1}

{1,11,1}

NIL

5 (3 5): {3,4,5}

(3 5):{3,0,1,4,5}

{5,4,5}

{0,1,0}

{0,1,2,0}

{0,10,0},{1,11,1},{5,4,5}

NIL

{1,11,1}

{1,11,1}

NIL

6 (3 6): {3,4,6}

(3 6):{3,0,1,4,6}

NIL

{0,1,0}

{0,1,2,0}

{0,10,0},{1,11,1}

NIL

{1,11,1}

{1,11,1}

NIL

7 (3 7):{3,7}

NIL NIL

8 (3 8):{3,4,8}

(3 8):{3,0,1,4,8}

{8,4,8}

{0,1,0}

{0,1,2,0}

{0,10,0},{1,11,1},{8,4,8}

NIL

{1,11,1}

{1,11,1}

NIL

9 (3 9):{3,9}

NIL NIL

10 (3 10):{3,0,10}

{0,1,0}

{0,1,2,0}

{0,10,0}

{1,11,1}

{1,11,1}

NIL

11 (3 11):{3,0,1,11}

{0,1,0}

{0,1,2,0}

{0,10,0},{1,11,1}

{1,11,1}

{1,11,1}

NIL

4.1.2 Transition Probabilities and the Mean Sojourn Times:



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Transition Probabilities:

qi,j(t): probability density function (p.d.f.) of the first passage time from a regenerative state i to

a regenerative state j or to a failed state j without visiting any other regenerative state in (0,t].

pi,j: steady state transition probability from a regenerative state i to a regenerative state j without

visiting any other regenerative state. pi,j = ; where denotes Laplace transformation

Table 4: Transition Probabilities

qi,j(t) pi,j=qi,j*(0)

q0,1= λ1e-(λ1+λ3+λ5)t

q0,3= λ3e-(λ1+λ3+λ5)t

q0,10= λ5e-(λ1+λ3+λ5)t

p0,1= λ1/λ1+ λ3+ λ5

p0,3= λ3/λ1+ λ3+ λ5

p0,10= λ5/λ1+ λ3+ λ5

q1,0=g1(t) e-(λ2+λ3+λ5)t

q1,2= λ2e-(λ2+λ3+λ5)t

G1 (t)

q1,4= λ3e-(λ2+λ3+λ5)t

G 1 (t)

q1,11= λ5e-(λ2+λ3+λ5)t

G 1 (t)

p1,0=g1*(λ2+ λ3+ λ5)

p1,2= λ2/ λ2+ λ3+ λ5{1-g1*( λ2+ λ3+ λ5)}

p1,4=λ3/ λ2+ λ3+ λ5{1-g1*( λ2+ λ3+ λ5)}

p1,11=λ5/ λ2+ λ3+ λ5{1-g1*( λ2+ λ3+ λ5)}

q2,0(t)= g2(t) p2,0=g2

*(0)

q3,0(t)= g3(t) e-(λ1+λ4+λ5)t

q3,4(t)= λ1e-(λ1+λ4+λ5)t

G3 (t)

q3,7(t)= λ4e-(λ1+λ4+λ5)t

G 3 (t)

q3,9(t)= λ5e-(λ1+λ4+λ5)t

G 3 (t)

p3,0=g3*( λ1+ λ4+ λ5)

p3,4=λ1/ λ1+ λ4+ λ5{1-g3*( λ1+ λ4+ λ5)}

p3,7=λ4/ λ1+ λ4+ λ5{1-g3*( λ1+ λ4+ λ5)}

p3,9=λ5/ λ1+ λ4+ λ5{1-g3*( λ1+ λ4+ λ5)}

q4,3(t)=g1(t) e-(λ2+λ4+λ5)t

q4,5(t)=λ4e-(λ2+λ4+λ5)t

G1 (t)

q4,6(t)=λ2e-(λ2+λ4+λ5)t

G1 (t)

q4,8(t)=λ5e-(λ2+λ4+λ5)t

G1 (t)

p4,3=g1*( λ2+ λ4+ λ5)

p4,5=λ4/ λ2+ λ4+ λ5{1-g1*( λ2+ λ4+ λ5)}

p4,6=λ2/ λ2+ λ4+ λ5{1-g1*( λ2+ λ4+ λ5)}

p4,8=λ5/ λ2+ λ4+ λ5{1-g1*( λ2+ λ4+ λ5)}

q5,4(t)=g4(t) p5,4=g4*(0)

q6,3(t)=g2(t) p6,3=g2*(0)

q7,3(t)=g4(t) p7,3=g4*(0)

q8,4(t)=g5(t) p8,4=g5*(0)

q9,3(t)=g5(t) p9,3=g5*(0)

q10,0(t)=g5(t) p10,0=g5*(0)

q11,1(t)=g5(t) p11,1=g5*(0)

