Fault Tree Analysis · Chapter 4

Worked Example

The method applied end-to-end on a concrete system, with numbers.

This is row 7 of the functional hazard assessment in the previous module: misleading attitude displayed on both flight displays, unannunciated, classified catastrophic, objective 2 × 10⁻⁹ per two-hour flight. The architecture proposed to meet it has three attitude sources with a two-out-of-three comparison, a comparison monitor in each display computer, and a standby indicator that this tree does not need. All rates are illustrative teaching values.

The inputs

Basic eventRateExposureWhere it comes from
Attitude source produces erroneous output1.5 × 10⁻⁵ per hour2 h flight, revealed at once by the votePrediction, split by the FMECA's mode ratio
Comparison monitor fails passive8.0 × 10⁻⁶ per hourlatent: 750 h scheduled testPrediction; the interval is the maintenance programme's
28 V bus transient corrupts the sources3.0 × 10⁻⁵ per hour2 h flightPrediction
A transient drives all three sources outside tolerance the same way0.01conditionalEngineering judgement, not data
Two erroneous outputs agree inside the comparator window0.05conditionalEngineering judgement, not data
Dispatch with one source inoperative0.02 of departuresconfigurationThe minimum equipment list

The agreement probability is the row to watch. It is a conditional probability in an inhibit gate, it is a judgement rather than a measurement, and by the end of this example it will turn out to matter more than any failure rate in the table.

The tree as first drawn

Three branches. Two of them are pairs of failures and the third is a single event, which is the finding no amount of arithmetic can soften.
Three branches. Two of them are pairs of failures and the third is a single event, which is the finding no amount of arithmetic can soften.

Branch A, two erroneous sources outvoting the healthy one. With three sources and a two-out-of-three comparison, one erroneous source is outvoted and annunciated. Two are not:

P(one source erroneous in a flight) = 1.5 × 10⁻⁵ × 2 = 3.0 × 10⁻⁵

3 pairs × (3.0 × 10⁻⁵)² = 2.7 × 10⁻⁹ per flight

conservatively assuming any two erroneous outputs agree closely enough to win the vote.

Branch B, one erroneous source hidden by a failed comparator. Only reachable when the aircraft is dispatched with a source inoperative, so the comparison is a two-way comparator with nothing to vote. The monitor is latent, tested every 750 hours, and must fail before the source for the error to go unannounced, which is the priority AND:

½ × 8.0 × 10⁻⁶ × 1.5 × 10⁻⁵ × 2 h × 750 h = 9.0 × 10⁻⁸ per MEL flight, per source

× 2 sources × 0.02 of departures = 3.6 × 10⁻⁹ per flight

Branch C, the common cause. All three sources are fed from one 28 V bus in this version of the design. A transient that drives all three outside tolerance in the same direction defeats the vote completely, and the 0.01 declared above is the judged probability that a transient does that rather than something the sources ride out:

3.0 × 10⁻⁵ × 2 h × 0.01 = 6.0 × 10⁻⁷ per flight

Total: 6.1 × 10⁻⁷ per flight, against an objective of 2 × 10⁻⁹. Three hundred times over.

The finding was not the number

The arithmetic says branch C dominates by more than two orders of magnitude, and a programme reading only the number would go looking for a better bus. Read the cut sets instead:

Cut setOrderPer flight
{bus transient}16.0 × 10⁻⁷
{source i erroneous, source j erroneous}, three of them29.0 × 10⁻¹⁰ each
{source erroneous, comparator latent}, two of them21.8 × 10⁻⁹ each, on MEL flights

The first row is a single event that produces a catastrophic failure condition, and the qualitative rule that governs it does not have a probability in it. A bus a hundred times better would still be an order-one cut set and still be non-compliant. The design has to change, and no amount of data improves it.

Three changes

ChangeKindEffect on the tree
Three sources on three separately protected suppliesArchitectureDeletes the order-one cut set; the residue is an order-three coincidence, negligible
Comparator proved at every power-up rather than every 750 hMaintenance and BITLatency falls from 750 h to one flight
Agreement probability estimated and stated, rather than assumed to be 1AnalysisBranch A falls by the factor that is argued, and the argument becomes visible

The first change is argued rather than calculated. Three independent supply faults in one flight is an order-three coincidence and falls out of the arithmetic, but that is only the independent part. What survives separation is the dependent fraction: shared routing, a shared protection design, one installation error repeated three times. That residue is what a beta-factor term would carry, and its value is settled by the zonal, particular-risk and common-mode analyses rather than by anything in this tree.

