Understand Dispatch Analysis
Purpose, scope, formulas and derivations for engine-control time-limited dispatch analysis.
Models, formulas and conditions of use
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.
Calculator input modes: individual fault inputs retain every fault and use Full Markov Equation 20 and its first-order Equation 20c. Averaged ST/LT inputs use two representative states, Full Markov Equation 18 and a first-order expansion of that equation. All four methods are available in each mode, subject to the first-order expansion domain. Averaged-state calculations do not reconstruct unprovided individual faults.
Understand the terms: HM, UC, X, individual faults, ST and LT
HM / HMC: mechanical and hydromechanical failures
HM refers to the mechanical/hydromechanical contribution. ARP5107C denotes its loss-of-thrust-control failure rate by λHMC. It is the sum of the rates of single mechanical or hydromechanical faults that lead to loss of thrust control. The input is the assessed rate for those consequences, not the rate of every mechanical fault regardless of its effect. See the definitions following Equation 1, p. 23.
UC: undetected / uncovered electrical and electronic failures
UC refers to the undetected/uncovered electrical/electronic fault contribution. λUC is the rate of those faults assumed to lead to loss of thrust control under the assessed architecture. It is separate from λHMC and from the additional electronic exits entered for ST and LT fault conditions. The combined baseline is:
X: the multiplier used to estimate the uncovered contribution
X is a dimensionless uncovered-fault multiplier. It is a fraction, not a failure rate: X = 0.02 means two percent. In Equation 2, it multiplies the summed electrical/electronic failure rates associated with loss-of-thrust-control-critical elements in both channels. That produces a failure rate λUC in events per hour.
The standard permits that sum to be approximated by the total ST plus total LT occurrence rates. This is the aggregate approximation implemented by the calculator:
For example, the Section 7.1.5 inputs are X = 0.02, λSTT = 3e-05 per hour, λLTT = 7e-05 per hour and λHMC = 4e-06 per hour. Therefore λUC = 0.02 × (3e-05 + 7e-05) = 2e-06 per hour, and λHM+UC = 4e-06 + 2e-06 = 6e-06 per hour.
Page 24 generally discusses X between 0 and 0.05, but does not assign the same uncovered percentage to every element. Its note explains that a detailed assessment normally sums uncovered rates element by element, such as the power supply, processor and sensors. The appropriate multiplier depends on architecture, coverage and channel operation. Thus 0.02 is an example value, not a universal coverage claim. X estimates UC; it does not scale the mechanical/hydromechanical rate.
In the generic Equation 20, λi means the occurrence rate into any one individual fault state. For a short-term fault it is λSTi; for a long-term fault it is λLTi. The index i identifies an individual fault, not an average. The source also writes these as λST(i) and λLT(i) in Equations 3, 8 and 9. Each ST sum and LT sum below runs over its own faults. λSTT and λLTT are totals across the applicable faults in both channels; the final T in STT/LTT denotes the total, not a repair time.
| Standard symbol | Meaning |
|---|---|
| λSTi; λLTi | Occurrence rates into an individual short-term or long-term fault state. |
| λSTT; λLTT | Totals of the respective individual occurrence rates, including both channels. |
| λSTi-LOTC; λLTi-LOTC | Electronic exit rates from the respective individual fault states into loss of thrust control; HM + UC excluded. |
| λST-LOTC-AVE; λLT-LOTC-AVE | Occurrence-weighted electronic exit rates for short-term and long-term operation, respectively. |
| λHMC; λUC; λHM+UC | Mechanical/hydromechanical rate; undetected/uncovered electrical/electronic rate; their combined rate. |
| X | Uncovered-fault multiplier used with the relevant electrical/electronic rate sum to estimate UC. |
| TST-REPAIR; TLT-REPAIR | Repair intervals. Their reciprocals are μST and μLT. |
| TFL; λFU-LOTC | Flight duration; fully operational electronic loss-of-thrust-control rate. |
| Fract(FU); Fract(ST); Fract(LT) | Fractions of total operating time spent fully operational, in short-term faults, and in long-term faults. |
| λi; μi; Ti | Generic individual-state notation in Equation 20. For ST use the ST fault rate, ST repair rate and ST interval; for LT use the corresponding LT quantities. |
ST and LT occurrence totals
The summation sign means add the displayed kind of term for every applicable individual fault. The subscript identifies the fault: ST1 is the first ST fault and LT1 is the first LT fault.
Short-term faults: total occurrence and weighted electronic exit
The occurrence rate tells you how often a particular ST fault is entered. Its electronic exit rate tells you how quickly an additional electronic failure can cause loss of thrust control while already in that fault state. Multiply these two rates for each ST fault, add the products, and divide by the total ST occurrence rate. Faults entered more often receive more weight. This is not an arithmetic average.
Long-term faults: total occurrence and weighted electronic exit
Repeat the same operation using LT faults only. The ST total is a sum of ST occurrence rates; the LT total is a sum of LT occurrence rates. The weighted ST and LT electronic exit rates describe operation after the first fault. They exclude HM + UC, which is accounted for separately.
Source-reading note: the printed Equation 8 numerator carries an LT label although its ST definition and explanatory prose identify the ST electronic exit. The ST formula above follows that definition and prose. This is an interpretation of the supplied text, not a publisher-issued correction.
The standard uses TST-REPAIR and TLT-REPAIR for the short-term and long-term repair intervals in its time-weighted-average equations; Equation 20c shortens them to TST and TLT. They refer to the same respective repair intervals. μST and μLT are repair rates, equal to their reciprocals.
For each individual fault, the combined loss-of-thrust-control exit contains the additional electronic exit plus the common HM + UC contribution. The standard uses these combined-rate symbols in Equation 20c.
Use of fault trees in dispatch analysis
ARP5107C Section 7.6 (p. 48) accepts fault trees for calculating failure rates while operating with faults, but stresses careful construction and simple reasonableness checks. A mathematically consistent tree can still give the wrong LOTC rate if it represents failed hardware without the FADEC’s functional response and reconfiguration.
The standard observes that Markov models tend to represent system functions, while fault trees tend to represent hardware. This is a practical warning, not a claim that fault trees are inherently inaccurate. Either method must represent the actual control behavior.
1. Start with the degraded configuration
Consider the standard’s dual-channel FADEC example. TLAA has already failed; TLAB remains healthy. Control may remain in Channel A using Channel B’s TLA through the cross-talk bus, or transfer to Channel B. The next-failure analysis starts from this operating configuration, not from full-up operation.