It can be easily verified that

p0,1+p0,3+p0,10=1 , p1,0+p1,2+p1,4+p1,11=1 ,p2,0=1 ,p3,0+p3,4+p3,7+p3,9= 1

p4,3+p4,5+p4,6+p4,8= 1 ,p5,4= 1 ,p6,3=1 ,p7,3=1 ,p8,4=1 ,p9,3=1 ,p10,0=1 ,p11,1=1

Mean Sojourn Times:

Ri(t) : reliability of the system at time t, given that the system in regenerative state i.

: mean sojourn time spent in state i, before visiting any other states;



34



Table 5: Mean Sojourn Times

Ri(t) µi=Ri*(0)

R0(t)= e-(λ1+λ3+λ5)t

µ0=1/λ1+ λ3+ λ5

R1(t)=e-(λ2+λ3+λ5)t

G 1 (t) µ1=1-g1*( λ2+ λ3+ λ5)/ λ2+ λ3+ λ5

R2(t)=G 2(t) µ2= - g2*’(0)

R3(t)=e-(λ1+λ4+λ5)t

G 3 (t) µ3=1-g3*( λ1+ λ4+ λ5)/ λ1+ λ4+ λ5

R4(t)=e-(λ2+λ4+λ5)t

G 1 (t) µ4=1-g1*( λ1+ λ4+ λ5)/ λ1+ λ4+ λ5

R5(t)= G 4(t) µ5= - g4*’(0)

R6(t)= G 2(t) µ6= - g2*’(0)

R7(t)= G4(t) µ7= - g4*’(0)

R8(t)= G5(t) µ8= - g5*’(0)

R9(t)= G 5(t) µ9= - g5*’(0)

R10(t)= G 5 (t) µ10= - g5*’(0)

R11(t)= G5 (t) µ11= - g5*’(0)

4.1.3 Evaluation of Parameters:

The mean time to system failure and all the key parameters of the system (under steady state

conditions) are evaluated, by applying Regenerative Point Graphical Technique (RPGT) and using ‘3’

as the base-state of the system as under. The transition probability factors of all the reachable states

from the base state ‘3’ are:

V3,0=[(3,0)]/[{1-(0,1,0)/1-(1,11,1)}(1-(1,11,1)){1-(0,1,2,0)/1-(1,11,1)}(1-(1,11,1)){1-(0,10,0)}]

=p30/[{1-(p1,11p11,1)-(p0,1p1,0)}{1-(p1,11p11,1)-(p0,1p1,2p2,0)}{1-(p0,10p10,0)}]

V3,1=[(3,0,1)]/[{1-(0,1,0)/1-(1,11,1)}{1-(1,11,1)}{1-(0,1,2,0)/1-(1,11,1)}{1-(1,11,1)}{1-(0,10,0)}

{1-(1,11,1)}]

=p3,0p0,1/[{1-(p1,11p11,1)-(p0,1p1,0)}{1-(p1,11p11,1)-(p0,1p1,2p2,0)}{1-(p0,10p10,0)}{1-(p1,11p11,1)}]

V3,2=[(3,0,1,2)]/[{1-(0,1,0/1-(1,11,1))}{1-(1,11,1)}{1-(0,1,2,0)/1-(1,11,1)}{1-(0,10,0)}{1-(1,11,1)}]

=p3,0 p0,1 p1,2/{1-(p1,11p11,1 –( p0,1 p1,0)}{1-(p1,11p11,1)-(p0,1 p1,2 p2,0 )}{1-( p0,10 p10,0 )}

{1-( p1,11 p11,1)}]

V3,3=[(3,0,1,4,3)]/[{1-(0,1,0)/1-(1,11,1)}{1-(1,11,1)}{1-(0,1,2,0)}{1-(0,10,0)}]+[(3,0,3)] /

[{1-(0,1,0)/1-(1,11,1)}{1-(1,11,1)}{1-(0,1,2,0)}{1-(0,10,0)}]+[(3,4,3)]+[(3,7,3)]+[(3,9,3)]

=p3,0p0,1 p1,4 p4,3 /[{1-(p1,11p11,1 –( p0,1 p1,0)}{(1-p0,1 p1,2 p2,0 )}{1-( p0,10 p10,0 )}]

