Measurements Meir Kalech Partially Based on slides of Brian Williams and Peter struss.
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Transcript of Measurements Meir Kalech Partially Based on slides of Brian Williams and Peter struss.
Outline Last lecture:
1. Justification-based TMS
2. Assumption-based TMS Consistency-based
diagnosis
Today’s lecture:
1. Generation of tests/probes
2. Measurement Selection
3. Probabilities of Diagnoses
Generation of tests/probes
Test: test vector that can be applied to the system assumption: the behavior of the component does not change
between tests
approaches to select the test that can discriminate between faults of different components (e.g. [Williams])
Probe: selection of the probe based on: predictions generated by each candidate on unknown
measurable points
cost/risk/benefits of the different tests/probes
fault probability of the various components
Generation of tests/probes (II)Approach based on entropy [deKleer, 87, 92]
A-priori probability of the faults (even a rough estimate)
Given set D1, D2, ... Dn of candidates to be discriminated
1. Generate predictions from each candidate
2. For each probe/test T, compute the a-posteriori probability p(Di|T(x)), for each possible outcome x of T
3. Select the test/probe for which the distribution p(Di|T(x)) has a minimal entropy (this is the test that on average best discriminates between the candidates)
A Motivating Example
Minimal diagnoses: {M1}, {A1}, {M2, M3}, {M2, A2} Where to measure next? X,Y, or Z? What measurement promises the most information? Which values do we expect?
Minimal diagnoses: {M1}, {A1}, {M2, M3}, {M2, A2} Where to measure next? X,Y, or Z? What measurement promises the most information? Which values do we expect?
10
12
X
Y
Z
M1
M2
M3
*
*
*
A1
A2
+
+
F
G
2
A
2
3
3
B CD
E
3
Outline Last lecture:
1. Justification-based TMS
2. Assumption-based TMS Consistency-based
diagnosis
Today’s lecture:
1. Generation of tests/probes
2. Measurement Selection
3. Probabilities of Diagnoses
Measurement Selection - Discriminating Variables
10
12
X
Y
Z
M1
M2
M3
*
*
*
A1
A2
+
+
F
G
2
A
2
3
3
B CD
E
3
Suppose: single faults are more likely than multiple faults Probes that help discriminating {M1} and {A1} are most valuable
Suppose: single faults are more likely than multiple faults Probes that help discriminating {M1} and {A1} are most valuable
?
Discriminating Variables - Inspect ATMS Labels!
6 {{M1}}{{}}
{{}}
{{}}
{{}}
{{}}
{ }12
{{ }}
{{ }}
4 {{M2, A1} {M3, A1, A2}}
{{M2}} 6
{{M1,A1}} 4
6 {{M3}}
8 {{M1, A1, A2}}
10
12
X
Y
Z
M1
M2
M3
*
*
*
A1
A2
+
+
F
G
2
A
2
3
3
B CD
E
3
Observations: Facts - not based on any assumption:Node has empty environment (as the only minimal one): always derivable
Note the difference: empty label node not derivable!
Observations: Facts - not based on any assumption:Node has empty environment (as the only minimal one): always derivable
Note the difference: empty label node not derivable!
{{M3, A2}{M2}} 6
?
Justification :{A,C}
Justification :{B,D}
Justification :{C,E}
Empty label
A1=10 and M1=6M2=4
A2=12 and M2=4M3=8
A2=12 and M3=6M2=6
A1=10 and M2=6M1=4
A1=10 and M2(depends on M3 and A2)=6M1=4
Fault Predictions
No fault models used Nevertheless, fault hypotheses make predictions! E.g. diagnosis {A1} implies OK(M1)
OK(M1) implies x=6
No fault models used Nevertheless, fault hypotheses make predictions! E.g. diagnosis {A1} implies OK(M1)
OK(M1) implies x=6
6 {{M1}}
{{}}
{{}}
{{}}
{{}}
{{}}
{ }12
{{ }}
{{ }}
4 {{M2, A1} {M3, A1, A2}}
{{M2}} 6
{{M1,A1}} 4
6 {{M3}}
8 {{M1, A1, A2}}
10
12
X
Y
Z
M1
M2
M3
*
*
*
A1
A2
+
+
F
G
2
A
2
3
3
B CD
E
3
{{M3, A2}{M2}} 6If we measure x and concludes x=6 then we can infer that A1 is the diagnosis rather than M1
If we measure x and concludes x=6 then we can infer that A1 is the diagnosis rather than M1
Predictions of Minimal Fault Localizations
ATMS Labels: ATMS Labels:
X = 4 {{M2, A1} {M3, A1, A2}} X = 6 {{M1}}Y = 6 {{M2} {M3, A2}} Y = 4 {{M1, A1}}Z = 6 {{M3} {M2, A2}} Z = 8 {{M1, A1, A2}}
X = 4 {{M2, A1} {M3, A1, A2}} X = 6 {{M1}}Y = 6 {{M2} {M3, A2}} Y = 4 {{M1, A1}}Z = 6 {{M3} {M2, A2}} Z = 8 {{M1, A1, A2}}
Minimal Fault Localization
Prediction X Y Z
{M1} 4 6 6 {A1} 6 6 6 {M2, A2} 6 4 8 {M2, M3} 6 4 8
Minimal Fault Localization
Prediction X Y Z
{M1} 4 6 6 {A1} 6 6 6 {M2, A2} 6 4 8 {M2, M3} 6 4 8
X 6: M1 is broken.
