Fault tree analysis (FTA) — top-down, AND - OR gates: Exercises

3.6.30 · D4 · Physics › Spacecraft Structures & Systems Engineering › Fault tree analysis (FTA) — top-down, AND - OR gates

Yeh page ek self-test ladder hai. Har problem pehle pen aur paper se karo, PHIR collapsible solution kholo. Difficulty dheere-dheere badhti hai — gates ko pehchanne se lekar poore trees banana tak. Yahan use kiye gaye har symbol ka base parent FTA note mein hai; agar koi formula anjaan lage, pehle woh note dobara padh lo.

Definition Do Gate Symbols (Shape Conventions)

Fault trees do standard shapes se draw ki jaati hain. Tumhe inhe shape se pehchanna aana chahiye, sirf colour se nahi:

OR gate — ek curved, pointed "shield" shape se draw hota hai (ek chevron jiska bottom edge curved ho). Mnemonic: curved bottom ek bade khule munh jaisa lagta hai jo koi bhi ek input nigal leta hai. Koi bhi ek input → output.
AND gate — ek flat-bottomed dome se draw hota hai (ek semicircle / "D" apni seedhi edge par leta hua). Mnemonic: flat base ek solid floor ki tarah hai jo tabhi tootega jab neeche ka har pillar dheela pade. Saari inputs → output.

Is page ke figures mein har gate box par words mein bhi label hai ("OR" / "AND") aur colour-coding bhi hai, isliye tumhe sirf colour par depend nahi karna padega. Use kiye gaye colours: OR = orange, AND = inner branches ke liye green aur top event ke liye red, basic events = blue. Har box ke andar word label hi authoritative cue hai.

Figure dekho: left panel (ek AND gate, flat-bottomed dome, labelled "AND") chote numbers ko multiply karta hai isliye output shrink hota hai (redundancy = achha); right panel (ek OR gate, curved shield, labelled "OR") vulnerabilities stack karta hai isliye output badhta hai (single points of failure = bura). Word labels padho, sirf colours nahi. Yeh mental image raho; neeche har answer bas yahi picture hai, baar baar repeat hoti hui.

Level 1 — Recognition

Exercise 1.1 (L1)

Ek subsystem tab fail hota hai jab ya toh fuse blow kare ya relay stick kare. Subsystem-failure node ke neeche in do basic events ko konsa gate connect karta hai — AND ya OR?

Recall Solution 1.1

Shabd "either… or…" ka matlab hai ki koi ek akela failure kaafi hai. Yeh OR gate hai (curved-shield shape). Koi redundancy nahi: ek akeli fault fatal hai. Answer: OR gate.

Exercise 1.2 (L1)

Do identical pumps coolant supply karte hain. Cooling system tabhi fail hota hai jab dono pumps fail ho jaayein. Konsa gate?

Recall Solution 1.2

"Dono ka fail hona zaroori" = saari inputs required = AND gate (flat-bottomed dome shape). Yeh redundancy hai: ek kaam karta pump system bacha leta hai. Answer: AND gate.

Exercise 1.3 (L1)

Ek basic event ki failure probability $P = 0.02$ hai. Uski survival probability kya hai (probability ki woh fail nahi hoga)?

Recall Solution 1.3

Survival = "the complement" = $1 - P$ . WHY: har event ya toh hota hai ya nahi hota, aur yeh dono possibilities $1$ mein sum honi chahiye, isliye "nahi hota" woh bacha hua hissa hai failure chance subtract karne ke baad. $1 - 0.02 = 0.98$ Answer: 0.98.

Level 2 — Application

Exercise 2.1 (L2)

Ek OR gate ke do independent inputs hain, $P_{1} = 0.1$ aur $P_{2} = 0.2$ . Output probability calculate karo.