+ p3,0p0,3/[{1-(p1,11p11,1 –( p0,1 p1,0)}{(1-p0,1 p1,2 p2,0 )}{1-( p0,10 p10,0 )}+ p3,4 p4,3 + p3,7 p7,3

+ p3,9 p9,3

V3,4=[(3,4)]+[(3,0,1,4)]/[{1-(0,1,0)/1-(1,11,1)}{1-(1,11,1)}{1-(0,1,2,0)/1-(1,11,1)}{1-(1,11,1)}

{1-(0,10,0)}{1-(1,11,1)}]

= p3,4 + p3,0p0,1 p1,4 /[{1-(p1,11p11,1 –( p0,1 p1,0)}{(1-(p1,11p11,1–( p0,1 p1,2 p2,0)}{1-( p0,10 p10,0 )}

{ 1-(p1,11p11,1)}]

V3,5=[(3,4,5)]/1-(5,4,5)+[(3,0,1,4,5)]/[{1-(0,1,0)/1-(1,11,1)}{1-(1,11,1)}{1-(0,1,2,0)/1-(1,11,1)}

{1-(1,11,1)}{1-(0,10,0)}{1-(1,11,1)}{1-(5,4,5)}

= p3,4 p4,5 /1- p5,4 p4,5 + p3,0p0,1 p1,4 p4,5 /[{1-(p1,11p11,1 –( p0,1 p1,0)}{(1-(p1,11p11,1–( p0,1 p1,2 p2,0)}

{1-( p0,10 p10,0 )}{1-(p1,11p11,1)}{1- p5,4 p4,5 )}]

V3,6=[(3,4,6)]+[(3,0,1,4,6)]/[{1-(0,1,0)/1-(1,11,1)}{1-(1,11,1)}{1-(0,1,2,0)/1-(1,11,1)}{1-(1,11,1)}

{1-(0,10,0)}{1-(1,11,1)}



35



= p3,4 p4,6 + p3,0p0,1 p1,4 p4,6 /[{1-(p1,11p11,1 –( p0,1 p1,0)}{1-(p1,11p11,1–( p0,1 p1,2 p2,0)}

{1-( p0,10 p10,0 )}{1-(p1,11p11,1)}]

V3,7 = (3,7) = p3,7

V3,8=[(3,4,8)]/[1-(8,4,8)]+[(3,0,1,4,8)]/[{(1-(0,1,0)/1-(1,11,1))}{1-(1,11,1)}

{(1-(0,1,2,0)/1-(1,11,1))}{1-(1,11,1)}{1-(0,10,0}{1-(1,11,1)}{1-(8,4,8,)}]

= p3,4 p4,8/{1- p8,4 p4,8 }+ p3,0p0,1 p1,4 p4,8 /[{1-(p1,11p11,1 –( p0,1 p1,0)}

{1-(p1,11p11,1–( p0,1 p1,2 p2,0)}{1-( p0,10 p10,0 )}{1-(p1,11p11,1)}{1- p8,4 p4,8)}]

V3,9= (3,9) = p3,9

V3,10=(3,0,10)/[{(1-(0,1,0)/1-(1,11,1))}{1-(1,11,1)}{(1-(0,1,2,0)/1-(1,11,1))}{1-(1,11,1)}

{1-(0,10,0}{1-(1,11,1)}

= p3,0 p0,10 /[{1-(p1,11p11,1 –( p0,1 p1,0)}{1-( p1,11p11,1–( p0,1 p1,2 p2,0)}{1-( p0,10 p10,0 )}{1-(p1,11p11,1)}]

V3,11=(3,0,1,11)/[{1-(0,1,0)/1-(1,11,1)}{1-(1,11,1)}{1-(0,1,2,0)/1-(1,11,1)}{1-(1,11,1)}{1-(0,10,0}

{1-(1,11,1)}]

=p3,0p0,1p1,11/[{1-(p1,11p11,1–(p0,1p1,0)}{1-( p1,11p11,1–( p0,1 p1,2 p2,0)}{1-( p0,10 p10,0 )}{1-(p1,11p11,1)}]

(a). MTSF(T0): From Fig.1, the regenerative un-failed states to which the system can transit(initial

state ‘0’), before entering any failed state are: i = 0,3,4 For ‘ ’ = ‘3’, MTSF is given by

