APEX ARP5107C — worked reference examples

Purpose and scope: engine-control time-limited dispatch

This tool supports time-limited dispatch (TLD) analysis of engine control systems, particularly full-authority digital engine control (FADEC). TLD assesses operation with detected faults awaiting repair. The tool calculates the fleet-average loss-of-thrust-control (LOTC) rate, compares it with a target, and explores how short-term (ST) and long-term (LT) analysis repair intervals affect that rate.

Methods: Full Markov, first-order and zero-order Markov approximations, and balanced time-weighted average. Inputs can retain individual faults or represent averaged ST/LT conditions.

Single- and two-fault states: this tool models a fully operational state, ST/LT single-fault conditions and LOTC. A second failure causing LOTC is represented by an exit from a single-fault state. Continued operation with two faults is not modeled explicitly. More detailed models can include those states; the repair-rate guideline helps assess when omitting them is reasonable.

Open and closed loops: ARP5107C calls a Markov model open-loop when there is no LOTC-to-full-up feedback. A closed-loop model adds simulated feedback for steady-state analysis. This tool uses closed-loop single-fault formulas; this feedback is mathematical, not an operational recovery from LOTC.

TWA conservatism and acceptability: ARP5107C Section 7.6 (p. 48) states that the time-weighted-average approach gives a conservative LOTC-rate estimate and is an acceptable method for completing the LOTC TLD analysis. This applies to both the original and revised time-fraction coefficients described in the standard.

Scope: engine-control-related LOTC. Model adequacy and permissible dispatch intervals require separate assessment; the calculation does not establish dispatch approval. Source: ARP5107C (September 2018), Sections 1–1.1 (p. 4), 7.1–7.2 (pp. 22–28), 7.2.2–7.2.4 (pp. 31–37) and Appendix G.

Two source datasets are worked in full below: Section 7.1.5 and the Section 7.4 Table 2 dataset at 1000 LT repair hours. All seven published Table 2 repair intervals belong to Example 2’s sensitivity analysis in the live calculator. The comparison table checks them all. Formula references are to supplied ARP5107C Revision C, September 2018. These are the implemented short-term and long-term single-fault methods; other standard methods are outside this calculator’s scope.

Source comparison checks

Reference case	Calculated method	Calculated rate per hour	Source reference value per hour
Example 1 — averaged ST/LT inputs, Section 7.1.5	Balanced TWA	9.454100652376514e-06	9.454100652376514e-06
Example 2 — individual fault inputs, Section 7.4 Table 2	Full Markov	7.810342390109076e-06	7.81e-06
Example 2 — individual fault inputs, Section 7.4 Table 2	Full Markov	8.74464913374253e-06	8.74e-06
Example 2 — individual fault inputs, Section 7.4 Table 2	Full Markov	9.532861613425565e-06	9.53e-06
Example 2 — individual fault inputs, Section 7.4 Table 2	Full Markov	1.0206049335624836e-05	1.021e-05
Example 2 — individual fault inputs, Section 7.4 Table 2	Full Markov	1.0787610056732499e-05	1.079e-05
Example 2 — individual fault inputs, Section 7.4 Table 2	Full Markov	1.1295004027916633e-05	1.13e-05
Example 2 — individual fault inputs, Section 7.4 Table 2	Full Markov	1.1741524463922093e-05	1.174e-05

Example 1 — averaged ST/LT inputs, Section 7.1.5

ARP5107C Section 7.1.5, page 27, Equation 11a

The standard leaves the LT repair interval to be determined. This example evaluates its published Equation 11a at an LT interval of 1000 hours; that evaluation interval is selected here, not a separate printed input.

Full Markov (averaged ST/LT states) — Equation 18

1. Averaged ST/LT inputs and combined baseline

Averaged ST/LT state model: one representative short-term state and one representative long-term state. The given totals and averages are used directly.

Hydromechanical and uncovered contributions — Equations 1–3, pages 23–24

X = 0.02; λ_HMC = 4e-06 per hour.

λ_UC ≈ X × (λ_STT + λ_LTT) = 0.02 × 1e-04 = 2e-06 per hour.

λ_HM+UC = λ_HMC + λ_UC = 4e-06 + 2e-06 = 6e-06 per hour.

The total occurrence rate uses the entered ST and LT rates for both channels. This uses the aggregate approximation described after Equation 2; the approximation assumes that the entered totals represent the relevant LOTC-critical electronic faults in both channels. Architecture-specific or element-level uncovered-rate assessments are outside this simplified input model.

ST representative state

λ_STT = 3e-05 per hour; λ_ST-LOTC-AVE = 6.5e-05 per hour; T_ST = 100 hours.

Combined exit = λ_ST-LOTC-AVE + λ_HM+UC = 6.5e-05 + 6e-06 = 7.1e-05 per hour.

µ_ST =1T_ST

Repair rate = 0.01 per hour.

LT representative state

λ_LTT = 7e-05 per hour; λ_LT-LOTC-AVE = 5e-05 per hour; T_LT = 1e+03 hours.

Combined exit = λ_LT-LOTC-AVE + λ_HM+UC = 5e-05 + 6e-06 = 5.6e-05 per hour.

µ_LT =1T_LT

Repair rate = 1e-03 per hour.

2. Formula for Full Markov (averaged ST/LT states)

Averaged ST/LT Full Markov — Equation 18, page 34

λ_LOTC =λ_HM+UC + λ_STT × (λ_ST-LOTC-AVE + λ_HM+UC)µ_ST + λ_ST-LOTC-AVE + λ_HM+UC + λ_LTT × (λ_LT-LOTC-AVE + λ_HM+UC)µ_LT + λ_LT-LOTC-AVE + λ_HM+UC1 + λ_STTµ_ST + λ_ST-LOTC-AVE + λ_HM+UC + λ_LTTµ_LT + λ_LT-LOTC-AVE + λ_HM+UC

3. Representative-state contributions

ST numerator and denominator contributions

Numerator contribution =3e-05 × 7.1e-050.01 + 7.1e-05

Numerator contribution = 2.115e-07 per hour.

Denominator contribution =3e-050.01 + 7.1e-05

Denominator contribution = 2.979e-03.

LT numerator and denominator contributions

Numerator contribution =7e-05 × 5.6e-051e-03 + 5.6e-05

Numerator contribution = 3.712e-06 per hour.

Denominator contribution =7e-051e-03 + 5.6e-05

Denominator contribution = 0.066.

4. Assemble Equation 18

λ_LOTC =6e-06 + 2.115e-07 + 3.712e-061 + 2.979e-03 + 0.066

λ_LOTC =9.924e-061.069

Result = 9.281e-06 per hour.

Inspect full-precision values used in this calculation

Parameter	Unrounded value
Combined HM + UC	6e-06
ST occurrence total	3e-05
ST averaged electronic exit	6.5e-05
ST repair interval	100
ST combined exit	7.099999999999999e-05
LT occurrence total	7e-05
LT averaged electronic exit	5e-05
LT repair interval	1e+03
LT combined exit	5.6000000000000006e-05
Equation 18 numerator	9.923619573753623e-06
Equation 18 denominator	1.0692667289517155
Equation 18 result	9.280770929328841e-06

First-order Markov (averaged ST/LT states) — First-order expansion of Equation 18

1. Averaged ST/LT inputs and combined baseline

Averaged ST/LT state model: one representative short-term state and one representative long-term state. The given totals and averages are used directly.

Hydromechanical and uncovered contributions — Equations 1–3, pages 23–24

X = 0.02; λ_HMC = 4e-06 per hour.

λ_UC ≈ X × (λ_STT + λ_LTT) = 0.02 × 1e-04 = 2e-06 per hour.

λ_HM+UC = λ_HMC + λ_UC = 4e-06 + 2e-06 = 6e-06 per hour.

ST representative state

λ_STT = 3e-05 per hour; λ_ST-LOTC-AVE = 6.5e-05 per hour; T_ST = 100 hours.

