Asset Management Using Failure
Forecasting and Statistical
Assessment of Diagnostic Testing
Miroslav M. Begovic
School of ECE, Georgia Institute of Technology
Fourth Annual Carnegie Mellon Conference on the Electricity Industry
FUTURE ENERGY SYSTEMS: EFFICIENCY, SECURITY, CONTROL
Pittsburgh, March 10-11, 2008
1
Background/Problem Statement
Most utilities have incomplete failure data on
various components, such as cables, switchgear,
etc.
Framework: a software-based algorithm to
estimate future failures based on data obtained
from incomplete failure information.
The optimal replacement rate needed to maintain
failures at a specified failure rate is sought.
Optimal failure rate estimation will yield the lowest
replacement and maintenance budget required to
meet the desired failure performance
2
Effect of Replacements on
Failure Rates
FAILURES
FAILURE
REDUCTION
INSTALLATION TIME
PARTIAL REPLACEMENT
TIME
3
Estimation and Control of Failures
in Composite Populations
ESTIMATED FAILURES
UPPER BOUND
(CONFIDENCE RANGE)
LOWER BOUND
INSTALLATION EVENTS
TIME
PARTIAL REPLACEMENT EVENTS
PROPOSE TO DEVELOP
SCHEDULE FOR
REPLACEMENT
EVENTS
4
Preliminaries
If p(t) is the pdf of the time to failure t of a single
component, the probability of that component failing
before time t is given by
P (t ) =
∫
t
0
p (u ) du
Assuming Weibull distribution for p(t) and population of N
identical components, the pdf of a component of that
population failing before time t is given by
−β
p0 (t ) = N βα t
β −1 − ( t / α ) N
β
e
,
t ≥ 0.
5
Preliminaries (cont.)
The expected value of T (the time to failure of a
single component) is
⎛1 ⎞
E (T ) = αΓ ⎜ + 1 ⎟
⎝β ⎠
The expected value of T (the time to failure of a
single component) in a population of N identical
components is
E (T0 ) =
α
N
1/ β
⎛1 ⎞
Γ ⎜ + 1⎟
⎝β ⎠
6
Failure Rates
The overall failure rate will be
f 0 (t ) =
p0 (t )
1 − P0 (t )
t≥0
,
In terms of the original α and N
f N (t ) = N βα t
−β
β −1
t≥0
,
Which will produce linear growth compared to the aging of
a single component
β
−β
p (t ) = Wei (α , β ) = βα t
f (t ) = βα t
−β
β −1
β −1
,
(
)
e
,
−
t
α
t≥0
t≥0
7
Formulation of the Failure
Model
f (t ) = X ⋅ K ⋅ (t − g )
b
Parameters K,b,g represent the description (model) of the failures
for population X (or N, if we deal with the discrete components)
When there are multiple populations Xi installed in years i = 0, 1, 2, …, k,
the cumulative failure model F(t,g,b,K) becomes a sum:
k
F (t , g , b, K ) = ∑ X i ⋅ K ⋅ (t − g − i )
b
i =0
8
Methodology for Identification of
Statistical Coefficients
160
140
140
120
120
number of failures
number of failures
Actual Failures and Curve Fit
Actual Failures
160
100
80
60
100
80
60
40
40
20
20
0
1960
1965
1970
1975
1980
year
1985
1990
1995
IDENTIFICATION OF {g,b,K}
STATISTICAL PARAMETERS
ESTIMATED FAILURES CURVE
0
1960
1965
1970
1975
1980
1985
1990
1995
year
CALCULATION OF ESTIMATED
FAILURES USING {g,b,K}
9
Failure Estimation
FAILURES
PAST
PRESENT
FUTURE
PROBABILITY OF
FAILURE ESTIMATE
…
t-1
t
t+1
YEAR
10
Test Data Set: New
Installation and Replacement
New Miles Installed and Replaced
600
installed
replaced
500
miles
400
300
200
100
0
1960
1965
1970
1975
1980
year
1985
1990
1995
2000
11
Distribution of Estimated Failures
Year 1 Projection of Frequency versus Number of Failures
25
Frequency
20
15
10
5
0
0
50
100
150
200
Number of Failures
250
300
350
400
300
350
400
300
350
400
Year 2 Projection of Frequency versus Number of Failures
25
Frequency
20
15
10
5
0
0
50
100
150
200
Number of Failures
250
Year 3 Projection of Frequency versus Number of Failures
25
Frequency
20
15
10
5
0
0
50
100
150
200
Number of Failures
250
Three year projection histograms for 1000 simulations.