2. Identify the failures that actually cause LOTC
Conceptual fault-tree drawing and algebra for the Section 7.6 example, prepared here for explanation; not a reproduced figure or a complete system fault tree from ARP5107C. AND requires both input events; OR requires any one input event. A and B identify the channels.
For this branch, the three two-event combinations are TLAA AND CPUB, TLAA AND PowerB, and TLAA AND TLAB. The first TLA fault alone does not cause LOTC in this example.
3. Check the cross-talk and remaining-TLA cases
| Next failure, with TLA_A already failed | Functional consequence in this example | LOTC? |
|---|---|---|
| Cross-talk bus | If Channel A loses access to Channel B’s TLA, control can transfer to healthy Channel B, which still has its own TLA, CPU and power supply. | No, with the stated transfer behavior |
| CPU_B | The remaining channel’s required processing capability is lost. | Yes, under the example’s assumptions |
| Power_B | The remaining channel’s required power supply is lost. | Yes, under the example’s assumptions |
| TLA_B | TLA_A is already failed. Losing TLA_B now removes the last healthy TLA signal; both TLA signals are failed. | Yes, under the example’s assumptions |
4. Condition on the known first fault before obtaining a rate
The event TLAA is already true in this degraded configuration. Do not multiply the conditional exit rate by the occurrence rate of TLAA again. That occurrence rate is a separate input to the dispatch model.
Here i denotes the individual TLAA-failed state. The component subscripts are explanatory labels for this example. The sum assumes distinct, non-overlapping LOTC-causing transition events and the behavior described above. For example, an intrinsic CPU failure event must not also count the CPU’s loss of operation caused by the separately counted power-supply failure. Common-cause events and additional relevant paths require their own treatment. The standard supplies no numerical rates for this example; the displayed rate expression is explanatory algebra, not an additional numbered source equation.
5. Use the correct dispatch-analysis input
Enter this conditional electronic LOTC exit rate for the relevant ST or LT fault: λSTi-LOTC or λLTi-LOTC. Enter the occurrence of the first fault separately as λSTi or λLTi. Follow the assessed repair category; this example does not assign one automatically.
The tool’s electronic exit input excludes HM + UC. The tool adds λHM+UC once to form the combined exit rate. Cross-talk is not added to the electronic sum for the functional scenario above. A top-event probability or a cut-set probability is not a per-hour conditional transition rate.
An exact Markov calculation cannot correct an incorrect exit-rate input. Likewise, the standard’s statement that TWA is conservative does not justify adding hardware combinations that do not cause LOTC or omitting combinations that do.
6. Apply simple reasonableness checks
- State which fault is already present, which channel has control, and which healthy functions remain.
- For each proposed next failure, trace the actual response after available channel transfer and reconfiguration.
- Verify that each cut set causes LOTC in that configuration; remove the cross-talk cut set only when the stated healthy-channel transfer is available.
- Check both channel directions and other degraded configurations when constructing the complete system analysis; this single branch is not the whole FADEC fault tree.
- Keep event probabilities, first-fault occurrence rates and conditional LOTC exit rates distinct. Check dependence, common causes and duplicated events before summing rates.
Reference: supplied SAE ARP5107C (September 2018), Section 7.6, p. 48. The fault-tree diagram, Boolean expansion and calculator-input mapping are explanatory derivations of its qualitative TLA example. They assume the channel-transfer behavior stated there, not every possible FADEC architecture.
Full Markov: individual faults and averaged ST/LT states
Individual-state formula in standard notation
In the generic Equation 20, λi means the occurrence rate into any one individual fault state. For a short-term fault it is λSTi; for a long-term fault it is λLTi. The index i identifies an individual fault, not an average. The source also writes these as λST(i) and λLT(i) in Equations 3, 8 and 9. Each ST sum and LT sum below runs over its own faults. λSTT and λLTT are totals across the applicable faults in both channels; the final T in STT/LTT denotes the total, not a repair time.
Averaged ST and LT formula
Equation 18 is an averaged ST/LT state model. Equation 20 retains every individual state. Because each individual exit appears inside a denominator, a weighted average generally cannot reconstruct the individual Full Markov result. They coincide under a common repair interval when all individual electronic exits within each ST or LT category equal the corresponding representative exit. Otherwise the averaged model needs its own justification; the calculator does not infer the individual result from means alone.
First-order Markov: individual constants and averaged ST/LT states
Individual-state expansion
In the generic Equation 20, λi means the occurrence rate into any one individual fault state. For a short-term fault it is λSTi; for a long-term fault it is λLTi. The index i identifies an individual fault, not an average. The source also writes these as λST(i) and λLT(i) in Equations 3, 8 and 9. Each ST sum and LT sum below runs over its own faults. λSTT and λLTT are totals across the applicable faults in both channels; the final T in STT/LTT denotes the total, not a repair time.
ST and LT coefficient form
ST coefficients
LT coefficients
First-order expansion of the averaged ST/LT model
Equation 20c needs sums involving squared individual combined exit rates: a1 and c1. The ST and LT averages do not determine those sums. The first-order expansion of Equation 18 below therefore describes the averaged ST/LT model, not a reconstruction of arbitrary individual states. This distinction is about the individual rates and their weighted spread; no additional grouping symbols are needed.
Zero-order Markov: individual faults and averaged ST/LT rates
Individual ST and LT sums
Averaged ST and LT formula
For a common interval within ST and within LT, the occurrence-weighted averages reproduce these zero-order individual sums. This identity does not mean zero-order is always equal to Full Markov.
Balanced time-weighted average: individual faults and averaged ST/LT rates
Individual ST and LT formula
Averaged ST and LT formula
When the single-fault model and the approximations are appropriate
Single-fault model requirement: Section 7.2.4, p. 37, supports using a single-fault model when the repair rate for each ST and LT fault state is approximately ten times or more than the maximum relevant failure rate into or out of those states. Check ST and LT separately. Include further failures from a single-fault state into a two-fault operating state when establishing the maximum. This condition supports neglecting such additional operating states; it is not an unconditional claim that any single-fault model is correct.
At a chosen sensitivity maximum, a longer repair interval gives a lower repair-rate ratio. Check the entered intervals and the limiting sensitivity intervals separately. Changing flight duration does not change the implemented Markov rates or repair-rate ratios.
How the equations are derived
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.
Each model below begins with its state balance, approximation step or operating-time accounting, then proceeds to the individual ST/LT formulas and the averaged ST/LT formulas. The summary defines every rate and repair symbol. There are no generic grouping indices or newly named mathematical states.