Combined exit = λ_ST-LOTC-AVE + λ_HM+UC = 6.5e-05 + 6e-06 = 7.1e-05 per hour.

µ_ST =1T_ST

Repair rate = 0.01 per hour.

LT representative state

λ_LTT = 7e-05 per hour; λ_LT-LOTC-AVE = 5e-05 per hour; T_LT = 1e+03 hours.

Combined exit = λ_LT-LOTC-AVE + λ_HM+UC = 5e-05 + 6e-06 = 5.6e-05 per hour.

µ_LT =1T_LT

Repair rate = 1e-03 per hour.

2. Formula for First-order Markov (averaged ST/LT states)

First-order expansion of the averaged Equation 18

λ_LOTC ≈λ_HM+UC + λ_STT × (λ_ST-LOTC-AVE + λ_HM+UC) × T_ST − λ_STT × (λ_ST-LOTC-AVE + λ_HM+UC)² × T_ST² + λ_LTT × (λ_LT-LOTC-AVE + λ_HM+UC) × T_LT − λ_LTT × (λ_LT-LOTC-AVE + λ_HM+UC)² × T_LT²1 + λ_STT × T_ST − λ_STT × (λ_ST-LOTC-AVE + λ_HM+UC) × T_ST² + λ_LTT × T_LT − λ_LTT × (λ_LT-LOTC-AVE + λ_HM+UC) × T_LT²

For ST and LT separately, combined exit × repair interval must be much smaller than one for useful accuracy; this implementation requires it below one. The expansion replaces each denominator term by its constant and linear terms. This approximates the two representative states of Equation 18. It is not the individual-state coefficient reconstruction from averaged rates.

3. Constants for the two representative states

These constants belong to the averaged-state model. Calculate them once when failure inputs are fixed; change ST/LT intervals without repeating the failure-rate products.

a₀ = λ_STT × (λ_ST-LOTC-AVE + λ_HM+UC) = 3e-05 × 7.1e-05 = 2.13e-09.

a₁ = λ_STT × (λ_ST-LOTC-AVE + λ_HM+UC)² = 3e-05 × 7.1e-05² = 1.512e-13.

b₀ = λ_STT = 3e-05 = 3e-05.

b₁ = a₀ = 2.13e-09 = 2.13e-09.

c₀ = λ_LTT × (λ_LT-LOTC-AVE + λ_HM+UC) = 7e-05 × 5.6e-05 = 3.92e-09.

c₁ = λ_LTT × (λ_LT-LOTC-AVE + λ_HM+UC)² = 7e-05 × 5.6e-05² = 2.195e-13.

d₀ = λ_LTT = 7e-05 = 7e-05.

d₁ = c₀ = 3.92e-09 = 3.92e-09.

4. Expansion condition and result

ST combined exit × interval = 7.1e-05 × 100 = 7.1e-03. It must be much smaller than one for useful approximation; the implemented domain requires it below one.

LT combined exit × interval = 5.6e-05 × 1e+03 = 0.056. It must be much smaller than one for useful approximation; the implemented domain requires it below one.

λ_LOTC ≈6e-06 + (2.13e-09 × 100 − 1.512e-13 × 100²) + (3.92e-09 × 1e+03 − 2.195e-13 × 1e+03²)1 + (3e-05 × 100 − 2.13e-09 × 100²) + (7e-05 × 1e+03 − 3.92e-09 × 1e+03²)

First-order result =9.912e-061.069

Result = 9.272e-06 per hour.

Inspect full-precision values used in this calculation

Parameter	Unrounded value
Combined HM + UC	6e-06
ST occurrence total	3e-05
ST averaged electronic exit	6.5e-05
ST repair interval	100
ST combined exit	7.099999999999999e-05
LT occurrence total	7e-05
LT averaged electronic exit	5e-05
LT repair interval	1e+03
LT combined exit	5.6000000000000006e-05
a₀	2.13e-09
a₁	1.5122999999999998e-13
b₀	3e-05
b₁	2.13e-09
c₀	3.92e-09
c₁	2.1952e-13
d₀	7e-05
d₁	3.92e-09
Averaged first-order numerator	9.9119677e-06
Averaged first-order denominator	1.0690587
Averaged first-order result	9.271677691786242e-06

Zero-order Markov (averaged ST/LT states) — Equation 20f

1. Averaged ST/LT inputs and combined baseline

Averaged ST/LT state model: one representative short-term state and one representative long-term state. The given totals and averages are used directly.

Hydromechanical and uncovered contributions — Equations 1–3, pages 23–24

X = 0.02; λ_HMC = 4e-06 per hour.

λ_UC ≈ X × (λ_STT + λ_LTT) = 0.02 × 1e-04 = 2e-06 per hour.

λ_HM+UC = λ_HMC + λ_UC = 4e-06 + 2e-06 = 6e-06 per hour.

ST representative state

λ_STT = 3e-05 per hour; λ_ST-LOTC-AVE = 6.5e-05 per hour; T_ST = 100 hours.

Combined exit = λ_ST-LOTC-AVE + λ_HM+UC = 6.5e-05 + 6e-06 = 7.1e-05 per hour.

µ_ST =1T_ST

Repair rate = 0.01 per hour.

LT representative state

λ_LTT = 7e-05 per hour; λ_LT-LOTC-AVE = 5e-05 per hour; T_LT = 1e+03 hours.

Combined exit = λ_LT-LOTC-AVE + λ_HM+UC = 5e-05 + 6e-06 = 5.6e-05 per hour.

µ_LT =1T_LT

Repair rate = 1e-03 per hour.

2. Formula for Zero-order Markov (averaged ST/LT states)

Zero-order averaged ST/LT equation

λ_LOTC ≈λ_HM+UC + λ_STT × T_ST × (λ_ST-LOTC-AVE + λ_HM+UC) + λ_LTT × T_LT × (λ_LT-LOTC-AVE + λ_HM+UC)1 + λ_STT × T_ST + λ_LTT × T_LT

3. Representative-state products and result

ST occurrence × interval × combined exit = 3e-05 × 100 × 7.1e-05 = 2.13e-07 per hour.

LT occurrence × interval × combined exit = 7e-05 × 1e+03 × 5.6e-05 = 3.92e-06 per hour.

λ_LOTC ≈6e-06 + 2.13e-07 + 3.92e-061 + 3e-03 + 0.07

Result = 9.444e-06 per hour.

Inspect full-precision values used in this calculation

Parameter	Unrounded value
Combined HM + UC	6e-06
ST occurrence total	3e-05
ST averaged electronic exit	6.5e-05
ST repair interval	100
ST combined exit	7.099999999999999e-05
LT occurrence total	7e-05
LT averaged electronic exit	5e-05
LT repair interval	1e+03
LT combined exit	5.6000000000000006e-05
Zero-order numerator	1.0133e-05
Zero-order denominator	1.073
Zero-order result	9.443616029822927e-06

Balanced time-weighted average (averaged ST/LT states) — Equations 6–9, 11

1. Averaged ST/LT inputs and combined baseline

Averaged ST/LT state model: one representative short-term state and one representative long-term state. The given totals and averages are used directly.

Hydromechanical and uncovered contributions — Equations 1–3, pages 23–24

X = 0.02; λ_HMC = 4e-06 per hour.

λ_UC ≈ X × (λ_STT + λ_LTT) = 0.02 × 1e-04 = 2e-06 per hour.

λ_HM+UC = λ_HMC + λ_UC = 4e-06 + 2e-06 = 6e-06 per hour.

ST representative state

λ_STT = 3e-05 per hour; λ_ST-LOTC-AVE = 6.5e-05 per hour; T_ST = 100 hours.

Combined exit = λ_ST-LOTC-AVE + λ_HM+UC = 6.5e-05 + 6e-06 = 7.1e-05 per hour.

µ_ST =1T_ST

Repair rate = 0.01 per hour.

