REPAIRABLESYSTEMS

1

AVAILABILITYENGINEERING

A Single Repairable ComponentWith Failure Rate λ and Repair Rate μ

The component may exist in one of Two States SO (Working) and S1 (Failing).It transits from SO to S1 by Failure and back from S1 to SO by Repair

SO

S1

λμ

The Concept of AVAILABILITY

STATE GRAPH

After time interval ΔtAt Present

SO S1

SO 1 - λ Δt λ Δt

S1 μ Δt 1 - μ Δt

TRANSITION MATRIX

It is required to evaluate Probabilities PO and P1 Of the component being in states SO and S1

TRANSITION EQUATIONS

After time interval ΔtAt Present

SO S1

SO 1 - λ Δt λ Δt

S1 μ Δt 1 - μ Δt

OO

OOO

OOO

OOO

OO

O

PdtdP

Pt

tPttPttPtPttPPttPtttP

PttPtttPPP

)()(

)1()(1)(1

1

1

1

Multiply both sides byte )(

KeeP

dteePd

eePdtd

ePeedtdP

ttO

ttO

ttO

tO

ttO

)()(

)()(

)()(

)()()(

)(

KeeP

dteePd

eePdtd

ePeedtdP

ttO

ttO

ttO

tO

ttO

)()(

)()(

)()(

)()()(

)(

tO KeP )(

)(

Then

, ,)()(

1 then )(

1

, 1 0

ThereforeKK

ThereforePtAt O

tO eP )(

)()(

At Steady state as t tends to infinity, the component(system) will be AVAILABLE with probability

APO

)(

After time interval ΔtAt Present

SO S1

SO 1 - λ Δt λ Δt

S1 μ Δt 1 - μ Δt

OO

OOO

OOO

OOO

OO

O

PdtdP

Pt

tPttPttPtPttPPttPtttP

PttPtttPPP

)()(

)1()(1)(1

1

1

1

As t tends to infinity, Po = Availability = const and dPo / d t =0, then

A

A λ=1/MTBF μ=1/MTTR Therefore

TimeDownMeanTimeUpMeanTimeUpMean

MTTRMTTFMTBFA

Another Approach toAVAILABILITY

Availability of A System with One Unit Standby

This system could exist in one of three states So , S1 and S2:

So state of system operating with probability Po

S1 state of system with the main component failed and switched to the standby unit to operate ( probability of being in this state P1,S2 State of having both units main and standby failed with probability P2

SOS1 S2

λ λ

μ2 μ

SO S1 S2So 1 - λ Δt λ Δt 0

S1 μ Δt 1 - λ Δt - μ Δt λ Δt

S2 0 2 μ Δt 1 – 2 μ Δt

Two Repair gangs are available

STATE GRAPH

TRANSITION MATRIX

TRANSITION EQUATIONS

)........(3....................2

)(2)()(

)(2)()()(21)(

..(2)..............................

before,shown as Therefore)(1

.....(1)..............................1

122

2122

2122

212

1

1

21

PPdtdP

tPtPt

tPttPttPttPtPttP

tPtttPttP

PPdtdP

PttPtttPPPP

OO

OO

O

SO S1 S2SO 1 - λ Δt λ Δt 0

S1 μ Δt 1 - λ Δt - μ Δt λ Δt

S2 0 2 μ Δt 1 – 2 μ Δt

Taking the Limit as the time tends to infinity, we find

)5(..........212

)4........(..........

2

212

11

O

OO

PPPP

PPPλP

.

Substitute (4) and (5) in (1), we find

22

2

2

2

222

1211

121.