Recall Solution 2.1

WHAT karna hai: OR formula apply karo. WHY: ek OR gate tab fire karta hai jab kam se kam ek input fail ho, aur "kam se kam ek" ko count karna sabse aasaan hai "koi fail nahi hota" ke complement ke roop mein. Chance ki koi nahi fail hota woh survival terms $(1 - P_{i})$ ka product hai, isliye output hai $1 - \prod (1 - P_{i})$ . $P_{out} = 1 - (1 - 0.1) (1 - 0.2) = 1 - (0.9) (0.8) = 1 - 0.72 = 0.28$ Answer: 0.28. Sanity check: dono inputs se bada ( $0.28 > 0.2$ ) — sahi hai, kyunki OR gate risk accumulate karta hai.

Exercise 2.2 (L2)

Ek AND gate ke teen independent inputs hain, har ek $P = 0.05$ ke saath. Output calculate karo.

Recall Solution 2.2

WHAT karna hai: AND formula apply karo. WHY: ek AND gate tabhi fire karta hai jab saari inputs ek saath fail hoon. Independent events ke liye kai cheezein ek saath hone ki probability unki individual probabilities ka product hai, isliye hum teeno inputs multiply karte hain. $P_{out} = 0.05 \times 0.05 \times 0.05 = 0.000125 = 1.25 \times 1 0^{- 4}$ Answer: $1.25 \times 1 0^{- 4}$ . Sanity check: kisi bhi single input se kaafi chota — sahi hai, AND risk suppress karta hai.

Exercise 2.3 (L2)

Ek OR gate ke teen independent inputs hain: $P_{1} = 0.001$ , $P_{2} = 0.01$ , $P_{3} = 0.005$ . Output calculate karo (yeh parent note ka "Solar Fail" branch hai).

Recall Solution 2.3

WHY OR formula phir se: solar branch tab marti hai jab deploy, degrade, ya tracking mein se koi ek fail ho — ek classic "at least one" sawaal — isliye hum $1$ lete hain minus teenon ke survive hone ka chance, yaani $1 - \prod (1 - P_{i})$ . $P_{out} = 1 - (1 - 0.001) (1 - 0.01) (1 - 0.005)$ $= 1 - (0.999) (0.99) (0.995)$ Step by step multiply karo: $0.999 \times 0.99 = 0.98901$ , phir $0.98901 \times 0.995 = 0.98406...$ $P_{out} = 1 - 0.98406 = 0.01594 \approx 0.016$ Answer: $\approx 0.016$ .

Level 3 — Analysis

Exercise 3.1 (L3)

Communication system (parent note se): Loss of Comm tab hoti hai jab (Transponder A AND Transponder B) fail hoon, OR high-gain antenna (HGA) fail ho. $P (A) = 0.01$ , $P (B) = 0.01$ , $P (HGA) = 0.05$ diya hai; $P_{Top}$ calculate karo aur batao konsa basic event risk dominate karta hai.

Recall Solution 3.1

Step 1 — inner AND, WHY: do transponders redundant hain, isliye branch tabhi fail hoti hai jab dono ek saath fail hoon → multiply karo. $P_{branch} = 0.01 \times 0.01 = 0.0001$ Step 2 — outer OR, WHY: top event tabhi fire karta hai jab ya toh transponder branch fail ho ya HGA fail ho — ek "at least one" combination — isliye $1 - \prod (1 - P_{i})$ use karo, jisme branch value ek input ki tarah feed hoti hai. $P_{Top} = 1 - (1 - 0.0001) (1 - 0.05) = 1 - (0.9999) (0.95)$ $= 1 - 0.949905 = 0.050095 \approx 0.0501$ Answer: $\approx 0.0501$ . Dominant event: HGA $0.05$ contribute karta hai jabki poori transponder branch sirf $0.0001$ contribute karti hai — HGA lagbhag $99.8%$ risk drive karta hai. Reliability effort antenna par lagani chahiye. Dekho Single Point Failure Review.