LT representative state

λ_LTT = 7e-05 per hour; λ_LT-LOTC-AVE = 5e-05 per hour; T_LT = 1e+03 hours.

Combined exit = λ_LT-LOTC-AVE + λ_HM+UC = 5e-05 + 6e-06 = 5.6e-05 per hour.

µ_LT =1T_LT

Repair rate = 1e-03 per hour.

2. Formula for Balanced time-weighted average (averaged ST/LT states)

ARP5107C, Section 7.1.4, page 26 — Equation 11

The standard writes the balanced time-weighted average as follows. HM + UC is outside the electronic contribution fraction.

λ_LOTC-AVE = λ_HM+UC +λ_FU-LOTC + (λ_STT × T_ST-REPAIR) × λ_ST-LOTC-AVE + (λ_LTT × T_LT-REPAIR) × λ_LT-LOTC-AVE1 + λ_STT × T_ST-REPAIR + λ_LTT × T_LT-REPAIR(11)

In simple terms: start with HM + UC. Add the electronic LOTC contribution from fully operational operation and from short-term and long-term fault operation, weighted by the time spent in each condition.

Time fractions — Equations 6a–6c, page 25

Fract(FU) =11 + λ_STT × T_ST-REPAIR + λ_LTT × T_LT-REPAIR(6a)

Fract(ST) =λ_STT × T_ST-REPAIR1 + λ_STT × T_ST-REPAIR + λ_LTT × T_LT-REPAIR(6b)

Fract(LT) =λ_LTT × T_LT-REPAIR1 + λ_STT × T_ST-REPAIR + λ_LTT × T_LT-REPAIR(6c)

These fractions add to one. The denominator accounts for time spent in all three operating conditions.

Fully operational electronic LOTC bound — Equation 7, page 25

λ_FU-LOTC ≤[(λ_STT + λ_LTT) / 2]² × T_FL(7)

Calculator choice: the calculator substitutes the right-hand side of Equation 7 for the fully operational electronic LOTC rate in Equation 11. The standard gives this as a conservative approximation for small failure-rate × flight-duration products. The result is therefore an estimate using this bound, not a separately supplied exact fully operational rate. Flight duration is T_FL.

Occurrence-weighted electronic exit rates — Equations 8 and 9, page 26

λ_ST-LOTC-AVE =∑[λ_ST(i) × λ_LT-LOTC(i)]∑[λ_ST(i)](8 — as printed)

λ_LT-LOTC-AVE =∑[λ_LT(i) × λ_LT-LOTC(i)]∑[λ_LT(i)](9)

Equation 8 interpretation: the printed numerator uses an LT exit label, but the definition immediately below it specifies the exit associated with that particular ST fault. The calculator follows that ST definition: sum each ST occurrence rate × its ST electronic exit rate, then divide by total ST occurrence rate. This is a disclosed interpretation, not a publisher-issued correction. Both weighted exit rates exclude HM + UC.

Terms: λ_STT and λ_LTT are total ST and LT occurrence rates; T_ST-REPAIR and T_LT-REPAIR are their repair intervals; λ_ST-LOTC-AVE and λ_LT-LOTC-AVE are occurrence-weighted electronic LOTC exit rates. All rates are per hour; all times are in hours.

3. Fractions and electronic contributions

Fract(FU) = 0.932.

Fract(ST) = 2.796e-03.

Fract(LT) = 0.065.

Equation 7 conservative fully operational electronic estimate = 1.125e-08 per hour.

λ_LOTC-AVE = 6e-06 +1.125e-08 + (3e-05 × 100 × 6.5e-05) + (7e-05 × 1e+03 × 5e-05)1.073

Result = 9.454e-06 per hour.

Inspect full-precision values used in this calculation

Parameter	Unrounded value
Combined HM + UC	6e-06
ST occurrence total	3e-05
ST averaged electronic exit	6.5e-05
ST repair interval	100
ST combined exit	7.099999999999999e-05
LT occurrence total	7e-05
LT averaged electronic exit	5e-05
LT repair interval	1e+03
LT combined exit	5.6000000000000006e-05
Fract(FU)	0.9319664492078286
Fract(ST)	2.7958993476234857e-03
Fract(LT)	0.06523765144454799
Fully operational estimate	1.1249999999999998e-08
Balanced denominator	1.073
Balanced time-weighted average	9.454100652376514e-06

Example 2 — individual fault inputs, Section 7.4 Table 2

ARP5107C Section 7.4, Table 2, pages 41–45

Table 2 provides the combined HM + UC baseline (7.6e-06 per hour), not separate HMC and X values. Its column D already includes HM + UC; this loader subtracts that baseline to populate the electronic-only exit fields, then the model adds it once. Flight duration 5 hours is an illustrative supplemental input: Table 2 does not specify it, and it does not affect the Markov result. The TWA result for this example is supplementary, not a published Table 2 benchmark.

Open a model to inspect its complete calculation. Each section repeats the inputs and calculates the parameters used by that model; no other model section needs to be opened. Substituted numbers are rounded for display; each model has a full-precision inspection table.

Full Markov — Equation 20

1. Inputs and notation

Combined hydromechanical and uncovered fault rate

λ_HM+UC

λ_HM+UC = 7.6e-06 per hour

Repair intervals

T_ST, T_LT

T_ST = 2.5e+02 h; T_LT = 1e+03 h

Flight duration

T_FL

T_FL = 5e+00 hours

The generic index i in Equation 20 runs over every individual fault. STi and LTj identify faults within the short-term and long-term groups. These are separate states, not weighted group substitutes.

Reference baseline: λ_HM+UC = 7.6e-06 per hour, as provided in Section 7.4/Table 2. Separate HMC, UC and X values are not specified there and are not inferred. The electronic-only exits have the baseline subtracted from the table’s combined column D.

2. Fault-level parameters

CPU (system) — ST1

Occurrence and electronic exit inputs

λ_ST1; λ_ST1,LOTC

λ_ST1 = 1.77e-05; λ_ST1,LOTC = 4.44e-05 per hour

Combined exit rate

λ_{ST1-LOTC+HM+UC} = λ_ST1,LOTC + λ_HM+UC

4.44e-05 + 7.6e-06 = 5.2e-05 per hour

Repair rate

µ_ST = 1 / T_ST

1 / 2.5e+02 = 4e-03 per hour

Equation 20 numerator term

λ_ST1 × λ_{ST1-LOTC+HM+UC} / (µ_ST + λ_{ST1-LOTC+HM+UC})

(1.77e-05 × 5.2e-05) / (4e-03 + 5.2e-05) = 2.27e-07 per hour

Equation 20 denominator term

λ_ST1 / (µ_ST + λ_{ST1-LOTC+HM+UC})

1.77e-05 / (4e-03 + 5.2e-05) = 4.37e-03

PMA — ST2

Occurrence and electronic exit inputs

λ_ST2; λ_ST2,LOTC

λ_ST2 = 1.5e-06; λ_ST2,LOTC = 4.42e-05 per hour

Combined exit rate

λ_{ST2-LOTC+HM+UC} = λ_ST2,LOTC + λ_HM+UC

4.42e-05 + 7.6e-06 = 5.18e-05 per hour

Repair rate

µ_ST = 1 / T_ST

1 / 2.5e+02 = 4e-03 per hour

Equation 20 numerator term

λ_ST2 × λ_{ST2-LOTC+HM+UC} / (µ_ST + λ_{ST2-LOTC+HM+UC})

(1.5e-06 × 5.18e-05) / (4e-03 + 5.18e-05) = 1.92e-08 per hour

Equation 20 denominator term

λ_ST2 / (µ_ST + λ_{ST2-LOTC+HM+UC})

