Worked examples — Fault tree analysis (FTA) — top-down, AND - OR gates

3.6.30 · D3 · Physics › Spacecraft Structures & Systems Engineering › Fault tree analysis (FTA) — top-down, AND - OR gates

Kuch bhi shuru karne se pehle, ek reminder ki symbols ka MATLAB kya hai, taaki hum unhe kabhi bina samjhe use na karein.

Number $(1-P_i)$ baar baar aata rehta hai — yeh event $i$ ki survival probability hai (woh chance ki yeh NAHI hoti). Yeh phrase apne dimaag mein rakho; yeh OR gate ka hero hai.

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Scenario matrix

Har FTA calculation jo tumhe kabhi milegi woh in cells mein se ek hai. Neeche ke examples us cell ke saath label kiye gaye hain jo unhe hit karti hai, aur milke yeh poora grid cover karte hain.

Cell	Case class	Kya mushkil banata hai
A	Single OR gate, chhote $P_i$	approximation vs exact
B	Single AND gate, chhote $P_i$	product kitna jaldi shrink karta hai
C	Degenerate input $P_i=0$	tree mein ek perfect part
D	Degenerate input $P_i=1$	tree mein ek guaranteed failure
E	Mixed tree (AND under OR)	order of operations, dominant term
F	Limiting behaviour: bahut saare identical OR inputs	$P_{\text{OR}}\to 1$
G	Common cause failure (dependent)	independence assumption toot jaati hai
H	Real-world word problem	English ko gates mein translate karo
I	Exam twist: "kaun sa improvement zyada help karta hai?"	sensitivity, dominant path
J	Badi probabilities ($P_i$ near 1)	sum-approximation blow up ho jaata hai

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Cell A — Single OR gate, chhoti probabilities

Forecast: Teen chhoti failures mein se koi bhi ek solar ko khatam kar sakti hai. Toh answer kisi bhi single input se bada hona chahiye lekin phir bhi chhota — guess karo "roughly sum, about $0.016$."

1. Har survival probability likho. Kyun yeh step? OR formula "kya yeh survive kiya?" se bana hai, toh pehle har failure ko uske opposite mein convert karo. $1-0.001=0.999,1-0.01=0.99,1-0.005=0.995$

2. Survivals multiply karo — woh chance ki TEENO healthy rahein. Kyun? Independent parts ka sab milke survive karna ek joint event hai, aur joint = product. $0.999\times 0.99\times 0.995=0.98405$

3. 1 se subtract karo. Kyun? "Kam se kam ek fail hua" "koi fail nahi hua" ka complement hai. $P_{\text{Solar}}=1-0.98405=0.01595\approx 0.016$

Verify: Lazy sum $0.001+0.01+0.005=0.016$ se compare karo. Exact answer $0.01595$ sum se thoda neeche hai — woh chhota gap woh double-counting hai jo sum ignore karta hai. Kyunki $P_i$ chhote hain, sum ≈ exact. Sanity: answer $0$ aur $1$ ke beech hai. ✓

Recall Exact OR answer hamesha inputs ke sum se ≤ kyun hota hai?

Sum "do ek saath fail" ke overlaps ko double-count karta hai ::: exact OR unhe subtract karta hai, toh exact ≤ sum, equality sirf tiny probabilities ke limit mein hoti hai.

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Cell B — Single AND gate, chhoti probabilities

Forecast: Dono tootne chahiye — ek se kahin zyada rare. Guess karo "$0.02$ se kaafi chhota."

1. Dono failure probabilities multiply karo. Kyun? AND = sab inputs ek saath fail = joint probability = product (independent banks ke liye). $P_{\text{Battery}}=0.02\times 0.02=0.0004$

Verify: $0.0004$, kisi bhi input ($0.02/0.0004=50$) se $50\times$ chhota hai. Yahi redundancy ka kaam hai — AND gate ne risk ko crush kar diya. Units: abhi bhi ek bare probability, dimensionless. ✓

Figure dekho: beech wali bar (teen inputs ka OR gate) yellow reference line se oopar hai jo ek single input (risk badh jaati hai) mark karti hai; daayein bar (do inputs ka AND gate) us line se kaafi neeche hai (risk collapse ho jaati hai). Same class ke chhote inputs, opposite behaviour — yahi is page ka poora point hai.

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Cell C — Degenerate input: ek perfect part ($P_i=0$)

Forecast: Zero-probability input OR mein kuch contribute nahi karta. Guess karo "answer sirf baaki doono ke OR tak gir jaata hai."

1. Perfect part ki survival $1-0=1$ hai. Kyun yeh step? Ek part jo kabhi fail nahi hota, hamesha survive karta hai.
2. Survivals multiply karo: Kyun? $1$ times kuch bhi use unchanged chhodta hai. $1\times 0.99\times 0.995=0.98505$
3. Subtract karo: $P=1-0.98505=0.01495$.