Exercise 3.2 (L3)

Usi tree ke liye, maano hum (a) $P (HGA)$ ko halve karke $0.025$ kar sakte hain, ya (b) har transponder ko halve karke $0.005$ kar sakte hain. Konsi change $P_{Top}$ zyada reduce karegi?

Recall Solution 3.2

Option (a): $P_{Top} = 1 - (1 - 0.0001) (1 - 0.025) = 1 - (0.9999) (0.975) = 1 - 0.9749025 = 0.0250975 \approx 0.0251$ . Option (b): inner AND $P_{branch} = 0.005 \times 0.005 = 0.000025$ ; phir $P_{Top} = 1 - (1 - 0.000025) (1 - 0.05) = 1 - (0.999975) (0.95) = 1 - 0.94997625 = 0.05002375 \approx 0.0500$ . Option (a) top ko $0.0501$ se $0.0251$ par le aata hai (bada faida); option (b) mein practically koi change nahi ( $0.0501 \to 0.0500$ ). Answer: HGA fix karo (option a). Pehle se redundant branch ko improve karna almost wasted effort hai.

Exercise 3.3 (L3)

Ek tree ka top event do branches ka OR hai: Branch-1 $= 0.02$ aur Branch-2 $= 0.03$ . Ek designer Branch-1 ki redundant copy add karta hai, us leg ko do identical $0.02$ events ke AND mein badal deta hai. Naya top probability aur fractional risk reduction calculate karo.

Recall Solution 3.3

Naya Branch-1 (do $0.02$ ka AND): $P_{branch} = 0.02 \times 0.02 = 0.0004$ . Purana top: $P_{Top} = 1 - (1 - 0.02) (1 - 0.03) = 1 - (0.98) (0.97) = 1 - 0.9506 = 0.0494$ . Naya top: $P_{Top} = 1 - (1 - 0.0004) (1 - 0.03) = 1 - (0.9996) (0.97) = 1 - 0.969612 = 0.030388 \approx 0.0304$ . Reduction: $(0.0494 - 0.0304) /0.0494 = 0.0190/0.0494 = 0.3846 \approx 38.5%$ . Answer: naya top $\approx 0.0304$ ; risk $\approx 38.5%$ cut hua.

Level 4 — Synthesis

Exercise 4.1 (L4)

Is description se ek fault tree banao aur quantify karo:

"Attitude control LOST hoti hai agar reaction wheels fail hoon AND thrusters fail hoon. Reaction wheels tab fail hote hain jab koi bhi teen wheels (W1, W2, W3) mein se koi fail ho. Thrusters tab fail hote hain jab fuel line clog ho OR dono thruster valves (V1, V2) fail hoon." $P (W_{i}) = 0.03$ har ek ke liye, $P (clog) = 0.004$ , $P (V 1) = P (V 2) = 0.1$ diya hai. $P_{Top}$ calculate karo.

Jo tree tumhe draw karni chahiye woh aisi dikhti hai — ise calculation ke saath saath study karo:

Figure mein, top box (flat-bottomed AND dome, labelled "AND", red) matlab dono major branches ko marna hoga; har OR box (curved shield, labelled "OR", orange) matlab koi bhi child us branch ko maar deta hai; inner AND box (labelled "AND", green) do valves ko combine karta hai. Har box apna word label carry karta hai, isliye labels follow karo — colours sirf secondary cue hain.