X = 6 : {A1} only single fault
Y or Z same for {A1}, {M1}
X best measurement.
X 6: M1 is broken.
X = 6 : {A1} only single fault
Y or Z same for {A1}, {M1}
X best measurement.
X=4 M1 is diagnosis, since it appears
only in x=6
Outline Last lecture:
1. Justification-based TMS
2. Assumption-based TMS Consistency-based
diagnosis
Today’s lecture:
1. Generation of tests/probes
2. Measurement Selection
3. Probabilities of Diagnoses
Probabilities of Diagnoses
Fault probability of component(type)s: pf
For instance, pf(Ci) = 0.01 for all Ci{A1, A2, M1, M2, M3} Normalization by = p(FaultLoc)
FaultLoc
Fault probability of component(type)s: pf
For instance, pf(Ci) = 0.01 for all Ci{A1, A2, M1, M2, M3} Normalization by = p(FaultLoc)
FaultLoc
10
12
X
Y
Z
M1
M2
M3
*
*
*
A1
A2
+
+
F
G
2
A
2
3
3
B CD
E
3
Probabilities of Diagnoses - Example
Assumption: independent faults Heuristic: minimal fault localizations only
Assumption: independent faults Heuristic: minimal fault localizations only
Minimal fault localization
p(FaultLoc)/
Prediction X Y Z
{M1} .495 4 6 6 {A1} .495 6 6 6 {M2, A2} .005 6 4 6 {M2, M3} .005 6 4 8
495.001.001.001.001.0
01.022
1
M
Entropy-based Measurement Proposal
Entropy of a Coin toss as a function of the probability of it coming up heads
The cost of locating a candidate with probability pi is log(1/pi) (binary search through 1/pi objects).
Meaning, needed cuts to find an object. Example: p(x)=1/25 the number of cuts in binary search will be
log(25) = 4.6 p(x)=1/2 the number of cuts in binary search will be log(2)
= 1
pi is the probability of Ci being actual candidate given a measurement outcome.
The Intuition Behind the Entropy
The cost of identifying the actual candidate, by the measurement is:
1. pi 0 occur infrequently, expensive to find pi log(1/pi) 0
2. pi 1 occur frequently, easy to find pi log(1/pi) 0
3. pi in between pi log(1/pi) 1
The Intuition Behind the Entropy
Go over through the possible candidates
The probability of candidate Ci to be faulted given an
assignment to the measurement
The cost of searching for this
probability
The expected entropy by measuring Xi is:
Intuition: the expected entropy of X = ∑ the probability of Xi * entropy of Xi
This formula is an approximation of the above:
The Intuition Behind the Entropy
Go over through the possible outcomes of
measurement Xi
The probability of measurement Xi to be
Vik
The entropy if Xi=Vik
m
This formula is an approximation of the above:
Where, Ui is the set of candidates which do not predict any value for xi
The goal is to find measurement xi that minimizes the above function
The Intuition Behind the Entropy
m
Entropy-based Measurement Proposal - Example
vari valik p(vari = valik)entropy
p(vari = valik)* log p (vari = valik)
X6
4
.505
.495-.99993
Y6
4
.990
.010-.0808
Z6
8
.995
.005-.0454
vari valik p(vari = valik)entropy
p(vari = valik)* log p (vari = valik)
X6
4
.505
.495-.99993
Y6
4
.990
.010-.0808
Z6
8
.995
.005-.0454
Proposal: Measure variable which minimizes the entropy: XProposal: Measure variable which minimizes the entropy: X
x=6 under the diagnoses: {A1}, {M2,A2}, {M2,M3}0.495+0.05+0.05=0.505
x=6 under the diagnoses: {A1}, {M2,A2}, {M2,M3}0.495+0.05+0.05=0.505
How to update the probability of a candidate?
Given measurement outcome xi=uik, the probability of
a candidate is computed via Bayes’ rule:
Meaning: the probability that Cl is the actual candidate
given the measurement xi=uik.
p(Cl) is known in advance.
Computing Posterior Probability
Normalization factor:The probability that xi = uik : the sum of theprobabilities of the Candidates consistent withthis measurment
How to compute p(xi=uik|Cl)? Three cases:
1. If the candidate Cl predicts the output xi=uik then p(xi=uik|
Cl)=1
2. If the candidate Cl predicts the output xi≠uik then p(xi=uik|
Cl)=0
3. If the candidate Cl predicts no output for xi then p(xi=uik|
Cl)=1/m (m is number of possible values for x)
Computing Posterior Probability
Example
Initial probability of failure in inverter is 0.01.
Assume the input a=1:
What is the best next measurement b or e?
Assume next measurement points on fault: Measuring closer to input produces less conflicts:
b=1 A is faulty
e=0 some components is faulty.
Example
On the other hand:
Measuring further away from the input is more likely to produce a discrepant value.
The large number of components the more likely that there is a fault.
the probability of finding a particular value outweighs the expected cost of isolating the candidate from a set.
the best next measurement is e.
Example
H(b) =
p(b = true | all diagnoses with observation a)
log p(b = true | all diagnoses with observation a) + p(b = false | all diagnoses with observation a)
log p(b = false | all diagnoses with observation a)