1.5e-06 / (4e-03 + 5.18e-05) = 3.7e-04

SBV A,B,C — LT1

Occurrence and electronic exit inputs

λ_LT1; λ_LT1,LOTC

λ_LT1 = 2.22e-05; λ_LT1,LOTC = 1.33e-05 per hour

Combined exit rate

λ_{LT1-LOTC+HM+UC} = λ_LT1,LOTC + λ_HM+UC

1.33e-05 + 7.6e-06 = 2.09e-05 per hour

Repair rate

µ_LT = 1 / T_LT

1 / 1e+03 = 1e-03 per hour

Equation 20 numerator term

λ_LT1 × λ_{LT1-LOTC+HM+UC} / (µ_LT + λ_{LT1-LOTC+HM+UC})

(2.22e-05 × 2.09e-05) / (1e-03 + 2.09e-05) = 4.54e-07 per hour

Equation 20 denominator term

λ_LT1 / (µ_LT + λ_{LT1-LOTC+HM+UC})

2.22e-05 / (1e-03 + 2.09e-05) = 2.17e-02

PS — LT2

Occurrence and electronic exit inputs

λ_LT2; λ_LT2,LOTC

λ_LT2 = 1.1e-05; λ_LT2,LOTC = 1.52e-05 per hour

Combined exit rate

λ_{LT2-LOTC+HM+UC} = λ_LT2,LOTC + λ_HM+UC

1.52e-05 + 7.6e-06 = 2.28e-05 per hour

Repair rate

µ_LT = 1 / T_LT

1 / 1e+03 = 1e-03 per hour

Equation 20 numerator term

λ_LT2 × λ_{LT2-LOTC+HM+UC} / (µ_LT + λ_{LT2-LOTC+HM+UC})

(1.1e-05 × 2.28e-05) / (1e-03 + 2.28e-05) = 2.45e-07 per hour

Equation 20 denominator term

λ_LT2 / (µ_LT + λ_{LT2-LOTC+HM+UC})

1.1e-05 / (1e-03 + 2.28e-05) = 1.08e-02

CCDL — LT3

Occurrence and electronic exit inputs

λ_LT3; λ_LT3,LOTC

λ_LT3 = 1.6e-06; λ_LT3,LOTC = 2.81e-05 per hour

Combined exit rate

λ_{LT3-LOTC+HM+UC} = λ_LT3,LOTC + λ_HM+UC

2.81e-05 + 7.6e-06 = 3.57e-05 per hour

Repair rate

µ_LT = 1 / T_LT

1 / 1e+03 = 1e-03 per hour

Equation 20 numerator term

λ_LT3 × λ_{LT3-LOTC+HM+UC} / (µ_LT + λ_{LT3-LOTC+HM+UC})

(1.6e-06 × 3.57e-05) / (1e-03 + 3.57e-05) = 5.52e-08 per hour

Equation 20 denominator term

λ_LT3 / (µ_LT + λ_{LT3-LOTC+HM+UC})

1.6e-06 / (1e-03 + 3.57e-05) = 1.54e-03

TLA — LT4

Occurrence and electronic exit inputs

λ_LT4; λ_LT4,LOTC

λ_LT4 = 5.4e-06; λ_LT4,LOTC = 1.23e-05 per hour

Combined exit rate

λ_{LT4-LOTC+HM+UC} = λ_LT4,LOTC + λ_HM+UC

1.23e-05 + 7.6e-06 = 1.99e-05 per hour

Repair rate

µ_LT = 1 / T_LT

1 / 1e+03 = 1e-03 per hour

Equation 20 numerator term

λ_LT4 × λ_{LT4-LOTC+HM+UC} / (µ_LT + λ_{LT4-LOTC+HM+UC})

(5.4e-06 × 1.99e-05) / (1e-03 + 1.99e-05) = 1.05e-07 per hour

Equation 20 denominator term

λ_LT4 / (µ_LT + λ_{LT4-LOTC+HM+UC})

5.4e-06 / (1e-03 + 1.99e-05) = 5.29e-03

CGV — LT5

Occurrence and electronic exit inputs

λ_LT5; λ_LT5,LOTC

λ_LT5 = 8.2e-06; λ_LT5,LOTC = 1.7e-05 per hour

Combined exit rate

λ_{LT5-LOTC+HM+UC} = λ_LT5,LOTC + λ_HM+UC

1.7e-05 + 7.6e-06 = 2.46e-05 per hour

Repair rate

µ_LT = 1 / T_LT

1 / 1e+03 = 1e-03 per hour

Equation 20 numerator term

λ_LT5 × λ_{LT5-LOTC+HM+UC} / (µ_LT + λ_{LT5-LOTC+HM+UC})

(8.2e-06 × 2.46e-05) / (1e-03 + 2.46e-05) = 1.97e-07 per hour

Equation 20 denominator term

λ_LT5 / (µ_LT + λ_{LT5-LOTC+HM+UC})

8.2e-06 / (1e-03 + 2.46e-05) = 8e-03

MV — LT6

Occurrence and electronic exit inputs

λ_LT6; λ_LT6,LOTC

λ_LT6 = 6.6e-06; λ_LT6,LOTC = 1.7e-05 per hour

Combined exit rate

λ_{LT6-LOTC+HM+UC} = λ_LT6,LOTC + λ_HM+UC

1.7e-05 + 7.6e-06 = 2.46e-05 per hour

Repair rate

µ_LT = 1 / T_LT

1 / 1e+03 = 1e-03 per hour

Equation 20 numerator term

λ_LT6 × λ_{LT6-LOTC+HM+UC} / (µ_LT + λ_{LT6-LOTC+HM+UC})

(6.6e-06 × 2.46e-05) / (1e-03 + 2.46e-05) = 1.58e-07 per hour

Equation 20 denominator term

λ_LT6 / (µ_LT + λ_{LT6-LOTC+HM+UC})

6.6e-06 / (1e-03 + 2.46e-05) = 6.44e-03

T2 — LT7

Occurrence and electronic exit inputs

λ_LT7; λ_LT7,LOTC

λ_LT7 = 3.8e-06; λ_LT7,LOTC = 1.16e-05 per hour

Combined exit rate

λ_{LT7-LOTC+HM+UC} = λ_LT7,LOTC + λ_HM+UC

1.16e-05 + 7.6e-06 = 1.92e-05 per hour

Repair rate

µ_LT = 1 / T_LT

1 / 1e+03 = 1e-03 per hour

Equation 20 numerator term

λ_LT7 × λ_{LT7-LOTC+HM+UC} / (µ_LT + λ_{LT7-LOTC+HM+UC})

(3.8e-06 × 1.92e-05) / (1e-03 + 1.92e-05) = 7.16e-08 per hour

Equation 20 denominator term

λ_LT7 / (µ_LT + λ_{LT7-LOTC+HM+UC})

3.8e-06 / (1e-03 + 1.92e-05) = 3.73e-03

4 SUMS — LT8

Occurrence and electronic exit inputs

λ_LT8; λ_LT8,LOTC

λ_LT8 = 9e-07; λ_LT8,LOTC = 6.1e-06 per hour

Combined exit rate

λ_{LT8-LOTC+HM+UC} = λ_LT8,LOTC + λ_HM+UC

6.1e-06 + 7.6e-06 = 1.37e-05 per hour

Repair rate

µ_LT = 1 / T_LT

1 / 1e+03 = 1e-03 per hour

Equation 20 numerator term

λ_LT8 × λ_{LT8-LOTC+HM+UC} / (µ_LT + λ_{LT8-LOTC+HM+UC})

(9e-07 × 1.37e-05) / (1e-03 + 1.37e-05) = 1.22e-08 per hour

Equation 20 denominator term

λ_LT8 / (µ_LT + λ_{LT8-LOTC+HM+UC})