Use of fault trees in dispatch analysis
ARP5107C Section 7.6 (p. 48) accepts fault trees for calculating failure rates while operating with faults, but stresses careful construction and simple reasonableness checks. A mathematically consistent tree can still give the wrong LOTC rate if it represents failed hardware without the FADEC’s functional response and reconfiguration.
The standard observes that Markov models tend to represent system functions, while fault trees tend to represent hardware. This is a practical warning, not a claim that fault trees are inherently inaccurate. Either method must represent the actual control behavior.
1. Start with the degraded configuration
Consider the standard’s dual-channel FADEC example. TLAA has already failed; TLAB remains healthy. Control may remain in Channel A using Channel B’s TLA through the cross-talk bus, or transfer to Channel B. The next-failure analysis starts from this operating configuration, not from full-up operation.
2. Identify the failures that actually cause LOTC
Conceptual fault-tree drawing and algebra for the Section 7.6 example, prepared here for explanation; not a reproduced figure or a complete system fault tree from ARP5107C. AND requires both input events; OR requires any one input event. A and B identify the channels.
For this branch, the three two-event combinations are TLAA AND CPUB, TLAA AND PowerB, and TLAA AND TLAB. The first TLA fault alone does not cause LOTC in this example.
3. Check the cross-talk and remaining-TLA cases
| Next failure, with TLA_A already failed | Functional consequence in this example | LOTC? |
|---|---|---|
| Cross-talk bus | If Channel A loses access to Channel B’s TLA, control can transfer to healthy Channel B, which still has its own TLA, CPU and power supply. | No, with the stated transfer behavior |
| CPU_B | The remaining channel’s required processing capability is lost. | Yes, under the example’s assumptions |
| Power_B | The remaining channel’s required power supply is lost. | Yes, under the example’s assumptions |
| TLA_B | TLA_A is already failed. Losing TLA_B now removes the last healthy TLA signal; both TLA signals are failed. | Yes, under the example’s assumptions |
4. Condition on the known first fault before obtaining a rate
The event TLAA is already true in this degraded configuration. Do not multiply the conditional exit rate by the occurrence rate of TLAA again. That occurrence rate is a separate input to the dispatch model.
Here i denotes the individual TLAA-failed state. The component subscripts are explanatory labels for this example. The sum assumes distinct, non-overlapping LOTC-causing transition events and the behavior described above. For example, an intrinsic CPU failure event must not also count the CPU’s loss of operation caused by the separately counted power-supply failure. Common-cause events and additional relevant paths require their own treatment. The standard supplies no numerical rates for this example; the displayed rate expression is explanatory algebra, not an additional numbered source equation.
5. Use the correct dispatch-analysis input
Enter this conditional electronic LOTC exit rate for the relevant ST or LT fault: λSTi-LOTC or λLTi-LOTC. Enter the occurrence of the first fault separately as λSTi or λLTi. Follow the assessed repair category; this example does not assign one automatically.
The tool’s electronic exit input excludes HM + UC. The tool adds λHM+UC once to form the combined exit rate. Cross-talk is not added to the electronic sum for the functional scenario above. A top-event probability or a cut-set probability is not a per-hour conditional transition rate.
An exact Markov calculation cannot correct an incorrect exit-rate input. Likewise, the standard’s statement that TWA is conservative does not justify adding hardware combinations that do not cause LOTC or omitting combinations that do.
6. Apply simple reasonableness checks
- State which fault is already present, which channel has control, and which healthy functions remain.
- For each proposed next failure, trace the actual response after available channel transfer and reconfiguration.
- Verify that each cut set causes LOTC in that configuration; remove the cross-talk cut set only when the stated healthy-channel transfer is available.
- Check both channel directions and other degraded configurations when constructing the complete system analysis; this single branch is not the whole FADEC fault tree.
- Keep event probabilities, first-fault occurrence rates and conditional LOTC exit rates distinct. Check dependence, common causes and duplicated events before summing rates.
Reference: supplied SAE ARP5107C (September 2018), Section 7.6, p. 48. The fault-tree diagram, Boolean expansion and calculator-input mapping are explanatory derivations of its qualitative TLA example. They assume the channel-transfer behavior stated there, not every possible FADEC architecture.
Derive Full Markov from probability balance: individual ST/LT and averaged ST/LT
Identify the states and follow their arrows
PFU is the probability of being fully operational. PSTi and PLTi are the probabilities of occupying individual short-term and long-term fault states. PLOTC is the probability of occupying the loss-of-thrust-control state. The arrows show the failure and repair transitions between them.
Individual ST and LT fault states Blue arrows enter an ST or LT fault state from fully operational operation. Green arrows repair the fault and return to fully operational operation. Red arrows lead to loss of thrust control. The dashed feedback arrow closes the mathematical model; it is not an operational repair instruction.
One ST fault and one LT fault are drawn for clarity. The ST and LT branches repeat for every entered individual fault; these two ellipses do not replace all faults with a single averaged state.
λSTi-LOTC+HM+UC =λSTi-LOTC + λHM+UCλLTi-LOTC+HM+UC =λLTi-LOTC + λHM+UCFor each individual fault, the combined loss-of-thrust-control exit contains the additional electronic exit plus the common HM + UC contribution. The standard uses these combined-rate symbols in Equation 20c.
Write probability flow as starting-state probability multiplied by transition rate
Probability flow = probability of occupying the starting state × transition rate. For example, λSTi PFU enters ST fault i because its arrow starts at the fully operational state. Both outgoing ST arrows start at the ST fault state, so their flows use PSTi.
Arrow Probability flow Meaning for the fault state FU to ST fault i λSTi PFU Incoming ST flow ST fault i to FU μST PSTi Outgoing ST repair flow ST fault i to LOTC λSTi-LOTC+HM+UC PSTi Outgoing ST loss-of-thrust-control flow FU to LT fault i λLTi PFU Incoming LT flow LT fault i to FU μLT PLTi Outgoing LT repair flow LT fault i to LOTC λLTi-LOTC+HM+UC PLTi Outgoing LT loss-of-thrust-control flow The same interpretation applies to LT. Add incoming flows and subtract outgoing flows to obtain the change of probability in each fault state:
dPSTidt =λSTi PFU − μST PSTi − λSTi-LOTC+HM+UC PSTidPLTidt =λLTi PFU − μLT PLTi − λLTi-LOTC+HM+UC PLTiThe derivative means change in state probability per unit time. Collecting the two outgoing terms gives the repair-plus-combined-exit expression used below.