O

O

OOO

P

P

PPP

This Result could be obtained directly by a method called HARMONIC BALANCE Apply Harmonic Balance to state So

SO

Expected IN [ µ P1 ] = Expected out [ λ Po ]

μ

S1λ

Similarly, apply Harmonic Balance to state S2 S1 S2

λ

2μ.2 21 PP .....(1)..............................121 PPPO

The System is AVAILABLE in case of being in either states So and S1

22

2

1

2222

1

A

PPPPA OOO

. 1 OPP

22

2

222

OP

In a metal cutting workshop, there are two different machines:Machine A: a mechanical cutting with capacity of 10 tons per hour. The machine has a constant failure rate of 3 failures per month and mean repair time of 5 daysMachine B: a gas cutting machine with capacity of 20 tons per hour . The machine has a constant failure rate of 4 failures per month and mean repair time of 3 days.Evaluate the EFFECTIVE CAPACITY of the workshop._____________________________________________________________________The Workshop could exist in one of the following statesSo Both machines are working. S1 Mechanical Cutting failed and gas Cutting worksS2 Gas Cutting failed and Mechanical Cutting worksS3 Both failed

S1S2

S3

λ m

μ mλ gRequired to find the FOUR probabilities, Po, P1, P2, P3

The first equation:Po + P1 + P2 + P3 = 1 ………………………….(1)The other THREE equation will be obtained by applying HARMONIC BALANCE to the three statesSo, S1, S2 as follows:

SO

μ m

λ mλ g

μ g

μ g103/30 465/30 3

gg

mm

)2...(..........21 PPP gmOgm Apply to So

Apply to S1 )3...(..........31 PPP gOmmg

Apply to S2 )4...(..........32 PPP mOgmg Omitting P3 from the two equations (3) and (4), we get

)5......(21 Oggmmggmmmg PPP Consider (2) and (5) to omit P2, we find

)6.(..........01

1

PP

PP

m

m

Ogmgmmmgmgm

Substitute in (2)

)7.........(02

20

PP

PPP

g

g

gmOgm

Substitute in (3)

)8.(..........3

30

Og

g

m

m

gOmm

mmg

PP

PPP

Effective Capacity = (20+10)*Po + 20*P1 + 10 * P2

tons95.201905.0*102381.0*204762.0*30

095.0 1905.0 2381.0 321 PPP

Substitute in (1), we get

i

ii

gmgmO

Og

g

m

mO

g

gO

m

mO

P

PPPP

)9.(..........1

1

1

Po + P1 + P2 + P3 =1……….(1)..01 PP

m

m

02 PPg

g

..3 Og

g

m

m PP

10 46 3

gg

mm

4762.02.04.05.01

1 4.0 5.0 0

Pgg

gm

m

m

Evaluation of the RELIABILITY of the above considered REPAIRABLE SYSTEM

)........(3....................

..(2)..............................

before, shown as Therefore

.....(1)..............................

122

2122

2122

212

1

1

21

2

)(2)()()(2)()()(21)(

)(11

PPdtdP

tPtPt

tPttPttPttPtPttP

tPtttPttP

PPdtdP

PttPtttPPPP

OO

OO

O

...(3)..............................2

...(2)..............................

...(1).......... ......................1

122

1

21

PPdtdP

PPdtdP

PPP

OO

O

Consider the following three equations in the three unknowns:

Apply Laplace Transform on the above three equations

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

)0(

1

2

*1

*22

*2

*1

**

*2

*1

*

PPPPPsP

PPPsPs

PPP

O

OOO

O

0

* dtePP stmm

Where, Is the Laplace Transform of Pm. Therefore,

..(6)..........2

5)1........(

)4..(..........1

*1

*2

*2

*1

**

*2

*1

*

PPsP

PPsPs

PPP

OO

O

Solving the above system of algebraic equations

At t = 0, PO =1 :

)(1

)(1

K

K

Therefore,

A

AtRtAs

etRtP

eeP

tO

ttO

)(

)()(

)(

)(

)()(

A is the AVAILABILITY λ=1/MTTF μ=1/MTTR

Therefore

TimeDownMeanTimeUpMeanTimeUpMean

MTTRMTTFMTTFA

KeeP ttO

)()(

)(