9e-07 / (1e-03 + 1.37e-05) = 8.88e-04

P3 — LT9

Occurrence and electronic exit inputs

λ_LT9; λ_LT9,LOTC

λ_LT9 = 1.1e-05; λ_LT9,LOTC = 1.52e-05 per hour

Combined exit rate

λ_{LT9-LOTC+HM+UC} = λ_LT9,LOTC + λ_HM+UC

1.52e-05 + 7.6e-06 = 2.28e-05 per hour

Repair rate

µ_LT = 1 / T_LT

1 / 1e+03 = 1e-03 per hour

Equation 20 numerator term

λ_LT9 × λ_{LT9-LOTC+HM+UC} / (µ_LT + λ_{LT9-LOTC+HM+UC})

(1.1e-05 × 2.28e-05) / (1e-03 + 2.28e-05) = 2.45e-07 per hour

Equation 20 denominator term

λ_LT9 / (µ_LT + λ_{LT9-LOTC+HM+UC})

1.1e-05 / (1e-03 + 2.28e-05) = 1.08e-02

3. Full Markov — Equation 20

ARP5107C Equation 20, page 35

λ_LOTC =λ_HM+UC + ∑[λ_i × λ_i-LOTC+HM+UC / (µ_i + λ_i-LOTC+HM+UC)]1 + ∑[λ_i / (µ_i + λ_i-LOTC+HM+UC)](20)

λ_i-LOTC+HM+UC = λ_i-LOTC + λ_HM+UC; µ_i = 1 / T_i. The index i runs over every individual fault. The combined exit includes HM + UC; the entered electronic exit excludes it.

Substitute every fault term

λ_LOTC =7.6e-06 + 2.27e-07 + 1.92e-08 + 4.54e-07 + 2.45e-07 + 5.52e-08 + 1.05e-07 + 1.97e-07 + 1.58e-07 + 7.16e-08 + 1.22e-08 + 2.45e-071 + 4.37e-03 + 3.7e-04 + 2.17e-02 + 1.08e-02 + 1.54e-03 + 5.29e-03 + 8e-03 + 6.44e-03 + 3.73e-03 + 8.88e-04 + 1.08e-02

Evaluate numerator and denominator

λ_LOTC =9.39e-061.07e+00

λ_LOTC = 8.74e-06 per hour

Inspect full-precision values used in this calculation

Parameter	Unrounded value
λ_HM+UC	7.6e-06
Equation 20 numerator	9.390825513639392e-06
Equation 20 denominator	1.0738939172989221
Equation 20 result	8.74464913374253e-06
λ_i=1	1.77e-05
λ_i=1,LOTC	4.4399999999999995e-05
λ_{i=1-LOTC+HM+UC}	5.2e-05
µ_i=1	4e-03
λ_i=2	1.5e-06
λ_i=2,LOTC	4.42e-05
λ_{i=2-LOTC+HM+UC}	5.18e-05
µ_i=2	4e-03
λ_i=3	2.22e-05
λ_i=3,LOTC	1.33e-05
λ_{i=3-LOTC+HM+UC}	2.09e-05
µ_i=3	1e-03
λ_i=4	1.1e-05
λ_i=4,LOTC	1.5199999999999998e-05
λ_{i=4-LOTC+HM+UC}	2.28e-05
µ_i=4	1e-03
λ_i=5	1.6e-06
λ_i=5,LOTC	2.81e-05
λ_{i=5-LOTC+HM+UC}	3.57e-05
µ_i=5	1e-03
λ_i=6	5.4e-06
λ_i=6,LOTC	1.2299999999999999e-05
λ_{i=6-LOTC+HM+UC}	1.99e-05
µ_i=6	1e-03
λ_i=7	8.2e-06
λ_i=7,LOTC	1.7e-05
λ_{i=7-LOTC+HM+UC}	2.46e-05
µ_i=7	1e-03
λ_i=8	6.6e-06
λ_i=8,LOTC	1.7e-05
λ_{i=8-LOTC+HM+UC}	2.46e-05
µ_i=8	1e-03
λ_i=9	3.8e-06
λ_i=9,LOTC	1.1599999999999999e-05
λ_{i=9-LOTC+HM+UC}	1.92e-05
µ_i=9	1e-03
λ_i=10	9e-07
λ_i=10,LOTC	6.099999999999999e-06
λ_{i=10-LOTC+HM+UC}	1.37e-05
µ_i=10	1e-03
λ_i=11	1.1e-05
λ_i=11,LOTC	1.5199999999999998e-05
λ_{i=11-LOTC+HM+UC}	2.28e-05
µ_i=11	1e-03

First-order Markov — Equation 20c

1. Inputs and notation

Combined hydromechanical and uncovered fault rate

λ_HM+UC

λ_HM+UC = 7.6e-06 per hour

Repair intervals

T_ST, T_LT

T_ST = 2.5e+02 h; T_LT = 1e+03 h

Flight duration

T_FL

T_FL = 5e+00 hours

µ_LT = 1 / T_LT

1 / 1e+03 = 1e-03 per hour

3. First-order Markov — Equation 20c

Calculate the coefficients below from the individual fault inputs. These coefficient definitions are given on page 36.

ST combined-exit coefficient

a₀ = ∑(λ_STi × λ_{STi-LOTC+HM+UC})

(1.77e-05 × 5.2e-05) + (1.5e-06 × 5.18e-05) = 9.98e-10 per hour²

ST squared-exit coefficient

a₁ = ∑(λ_STi × λ_{STi-LOTC+HM+UC}²)

(1.77e-05 × 5.2e-05²) + (1.5e-06 × 5.18e-05²) = 5.19e-14 per hour³

ST occurrence coefficient

b₀ = ∑(λ_STi)

1.77e-05 + 1.5e-06 = 1.92e-05 per hour

ST denominator correction coefficient

b₁ = a₀

9.98e-10 = 9.98e-10 per hour²

LT combined-exit coefficient

c₀ = ∑(λ_LTi × λ_{LTi-LOTC+HM+UC})

(2.22e-05 × 2.09e-05) + (1.1e-05 × 2.28e-05) + (1.6e-06 × 3.57e-05) + (5.4e-06 × 1.99e-05) + (8.2e-06 × 2.46e-05) + (6.6e-06 × 2.46e-05) + (3.8e-06 × 1.92e-05) + (9e-07 × 1.37e-05) + (1.1e-05 × 2.28e-05) = 1.58e-09 per hour²

LT squared-exit coefficient

c₁ = ∑(λ_LTi × λ_{LTi-LOTC+HM+UC}²)

(2.22e-05 × 2.09e-05²) + (1.1e-05 × 2.28e-05²) + (1.6e-06 × 3.57e-05²) + (5.4e-06 × 1.99e-05²) + (8.2e-06 × 2.46e-05²) + (6.6e-06 × 2.46e-05²) + (3.8e-06 × 1.92e-05²) + (9e-07 × 1.37e-05²) + (1.1e-05 × 2.28e-05²) = 3.58e-14 per hour³

LT occurrence coefficient

d₀ = ∑(λ_LTi)

2.22e-05 + 1.1e-05 + 1.6e-06 + 5.4e-06 + 8.2e-06 + 6.6e-06 + 3.8e-06 + 9e-07 + 1.1e-05 = 7.07e-05 per hour

LT denominator correction coefficient

d₁ = c₀

1.58e-09 = 1.58e-09 per hour²

ARP5107C Equation 20c, page 36

λ_LOTC ≈λ_HM+UC + a₀ × T_ST − a₁ × T_ST² + c₀ × T_LT − c₁ × T_LT²1 + b₀ × T_ST − b₁ × T_ST² + d₀ × T_LT − d₁ × T_LT²(20c)

a₀ = ∑(λ_STi × λ_{STi-LOTC+HM+UC}); a₁ = ∑(λ_STi × λ_{STi-LOTC+HM+UC}²); b₀ = ∑λ_STi; b₁ = a₀.

c₀ = ∑(λ_LTi × λ_{LTi-LOTC+HM+UC}); c₁ = ∑(λ_LTi × λ_{LTi-LOTC+HM+UC}²); d₀ = ∑λ_LTi; d₁ = c₀.

The combined exits include HM + UC. The first correction is retained in both numerator and denominator.