Identify the standard ST and LT transition symbols
In the generic Equation 20, λi means the occurrence rate into any one individual fault state. For a short-term fault it is λSTi; for a long-term fault it is λLTi. The index i identifies an individual fault, not an average. The source also writes these as λST(i) and λLT(i) in Equations 3, 8 and 9. Each ST sum and LT sum below runs over its own faults. λSTT and λLTT are totals across the applicable faults in both channels; the final T in STT/LTT denotes the total, not a repair time.
λSTi-LOTC+HM+UC =λSTi-LOTC + λHM+UCλLTi-LOTC+HM+UC =λLTi-LOTC + λHM+UCFor each individual fault, the combined loss-of-thrust-control exit contains the additional electronic exit plus the common HM + UC contribution. The standard uses these combined-rate symbols in Equation 20c.
The standard uses TST-REPAIR and TLT-REPAIR for the short-term and long-term repair intervals in its time-weighted-average equations; Equation 20c shortens them to TST and TLT. They refer to the same respective repair intervals. μST and μLT are repair rates, equal to their reciprocals.
μST =1TST-REPAIRμLT =1TLT-REPAIRUse the fully operational state, individual ST states, individual LT states and the loss-of-thrust-control state. A closed-loop feedback arrow returns from loss of thrust control to fully operational operation; this is a mathematical closure, not an operational repair instruction. Sections 7.2.2.2-7.2.2.4, pp. 33-35, derive the closed-loop result.
Write probability-flow equations before taking the steady solution
dPSTidt =λSTi PFU − (μST + λSTi-LOTC+HM+UC) PSTidPLTidt =λLTi PFU − (μLT + λLTi-LOTC+HM+UC) PLTidPFUdt =− (λHM+UC + ∑i λSTi + ∑i λLTi) PFU + ∑i μST PSTi + ∑i μLT PLTi + μFB PLOTCdPLOTCdt =λHM+UC PFU + ∑i λSTi-LOTC+HM+UC PSTi + ∑i λLTi-LOTC+HM+UC PLTi − μFB PLOTCEach derivative is incoming probability flow minus outgoing probability flow. The feedback rate μFB closes the mathematical model. Add all four kinds of equations: every transition cancels, giving conservation of total probability. For the steady closed-loop solution, set the derivatives to zero; the individual ST and LT equations then give the balances in the next step.
Take the steady-state limit, then rearrange each balance
Equation 20 is the steady closed-loop result. At steady state the probabilities of occupying the states no longer change, so dPdt = 0. This does not mean failures or repairs have stopped: their probability flows continue and balance. Applying that condition to the ST equation gives zero equals inflow minus outflow. Move the outflow to the other side and divide by its total exit rate:
(μST + λSTi-LOTC+HM+UC) PSTi =λSTi PFUThe ST state-probability expression below follows from this rearrangement; the LT expression follows by exactly the same steps.
0 =λSTi PFU − (μST + λSTi-LOTC+HM+UC) PSTiPSTi =λSTi PFUμST + λSTi-LOTC+HM+UC0 =λLTi PFU − (μLT + λLTi-LOTC+HM+UC) PLTiPLTi =λLTi PFUμLT + λLTi-LOTC+HM+UCOccurrence supplies probability into each fault state. Repair and the combined loss-of-thrust-control exit remove it. At steady state those flows balance.
Derive the loss-of-thrust-control rate from the hazard-rate definition
For an absorbing failure state with no return transition, the probability of not having failed is 1 − PLOTC. The probability entering the failure state per unit time is dPLOTCdt. The conditional rate of failure is therefore:
λLOTC =dPLOTCdt1 − PLOTCAbsorbing-state hazard definitionEquation 13, p. 31, uses the more general definition for a repairable or feedback-closed LOTC state: incoming probability flow divided by the probability of not occupying that state. Its note explains that the incoming flow is not the same as the net derivative when probability also leaves LOTC.
λLOTC =probability flow entering the LOTC state1 − PLOTCARP5107C Equation 13dPLOTCdt =incoming LOTC probability flow − μFB PLOTCλLOTC =dPLOTCdt + μFB PLOTC1 − PLOTCSame definition expressed using the net derivative and feedback outflowThus the numerator in the next equation is obtained by adding all incoming arrows to LOTC: the direct HM + UC arrow from FU, the combined exits from individual ST faults, and the combined exits from individual LT faults. At steady state dPLOTCdt = 0 because incoming and feedback flows balance; that does not make the incoming failure rate zero.
λLOTC =λHM+UC PFU + ∑i λSTi-LOTC+HM+UC PSTi + ∑i λLTi-LOTC+HM+UC PLTi1 − PLOTCEquation 13 applied to individual ST/LT states1 − PLOTC =PFU + ∑i PSTi + ∑i PLTiThe numerator is probability flow into loss of thrust control. The denominator is the probability of not being in that state. For the repairable closed-loop failure state, do not replace the incoming flow by its net probability slope.
Substitute both sets of state balances
Replace each individual fault-state probability by its expression above. Every numerator and denominator term contains the common factor PFU. Divide both by that factor, producing:
λLOTC =λHM+UC + ∑i λSTi λSTi-LOTC+HM+UCμST + λSTi-LOTC+HM+UC + ∑i λLTi λLTi-LOTC+HM+UCμLT + λLTi-LOTC+HM+UC1 + ∑i λSTiμST + λSTi-LOTC+HM+UC + ∑i λLTiμLT + λLTi-LOTC+HM+UCEquation 20 expanded into ST and LT sumsλLOTC =λHM+UC + ∑i λi λi-LOTC+HM+UCμi + λi-LOTC+HM+UC1 + ∑i λiμi + λi-LOTC+HM+UCEquation 20, p. 35The generic sum in Equation 20 is exactly the combined ST and LT sums. No feedback rate remains in this expression.
Derive the averaged ST/LT model separately
Averaged ST and LT states Blue arrows enter an ST or LT fault state from fully operational operation. Green arrows repair the fault and return to fully operational operation. Red arrows lead to loss of thrust control. The dashed feedback arrow closes the mathematical model; it is not an operational repair instruction.