The second change is the one worth dwelling on, because it cost almost nothing:

½ × 8.0 × 10⁻⁶ × 1.5 × 10⁻⁵ × 2 h × 2 h = 2.4 × 10⁻¹⁰ per source

× 2 sources × 0.02 = 9.6 × 10⁻¹² per flight

A 375-fold reduction from a test interval, with no new hardware. That is the routine result of taking latency seriously, and it is why the exposure column of a fault tree is worth more attention than the rate column.

The tree after the redesign

The same top event with the special gates the logic actually needs: a voting gate for the triplex comparison, an inhibit carrying the agreement probability, a house event for the dispatch configuration, and a priority AND for the monitor that has to fail first.
The same top event with the special gates the logic actually needs: a voting gate for the triplex comparison, an inhibit carrying the agreement probability, a house event for the dispatch configuration, and a priority AND for the monitor that has to fail first.
BranchPer flight
Two erroneous sources, agreeing1.4 × 10⁻¹⁰
Erroneous source hidden by a failed comparator, MEL flights only9.6 × 10⁻¹²
Two erroneous sources agreeing on an MEL flight9.0 × 10⁻¹³
Common cause through the suppliesnegligible, order three
Top event1.5 × 10⁻¹⁰ per flight, against 2 × 10⁻⁹

Unrounded, branch A is 2.7 × 10⁻⁹ × 0.05 = 1.35 × 10⁻¹⁰, and the third row is the same coincidence on the two sources left after an MEL dispatch: 9.0 × 10⁻¹⁰ × 0.05 × 0.02 = 9.0 × 10⁻¹³. The three quantified rows sum to 1.35 × 10⁻¹⁰ + 9.6 × 10⁻¹² + 9.0 × 10⁻¹³ = 1.455 × 10⁻¹⁰, so 1.455 × 10⁻¹⁰ ÷ 2 h = 7.3 × 10⁻¹¹ per flight hour: comfortably inside the catastrophic band, with better than a factor of thirteen in hand.

The FHA recorded a twenty-second at-risk window against this failure condition. That window does not change the budget, which is per flight either way. It changes the rate an item supplier can be given, because an item whose failure only matters inside the window is at risk for twenty seconds rather than two hours. This tree is quantified over the whole flight, which is the conservative direction, and the window is left where it belongs: in the supplier requirement rather than in the compliance total.

Where the number actually comes from

Contribution after the redesign. One class of cut set carries almost everything, and it is the class whose probability depends on a judgement rather than on a measured rate.
Contribution after the redesign. One class of cut set carries almost everything, and it is the class whose probability depends on a judgement rather than on a measured rate.
Cut set classSharePer flight
Two sources erroneous and agreeing92.8%1.4 × 10⁻¹⁰
Source erroneous with the comparator latent, MEL only6.6%9.6 × 10⁻¹²
Two sources erroneous on an MEL flight0.6%9.0 × 10⁻¹³

Every surviving cut set contains an erroneous attitude source, so that mode's Fussell-Vesely importance is 1.00: nothing else in the model can produce the top event on its own terms. The comparator's is 0.066, under seven per cent, which is worth knowing before anybody spends money making it better.

The assumption the compliance case rests on

The top event against the agreement probability. At the value assumed the design passes with margin; at the conservative value it fails. Nothing in the failure rate data decides which is right.
The top event against the agreement probability. At the value assumed the design passes with margin; at the conservative value it fails. Nothing in the failure rate data decides which is right.
P(two erroneous outputs agree)Top event, per flightVerdict
0.013.7 × 10⁻¹¹compliant
0.05, assumed1.5 × 10⁻¹⁰compliant, 13× margin
0.25.5 × 10⁻¹⁰compliant, 3.6× margin
0.51.4 × 10⁻⁹marginal
1.0, fully conservative2.7 × 10⁻⁹not compliant

This is the honest ending of most real fault trees. The failure rates are traceable, the logic is checkable, the arithmetic is elementary, and the answer swings by two orders of magnitude on one conditional probability that is an argument rather than a measurement. Reporting that number, and the reasoning behind it, is not a caveat: it is the main result. A tree that buries it in a spreadsheet cell has published a conclusion whose foundation nobody can examine.

What the analysis leaves behind

OutputOwner
1.5 × 10⁻¹⁰ per flight against a 2 × 10⁻⁹ objective, with its cut setsThe certification file
Three separately protected supplies, one per sourceThe architecture, as a safety requirement
Power-up test of each comparison monitorBuilt-in test specification, and testability
An agreement probability of 0.05, with the study that supports itAn open item until the study exists
Source erroneous-output rate 1.5 × 10⁻⁵ per hourThe supplier requirement, and what field data will grade

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