Substitute grouped coefficients

λ_LOTC ≈7.6e-06 + 9.98e-10 × 2.5e+02 − 5.19e-14 × 2.5e+02² + 1.58e-09 × 1e+03 − 3.58e-14 × 1e+03²1 + 1.92e-05 × 2.5e+02 − 9.98e-10 × 2.5e+02² + 7.07e-05 × 1e+03 − 1.58e-09 × 1e+03²

Evaluate first-order numerator and denominator

λ_LOTC ≈9.39e-061.07e+00

λ_LOTC ≈ 8.74e-06 per hour

Inspect full-precision values used in this calculation

Parameter	Unrounded value
λ_HM+UC	7.6e-06
Equation 20c numerator	9.38997472525e-06
Equation 20c denominator	1.07385808875
Equation 20c result	8.744148620401216e-06
λ_i=1	1.77e-05
λ_i=1,LOTC	4.4399999999999995e-05
λ_{i=1-LOTC+HM+UC}	5.2e-05
µ_i=1	4e-03
λ_i=2	1.5e-06
λ_i=2,LOTC	4.42e-05
λ_{i=2-LOTC+HM+UC}	5.18e-05
µ_i=2	4e-03
λ_i=3	2.22e-05
λ_i=3,LOTC	1.33e-05
λ_{i=3-LOTC+HM+UC}	2.09e-05
µ_i=3	1e-03
λ_i=4	1.1e-05
λ_i=4,LOTC	1.5199999999999998e-05
λ_{i=4-LOTC+HM+UC}	2.28e-05
µ_i=4	1e-03
λ_i=5	1.6e-06
λ_i=5,LOTC	2.81e-05
λ_{i=5-LOTC+HM+UC}	3.57e-05
µ_i=5	1e-03
λ_i=6	5.4e-06
λ_i=6,LOTC	1.2299999999999999e-05
λ_{i=6-LOTC+HM+UC}	1.99e-05
µ_i=6	1e-03
λ_i=7	8.2e-06
λ_i=7,LOTC	1.7e-05
λ_{i=7-LOTC+HM+UC}	2.46e-05
µ_i=7	1e-03
λ_i=8	6.6e-06
λ_i=8,LOTC	1.7e-05
λ_{i=8-LOTC+HM+UC}	2.46e-05
µ_i=8	1e-03
λ_i=9	3.8e-06
λ_i=9,LOTC	1.1599999999999999e-05
λ_{i=9-LOTC+HM+UC}	1.92e-05
µ_i=9	1e-03
λ_i=10	9e-07
λ_i=10,LOTC	6.099999999999999e-06
λ_{i=10-LOTC+HM+UC}	1.37e-05
µ_i=10	1e-03
λ_i=11	1.1e-05
λ_i=11,LOTC	1.5199999999999998e-05
λ_{i=11-LOTC+HM+UC}	2.28e-05
µ_i=11	1e-03
a₀	9.980999999999999e-10
a₁	5.188565999999999e-14
b₀	1.92e-05
b₁	9.980999999999999e-10
c₀	1.5795300000000002e-09
c₁	3.5837421e-14
d₀	7.07e-05
d₁	1.5795300000000002e-09

Zero-order Markov — Equation 20f

1. Inputs and notation

Combined hydromechanical and uncovered fault rate

λ_HM+UC

λ_HM+UC = 7.6e-06 per hour

Repair intervals

T_ST, T_LT

T_ST = 2.5e+02 h; T_LT = 1e+03 h

Flight duration

T_FL

T_FL = 5e+00 hours

µ_LT = 1 / T_LT

1 / 1e+03 = 1e-03 per hour

Group sums used by Equation 20f

∑λ_STi = 1.77e-05 + 1.5e-06 = 1.92e-05 per hour.

∑[λ_STi × λ_{STi-LOTC+HM+UC}] = (1.77e-05 × 5.2e-05) + (1.5e-06 × 5.18e-05) = 9.980999999999999e-10 per hour².

∑λ_LTi = 2.22e-05 + 1.1e-05 + 1.6e-06 + 5.4e-06 + 8.2e-06 + 6.6e-06 + 3.8e-06 + 9e-07 + 1.1e-05 = 7.07e-05 per hour.

∑[λ_LTi × λ_{LTi-LOTC+HM+UC}] = (2.22e-05 × 2.09e-05) + (1.1e-05 × 2.28e-05) + (1.6e-06 × 3.57e-05) + (5.4e-06 × 1.99e-05) + (8.2e-06 × 2.46e-05) + (6.6e-06 × 2.46e-05) + (3.8e-06 × 1.92e-05) + (9e-07 × 1.37e-05) + (1.1e-05 × 2.28e-05) = 1.5795300000000002e-09 per hour².

3. Zero-order Markov — Equation 20f

ARP5107C Equation 20f, page 37

λ_LOTC ≈λ_HM+UC + T_ST × ∑[λ_STi × λ_{STi-LOTC+HM+UC}] + T_LT × ∑[λ_LTi × λ_{LTi-LOTC+HM+UC}]1 + T_ST × ∑[λ_STi] + T_LT × ∑[λ_LTi](20f)

The combined exit rates in these sums include HM + UC. The calculation is algebraically equivalent to HM + UC plus the electronic contribution divided by the denominator. Averaged mode supplies the occurrence-weighted sums; it cannot recover individual full or first-order Markov values.

Substitute group sums

λ_LOTC ≈7.6e-06 + 2.5e+02 × 9.98e-10 + 1e+03 × 1.58e-091 + 2.5e+02 × 1.92e-05 + 1e+03 × 7.07e-05

Evaluate zero-order numerator and denominator

λ_LOTC ≈9.43e-061.08e+00

λ_LOTC ≈ 8.77e-06 per hour

Inspect full-precision values used in this calculation

Parameter	Unrounded value
λ_HM+UC	7.6e-06
Equation 20f numerator	9.429055e-06
Equation 20f denominator	1.0755
Equation 20f result	8.767136215713621e-06
λ_i=1	1.77e-05
λ_i=1,LOTC	4.4399999999999995e-05
λ_{i=1-LOTC+HM+UC}	5.2e-05
µ_i=1	4e-03
λ_i=2	1.5e-06
λ_i=2,LOTC	4.42e-05
λ_{i=2-LOTC+HM+UC}	5.18e-05
µ_i=2	4e-03
λ_i=3	2.22e-05
λ_i=3,LOTC	1.33e-05
λ_{i=3-LOTC+HM+UC}	2.09e-05
µ_i=3	1e-03
λ_i=4	1.1e-05
λ_i=4,LOTC	1.5199999999999998e-05
λ_{i=4-LOTC+HM+UC}	2.28e-05
µ_i=4	1e-03
λ_i=5	1.6e-06
λ_i=5,LOTC	2.81e-05
λ_{i=5-LOTC+HM+UC}	3.57e-05
µ_i=5	1e-03
λ_i=6	5.4e-06
λ_i=6,LOTC	1.2299999999999999e-05
λ_{i=6-LOTC+HM+UC}	1.99e-05
µ_i=6	1e-03
λ_i=7	8.2e-06
λ_i=7,LOTC	1.7e-05
λ_{i=7-LOTC+HM+UC}	2.46e-05
µ_i=7	1e-03
λ_i=8	6.6e-06
λ_i=8,LOTC	1.7e-05
λ_{i=8-LOTC+HM+UC}	2.46e-05
µ_i=8	1e-03
λ_i=9	3.8e-06
λ_i=9,LOTC	1.1599999999999999e-05
λ_{i=9-LOTC+HM+UC}	1.92e-05
µ_i=9	1e-03
λ_i=10	9e-07
λ_i=10,LOTC	6.099999999999999e-06
λ_{i=10-LOTC+HM+UC}	1.37e-05
µ_i=10	1e-03
λ_i=11	1.1e-05
λ_i=11,LOTC	1.5199999999999998e-05
λ_{i=11-LOTC+HM+UC}	2.28e-05
µ_i=11	1e-03

Balanced time-weighted average — Equation 11

1. Inputs and notation

Combined hydromechanical and uncovered fault rate

λ_HM+UC

λ_HM+UC = 7.6e-06 per hour

Repair intervals

T_ST, T_LT

T_ST = 2.5e+02 h; T_LT = 1e+03 h

Flight duration

T_FL

T_FL = 5e+00 hours

The generic index i in Equation 20 runs over every individual fault. STi and LTj identify faults within the short-term and long-term groups. These are separate states, not weighted group substitutes.