Use the same inflow-minus-outflow rule, now with total occurrence into the representative ST state and total occurrence into the representative LT state:
dPST / dt =λSTT PFU − (μST + λST-LOTC-AVE + λHM+UC) PSTdPLT / dt =λLTT PFU − (μLT + λLT-LOTC-AVE + λHM+UC) PLTSet these derivatives to zero for the steady solution. Move each outflow term to the other side, then divide by its repair-plus-combined-exit rate:
PST =λSTT PFUμST + λST-LOTC-AVE + λHM+UCPLT =λLTT PFUμLT + λLT-LOTC-AVE + λHM+UCUse the same probability-flow definition with one representative ST state and one representative LT state. Substitute these two balances and divide numerator and denominator by PFU:
λLOTC =λHM+UC + λSTT (λST-LOTC-AVE + λHM+UC)μST + λST-LOTC-AVE + λHM+UC + λLTT (λLT-LOTC-AVE + λHM+UC)μLT + λLT-LOTC-AVE + λHM+UC1 + λSTTμST + λST-LOTC-AVE + λHM+UC + λLTTμLT + λLT-LOTC-AVE + λHM+UCEquation 18, p. 34: ST and LT averaged-state modelEquation 18 is an averaged ST/LT state model. Equation 20 retains every individual state. Because each individual exit appears inside a denominator, a weighted average generally cannot reconstruct the individual Full Markov result. They coincide under a common repair interval when all individual electronic exits within each ST or LT category equal the corresponding representative exit. Otherwise the averaged model needs its own justification; the calculator does not infer the individual result from means alone.
Single-fault model requirement: Section 7.2.4, p. 37, supports using a single-fault model when the repair rate for each ST and LT fault state is approximately ten times or more than the maximum relevant failure rate into or out of those states. Check ST and LT separately. Include further failures from a single-fault state into a two-fault operating state when establishing the maximum. This condition supports neglecting such additional operating states; it is not an unconditional claim that any single-fault model is correct.
Derive first-order Markov: reciprocal expansion and ST/LT constants
Start from Equation 20, retaining the individual fault terms
Equation 20 comes from the continuous-time Markov inflow-minus-outflow equations and the steady closed-loop balance. Read that derivation from state probabilities and transition flows before applying the approximation below.
λLOTC =λHM+UC + ∑i λi λi-LOTC+HM+UCμi + λi-LOTC+HM+UC1 + ∑i λiμi + λi-LOTC+HM+UCEquation 20, p. 35In the generic Equation 20, λi means the occurrence rate into any one individual fault state. For a short-term fault it is λSTi; for a long-term fault it is λLTi. The index i identifies an individual fault, not an average. The source also writes these as λST(i) and λLT(i) in Equations 3, 8 and 9. Each ST sum and LT sum below runs over its own faults. λSTT and λLTT are totals across the applicable faults in both channels; the final T in STT/LTT denotes the total, not a repair time.
λiμi + λi-LOTC+HM+UC =λi Ti1 + λi-LOTC+HM+UC TiMultiply numerator and denominator of each state term by its repair interval. The printed Equation 20a denominator omits this repair-interval factor in its numerator; the consistent form above follows directly from Equation 20. This is an algebraic interpretation, not a publisher-issued correction.
Expand the reciprocal, showing the remainder
11 + λi-LOTC+HM+UC Ti =1 − λi-LOTC+HM+UC Ti + (λi-LOTC+HM+UC Ti)21 + λi-LOTC+HM+UC TiFirst-order retains the constant and linear terms and drops the displayed remainder. Apply this replacement in both the numerator state sum and the normalization denominator. Small combined-exit-rate times repair interval makes the omitted contribution small.
Substitute the expansion in both sums
λLOTC ≈λHM+UC + ∑i [λi λi-LOTC+HM+UC Ti − λi λi-LOTC+HM+UC2 Ti2]1 + ∑i [λi Ti − λi λi-LOTC+HM+UC Ti2]Equation 20b, p. 36The numerator correction contains the square of the individual combined exit and the square of its interval. The denominator correction contains one combined exit factor and the square of its interval.
Separate ST and LT and calculate the standard constants
ST coefficients
a0 =∑i λSTi λSTi-LOTC+HM+UCa1 =∑i λSTi λSTi-LOTC+HM+UC2b0 =∑i λSTib1 =a0LT coefficients
c0 =∑i λLTi λLTi-LOTC+HM+UCc1 =∑i λLTi λLTi-LOTC+HM+UC2d0 =∑i λLTid1 =c0All ST states use the ST interval and all LT states use the LT interval. Move those interval factors outside the respective sums, leaving the constants above:
λLOTC ≈λHM+UC + a0 TST − a1 TST2 + c0 TLT − c1 TLT21 + b0 TST − b1 TST2 + d0 TLT − d1 TLT2Equation 20c, p. 36This step only rearranges the first-order expression. The constants contain failure-rate data only: calculate them once when that data is fixed, then evaluate new ST/LT intervals without repeating the individual sums. This is the computational benefit discussed in Section 7.3.
Derive the averaged ST/LT version without pretending it is the same individual model
λLOTC =λHM+UC + λSTT (λST-LOTC-AVE + λHM+UC)μST + λST-LOTC-AVE + λHM+UC + λLTT (λLT-LOTC-AVE + λHM+UC)μLT + λLT-LOTC-AVE + λHM+UC1 + λSTTμST + λST-LOTC-AVE + λHM+UC + λLTTμLT + λLT-LOTC-AVE + λHM+UCEquation 18, p. 34: ST and LT averaged-state modelFor ST, replace the reciprocal of [1 + (λST-LOTC-AVE + λHM+UC) TST] by 1 − (λST-LOTC-AVE + λHM+UC) TST. For LT, make the corresponding replacement using λLT-LOTC-AVE and TLT. Substitute into both numerator and denominator:
λLOTC ≈λHM+UC + λSTT (λST-LOTC-AVE + λHM+UC) TST − λSTT (λST-LOTC-AVE + λHM+UC)2 TST2 + λLTT (λLT-LOTC-AVE + λHM+UC) TLT − λLTT (λLT-LOTC-AVE + λHM+UC)2 TLT21 + λSTT TST − λSTT (λST-LOTC-AVE + λHM+UC) TST2 + λLTT TLT − λLTT (λLT-LOTC-AVE + λHM+UC) TLT2First-order expansion of Equation 18; derived, not a new numbered source equationEquation 20c needs sums involving squared individual combined exit rates: a1 and c1. The ST and LT averages do not determine those sums. The first-order expansion of Equation 18 below therefore describes the averaged ST/LT model, not a reconstruction of arbitrary individual states. This distinction is about the individual rates and their weighted spread; no additional grouping symbols are needed.