Verify: Cell A ($0.01595$) se compare karo. Deployment risk hatane se total $0.01595-0.01495=0.001$ gira — essentially woh $0.001$ deployment probability. OR gate mein ek perfect input invisible hota hai — bilkul wahi jo hum expect karte hain. ✓

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Cell D — Degenerate input: ek guaranteed failure ($P_i=1$)

Forecast: OR gate par ek guaranteed input OR ko guaranteed fire karta hai. Guess karo "$P_{\text{Top}}=1$."

1. AND branch: $0.01\times 0.01=0.0001$ (Kyun? dono transponders ko marna chahiye.)
2. HGA ki survival: $1-1=0$. Kyun yeh step? Ek part jo hamesha fail hota hai kabhi survive nahi karta.
3. OR formula: $P=1-(1-0.0001)(1-1)=1-(0.9999)(0)=1-0=1$.

Verify: $P_{\text{Top}}=1$ matlab certain failure — correct, kyunki HGA akela (ek OR input) ek single point of failure hai jo pehle se hi broken hai. $P=1$ input ek OR gate ko saturate karta hai. (AND gate mein yeh invisible hota: $1\times x=x$ — phir se bilkul Cell C ke opposite.) ✓

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Cell E — Mixed tree (AND nested under OR)

Forecast: Do redundant transponders AND branch ko tiny banate hain; akela HGA exposed hai. Guess karo "HGA dominate karta hai, answer ≈ $0.05$."

1. Pehle sabse deep gate solve karo (AND). Kyun? Tum ek gate evaluate nahi kar sakte jab tak uske inputs numbers na hon; andar se bahar kaam karo. $P_{\text{AND}}=0.01\times 0.01=0.0001$
2. Usse OR gate mein HGA ke saath feed karo. Kyun? AND result ab OR ka sirf ek input hai. $P_{\text{Top}}=1-(1-0.0001)(1-0.05)$
3. Survivals compute karo aur multiply karo: $0.9999\times 0.95=0.949905$.
4. Subtract karo: $P_{\text{Top}}=1-0.949905=0.050095\approx 0.0501$.

Verify: HGA $0.05$ contribute karta hai; poora answer $0.0501$ hai. Toh redundant transponders sirf $0.0001$ add karte hain — HGA ≈ $99.8\%$ risk drive karta hai ($0.05/0.0501$). Forecast se match karta hai. Yahi FTA ka payoff hai: yeh single point of failure par ek ungli point karta hai. ✓

Figure top-down padho: pink Loss of Comm box ek yellow OR junction ke upar hai jiske do branches hain. Baayein blue branch do transponders ka AND hai ($0.0001$); daayein pink branch akela HGA hai ($0.05$). Yellow arrow HGA branch ki taraf point karta hai — woh dominant OR input hai jahan almost saara risk rehta hai, aur yeh woh box hai jise ek designer pehle attack karna chahiye.

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Cell F — Limiting behaviour: bahut saare identical OR inputs

Is example se pehle humhe ek aur tool chahiye, aur hum use zero se build karte hain.

Forecast: Chhote $0.001$ risks bhi, OR gate mein baar baar pile hote hain, upar creep karne chahiye. Guess karo "107 of them kuch percent dete hain; infinitely many → certain failure."

1. General formula: sab identical hain, toh $P_{\text{OR}}=1-(1-0.001{)}^n=1-0.999^n$. Kyun? OR formula $n$ survival factors $(1-P_i)$ multiply karta hai; jab har factor same number $0.999$ ho, use $n$ baar multiply karna hi use $n$th power tak raise karna hai.
2. $n=1$: $1-0.999=0.001$. (Ek mechanism = uska apna risk.)
3. $n=107$: $1-0.999^{107}=1-0.89848=0.10152\approx 0.102$.
4. $n\to \infty$: $0.999^n\to 0$, toh $P_{\text{OR}}\to 1$.

Verify: $0.999^{107}$ logarithm trick se check karo jo abhi define ki. Logs kyun? $0.999$ ko by hand 107 baar multiply karna brutal hai, lekin power-slides-out rule use ek multiplication mein badal deta hai: ${\log }_{10}(0.999^{107})=107\cdot {\log }_{10}(0.999)$. Kyunki ${\log }_{10}(0.999)\approx -0.000434$ (1 se thoda neeche wale number ka log ek chhota negative hota hai), exponent $107\times (-0.000434)\approx -0.04648$ hai, aur antilog $10^{-0.04648}\approx 0.8985$ hai, jo $P\approx 0.1015$ deta hai. ✓ Lesson: bahut saare parts par ek OR gate ek slow poison hai — isliye JWST ne per-mechanism reliability $0.999+$ ki maang ki; $0.99$ each se, $1-0.99^{107}\approx 0.66$, ek coin-flip mission. Reliability Block Diagrams mein compare karo — failures ka OR gate survivals ki ek series chain hai.