2. Fault-level parameters

CPU (system) — ST1

Occurrence and electronic exit inputs

λ_ST1; λ_ST1,LOTC

λ_ST1 = 1.77e-05; λ_ST1,LOTC = 4.44e-05 per hour

PMA — ST2

Occurrence and electronic exit inputs

λ_ST2; λ_ST2,LOTC

λ_ST2 = 1.5e-06; λ_ST2,LOTC = 4.42e-05 per hour

SBV A,B,C — LT1

Occurrence and electronic exit inputs

λ_LT1; λ_LT1,LOTC

λ_LT1 = 2.22e-05; λ_LT1,LOTC = 1.33e-05 per hour

PS — LT2

Occurrence and electronic exit inputs

λ_LT2; λ_LT2,LOTC

λ_LT2 = 1.1e-05; λ_LT2,LOTC = 1.52e-05 per hour

CCDL — LT3

Occurrence and electronic exit inputs

λ_LT3; λ_LT3,LOTC

λ_LT3 = 1.6e-06; λ_LT3,LOTC = 2.81e-05 per hour

TLA — LT4

Occurrence and electronic exit inputs

λ_LT4; λ_LT4,LOTC

λ_LT4 = 5.4e-06; λ_LT4,LOTC = 1.23e-05 per hour

CGV — LT5

Occurrence and electronic exit inputs

λ_LT5; λ_LT5,LOTC

λ_LT5 = 8.2e-06; λ_LT5,LOTC = 1.7e-05 per hour

MV — LT6

Occurrence and electronic exit inputs

λ_LT6; λ_LT6,LOTC

λ_LT6 = 6.6e-06; λ_LT6,LOTC = 1.7e-05 per hour

T2 — LT7

Occurrence and electronic exit inputs

λ_LT7; λ_LT7,LOTC

λ_LT7 = 3.8e-06; λ_LT7,LOTC = 1.16e-05 per hour

4 SUMS — LT8

Occurrence and electronic exit inputs

λ_LT8; λ_LT8,LOTC

λ_LT8 = 9e-07; λ_LT8,LOTC = 6.1e-06 per hour

P3 — LT9

Occurrence and electronic exit inputs

λ_LT9; λ_LT9,LOTC

λ_LT9 = 1.1e-05; λ_LT9,LOTC = 1.52e-05 per hour

3. Balanced time-weighted average — Equations 6–9 and 11

ARP5107C, Section 7.1.4, page 26 — Equation 11

The standard writes the balanced time-weighted average as follows. HM + UC is outside the electronic contribution fraction.

λ_LOTC-AVE = λ_HM+UC +λ_FU-LOTC + (λ_STT × T_ST-REPAIR) × λ_ST-LOTC-AVE + (λ_LTT × T_LT-REPAIR) × λ_LT-LOTC-AVE1 + λ_STT × T_ST-REPAIR + λ_LTT × T_LT-REPAIR(11)

Time fractions — Equations 6a–6c, page 25

Fract(FU) =11 + λ_STT × T_ST-REPAIR + λ_LTT × T_LT-REPAIR(6a)

Fract(ST) =λ_STT × T_ST-REPAIR1 + λ_STT × T_ST-REPAIR + λ_LTT × T_LT-REPAIR(6b)

Fract(LT) =λ_LTT × T_LT-REPAIR1 + λ_STT × T_ST-REPAIR + λ_LTT × T_LT-REPAIR(6c)

These fractions add to one. The denominator accounts for time spent in all three operating conditions.

Fully operational electronic LOTC bound — Equation 7, page 25

λ_FU-LOTC ≤[(λ_STT + λ_LTT) / 2]² × T_FL(7)

Occurrence-weighted electronic exit rates — Equations 8 and 9, page 26

λ_ST-LOTC-AVE =∑[λ_ST(i) × λ_LT-LOTC(i)]∑[λ_ST(i)](8 — as printed)

λ_LT-LOTC-AVE =∑[λ_LT(i) × λ_LT-LOTC(i)]∑[λ_LT(i)](9)

Total ST occurrence rate

λ_STT = ∑λ_ST(i)

1.77e-05 + 1.5e-06 = 1.92e-05 per hour

ST weighted electronic exit rate

λ_ST-LOTC-AVE =(1.77e-05 × 4.44e-05) + (1.5e-06 × 4.42e-05)1.92e-05

λ_ST-LOTC-AVE = 4.44e-05 per hour

Total LT occurrence rate

λ_LTT = ∑λ_LT(i)

2.22e-05 + 1.1e-05 + 1.6e-06 + 5.4e-06 + 8.2e-06 + 6.6e-06 + 3.8e-06 + 9e-07 + 1.1e-05 = 7.07e-05 per hour

LT weighted electronic exit rate

λ_LT-LOTC-AVE =(2.22e-05 × 1.33e-05) + (1.1e-05 × 1.52e-05) + (1.6e-06 × 2.81e-05) + (5.4e-06 × 1.23e-05) + (8.2e-06 × 1.7e-05) + (6.6e-06 × 1.7e-05) + (3.8e-06 × 1.16e-05) + (9e-07 × 6.1e-06) + (1.1e-05 × 1.52e-05)7.07e-05

λ_LT-LOTC-AVE = 1.47e-05 per hour

Full-up time fraction — Equation 6a

Fract(FU) =11 + 1.92e-05 × 2.5e+02 + 7.07e-05 × 1e+03

Evaluate Fract(FU)

Fract(FU) = 1 / (1 + λSTT × TST-REPAIR + λLTT × TLT-REPAIR)

1 / 1.08e+00 = 9.3e-01

Short-term time fraction — Equation 6b

Fract(ST) = (λSTT × TST-REPAIR) / (1 + λSTT × TST-REPAIR + λLTT × TLT-REPAIR)

(1.92e-05 × 2.5e+02) / 1.08e+00 = 4.46e-03

Long-term time fraction — Equation 6c

Fract(LT) = (λLTT × TLT-REPAIR) / (1 + λSTT × TST-REPAIR + λLTT × TLT-REPAIR)

(7.07e-05 × 1e+03) / 1.08e+00 = 6.57e-02

Check conservation — Equation 5

Fract(FU) + Fract(ST) + Fract(LT) = 1

9.3e-01 + 4.46e-03 + 6.57e-02 = 1e+00

Conservative full-up electronic rate substituted from Equation 7

λ_FU-LOTC ≤ [( λ_STT + λ_LTT ) / 2]² × T_FL

[(1.92e-05 + 7.07e-05) / 2]² × 5e+00 = 1.01e-08 per hour

Equation 11 — numerical substitution

λ_LOTC-AVE = 7.6e-06 +1.01e-08 + (1.92e-05 × 2.5e+02) × 4.44e-05 + (7.07e-05 × 1e+03) × 1.47e-051 + 1.92e-05 × 2.5e+02 + 7.07e-05 × 1e+03

Equation 11 — evaluate the electronic numerator and denominator

λ_LOTC-AVE = 7.6e-06 +1.27e-061.08e+00

Balanced time-weighted average result

λ_LOTC-AVE = λ_HM+UC + Fract(FU) × λ_FU-LOTC + Fract(ST) × λ_ST-LOTC-AVE + Fract(LT) × λ_LT-LOTC-AVE