12
Failure Estimation and
Confidence Intervals
95% Confidence Interval
50% Confidence Interval
250
180
160
200
140
No. of Failures
No. of Failures
120
100
80
150
100
60
40
50
20
0
1970
1975
1980
1985
Year
25% Confidence Interval.
1990
1995
0
1980
1985
1990
1995
Year
95% confidence interval .
13
Bottom Line: 1-Year Replacement
Upper Bound
Future Year 1 Replacement Confidence
600
replacement (miles)
550
500
450
400
350
300
50
55
60
65
70
75
80
confidence (%)
85
90
95
100
14
Asset Management
Objectives
Objectives
Follow-Up
Follow-Up
Maintenance
Maintenance
Scheduling
Scheduling
Planning
Planning
Asset
Asset
Management
Management
Analysis
Analysis
Budgeting
Budgeting
Data
DataLogging
Logging
Source: Joshua Perkel, NEETRAC
15
Cost Analysis
LOAD VOLTAGE
Refurbishment
Repair
Preventive Replacement
Diagnostic Testing
Replacement on Failure
Cost of Outage
DURATION OF
OUTAGE
OCCURRENCE
OF OUTAGE
TIME
COST OF OUTAGE
COST OF ENERGY
NOT SERVED
INITIAL COST OF
OUTAGE
Hard to Assess the Cost of
Reliability for Utilities
16
Motivation
Diagnostic tests look for symptoms of
degradation, not failures.
Symptoms are difficult to relate to future failures
unless they are in the extremes.
Probability
“Good”
?
“Bad”
No Failure
Failure
Diagnostic Measurement
17
Diagnostics – where to look
FRACTION OF CABLE
FRACTION OF CABLE
POPULATION TESTED
POPULATION TESTED
POSITIVE FOR
POSITIVE FOR
REPLACEMENT
REPLACEMENT
TESTED CABLE
TESTED CABLE
POPULATION
POPULATION
TOTAL CABLE
TOTAL CABLE
POPULATION
POPULATION
18
Passive Approach – do nothing
Cost Parameters:
A = Failure rates [failures/miles/year]
B = Average length of segment [ft]
C = Average cost of replacement cable [$/ft]
D = Average cost of repair [$/ft]
E = Length of population [miles]
F = Average Momentary Cost of Outage [$/failure]
G = Average Cost of Energy Not Served [$/min/failure]
TF = Average Duration of Outage [min]
Cost = A ⋅ E ⋅ [ (C + D ) ⋅ B + F + G ⋅ TF ] [$ / yr ]
19
Passive Cost
AVOIDABLE COST
Cost = A ⋅ E ⋅ [ (C + D) ⋅ B + F + G ⋅ TF ] [$ / yr ]
AVERAGE COST OF
ENERGY NOT SERVED
PER FAILURE
TOTAL NUMBER OF
FAILURES PER YEAR
COST OF REPLACEMENT
(CABLE + WORK)
AVERAGE COST OF
MOMENTARY
INTERRUPTION
20
SoE1
Failure Management
Avoidable cost can be reduced by replacing
suspect cable segments in an efficient way
before they fail
Need to know how many
failures are anticipated –
failure forecasting
Need to know
which segments to replace
„how accurate the identification
„
21
Slide 21
SoE1
Figure Hard to read
School of ECE, 1/29/2007
Failure Management
Cost Parameters:
B = Average length of segment [ft]
C = Average cost of replacement [$/ft]
D = Average cost of repair [$/ft]
I = Total number of tested segments [ ]
J = Cost of diagnostic test per day [$/day]
K = Number of segments tested per day [day-1]
ξ = Fraction of the tested segments to be replaced – ratio
red to green areas
Cost = (C + D) ⋅ B ⋅ ξ ⋅ I + I ⋅ J / K [$ / yr ]
22
Cost of Failure Management
Cost = (C + D) ⋅ B ⋅ ξ ⋅ I + I ⋅ J / K [$ / yr ]
COST OF CABLE
(REPLACEMENT AND
WORK) PER SEGMENT
NUMBER OF SEGMENTS
NEEDING REPLACEMENT
COST OF DIAGNOSTIC
TEST PER SEGMENT
TOTAL NUMBER OF
TESTED SEGMENTS
23
Observations
Savings are achieved from small ξ
values (only segments correctly
diagnosed as bad are replaced)
Diagnostic tests add to the cost
It is not practical to test every segment
every year (cost would be too high)
How to determine which segments to
test?
