Transcript DESIGN FOR RELIABILITY

Reliability Prediction
A Quest for Reliable Parameters
By
Yair Shai
1
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.
2
Promo :
Description of Parameters
t1
t2
t3
t4
 Failure Event of an Item
t5
t
=
3
=
i
i
r
t
Non Repairable Items:
Mean Time To Failure
time
r =Number of Failures
Repairable Items:
Mean Time Between Failures
.......
i
r
i
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
TIME
DOWN
4
Critical Failures
Moving towards System Design
A System Failure resulting in (temporary or
permanent) Mission Termination.
X
X
COMPUTER
COMPUTER
SUBSYSTEM
A simple
configuration of
parallel hot
Redundancy.
A Failure: any computer failure
A Critical Failure: two computers failed
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Critical Failures
A 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
Switch control
UNIT A
ANTENA
UNIT B
ANTENA
CPU
POWER
SUPPLY
4 CHANNEL
RECEVER
CONTROLER
UNIT C
sw
ANTENA
CPU
POWER
SUPPLY
UNIT D
POWER
SUPPLY
ANTENA
2/4
Operational Demand: At least two receiver units
and one antenna should work to operate the system.
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From System Design to
Reliability Model
CPU
ANT
B
ANT
PS1
CONT
x
CPU
A
PS2
PS1
x
Is this a Critical Failure ?
sw
x C
ANT
D
ANT
2/4
Serial model : Rs = R1x R2
Parallel model : Rs = 1- (1-R1)x(1-R2)
K out of N model : Rs = Binomial Solution
8
From RBD Logic Diagram
to Reliability Function
Simple mathematical manipulation:
Rsys(t)= f( serial / parallel / K out of N)
Classic parameter evaluation:

WARNING !!!
0
Is this realistic
?
MTTF
MTBF   Rsys
Rsys(t )dt
dt
MTBCF
MTTCF
After[ each
S.Zacks,
repair
Springer-Verlag
of a critical failure
1991,-Introduction
The whole system
To
returns
Reliability
to status
Analysis,
“As Good
ParAs
3.5]
New”.
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MTBCF vs. MTTCF
A New Interpretation
Common practice interpretation:
First
MTBCF = MTTCF = MTTCFF
Each repair “Resets” the time count to idle status
(or) Each failure is the first failure.
Realistic interpretation:
MTBCF = MTTCF
Only failed Items which cause the failure are repaired
to idle. All other components keep on aging.
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Presentation I
Simple 3 aging components serial system model
A
B
3
2
HAD WE
KNOWN
THE FUTURE…
C
2
1
2 3
1
1
2
1
13
3
2
TTCF
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A
B
C
Presentation II
Simple 3 aging components serial system model
A
B
1
1
C
2
4
3
2
1
HAD WE
KNOWN
THE FUTURE…
2
A
B
C
3
3
4
TBCF
12
Presentation III
Simple 3 aging components serial system model
A
B
C
2
1
A
B
C
4
3
2
1
1
HAD WE
KNOWN
THE FUTURE…
3
2
3
4
TBCF
MTBCF < MTTCF
3
2
2
1
2 3
1
1
2
1
13
3
2
TTCF
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A
B
C
Simulation Method
MONTE – CARLO
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MIN (X1,N X2,N X3,N)
_________________
1 N
  mini
N i 1
MIN (X1,1 X2,1 X3,1)
MIN (X1,2 Δ1,2 Δ2,2)
…………………….
N=100,000 SETS
MIN (X1,1 X2,1 X3,1)
MIN (X1,2 X2,2 X3,2)
…………………….
N=100,000 SETS
MATHCAD
MIN (X1,N Δ1,N Δ2,N)
_________________
1 N
  mini
N i 1
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:
X
MTBCF = 780 hour
-40%
ATTENTION !!! How was it CALCULATED ????
Is this MTBCF or MTTCF ????
“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 2
Two redundant subsystems in series – each
component is exponentially distributed
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Constant
failure rate
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serial
Constant
failure rate
parallel
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A Comment about
Asymptotic Availability
(*)
E{TTF }
E{TBF }
A 

E{TTF }  E{TTR} E{TBF }  E{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
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