Recall Solution 4.1

Step 1 — Wheels branch (3 ka OR), WHY: koi bhi ek wheel fail hone par branch jaati hai → "at least one" → $1 - \prod (1 - P)$ . $P_{branch,W} = 1 - (1 - 0.03)^{3} = 1 - 0.9 7^{3} = 1 - 0.912673 = 0.087327$ Step 2 — Valves branch (2 ka AND), WHY: dono valves ek saath fail hone chahiye → multiply karo. $P_{branch,V} = 0.1 \times 0.1 = 0.01$ . Step 3 — Thruster branch (clog aur valves-branch ka OR), WHY: thrusters tab maarte hain jab clog ya valve pair fail ho → "at least one" → $1 - \prod (1 - P)$ . $P_{branch,T} = 1 - (1 - 0.004) (1 - 0.01) = 1 - (0.996) (0.99) = 1 - 0.98604 = 0.01396$ Step 4 — Top (wheels aur thrusters ka AND), WHY: attitude tabhi jaati hai jab dono independent branches ek saath fail hoon → multiply karo. $P_{Top} = P_{branch,W} \times P_{branch,T} = 0.087327 \times 0.01396 = 0.0012191 \approx 1.22 \times 1 0^{- 3}$ Answer: $\approx 1.22 \times 1 0^{- 3}$ . Design ki samajhdaari note karo: top AND gate do independent, moderately unreliable branches ko saath baandhta hai, mission risk ko lagbhag ek in 820 tak neeche kheench deta hai. Dekho Spacecraft Redundancy Architectures.

Exercise 4.2 (L4)

4.1 ke tree ke liye, minimal cut sets list karo — basic events ke sabse chote combinations jo akele top event cause karte hain. (Cut set events ka woh set hai jinke ek saath hone par top event guaranteed hota hai; yeh minimal hota hai agar koi bhi member remove karne par woh toot jaata hai.)

Recall Solution 4.2

Top AND(Wheels, Thrusters) hai. Isliye har cut set mein (ek wheel-failure) aur (ek thruster-side failure) honi chahiye.

Wheels branch OR(W1,W2,W3) hai → koi bhi akela $W_{i}$ wheels fail karta hai.
Thrusters branch OR(clog, AND(V1,V2)) hai → ya "clog" akela, ya "V1 aur V2 ek saath".

Combine karne par, minimal cut sets hain: ${W_{1}, clog}, {W_{2}, clog}, {W_{3}, clog},$ ${W_{1}, V_{1}, V_{2}}, {W_{2}, V_{1}, V_{2}}, {W_{3}, V_{1}, V_{2}}$ Yaani 3 two-element cut sets aur 3 three-element cut sets = kul 6 minimal cut sets. Formal algorithm ke liye dekho Minimum Cut Sets.

Level 5 — Mastery

Exercise 5.1 (L5) — Common Cause Failure

Do transponders A aur B AND-redundancy mein hain, har ek $P = 0.01$ ke saath, isliye naively $P_{branch} = 0.0001$ . Lekin dono ek hi power regulator se draw karte hain, jo $P_{R} = 0.008$ se fail hota hai aur dono transponders ko guaranteed maar deta hai. Ise sahi se model karo aur true branch probability calculate karo. Common cause ignore karne se risk kitne factor se understate hota hai?

Recall Solution 5.1

WHY naive number galat hai: $P_{branch} = P_{A} P_{B}$ independence assume karta hai, lekin shared regulator A aur B ko saath fail karata hai jab bhi woh maarta hai. Yeh shared path ek common cause failure (CCF) hai. Sahi model, WHY OR: transponder branch tab fail hoti hai jab (A fail AND B independently fail) OR (regulator fail) — do independent contributors ka "at least one" combination — isliye hum unhe OR karte hain $1 - \prod (1 - P)$ se: $P_{branch} = 1 - (1 - P_{A B}) (1 - P_{R}) = 1 - (1 - 0.0001) (1 - 0.008)$ $= 1 - (0.9999) (0.992) = 1 - 0.99180 = 0.008199 \approx 8.20 \times 1 0^{- 3}$ Understatement factor: $\frac{0.008199}{0.0001} = 81.99 \approx 82$ . Answer: true branch probability $\approx 8.20 \times 1 0^{- 3}$ ; naive estimate $\approx 82 \times$ zyada optimistic hai. Regulator, redundancy nahi, ab dominate karta hai — yahi Common Cause Failure Analysis ka poora point hai.