MTSF (T0): =

MTSF(T0)=[(3,0)µ0+(3,3)µ3+(3,4)µ4]÷[1- {(3,4,3)+(3,4,6,3)+(3,0,1,4,3)+(3,0,1,4,6,3)}]

=[ p3,0 µ0+ p3,3 µ3+p3,4 µ4 ]÷[1-{p3,4 p4,3+ p3,4 p4,6 p6,3+ p3,0 p0,1 p1,4 p4,3+ p3,0 p0,1

p1,4 p4,6 p6,3 }]

=[p3,0 µ0+p3,1 µ1+µ3+p3,4 µ4 ]÷[1-{p3,4 p4,3+ p3,4 p4,6 p6,3+ p3,0 p0,1 p1,4 p4,3+ p3,0 p0,1

p1,4 p4,6 p6,3 }]

(b). Availability of the system(A0): From Fig.1, the regenerative states, at which the system is

available are: j = 0,1,3,4 and the regenerative states are i = 0 to 11. For ‘ ’ = ‘3’, the total fraction of

time for which the system remains available is given by

A0 = [V3,0f0 µ0+ V3,1f1 µ1 + V3,3f3µ3+ V3,4f4µ4] ÷ [V3,0 µ01+ V3,1µ11+ V3,2 µ21+ V3,3

µ31+ V3,4µ41+ V3,5 µ51+ V3,6 µ61+ V3,7 µ71 + V3,8µ81+ V3,9 µ91+ V3,10 µ101+ V3,11µ111 ]

Where { fj=1 for all j}

{ µj1= µj for all j}

(c). Busy period of the Server (B0): From Fig.1, the regenerative states where Server is busy while

doing repairs are: j = 1, 2, 3, 4, 5,6,7,8,9,10,11; and the regenerative states are: i = 0 to 11. For ‘ ’ =

‘3’, the total fraction of time for which the Server remains busy is given by

=



36



=

= [V3,0f0 µ0+ V3,1f1 µ1 + V3,3f3µ3+ V3,4f4µ4] ÷ [V3,0 µ01+

V3,1µ1

1+ V3,2 µ2

1+ V3,3 µ3

1+ V3,4µ4

1+ V3,5 µ5

1+ V3,6

µ61+ V3,7 µ7

1 + V3,8µ8

1+ V3,9 µ9

1+ V3,10 µ10

1+ V3,11µ11

1 ]

Where { fj=1 for all j}{ µj1= µj for all j}

(c). Busy period of the Server (B0): From Fig.1, the regenerative states where Server is busy while doing repairs

are: j = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11; and the regenerative states are: i = 0 to 11. For ‘’ = ‘3’, the total fraction

of time for which the Server remains busy is

=

=

(d). Expected number of Server’s visits: From Fig.1, the Regenerative States where the Server visits

(afresh) for repairs of the system are: j=1,3,10; the Regenerative States are: i = 0 to 11. For ‘’ = ‘3’,

the Expected number of Server’s Visits per unit time is given by

V1= =

V1 = [V3,1 +V3,3+ V3,10] ÷ [V3,0 µ01 + V3,1 µ11 + V3,2 µ21 + V3,3 µ31+ V3,4 µ41 + V3,5 µ51

+V3,6 µ61 + V3,7 µ71 + V3,8 µ81 + V3,9 µ91+ V3,10 µ101 + V3,11 µ111 ]

5 Particular Case

Let us take;

g1(t)= ω1e-ω

1t, g2(t)= ω2e

-ω2

t, g3(t)= ω3e

-ω3

t, g4(t)= ω4e

-ω4

t, g5(t)= ω5e

-ω5

t

we have,

p0,1= λ1/ λ1+λ3+ λ5, p0,3= λ3/ λ1+λ3+ λ5, p0,10 = λ5/ λ1+λ3+ λ5, p1,0= ω1/ ω1+

λ2+λ3+ λ5, p1,2= λ2 /ω1+ λ2+λ3+ λ5, p1,4= λ3 /ω1+ λ2+λ3+ λ5, p1,11= λ5/ω1+ λ2+λ3+ λ5, p2,0=1,

p3,0= ω3/ ω3+ λ1+λ4+ λ5, p3,4= λ1/ω3+ λ1+λ4+ λ5, p3,7= λ4/ω3+ λ1+λ4+ λ5, p3,9== λ5/ω3+ λ1+λ4+ λ5 ,

p4,3= ω1/ω1+ λ2+λ4+ λ5, p4,5= λ4/ ω1+ λ2+λ4+ λ5 , p4,6= λ2/ ω1+ λ2+λ4+ λ5, p4,8= λ5/ ω1+ λ2+λ4+

