systems reliabilty 1. systems are basically built of different components and /or subsystems. for...

SYSTEMS RELIABILTY

1

SYSTEMS are basically built of different components and /or subsystems.For each component, there is an assigned role in the system performance.Take the motorcar as an illustrating example. The car is a system composed of:• Prime mover• Transmission subsystem• Car body• Braking subsystem

The transmission subsystem is composed of: Clutch, Transmission shaft, Differential gear, wheel axes and wheels. All these components are connected in SERIES

The Braking system is composed mainly of two Alternative subsystems:• Hydraulic leg actuated Brakes• Mechanical hand operated brakes.• The TWO subsystem are working in PARALLEL

Electric power generating systems may consist of N generating sets,For purposes of Reliability Increase,Out of them M generating sets (M<N)are sufficient to provide the necessary power

GENERALLY,

SYSTEMS may be built in one of the following configurations:

SERIES

PARALLEL

(With HOT REDUNDANCY)

Parallel Of Series

Series Of Parallel

M Out Of N (MOON)

K Consecutive Out N: Failed

2/4

2 Consecutive Out 8: Failed

System Fails

System Survives

SYSTEMS IN SERIES

The SYSTEM is considered failed if ONE of the components failedThe SYSTEM is considered Working ONLY IF ALL components are working

A B C

If A, B and C are Events of having components A, B and C working, then System RELIABILITY is given by:

CBAPRS

Assuming that the state of any of the components is independent of the others

CBAS

S

RRRRCPBPAPR

)()()(

Generally, if there N components are connected in SERIES, the System Reliability will be

N

KKS RR

1

SYSTEMS IN PARALLEL

The SYSTEM is considered failed if ALL components failed

A

B

C

Generally, if there N components are connected in PARALLEL, the System Reliability will be

N

KK

N

KKS RFR

11

111

CBACBAS

CBAS

RRRFFFRFFFF

11111

Then

A special case of having TWO Identical components

2

22

2

)21(111

RRR

RRRR

S

S

M Out Of N (MOON)

2/4

A

B

C

D

DCBADCBA

DCBADCBADCBA

DCBACBAADCBA

DCBADCBADCBAS

RRFFRFRFFRRFRFFRFRFRFFRRRRRFRRFRRFRRFRRRRRRRR

The System operates if ALL the four components operate A B C D =1 Alternative OR THREE OUT OF FOUR operate A B C D’ ABD C’ ACD B’ BCD A’ =4 AlternativesOR TWO OUT OF FOUR operate AB C’D’ AC B’D’ AD B’C’ BC A’D’ BD A’ C’ CD A’B’ =6 Alternatives

The Important Special Case, when ALL components are IDENTICAL2234 64 FRFRRRS

And Generally Having M Out Of N

N

MK

KNKNKS FRCR

6!2 !2

!4 4!1!3!4

)!(!! 4

243

CC

mnmnC n

m

STANDBY SYSTEMSc

Main Unit

STANDBY Unit

STANDBY Unit)( MXPRS

M = Number of Standby Components

`X is a Discrete Random Variable distributed according to POISSON’s Distribution

...!3!2

1!

32 ttteXteR t

MX

oX

Xt

S

1...

!3!3

!2!211

0

MRdtMTTF

Systems in SERIES OF PARALLELA

B

C

D

Consider first the PARALLEL and then the SERIES,, we get

DCBAS FFFFR 11

Generally,

N

G

M

kKS

M

kKG

N

GGS

G

G

G

G

G

G

FR

FRRR

1 1

11

1

1

N is the number of PARALLEL Groups connected in SERIESMG is the number of components in Parallel in the G th Group

For COMPLEX SYSTEMS, The RELIABILITY and MTTF Cannot be obtained in a CLOSED FORM as obtained for the Previous configurations.

