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Reliability PredictionA Quest for Reliable Parameters

By

Yair Shai

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Goals

• Compare the MTBCF & MTTCF parameters in view of complex systems engineering.

• Failure repair policy as the backbone for realistic MTBCF calculation.

• Motivation for modification of the technical specification requirements.

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Promo : Description of Parameters

r

ti

i

timet1 t2 t3 t4 t5 .......

Failure Event of an Item

Repairable Items:

Mean Time Between Failures =

Non Repairable Items:

Mean Time To Failure =

r

ti

ir =Number of Failures

Seman

tics

?

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MTBF = MTTF?? An assumption:

Failed item returns to “As Good As New” status after repair or renewal.

note: Time To Repair is not considered.

UP

DOWN

TIME

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Critical FailuresMoving towards System Design

A System Failure resulting in (temporary or permanent) Mission Termination.

COMPUTER

COMPUTER

SUBSYSTEM

A simple configuration of parallel hot Redundancy.

A Failure: any computer failure

A Critical Failure: two computers failed

XX

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Critical FailuresA clue for Design Architecture

MTBCF

Mean Time Between Critical Failures

MTTCF

Mean Time To Critical Failure

SAME? Remember the assumptions

Determining the failure repair policy: COLD REPAIR

No time for repair actions during the mission

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Functional System Design

Operational Demand: At least two receiver units and one antenna should work to operate the system.

CPU

CPUPOWER

SUPPLY

POWER SUPPLY 4 CHANNEL

RECEVERCONTROLER

UNIT A

UNIT B

UNIT C

UNIT D

POWER SUPPLY

ANTENA

ANTENA

ANTENA

ANTENA

2 / 4

sw

Switch control

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From System Design to Reliability Model

Serial model : Rs = R1x R2

Parallel model : Rs = 1- (1-R1)x(1-R2)

K out of N model : Rs = Binomial Solution

CPU PS1

CPU PS1

CONT PS2

A

B

C

D

2 / 4

xx

x

ANT

ANT

ANT

ANT

sw

Is this a Critical Failure ?

INDEPENDENT

BLOCKS

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From RBD Logic Diagram to Reliability Function

Rsys(t)= f( serial / parallel / K out of N)Classic parameter evaluation:

0

)( dttRsysMTBFWARNING!!!

After each repair of a critical failure - The whole system returns to status “As Good As New”.

Is this realistic?

Simple mathematical manipulation:

MTTCF 0

( )MTTF Rsys t dt

MTBCF

[ S.Zacks, Springer-Verlag 1991, Introduction To Reliability Analysis, Par 3.5]

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Realistic interpretation:

MTBCF = MTTCF

Only failed Items which cause the failure are repaired to idle. All other components keep on aging.

MTBCF vs. MTTCFA New Interpretation

Common practice interpretation:

MTBCF = MTTCF = MTTCFF

Each repair “Resets” the time count to idle status (or) Each failure is the first failure.

First

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Presentation I

TTCF

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1

231

3 12

1

231

2

A B CHAD WE KNOWN

THE FUTURE…

Simple 3 aging components serial system model

AB

C

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Presentation IISimple 3 aging components serial system model

A B C

TBCF

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1

32

2

323

4

HAD WE KNOWN

THE FUTURE…

4 AB

C

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Presentation III Simple 3 aging components serial system model

A B C

AB

C

TBCF

11

1

32

2

323

4

AB

C

TTCF

23

1

231

3 12

HAD WE KNOWN

THE FUTURE…

1

4

231

2

MTBCF < MTTCF

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Simulation MethodMONTE – CARLO

MATHCAD

N=

100,

000

SE

TS

MIN (X1,1 X2,1 X3,1)

MIN (X1,2 X2,2 X3,2)

.……

……

……

……

MIN (X1,N X2,N X3,N)_________________

N

iiN 1

min1

N=

100,

000

SE

TS

MIN (X1,1 X2,1 X3,1)

MIN (X1,2 Δ1,2 Δ2,2)

.……

……

……

……

MIN (X1,N Δ1,N Δ2,N)_________________

N

iiN 1

min1

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How “BIG” is the Difference?

1. Depends on the System Architecture.

2. Depends on the Time-To-Failure distribution of each component.

3. The difference in a specific complex electronic system was found to be ~40%

Note: True in redundant systems even when all components have constant failure rates.

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Why Does It Matter? Suppose a specification demand for a system’s reliability :

MTBCF = 600 hour

Suppose the manufacturer prediction of the parameter:

MTBCF = 780 hour

ATTENTION !!! How was it CALCULATED ????

Is this MTBCF or MTTCF ????

X -40%

“Real” MTBCF = 480 < 600 (spec)

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Example 1

Aging serial system – each component is weibull distributed

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התפלגות ווייבול זהה לכל הפריטים

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התפלגות ווייבול זהה לכל הפריטים

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התפלגות ווייבול זהה לכל הפריטים

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התפלגות ווייבול זהה לכל הפריטים

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Example 2Two redundant subsystems in series –

each component is exponentially distributed

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Constant failure rate

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serial

parallel

Constant failure rate

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A Comment about Asymptotic Availability

{ } { }

{ } { } { } { }

E TTF E T FA

E TTF E TTR E T F E

B

B TTR

(*)

(*) [ S.Zacks, Springer-Verlag 1991, Introduction To Reliability Analysis, Par 4.3]

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Repair policies

1. “Hot repair” is allowed for redundant components.

2. All components are renewed on every failure event.

3. All failed components are renewed on every failure event.

4. Failed components are renewed only in blocks which caused the system failure.

5. Failed subsystems are only partially renewed.

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Conclusions

• System configuration and distribution of components determine the gap.

• Repair policy should be specified in advance to determine calculation method.

• Flexible software solutions are needed to simulate real MTBCF for a given RBD.

• Predict MTBCF not MTTCF