λ5, p5,4 =1, p6,3 =1, p7,3 = 1, p8,4 = 1, p9,3 = 1, p10,0 = 1, p11,1 = 1

µ0 = 1/λ1+λ3+λ5, µ1= 1/ω1+λ2+λ3+λ5, µ2= 1/ω2, µ3=1/ω3+λ1+λ4+λ5, µ4=1/ω1+λ1+λ4+λ5, µ5=1/ω4, µ6=1/ω2 ,

µ7=1/ω4 , µ8=1/ω5 , µ9=1/ω5 , µ10=1/ω5 , µ11=1/ω5

By using these results, we get the following:

MTSF(T0)= [ω3(ω1+λ1+λ4+ λ5)+(ω1+ λ1+λ4+ λ5)( λ1+λ3+ λ5)+ λ1(λ1+λ3+ λ5)](ω1+ λ2+λ3+ λ5)(ω1+ λ2+λ4+ λ5)

÷ [λ1 ω1(ω1+ λ2+λ3+ λ5)( λ1+λ3+ λ5)+ λ1 λ2(ω1+ λ2+λ3+ λ5)( λ1+λ3+ λ5)+ ω1ω3 λ1 λ3+ ω3 λ1 λ2 λ3] (ω1+ λ1+λ4+

λ5)



37



Availability (A0) = [[(ω1+λ2+λ3+ λ5)( λ1+λ3+λ5)-λ5(λ1+λ3+ λ5)- λ1ω1]

[(ω1+λ2+λ3+λ5)(λ1+λ3+λ5)-λ5(λ1+λ3+λ5)-λ1λ2](λ1+λ3)[ω2λ1(ω1+λ2+λ3)

+(ω3+λ1+λ4+λ5){λ1ω1+(λ4+λ5)(ω1+λ2+λ4+λ5)}]+ω3λ3(ω1+λ2+λ3+λ5)(λ1+λ3+λ5)2

[ω1ω2λ1+ω2(ω1+λ2+λ3+λ5)(ω1+λ2+λ4+λ5)+ω2λ1(ω1+λ2+λ3+λ5)]+[ω2(ω1+λ2+λ3)+ω2λ1+λ1λ2

(ω3+λ1+λ4+λ5)](ω3+λ1+λ4+λ5)((ω1+λ2+λ3+λ5)](ω1+λ2+λ4)((ω1+λ2+λ5)÷[(ω1+λ2+λ3+λ5)

(λ1+λ3+λ5)-λ5(λ1+λ3+λ5)-λ1ω1][(ω1+λ2+λ3+λ5)(λ1+λ3+λ5)-λ5(λ1+λ3+λ5)-λ1λ2] (λ1+λ3)[{(ω1+λ2+λ3)

{λ1λ4ω2(ω1+λ2+λ4+λ5)+λ1λ2ω4(ω1+λ2+λ5)}ω5(ω1+λ2+λ4)+λ5ω2ω4(ω1+λ2+λ5)

(ω1+λ2+λ4+λ5)}+ω2{λ1(ω1+λ2+λ3)+(ω3+λ1+λ4+λ5)λ1ω1+(λ4+λ5)(ω1+λ2+λ4+λ5)}]

+ω3λ3(ω1+λ2+λ3+λ5)(λ1+λ3+λ5)2[{ω1λ1+(ω1+λ2+λ3+λ5)(ω1+λ2+λ4+λ5)}ω2+ω2λ1(ω1+λ2+λ3+λ5)]

(ω1+λ2+λ4)(ω1+λ2+λ5)+[ω2(ω1+λ2+λ3)+ω2λ1+λ1λ2(ω3+λ1+λ4+λ5)(ω1+λ2+λ4+λ5)(ω1+λ2+λ4)

(ω1+λ2+λ5)+[(λ4ω5+λ5ω4)ω2(ω1+λ2+λ5)(ω1+λ2+λ4)(ω1+λ2+λ4+λ5)](ω3+λ1+λ4+λ5)ω4ω5+ω3)

(ω1+λ2+λ3+λ5)2(λ1+λ3+λ5)

2(ω1+λ2+λ4)[λ1λ3ω5{λ4ω2(ω1+λ2+λ4+λ5)+λ2ω4(ω1+λ2+λ5)

+λ5ω2ω4{λ1+(ω1+λ2+λ3+λ5)}(ω1+λ2+λ5)(ω1+λ2+λ4+λ5)