Therefore, Special Methods will be applied as will be shown later

Complex Systems

Cut Sets

1,2 3,4 1,5,3

2,4,5

1 2 3 4 1 5 3 2 4 5 F .05 .1 .15 .2 .05 .15 .25 .1 .2 .25

F cut sets

.005 .03

.00188 .005

R cut sets

0.995 0.97 0.99812 0.995

R system

0.9582

MINIMUM CUTSETS METHOD

Systems in PARALLEL OF SERIES A

B

C

DConsider first the SERIES and then the PARALLEL,, we get

DBCAS RRRRR 111

Generally

N

G

M

KKS

N

G

M

KKGGS

G

G

G

G

G

G

RR

RFFR

1 1

1 1

11

11

N is the number of PARALLEL BranchesMG is the number of components in series in the G th branch

K Consecutive Out N: Failed

2 Consecutive Out 6: Failed

R is the component ReliabilityRS is the system Reliability N=6 the total number of components

K =2 The number of consecutive components If Failed, the system fails

2

1

0

1 )1(,2,

N

J

JNJJNJS RRCNRR

3

0

67 )1(6,2,J

JJJJS RRCRR

334

3425

2

561

6

)1()1(

)1(6,2,

RRCRRC

RRCRRRS

RELIABILITY & MTTF FOR

SYSTEMS WITH COMPONENTS HAVING

CONSTANT FAILURE RATE

CONSTANT FAILURE RATE (CFR)

Given a System composed of N components connected in SeriesThe Reliability of the K th component is given by

tK

KeR The System Reliability:

N

KK

K

tN

K

tS eeR 1

1

The Mean Time To Failure MTTF of the SYSTEM

N

KK

S

t

SS

MTTF

dtedtRMTTF

N

KK

1

0

0

1

1

As the number of components in series increases, MTTF of the system DECREASES

SERIES

CONSTANT FAILURE RATE (CFR) PARALLELGiven a System composed of N components connected in ParallelThe Reliability of the K th component is given by t

KKeR

The System Reliability:

N

K

tN

KKS

KeRR11

1111

The Mean Time To Failure MTTF of the SYSTEM

dtedtRMTTFN

K

tSS

K

0 10

11

Example: Take =2 and the components are identical with the same failure rate λ

tttS eeeR 22

211

5.1212

20

2

0

dteedtRMTTF ttSS

Two Identical ComponentsWith Failure Rate λ =0.01 (MTTF=100 hrs)

R=0.9

Configuration IN SERIES IN PARALLEL

MTTFS

1 / 2λ 1.5 / λ

50 hrs 150 hrs

RSR2

0.81

2R - R2

0.99

0 30 60 90 120

150

180

220

280

340

400

460

520

00.0020.0040.0060.008

0.010.012

Time

Two components each with λ=0.01 are put in Parallel

λ=0.01

λ=0.01

00.20.40.60.8

1

RS

R

ttS

t

eeR

eR

22

t

t

S

ttS

t

eeh

eetf

hetf

212

12)(

)(

Each component ofThe FOUR:Λ=0.02, R=0.95Find System Hazard RateAnd Reliability

9905.0)95.0(1122 SR

222 ttS eeR

2211 tS eR

121144

2

0

4

0

3

0

2

0

22

0

dtedtedte

eedtRMTTF

ttt

ttS

995.0)95.01(122 SR

432

11

0

4

0

2

0

22

0

dtedte

dtedtRMTTF

tt

tS

1) Unit reliability R = 10 / (10 + t) t in years. How many units in parallel are required to achieve a reliability of 0.98 in 5 years? If there is an additional common mode failure rate of 0.002 as a result of environmental factors. How many units in this case?

2) A natural gas distribution network contains FIVE shut-off valves. Valves 1 – 4 have probability of 0.02 of failing open and a probability of 0. 15 failing short. Valve 5 has probability of 0.05 of failing openand a probability of 0.2 failing short . Find system reliability.

4

5

11

2

3

N

m m1

11

N

Km m11

N

m m1

11

N

Km m11