Single-fault model requirement: Section 7.2.4, p. 37, supports using a single-fault model when the repair rate for each ST and LT fault state is approximately ten times or more than the maximum relevant failure rate into or out of those states. Check ST and LT separately. Include further failures from a single-fault state into a two-fault operating state when establishing the maximum. This condition supports neglecting such additional operating states; it is not an unconditional claim that any single-fault model is correct.
Derive zero-order: individual approximation, then averaged ST/LT form
A. Derivation using individual fault rates
Start from Full Markov with individual fault rates
In the generic Equation 20, λi means the occurrence rate into any one individual fault state. For a short-term fault it is λSTi; for a long-term fault it is λLTi. The index i identifies an individual fault, not an average. The source also writes these as λST(i) and λLT(i) in Equations 3, 8 and 9. Each ST sum and LT sum below runs over its own faults. λSTT and λLTT are totals across the applicable faults in both channels; the final T in STT/LTT denotes the total, not a repair time.
λSTi-LOTC+HM+UC =λSTi-LOTC + λHM+UCλLTi-LOTC+HM+UC =λLTi-LOTC + λHM+UCFor each individual fault, the combined loss-of-thrust-control exit contains the additional electronic exit plus the common HM + UC contribution. The standard uses these combined-rate symbols in Equation 20c.
λLOTC =λHM+UC + ∑i λi λi-LOTC+HM+UCμi + λi-LOTC+HM+UC1 + ∑i λiμi + λi-LOTC+HM+UCEquation 20, p. 35At this stage every occurrence and combined exit belongs to an individual fault state. No ST or LT average has been substituted.
Make the zero-order approximation here: neglect the combined exit beside the repair rate
The individual state denominator is repair rate plus combined exit rate. Under the condition above, the repair rate dominates. The approximation is therefore:
μST + λSTi-LOTC+HM+UC ≈μSTμLT + λLTi-LOTC+HM+UC ≈μLTApply that replacement to both occurrences of the state denominator in Equation 20. For the generic individual state, the two replacements are:
λi λi-LOTC+HM+UCμi + λi-LOTC+HM+UC ≈λi λi-LOTC+HM+UCμiλiμi + λi-LOTC+HM+UC ≈λiμiThis denominator replacement is the zero-order approximation. The subsequent substitutions and separation into ST and LT are algebraic rearrangements of that approximated equation.
Substitute those two replacements into the complete rate equation
λLOTC ≈λHM+UC + ∑i λi λi-LOTC+HM+UCμi1 + ∑i λiμiEquation 20d, p. 36The normalization denominator is still present. Only the combined-exit part inside each state denominator has been neglected; the occurrence contributions to the overall denominator remain.
Replace reciprocal repair rates by repair intervals
1μi =TiλLOTC ≈λHM+UC + ∑i λi λi-LOTC+HM+UC Ti1 + ∑i λi TiEquation 20e, p. 37This is Equation 20e. It is still an individual-fault equation; the index i runs over the ST and LT faults.
Separate the individual ST and LT sums
All ST faults use the ST repair interval, and all LT faults use the LT repair interval. Take those common intervals outside their respective sums:
λLOTC ≈λHM+UC + TST ∑i λSTi λSTi-LOTC+HM+UC + TLT ∑i λLTi λLTi-LOTC+HM+UC1 + TST ∑i λSTi + TLT ∑i λLTiEquation 20f, p. 37Individual-rate derivation complete. Equation 20f above can be evaluated directly from the individual occurrence and electronic exit inputs. Averaging is not a step required to obtain it.
Compare the individual zero-order result with balanced time-weighted average
Use the same individual occurrence rates, electronic exits, HM + UC baseline, and common ST/LT repair intervals in both methods. The balanced time-weighted-average model is the Equation 6/11 form implemented here.
Zero-order Markov
λLOTC ≈ λHM+UC +TST-REPAIR ∑i λSTi λSTi-LOTC + TLT-REPAIR ∑i λLTi λLTi-LOTC1 + TST-REPAIR ∑i λSTi + TLT-REPAIR ∑i λLTiIndividual zero-order result with the common baseline separatedBalanced time-weighted average
λLOTC-AVE = λHM+UC +λFU-LOTC + TST-REPAIR ∑i λSTi λSTi-LOTC + TLT-REPAIR ∑i λLTi λLTi-LOTC1 + TST-REPAIR ∑i λSTi + TLT-REPAIR ∑i λLTiEquation 11 with individual ST and LT sums substitutedSubtract the zero-order expression from the balanced time-weighted-average expression. The HM + UC baseline and both degraded-state electronic sums cancel:
λLOTC-AVE (balanced time-weighted average) − λLOTC (zero-order Markov) =λFU-LOTC1 + TST-REPAIR ∑i λSTi + TLT-REPAIR ∑i λLTi=Fract(FU) λFU-LOTCThe difference is the fully operational electronic contribution multiplied by the fully operational time fraction. It is not the unweighted fully operational rate. For the same inputs, the two expressions coincide if that electronic contribution is zero. This identity concerns the balanced time-weighted average and zero-order Markov, not Full Markov. The optional averaged ST/LT rewriting is derived separately in section B.
B. Separate rewriting using averaged ST and LT rates
Define the ST and LT totals and weighted electronic exits
This section starts from the individual zero-order result already derived. Its purpose is to rewrite that result when ST and LT totals and occurrence-weighted electronic averages are the available inputs. It does not introduce another zero-order approximation.
λSTT =∑i λSTiEquation 3a, p. 24λLTT =∑i λLTiEquation 3b, p. 24λSTT =λST1 + λST2 + …λLTT =λLT1 + λLT2 + …The summation sign means add the displayed kind of term for every applicable individual fault. The subscript identifies the fault: ST1 is the first ST fault and LT1 is the first LT fault.
Short-term faults: total occurrence and weighted electronic exit
λST-LOTC-AVE =∑i λSTi λSTi-LOTC∑i λSTiEquation 8 definition and prose, p. 26The occurrence rate tells you how often a particular ST fault is entered. Its electronic exit rate tells you how quickly an additional electronic failure can cause loss of thrust control while already in that fault state. Multiply these two rates for each ST fault, add the products, and divide by the total ST occurrence rate. Faults entered more often receive more weight. This is not an arithmetic average.
Long-term faults: total occurrence and weighted electronic exit
λLT-LOTC-AVE =∑i λLTi λLTi-LOTC∑i λLTiEquation 9, p. 26Repeat the same operation using LT faults only. The ST total is a sum of ST occurrence rates; the LT total is a sum of LT occurrence rates. The weighted ST and LT electronic exit rates describe operation after the first fault. They exclude HM + UC, which is accounted for separately.