Busy period of the Server (B0 )=

[ω2ω3ω4λ1λ3(ω1+λ2+λ5)(ω1+λ2+λ4)(ω1+λ2+λ4+λ5)(λ3ω5+λ5(ω1+λ1+λ4+λ5))+ω3ω5λ1λ3

(ω1+λ1+λ4+λ5)(ω1+λ2+λ4)(ω2λ4(ω1+λ2+λ4+λ5)+ω4λ2(ω1+λ2+λ5))+ω2ω3ω4λ5(ω1+λ1+λ4+λ5)

(ω1+λ2+λ5)(ω1+λ2+λ4+λ5)(λ1λ3+ω1+λ2+λ4)+ω3(ω1+λ2+λ4+λ5){ω2λ1+λ1λ2(ω1+λ1+λ4+λ5)}]

(ω1+λ2+λ3+λ5)2(λ1+λ3+λ5)

2(ω3+λ1+λ4+λ5){(ω1+λ2+λ3+λ5)(λ1+λ3+λ5)-λ1λ2}+[λ1ω5(ω1+λ2+λ4)

{(ω1+λ2+λ5)(ω1+λ2+λ4+λ5)ω2ω4+λ4ω2(ω1+λ2+λ4+λ5)(ω1+λ1+λ4+λ5)+λ2ω4(ω1+λ1+λ4+λ5)

(ω1+λ2+λ5)}+ω2(ω1+λ1+λ4+λ5)(ω1+λ2+λ5)(ω1+λ2+λ4+λ5){λ4ω5(ω1+λ2+λ4)+λ5ω4

(λ1+ω1+λ2+λ4)}(ω3+λ1+λ4+λ5)(ω1+λ2+λ4+λ5)+{λ1ω1+(λ4+λ5)(ω1+λ2+λ4+λ5)}ω2] {(ω1+λ2+λ3+λ5)(λ1+λ3+λ5)-

λ5(λ1+λ3+λ5)-λ1ω1}{(ω1+λ2+λ3+λ5)(λ1+λ3+λ5)-λ5(λ1+λ3+λ5)-λ1λ2}

{ω1+λ2+λ3}{λ1+λ3}{(ω1+λ2+λ3+λ5)(λ1+λ3+λ5)-λ1λ2}÷[ω2ω3ω4λ1λ3(ω1+λ2+λ5)(ω1+λ2+λ4)

(ω1+λ2+λ4+λ5)(λ3ω5+λ5(ω1+λ1+λ4+λ5))+ω3ω5λ1λ3(ω1+λ1+λ4+λ5)(ω1+λ2+λ4)

(ω2λ4(ω1+λ2+λ4+λ5)+ω4λ2(ω1+λ2+λ5))+ω2ω3ω4λ5(ω1+λ1+λ4+λ5)(ω1+λ2+λ5)(ω1+λ2+λ4+λ5)

(λ1λ3+ω1+λ2+λ4)+ω3(ω1+λ2+λ4+λ5){(ω1+λ2+λ3)ω2+ω2λ1+λ1λ2(ω1+λ1+λ4+λ5)}](ω1+λ2+λ3+λ5)2

(λ1+λ3+λ5)2(ω3+λ1+λ4+λ5){(ω1+λ2+λ3+λ5)(λ1+λ3+λ5)-λ1λ2}+[λ1ω5(ω1+λ2+λ4){(ω1+λ2+λ5)

(ω1+λ2+λ4+λ5)ω2ω4+λ4ω2(ω1+λ2+λ4+λ5)(ω1+λ1+λ4+λ5)+λ2ω4(ω1+λ1+λ4+λ5)(ω1+λ2+λ5)}+ω2

(ω1+λ1+λ4+λ5)(ω1+λ2+λ5)(ω1+λ2+λ4+λ5){λ4ω5(ω1+λ2+λ4)+ λ5ω4(λ1+ω1+λ2+λ4)}(ω3+λ1+λ4+λ5)

(ω1+λ2+λ4+λ5)+{λ1ω1+(λ4+λ5)(ω1+λ2+λ4+λ5)}ω2]{(ω1+λ2+λ3+λ5)(λ1+λ3+λ5)-λ5(λ1+λ3+λ5)-

λ1ω1}{(ω1+λ2+λ3+λ5)(λ1+λ3+λ5)-λ5(λ1+λ3+λ5)-λ1λ2}{ω1+λ2+λ3}{λ1+λ3}{(ω1+λ2+λ3+λ5) (λ1+λ3+λ5)-λ1λ2}