24
Diagnostics: Issues
UNEXPECTED
FAILURES
FRACTION OF CABLE
FRACTION OF CABLE
POPULATION TESTED
POPULATION TESTED
POSITIVE FOR
POSITIVE FOR
REPLACEMENT
REPLACEMENT
TESTED CABLE
TESTED CABLE
POPULATION
POPULATION
TOTAL CABLE
TOTAL CABLE
POPULATION
POPULATION
If number of replaced segments is smaller than
failure forecast, unexpected failures will likely occur
Even if the number of replaced segments is equal to
the failure forecast, there may (and probably will) be
unexpected failures in the untested population
25
Diagnostics: Issues (2)
UNEXPECTED
FAILURES
FRACTION OF CABLE
FRACTION OF CABLE
POPULATION TESTED
POPULATION TESTED
POSITIVE FOR
POSITIVE FOR
REPLACEMENT
REPLACEMENT
TESTED CABLE
TESTED CABLE
POPULATION
POPULATION
TOTAL CABLE
TOTAL CABLE
POPULATION
POPULATION
If population to be tested is poorly chosen,
the benefits of the diagnostic test are lost
26
Diagnostic Accuracy
UNEXPECTED
FAILURES
FRACTION OF CABLE
FRACTION OF CABLE
POPULATION TESTED
POPULATION TESTED
POSITIVE FOR
POSITIVE FOR
REPLACEMENT
REPLACEMENT
TESTED CABLE
TESTED CABLE
POPULATION
POPULATION
TOTAL CABLE
TOTAL CABLE
POPULATION
POPULATION
P (T < 0 G ) = 1 − ε
P (T > 0 G ) = ε
P (T > 0 B ) = 1 − ε
P (T < 0 B ) = ε
27
Diagnostic Accuracy (2)
P(G T < 0) =
P(T < 0 G ) P(G )
P(T < 0 G ) P(G ) + P(T < 0 B) P( B)
=
1⋅ P(G )
P(G )
=
=
=1
1 ⋅ P(G ) + 0 ⋅ P( B) P(G )
If diagnostic accuracy is perfect, testing will identify
all the good (G) and bad (B) tested components
28
Diagnostic Accuracy (3)
P(T < 0 G ) P(G )
=
P(G T < 0) =
P(T < 0 G ) P(G ) + P(T < 0 B ) P ( B)
0.5 ⋅ P(G )
0.5 ⋅ P(G )
=
=
= P(G )
0.5 ⋅ P(G ) + 0.5 ⋅ P( B)
0.5
If diagnostic accuracy is bad, testing will identify
all the good (G) and bad (B) tested components only
as their proportions in the tested population
29
Replacement Cost
Population: 100 miles
Diagnostic Cost: $6k/mile
Cycle: 6 years
Cost per annum: $100k/yr
Failure rate: 30 failures/100 mi/yr
Replacement on failure:
uniformly distributed [$5k, $10k]
Replacement on failure:
uniformly distributed [$5k, $10k]
Cable cost: uniformly distributed
between [$27,$33]
Diagnostic accuracy: uniformly
distributed in [80%,100%]
Forecast: Total Cost {$/yr]
100,000 Trials
Frequency Chart
100,000 Displayed
.072
7192
.054
.036
.018
Mean =493,884.48
.000
100,000.00
375,000.00
650,000.00
$/yr
0
925,000.00
1,200,000.00
30
Replacement Cost
DISTRIBUTION
DISTRIBUTIONOF
OFTOTAL
TOTALCOST
COST
1.2E+06
1.2E+06
1.0E+06
1.0E+06
$/YEAR
$/YEAR
8.0E+05
8.0E+05
6.0E+05
6.0E+05
4.0E+05
4.0E+05
2.0E+05
2.0E+05
0.0E+00
0.0E+00
[80%
[80%,100%
,100%] ]
0%
0%
[90%
[95%
[90%,100%
,100%] ]
[95%,100%
,100%] ]
DIAGNOSTIC
DIAGNOSTICACCURACY
ACCURACYDISTRIBUTION
DISTRIBUTION
10%
10%
25%
25%
50%
50%
75%
75%
90%
90%
[100%
[100%,100%
,100%] ]
100%
100%
31
Cost Benefit of Diagnostics
Cost [$]
BENEFIT
Consequence
LOSS
Alternate
Program 1
Maintenance
Alternate
Program 2
Diagnostic
Selection
Source: Joshua Perkel, NEETRAC
32
Conclusions
Failure Forecasting algorithm provides some guidance
under the assumption that oldest population of equipment is
the most prone to failures
Diagnostic testing may provide better targeting of the
candidates for replacement, but at a cost (both due to the
procedure and its limited accuracy)
Analysis of different scenarios of desired failure
performance assist in formulating optimal strategies
Circumstances may significantly influence the cost of
diagnostic testing
33