Source-reading note: the printed Equation 8 numerator carries an LT label although its ST definition and explanatory prose identify the ST electronic exit. The ST formula above follows that definition and prose. This is an interpretation of the supplied text, not a publisher-issued correction.
Multiply each weighted-average definition by its occurrence total
∑i λSTi λSTi-LOTC =λSTT λST-LOTC-AVE∑i λLTi λLTi-LOTC =λLTT λLT-LOTC-AVENow use the standard combined exit definitions. Expand the ST sum and the LT sum separately:
∑i λSTi λSTi-LOTC+HM+UC =∑i λSTi (λSTi-LOTC + λHM+UC)=λSTT λST-LOTC-AVE + λHM+UC λSTT = λSTT (λST-LOTC-AVE + λHM+UC)∑i λLTi λLTi-LOTC+HM+UC =∑i λLTi (λLTi-LOTC + λHM+UC)=λLTT λLT-LOTC-AVE + λHM+UC λLTT = λLTT (λLT-LOTC-AVE + λHM+UC)Substitute those identities into the individual Equation 20f
λLOTC ≈λHM+UC + λSTT TST-REPAIR (λST-LOTC-AVE + λHM+UC) + λLTT TLT-REPAIR (λLT-LOTC-AVE + λHM+UC)1 + λSTT TST-REPAIR + λLTT TLT-REPAIREquation 20f rewritten with ST and LT occurrence-weighted averagesThis averaged-input expression is the same zero-order result when every ST fault shares the ST interval and every LT fault shares the LT interval. The weighted-average identities preserve the individual sums. They do not prove that an averaged-state Full Markov model equals the individual-state Full Markov model.
Separate the common HM + UC contribution
Expand the numerator. Its HM + UC terms are:
λHM+UC + λHM+UC λSTT TST-REPAIR + λHM+UC λLTT TLT-REPAIR =λHM+UC [1 + λSTT TST-REPAIR + λLTT TLT-REPAIR]The bracket is exactly the overall denominator. Dividing cancels that bracket from the HM + UC contribution, leaving:
λLOTC ≈ λHM+UC +λSTT TST-REPAIR λST-LOTC-AVE + λLTT TLT-REPAIR λLT-LOTC-AVE1 + λSTT TST-REPAIR + λLTT TLT-REPAIRSame zero-order expression with HM + UC separatedThis final form keeps HM + UC once outside the electronic contribution. The electronic-only ST/LT exit fields must therefore exclude that same baseline.
The same comparison using averaged ST and LT inputs
For the averaged-input form, the electronic ST and LT numerators are the same in the zero-order result above and Equation 11. Equation 11 additionally contains the fully operational electronic term:
λLOTC-AVE = λHM+UC +λFU-LOTC + (λSTT TST-REPAIR) λST-LOTC-AVE + (λLTT TLT-REPAIR) λLT-LOTC-AVE1 + λSTT TST-REPAIR + λLTT TLT-REPAIREquation 11, p. 26λLOTC-AVE (balanced time-weighted average) − λLOTC (zero-order Markov) =λFU-LOTC1 + λSTT TST-REPAIR + λLTT TLT-REPAIR=Fract(FU) λFU-LOTCBoth methods use the same balanced denominator. In this calculator, the Equation 7 upper bound is substituted for the fully operational electronic rate when forming the time-weighted average; the difference shown therefore uses that selected conservative estimate. It is not an extra term to add to the Markov equations.
Single-fault model requirement: Section 7.2.4, p. 37, supports using a single-fault model when the repair rate for each ST and LT fault state is approximately ten times or more than the maximum relevant failure rate into or out of those states. Check ST and LT separately. Include further failures from a single-fault state into a two-fault operating state when establishing the maximum. This condition supports neglecting such additional operating states; it is not an unconditional claim that any single-fault model is correct.
Derive balanced time-weighted average from occurrence counts and operating time
Start with fully operational exposure, without choosing an observation duration
Consider the total operating time. A fraction Fract(FU) of that time is spent fully operational. The expected number of first ST or LT faults is based on that fully operational exposure. No particular observation duration or choice of time unit is needed for this algebra.
Fully operational exposure =Fract(FU) × total operating timeDerive the short-term fraction from occurrence and repair interval
Expected number of ST fault entries =λSTT × Fract(FU) × total operating timeExpected time operating with ST faults =λSTT × Fract(FU) × total operating time × TST-REPAIRDivide the time operating with ST faults by the total operating time. The same total time appears in numerator and denominator and cancels:
Fract(ST) =λSTT × Fract(FU) × total operating time × TST-REPAIRtotal operating timeFract(ST) =Fract(FU) λSTT TST-REPAIREquation 4a, p. 24Thus the ST time fraction is the fully operational time fraction multiplied by the ST occurrence rate and its repair interval. Rate multiplied by interval is dimensionless when their time units are consistent.
Derive the long-term fraction in the same way
Expected number of LT fault entries =λLTT × Fract(FU) × total operating timeExpected time operating with LT faults =λLTT × Fract(FU) × total operating time × TLT-REPAIRFract(LT) =λLTT × Fract(FU) × total operating time × TLT-REPAIRtotal operating timeFract(LT) =Fract(FU) λLTT TLT-REPAIREquation 4b, p. 25Apply conservation of the whole operating record
Fract(FU) + Fract(ST) + Fract(LT) =1Equation 5Substitute the fractions derived from the occurrence counting, rather than assuming that fully operational operation fills the whole record:
Fract(FU) + Fract(FU) λSTT TST-REPAIR + Fract(FU) λLTT TLT-REPAIR =1Fract(FU) [1 + λSTT TST-REPAIR + λLTT TLT-REPAIR] =1Now divide by the bracket to solve for Fract(FU). Substitute that result back into Equations 4a and 4b to obtain Fract(ST) and Fract(LT):
Fract(FU) =11 + λSTT TST-REPAIR + λLTT TLT-REPAIREquation 6aFract(ST) =λSTT TST-REPAIR1 + λSTT TST-REPAIR + λLTT TLT-REPAIREquation 6bFract(LT) =λLTT TLT-REPAIR1 + λSTT TST-REPAIR + λLTT TLT-REPAIREquation 6cAdding these three fractions returns one because their numerators add to the common denominator. The fully operational exposure adjustment is the reason these are called the balanced fractions.