Number of server’s visits (V0) =

[ω3(λ1+λ3+λ5)2(ω1+λ2+λ3+λ5)[(ω1+λ2+λ3+λ5)

2(λ1+λ5){(ω1+λ2+λ3+λ5)(λ1+λ3+λ5)-λ1λ2}+λ3

{ω1λ1+(ω1+λ2+λ3+λ5)(ω1+λ2+λ4+λ5)}(ω1+λ2+λ3){(ω1+λ2+λ3+λ5)(λ1+λ3+λ5)-λ5(λ1+λ3+λ5)-

λ1λ2}]+[{ω1λ1+(λ4+λ5)(ω1+λ2+λ4+λ5)}{(ω1+λ2+λ3+λ5)(λ1+λ3+λ5)-λ5(λ1+λ3+λ5)-λ1λ2}{λ1+λ3}

{ω1+λ2+λ3}{(ω1+λ2+λ3+λ5)(λ1+λ3+λ5)-λ5(λ1+λ3+λ5)-λ1ω1}{(ω1+λ2+λ3+λ5)(λ1+λ3+λ5)-λ1λ2}]]

(ω3+λ1+λ4+λ5)(ω1+λ1+λ4+λ5)(ω1+λ2+λ5)(ω1+λ2+λ4+λ5)(ω1+λ2+λ4)ω2ω4ω5÷

[ω2ω3ω4λ1λ3(ω1+λ2+λ5)(ω1+λ2+λ4)(ω1+λ2+λ4+λ5)(λ3ω5+λ5(ω1+λ1+λ4+λ5))+ω3ω5λ1λ3

(ω1+λ1+λ4+λ5)(ω1+λ2+λ4)(ω2λ4(ω1+λ2+λ4+λ5)+ω4λ2(ω1+λ2+λ5))+ω2ω3ω4λ5(ω1+λ1+λ4+λ5)

(ω1+λ2+λ5)(ω1+λ2+λ4+λ5)(λ1λ3+ω1+λ2+λ4)+ω3(ω1+λ2+λ4+λ5){(ω1+λ2+λ3)ω2+ω2λ1+λ1λ2

(ω1+λ1+λ4+λ5)}](ω1+λ2+λ3+λ5)2(λ1+λ3+λ5)

2(ω3+λ1+λ4+λ5){(ω1+λ2+λ3+λ5)(λ1+λ3+λ5)-

λ1λ2}+[λ1ω5(ω1+λ2+λ4){(ω1+λ2+λ5)(ω1+λ2+λ4+λ5)ω2ω4+λ4ω2(ω1+λ2+λ4+λ5)(ω1+λ1+λ4+λ5)

+λ2ω4(ω1+λ1+λ4+λ5)(ω1+λ2+λ5)}+ω2(ω1+λ1+λ4+λ5)(ω1+λ2+λ5)(ω1+λ2+λ4+λ5){λ4ω5(ω1+λ2+λ4)+

λ5ω4(λ1+ω1+λ2+λ4)}(ω3+λ1+λ4+λ5)(ω1+λ2+λ4+λ5)+{λ1ω1+(λ4+λ5)(ω1+λ2+λ4+λ5)}ω2]{(ω1+λ2+λ3+λ5)(λ1+λ3+λ5



38



)-λ5(λ1+λ3+λ5)-λ1ω1}{(ω1+λ2+λ3+λ5)(λ1+λ3+λ5)-λ5(λ1+λ3+λ5)-λ1λ2}{ω1+λ2+λ3}{

λ1+λ3}{(ω1+λ2+λ3+λ5)(λ1+λ3+λ5)-λ1λ2}

6 Analytical Discussions: The following tables, graphs and conclusions are obtained for:

λ1 = λ2= λ3= λ4= λ5= λ , ω1= ω2= ω3= ω4= ω4 = ω

(A) MTSF vs. Repair Rate:

The MTSF of the system is calculated for different values of the Failure Rate (λ) by taking λ=0.0005,

0.0006, 0.0007, 0.0008, 0.0009 and 0.0010 and for different values of the Repair Rate(ω) by taking

ω=0.80,0.85,0.90,0.95 and 1.00 .The data so obtained are shown in Table 6 and Graphically in fig.2

Table: 6: The data obtained from evaluation

λ T0

(ω=0.80)

T0 (ω=0.85) T0 (ω=0.90) T0 (ω=0.95) T0 (ω=1.00)

0.0005 809178.686 814732.000 861805.426 942141.816 1006014.270

0.0006 558512.905 591508.764 631306.077 699458.715 751551.648

0.0007 379730.524 479285.508 578840.492 678506.054 714567.8903

0.0008 252948.263 365715.9791 469245.789 56782.5698 699456.0987

0.0009 126150.052 240046.1222 345367.879 456784.4532 56345.6543

0.001 112345.890 127676.5936 23462.9876 32451.5642 44673.6754

Table 6 shows the behaviour of the MTSF(T0) vs. the repair rate(ω) of the unit of the system for

different values of the failure rate (λ).It is concluded that MTSF increases with increase in the values

of the repair rate(ω).