7.6e-06 + 1.18e-06 = 8.78e-06 per engine flight hour

Inspect full-precision values used in this calculation

Parameter	Unrounded value
λ_HM+UC	7.6e-06
Equation 7 upper-bound estimate	1.0102512499999996e-08
Equation 11 electronic numerator	1.2653575125e-06
Equation 11 denominator	1.0755
Equation 11 result	8.776529532775452e-06
λ_i=1	1.77e-05
λ_i=1,LOTC	4.4399999999999995e-05
λ_i=2	1.5e-06
λ_i=2,LOTC	4.42e-05
λ_i=3	2.22e-05
λ_i=3,LOTC	1.33e-05
λ_i=4	1.1e-05
λ_i=4,LOTC	1.5199999999999998e-05
λ_i=5	1.6e-06
λ_i=5,LOTC	2.81e-05
λ_i=6	5.4e-06
λ_i=6,LOTC	1.2299999999999999e-05
λ_i=7	8.2e-06
λ_i=7,LOTC	1.7e-05
λ_i=8	6.6e-06
λ_i=8,LOTC	1.7e-05
λ_i=9	3.8e-06
λ_i=9,LOTC	1.1599999999999999e-05
λ_i=10	9e-07
λ_i=10,LOTC	6.099999999999999e-06
λ_i=11	1.1e-05
λ_i=11,LOTC	1.5199999999999998e-05
Fract(FU)	0.9298000929800094
Fract(ST)	4.463040446304045e-03
Fract(LT)	0.06573686657368666

Comparison and applicability checks

7. Comparison and applicability

Method differences are numerical comparisons to Equation 20, not measured model errors or certification confidence.

Which Markov method do these checks support?

Main-input operating point: short-term repair interval = 2.50e+2 hours; long-term repair interval = 1.00e+3 hours. The checks below use these entered intervals, independently of the sensitivity range.

Check or method	Relevant quantity at the main inputs	Meaning for method selection
Single-fault model (shared by all three Markov equations)	ST: 76.92 times; LT: 28.01 times	The entered rates support using the single-fault model under the approximately ten-times guideline in Section 7.2.4. Confirm that omitted transitions, including further failures into two-fault operating states, would not reduce these ratios below ten.
Full Markov - Equation 20	ST: 76.92 times; LT: 28.01 times	Direct numerical reference within the specified single-fault model. No high-repair-rate denominator approximation is used; the single-fault model still needs justification.
First-order Markov - Equation 20c	ST: 76.92 times; LT: 28.01 times	Within the implemented expansion domain. Section 7.3 prefers this approximation for closer agreement with Full Markov when varying repair intervals. The ten-times ratio is not a mandatory switch to zero-order; inspect the ST/LT factor errors below.
Zero-order Markov - Equation 20f	ST: 76.92 times; LT: 28.01 times	The known combined-exit ratios meet the approximately ten-times simplification check in Section 7.2.2.6. Zero-order is an available simplification, not a required choice or a separate proof of single-fault adequacy.

Two different denominators: the single-fault check uses the maximum known failure rate into or out of the individual fault states in each group. Individual-mode approximation checks use the maximum individual electronic exit plus HM + UC, giving the lowest ratio in each repair category. Averaged-mode approximation checks use each representative ST/LT exit plus HM + UC; they do not identify individual maxima.

Model recommendation

Zero-order Markov, Equation 20f, is also supported by the known combined-exit ratios at the entered repair intervals (Section 7.2.2.6). It is an available simplification. Preferred approximation: first-order Markov, Equation 20c (Section 7.3), when closer numerical agreement with Full Markov is desired. The entered rates support using the single-fault model under Section 7.2.4, provided omitted transitions, including those into two-fault operating states, do not reduce the repair-rate ratios below ten. Use full Markov Equation 20 as the direct reference within this same model.

Single-fault model check — Section 7.2.4, page 37

Formula: repair rate ÷ maximum failure rate into or out of the fault states ≥ 10. Repair rate = one ÷ repair interval.

Short-term (ST)

Calculated ratio: 4.00e-3 per hour ÷ 5.20e-5 per hour = 76.92 times. Threshold: 10 times.

The entered rates meet the ten-times guideline in Section 7.2.4 and support using a single-fault model for these modeled transitions. Confirm that no omitted failure transition, including a further failure from a single-fault state into a two-fault operating state, has a higher rate that would reduce this ratio below ten.

Long-term (LT)

Calculated ratio: 1.00e-3 per hour ÷ 3.57e-5 per hour = 28.01 times. Threshold: 10 times.

Read the standard wording and why this check matters

“approximately 10 times (or more) greater than the maximum failure rate into or out of those fault states.”

Source: supplied ARP5107C, Section 7.2.4, page 37. This supports neglecting additional operating fault states when repair makes a second fault sufficiently unlikely. The section describes reasonable accuracy in approximately 5% terms under its modeling discussion; this is not an independently established error bound for your aircraft or for the zero-order equation.

A maximum calculated from the entered fault rows covers only the modeled occurrence and direct combined exit rates. Relevant transitions to additional operating fault states can be larger. The tool cannot infer additional transitions that have not been modeled.

Zero-order approximation check — Section 7.2.2.6, pages 36–37

Formula: repair rate ÷ maximum combined exit rate ≥ 10, where combined exit rate = electronic loss-of-thrust-control exit rate + HM + UC rate. Equivalently, maximum combined exit rate × repair interval ≤ 0.10.

Short-term (ST)

Calculated ratio: 4.00e-3 per hour ÷ 5.20e-5 per hour = 76.92 times.

The modeled individual combined exit rates meet this numerical approximation check.

Long-term (LT)

Calculated ratio: 1.00e-3 per hour ÷ 3.57e-5 per hour = 28.01 times.

The modeled individual combined exit rates meet this numerical approximation check.

Approximation factor errors for your entered ST and LT intervals

The ratio here is repair rate divided by the modeled combined exit rate (electronic loss-of-thrust-control exit rate + HM + UC). Multiplying that exit rate by its repair interval gives exactly one divided by this ratio. It is the same information, not an additional suitability test.

Zero-order factor error (%) = 100 / ratio.
First-order factor error (%) = 100 / (ratio × ratio). These percentages are the magnitudes of the errors in the replaced reciprocal factor, relative to its Full Markov value. Full Markov uses 1 / (1 + 1 / ratio); zero-order replaces this factor with 1; first-order replaces it with 1 - 1 / ratio.

Repair group	Entered repair interval (hours)	Limiting fault state	Repair / combined-exit ratio	Zero-order factor error (%)	First-order factor error (%)
ST	2.50e+2	CPU (system)	76.92 times	1.30%	0.02%
LT	1.00e+3	CCDL	28.01 times	3.57%	0.13%

Both ST and LT use these approximations. Individual mode shows the largest factor-error magnitude among the entered faults; averaged mode shows the factor error for each representative ST/LT state. The rows use the main-input repair intervals, not the sensitivity maximum.

These are factor errors, not the total loss-of-thrust-control-rate discrepancy: the final rate also combines occurrence rates, state weights, HM + UC and the normalization denominator.

Read the approximation guidance and preferred method

For your entered repair intervals: the known ratios also support zero-order approximation. You may use this simplification, subject to the model assumptions. First-order is preferred when closer agreement with Full Markov is desired.

Section 7.2.2.6 describes repair rates “approximately 10 times greater” than the combined loss-of-thrust-control and HM + UC exit rates when simplifying Equations 19 and 20 to the zero-order form. Page 37 notes that the 1,000-hour discussion is relative, not an absolute limit: the rate ratios govern applicability.

“The preferred approach is to use the first order approximation given by Equation 20c.”

Source: ARP5107C, Section 7.3, page 37. The preference concerns convenient approximation of full Markov, especially when varying repair times. Full Equation 20 is the individual-state reference; Equation 18 is the averaged ST/LT-state reference. Balanced time-weighted average includes its separate conservative fully operational contribution, so it is not always numerically equivalent.