Figure mein curve $P=1-0.999^n$ hai jo OR inputs $n$ ki number ke saath climb karti hai. Yellow dot $n=1$ mark karta hai (risk $=0.001$, floor se barely upar); pink dot $n=107$ mark karta hai (risk $\approx 0.102$ tak jump kar gaya); aur upar dotted line ceiling $P\to 1$ dikhati hai jiske taraf curve $n\to \infty$ jaane par creep karti hai. Dekho kaise ek "negligible" per-part risk ek real threat ban jaata hai jab tum unhe kaafi saare OR ke neeche stack karo.

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Cell G — Common cause failure (dependence formula tod deti hai)

Forecast: Naive AND ne $0.0001$ kaha tha. Lekin ek shared cause ek "dono ek saath fail" shortcut banata hai. Guess karo "$0.0001$ se kaafi bada — regulator dominate karta hai."

1. Naive product kyun galat hai. Formula $P_A\times P_B$ independence assume karta hai. Common cause use violate karta hai: ek event dono ko ek saath flip karta hai. Multiply karna pretend karta hai ki yeh kabhi nahi hota.
2. Har transponder ka total risk shared + independent mein split karo. Har transponder ki total $P=0.01$ shared regulator ($0.008$) PLUS uske apne independent modes se bani hai. Toh independent-only part hai $P^{\prime }=0.01-0.008=0.002$. Subtract kyun? $0.008$ pehle se shared term mein counted hai; agar hum use $P^{\prime }$ mein bhi rakhen toh double-count ho jaayega.
   • Dono apne khud ke independent causes se fail hote hain: $0.002\times 0.002=0.000004$.
3. Shared cause ko independent double-failure ke saath OR karo. Kyun OR? Dono transponders ko dead karne ke liye koi bhi route kaam karta hai: regulator ka marna, OR dono independently marna. $P_{\text{both}}=1-(1-0.008)(1-0.000004)=1-(0.992)(0.999996)$
4. Compute karo: $0.992\times 0.999996=0.99199603$, toh $P_{\text{both}}=0.00800397\approx 0.0080$.

Verify: Sahi answer $\approx 0.008004$ vs naive $0.0001$ — shared regulator "redundant" pair ke dono fail hone ki likelihood ko 80× zyada banata hai ($0.008004/0.0001\approx 80$). Independent double-failure ($0.000004$) shared $0.008$ ke samne bilkul negligible hai; regulator HI risk hai. Redundancy ek illusion hai jab tak tum do transponders ko alag power nahi dete. Boolean Algebra and Logic Design dekho ki hum regulator ko apne basic event ke roop mein kyun add karte hain bajaye usse hide karne ke. ✓

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Cell H — Real-world word problem (English → gates translate karo)

Forecast: Teen redundant wheels → tiny AND. Akela star tracker → exposed OR input. Guess karo "≈ $0.02$, tracker-dominated."

1. English translate karo. "Kam se kam ek wheel chahiye" = "sirf tabhi lost hoti hai jab sab fail ho jaayein" = teen wheels ka AND. "Star tracker mein backup nahi" = top par OR input. Kyun yeh step? Gates redundancy words se aate hain: all/both → AND, any/either → OR.
2. Wheel AND gate: $0.03^3=0.000027$. Kyun cube? Teen independent wheels ka sab fail hona product $0.03\times 0.03\times 0.03$ hai.
3. Top OR gate: $P=1-(1-0.000027)(1-0.02)$. Kyun? Wheel-AND result ab star tracker ke saath ek OR input hai.
4. Survivals compute karo aur multiply karo: $(0.999973)(0.98)=0.97997354$, toh $P=1-0.97997354=0.02002646\approx 0.0200$.

Verify: Answer $0.0200$ vs star-tracker akela $0.02$ — teen wheels sirf $0.000027$ add karte hain. Agar ek designer paise kharcha karna chahta ho, ek doosra star tracker add karna (usse AND branch banana, $0.02\times 0.02=0.0004$) top risk ko kaafi zyada slash karega banisbat ek chauthe wheel ke. Spacecraft Redundancy Architectures intuition se match karta hai. ✓

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Cell I — Exam twist: "kaun sa ek improvement sabse zyada help karta hai?"

Forecast: HGA dominant term hai, toh use fix karna bahut bada farak karna chahiye. Guess karo "(a)."