0

200000

400000

600000

800000

1000000

1200000

0.8 0.85 0.9 0.95 1

MTS

F(T0)

REPAIR RATE(ω)

MTSF GRAPH

T0(λ=0.0005)

T0((λ=0.0007)

Figure 2: MTSF Graph

Further it can be concluded from the fig.3 that values of MTSF( T0) shows the expected trend for

different values of failure rate(λ=0.0005 & 0.0006),as T0 increases in the values of repair rate(ω).

(B)Availability(A1) vs. the Repair Rate(ω) &failre rate :



39



The Availability of the system is calculated for different values of the failure rate (λ) by taking λ

=0.0005, 0.0006, 0.0007, 0.0008, 0.0009 and 0.001 and for different values of the repair rate (ω) by

taking ω =0.80, 0.85, 0.90, 0.95 and1.0.The data so obtained are shown in table 7 and graphically in

fig.3 Table 7: The behaviour of the Availability (A0) vs. the Repair rate (ω) of the unit of the system for different values of

the failure rate (λ).

λ A0 (ω=0.80) A0 (ω=0.85) A0 (ω=0.90) A0 (ω=0.95) A0 (ω=1.00)

0.0005 0.4147411 0.5278114 0.66573563 0.7800141 0.99308441

0.0006 0.41422548 0.52755351 0.66420895 0.77848596 0.99157184

0.0007 0.41271037 0.52656525 0.66268337 0.77696039 0.99004626

0.0008 0.40919573 0.52431836 0.66121781 0.77544480 0.98852072

0.0009 0.40968062 0.5223305 0.65975255 0.77394253 0.98700848

0.001 0.40816598 0.52008361 0.62570068 0.76789092 0.98548293

Table shows the behaviour of the Availability (A0) vs. the Repair rate (ω) of the unit of the system for

different values of the failure rate (λ).It is concluded that Availability (A0) increases with increase in

the values of the Repair rate (ω) and decreases with the increase in failure rates.

Figure 3: Availability Graph

Further it can be concluded from the fig.3 that values of availability( A0) shows the expected trend

for different values of failure rate(λ=0.0005 & 0.0007),as A0 increases in the values of repair rate(ω).

6 Conclusion:

From the Graphs and Tables, we see that as the Repair Rate (ω) increases, Availability of the

system increases, which would be. The Regenerative-Point Graphical Technique is useful for

the evaluation of the parameters in a simple way, without writing any state equation and

without doing any lengthy and cumbersome calculations. This study can be extended to time

dependent case. As it is easy for the management to control repair rates in comparison to



40



failure rates, fixing the target of availability repair rates can be determined and managed by

having efficient server.

7 References:

[1] Gupta, V. K. Singh & Vanita (2010), ‘ The New Concept of Base State in the Reliability Analysis’; Journal

of Mathematics and System Science (JMASS), Vol. 6, No. 2.

[2] Jindal, G. (2012), ‘Behavior and Availability Analysis of Some Industrial Systems’; Ph.D. Thesis, JJTU

(Jhunjhunu).

[3] Goel, P. & Singh, J. (1995), ‘Availability Analysis of a Stand-by Complex System Having Imperfect

Switch.’ Microelectronics and Relief, Vol. 35, No. 2 pp 285-288.

[4] Gupta, V. K. & Singh, J. (2007), ‘A New Approach for Availability Analysis’; IEEF transitions on

reliability, Tr 2007-107.

[5] Gupta, V. K (2008), ‘Behavior and Profit Analysis of Some Process Industries’; Ph.D. Thesis, NIT

Kurukshetra (India).

[6] Nidhi, Goel & Garg discussed Availability Analysis of a Soap Industry Using Regenerative Point Graphical

Technique (RPGT) Vol. 1 Issue 2, October, 2013 IJIFR, ISSN: 2347-1697.

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