Derive the individual ST and LT time contributions before averaging
The same counting works for one particular fault. Replace the ST total occurrence by its individual occurrence, and do the corresponding operation for an LT fault:
Fraction of the record in ST fault i =Fract(FU) λSTi TST-REPAIRFraction of the record in LT fault i =Fract(FU) λLTi TLT-REPAIRMultiply each individual fraction by its own electronic loss-of-thrust-control exit rate and sum the ST contributions and LT contributions separately. Adding the fully operational electronic contribution and the HM + UC baseline gives:
λLOTC-AVE = λHM+UC +λFU-LOTC + TST-REPAIR ∑i λSTi λSTi-LOTC + TLT-REPAIR ∑i λLTi λLTi-LOTC1 + TST-REPAIR ∑i λSTi + TLT-REPAIR ∑i λLTiEquation 11 with individual ST and LT sums substitutedDerive the ST and LT electronic averages from those weighted sums
Short-term faults: total occurrence and weighted electronic exit
λST-LOTC-AVE =∑i λSTi λSTi-LOTC∑i λSTiEquation 8 definition and prose, p. 26The occurrence rate tells you how often a particular ST fault is entered. Its electronic exit rate tells you how quickly an additional electronic failure can cause loss of thrust control while already in that fault state. Multiply these two rates for each ST fault, add the products, and divide by the total ST occurrence rate. Faults entered more often receive more weight. This is not an arithmetic average.
Long-term faults: total occurrence and weighted electronic exit
λLT-LOTC-AVE =∑i λLTi λLTi-LOTC∑i λLTiEquation 9, p. 26Repeat the same operation using LT faults only. The ST total is a sum of ST occurrence rates; the LT total is a sum of LT occurrence rates. The weighted ST and LT electronic exit rates describe operation after the first fault. They exclude HM + UC, which is accounted for separately.
Source-reading note: the printed Equation 8 numerator carries an LT label although its ST definition and explanatory prose identify the ST electronic exit. The ST formula above follows that definition and prose. This is an interpretation of the supplied text, not a publisher-issued correction.
Each ST electronic numerator sum equals λSTT λST-LOTC-AVE. Each LT electronic numerator sum equals λLTT λLT-LOTC-AVE. Substitute those two identities into the individual expression:
λLOTC-AVE = λHM+UC +λFU-LOTC + (λSTT TST-REPAIR) λST-LOTC-AVE + (λLTT TLT-REPAIR) λLT-LOTC-AVE1 + λSTT TST-REPAIR + λLTT TLT-REPAIREquation 11, p. 26This is Equation 11. It shows why occurrence weighting is needed and why the same ST interval and the same LT interval are assumed within their respective categories.
Explain and include the fully operational electronic term
Section 7.1.1 considers two independent electronic failures during one flight, one affecting each redundant channel. For small channel-rate-times-flight-duration products, Equation 7 gives a conservative upper bound, using half the both-channel total as the channel-rate estimate:
λFU-LOTC ≤[λSTT + λLTT2]2 TFLEquation 7, p. 25: conservative upper boundThe bound assumes channel faults conservatively count as channel loss. It is not an equality valid for every architecture. The calculator uses the right-hand side as its conservative fully operational electronic estimate. This term is multiplied by Fract(FU), so it enters the same balanced fraction as the other electronic contributions.
Explain where HM + UC enters, and assemble the result
HM / HMC: mechanical and hydromechanical failures
HM refers to the mechanical/hydromechanical contribution. ARP5107C denotes its loss-of-thrust-control failure rate by λHMC. It is the sum of the rates of single mechanical or hydromechanical faults that lead to loss of thrust control. The input is the assessed rate for those consequences, not the rate of every mechanical fault regardless of its effect. See the definitions following Equation 1, p. 23.
UC: undetected / uncovered electrical and electronic failures
UC refers to the undetected/uncovered electrical/electronic fault contribution. λUC is the rate of those faults assumed to lead to loss of thrust control under the assessed architecture. It is separate from λHMC and from the additional electronic exits entered for ST and LT fault conditions. The combined baseline is:
λHM+UC =λHMC + λUCDefinitions following Equation 1, p. 23X: the multiplier used to estimate the uncovered contribution
X is a dimensionless uncovered-fault multiplier. It is a fraction, not a failure rate: X = 0.02 means two percent. In Equation 2, it multiplies the summed electrical/electronic failure rates associated with loss-of-thrust-control-critical elements in both channels. That produces a failure rate λUC in events per hour.
λUC =X × [sum of both-channel electrical/electronic failure rates associated with loss-of-thrust-control-critical elements]Equation 2, p. 24The standard permits that sum to be approximated by the total ST plus total LT occurrence rates. This is the aggregate approximation implemented by the calculator:
λUC ≈X (λSTT + λLTT)λHM+UC ≈λHMC + X (λSTT + λLTT)For example, the Section 7.1.5 inputs are X = 0.02, λSTT = 3e-05 per hour, λLTT = 7e-05 per hour and λHMC = 4e-06 per hour. Therefore λUC = 0.02 × (3e-05 + 7e-05) = 2e-06 per hour, and λHM+UC = 4e-06 + 2e-06 = 6e-06 per hour.
Page 24 generally discusses X between 0 and 0.05, but does not assign the same uncovered percentage to every element. Its note explains that a detailed assessment normally sums uncovered rates element by element, such as the power supply, processor and sensors. The appropriate multiplier depends on architecture, coverage and channel operation. Thus 0.02 is an example value, not a universal coverage claim. X estimates UC; it does not scale the mechanical/hydromechanical rate.
The mechanical/hydromechanical and uncovered contributions are common to the operating conditions. Weighting a common baseline by fractions that sum to one leaves that baseline once. Hence it is outside the electronic fraction in Equation 11:
λLOTC-AVE = λHM+UC +λFU-LOTC + (λSTT TST-REPAIR) λST-LOTC-AVE + (λLTT TLT-REPAIR) λLT-LOTC-AVE1 + λSTT TST-REPAIR + λLTT TLT-REPAIREquation 11, p. 26Do not add the same baseline again inside the electronic-only ST/LT exits. Do not add the separate fully operational bound to the Markov equations: their model expressions are different.
Single-fault model requirement: Section 7.2.4, p. 37, supports using a single-fault model when the repair rate for each ST and LT fault state is approximately ten times or more than the maximum relevant failure rate into or out of those states. Check ST and LT separately. Include further failures from a single-fault state into a two-fault operating state when establishing the maximum. This condition supports neglecting such additional operating states; it is not an unconditional claim that any single-fault model is correct.