1. Option (a): HGA do ka AND ban jaata hai. Naya HGA branch $=0.05\times 0.05=0.0025$. Kyun? Do identical antennas ka dono fail hona zaroori hai. $P_{\text{Top}}^{(a)}=1-(1-0.0001)(1-0.0025)=1-(0.9999)(0.9975)=1-0.99740025=0.00259975\approx 0.0026$
2. Option (b): transponders improve ho jaate hain. AND branch $=0.001\times 0.001=10^{-6}$. $P_{\text{Top}}^{(b)}=1-(1-10^{-6})(1-0.05)=1-(0.999999)(0.95)=1-0.94999905=0.05000095\approx 0.0500$
3. Compare karo. (a) risk ko $0.0501$ se $0.0026$ tak girata hai ($19\times$ improvement). (b) barely move karta hai — $0.0501\to 0.0500$.

Verify: Option (a) overwhelmingly jeetta hai, forecast confirm hota hai. Exam lesson: hamesha sabse pehle dominant OR input attack karo — kisi aisi term ko improve karna jo pehle se negligible hai (transponders) budget waste karta hai. Minimum Cut Sets ke saath dominant paths dhundo pehle ek bhi rupaya kharcha karne se. ✓

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Cell J — Badi probabilities ($P_i$ near 1)

Forecast: Itne bade numbers ke saath, Cell A se "sum ≈ OR" trick toot jaani chahiye — sum $0.6+0.5+0.4=1.5$ already 1 se upar hai, jo probability ke liye impossible hai. Guess karo "sahi OR high hai lekin 1 se neeche rahega; AND ek modest fraction hoga."

1. Survival probabilities. Kyun? OR formula abhi bhi "kya yeh survive kiya?" par chalta hai, chahe failure risks kitne bhi bade hon. $1-0.6=0.4,1-0.5=0.5,1-0.4=0.6$
2. Survivals multiply karo — chance ki teeno survive karein. Kyun? All-survive abhi bhi ek joint event = product hai. $0.4\times 0.5\times 0.6=0.12$
3. OR result: $P_{\text{OR}}=1-0.12=0.88$.
4. AND result: failure probabilities directly multiply karo. Kyun? AND = sab ek saath fail = $P_i$ ka product. $P_{\text{AND}}=0.6\times 0.5\times 0.4=0.12$

Verify: Lazy sum $1.5$ deta hai — ek bakwaas "probability" jo 1 se upar hai, toh chhote-number shortcut yahan completely fail ho gaya hai; sirf exact formula $1-\prod (1-P_i)=0.88$ trustworthy hai. Sanity: $0.88$, $[0,1]$ mein hai. ✓ Gaur karo OR ($0.88$) AND ($0.12$) se kaafi bada hai — Cell B se wahi gap-between-gates, lekin ab dramatic kyunki inputs huge hain. Lesson: "OR ≈ sum" rule sirf tab safe hai jab har $P_i$ chhota ho; $P=1$ ke paas tum ZAROORI HO ki survival-product form use karo.

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Spacecraft Design se Connections

In cells ke aage, FTA directly Monte Carlo Reliability Simulation (jab probabilities khud uncertain hoon), Failure Modes and Effects Analysis (FMEA) (bottom-up complement), aur Single Point Failure Review (jo exactly woh OR inputs dhundta hai jo Cells E, I, aur J mein dominate kiye) mein feed karta hai.

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Active recall

Recall OR gate mein,

$P_i=0$ input kya karta hai, aur $P_i=1$ ke baare mein? $P_i=0$ invisible hai (survival $=1$, kuch nahi badalta). $P_i=1$ gate ko $1$ tak saturate karta hai (certain failure) ::: AND gate ke opposite, jahan $0$ gate की failure ko $0$ tak force karta hai (us branch ke liye guaranteed success) aur $1$ invisible hota hai.

Recall Do "redundant" units ek power supply share karte hain.

$P_A\times P_B$ kyun galat hai? Woh formula independence assume karta hai; shared supply ek common cause hai jo dono ko ek saath fail karta hai ::: use ek alag basic event ke roop mein model karo jo independent double-failure ke saath OR-ed ho — aur yaad raho ki square karne se pehle shared risk ko har unit ke total se subtract karo.

Recall Mixed tree diya ho toh, kaun sa gate pehle evaluate karte ho?

Sabse deep gate (inputs ke numbers hone zaroori hain pehle gate compute ho sake) ::: andar se bahar, bottom-up kaam karo.

Recall "OR ≈ inputs ka sum" shortcut kab safe hai?

Sirf jab har $P_i$ chhota ho ::: $P=1$ ke paas sum 1 se upar ja sakta hai (impossible), toh tumhe exact survival-product form $1-\prod (1-P_i)$ use karni padegi.