Exercise 5.2 (L5) — Design to a Requirement

Ek single-string branch mein do basic events ka OR hai, $P_{1} = 0.04$ aur $P_{2} = 0.04$ . Requirement hai $P_{branch} \leq 1 \times 1 0^{- 3}$ . Agar tum poori branch ko $k$ -fold redundant karo (us OR-branch ki $k$ independent identical copies ka AND), toh requirement meet karne wala sabse chota $k$ kya hai?

Recall Solution 5.2

Single branch value, WHY OR: branch tab fail hoti hai jab koi bhi basic event fire kare → "at least one" → $q = 1 - (1 - 0.04)^{2} = 1 - 0.9216 = 0.0784$ . $k$ -fold AND of identical branches, WHY multiply: $k$ independent copies ke saath poori cheez tabhi fail hoti hai jab saari copies ek saath fail hoon → $P_{branch} = q^{k} = 0.078 4^{k}$ . Require $0.078 4^{k} \leq 1 0^{- 3}$ . Logarithm lo (log exponentiation undo karta hai, $k$ solve karne deta hai): $k \geq \frac{l n ( 1 0 ^{- 3} )}{l n ( 0.0784 )} = \frac{- 6.9078}{- 2.5459} = 2.713$ Kyunki $k$ poora number hona chahiye, upar round karo: $k = 3$ . Verify karo: $0.078 4^{3} = 4.82 \times 1 0^{- 4} \leq 1 0^{- 3}$ ✓ (jabki $0.078 4^{2} = 6.15 \times 1 0^{- 3} > 1 0^{- 3}$ ✗). Answer: $k = 3$ (triple redundancy).

Exercise 5.3 (L5) — Full System Roll-up

Ek power system: Top = AND(Solar, Battery). Solar = OR(deploy $0.001$ , degrade $0.01$ , tracking $0.005$ ). Battery = AND(cell1 $0.02$ , cell2 $0.02$ ). $P_{Top}$ calculate karo, phir decide karo: top risk half karne ke liye, kya teesra battery cell add karna better hai (AND) ya ek redundant solar array add karna (Solar ko do identical solar OR-branches ke AND mein banana)?

Recall Solution 5.3

Baseline. Solar branch, WHY OR: deploy/degrade/tracking mein se koi bhi solar kho deta hai → $1 - \prod (1 - P)$ : $P_{branch,Solar} = 1 - (0.999) (0.99) (0.995) = 0.015940$ Battery branch, WHY AND: dono cells ka marna zaroori → multiply karo: $P_{branch,Battery} = 0.02 \times 0.02 = 0.0004$ Top, WHY AND: power tabhi jaati hai jab dono solar aur battery branches ek saath fail hoon → multiply karo: $P_{Top} = 0.015940 \times 0.0004 = 6.376 \times 1 0^{- 6}$ Option A — teesra battery cell (teen $0.02$ ka AND), WHY: teesra redundant cell matlab teeno ka fail hona zaroori → $P_{branch,Battery} = 0.0 2^{3} = 8 \times 1 0^{- 6}$ . $P_{Top}^{A} = 0.015940 \times 8 \times 1 0^{- 6} = 1.2752 \times 1 0^{- 7}$ Option B — redundant solar (do identical $0.015940$ branches ka AND), WHY: do independent solar arrays dono ko fail hona zaroori → branch value square karo: $P_{branch,Solar}^{B} = 0.01594 0^{2} = 2.5408 \times 1 0^{- 4}$ $P_{Top}^{B} = 2.5408 \times 1 0^{- 4} \times 0.0004 = 1.0163 \times 1 0^{- 7}$ Comparison. Baseline ka half hai $\frac{1}{2} (6.376 \times 1 0^{- 6}) = 3.188 \times 1 0^{- 6}$ . Dono options us threshold se kaafi neeche hain: Option A deta hai $1.28 \times 1 0^{- 7}$ aur Option B deta hai $1.02 \times 1 0^{- 7}$ — har ek roughly "half" goal ke $25 \times$ neeche, isliye dono requirement easily satisfy karte hain. Option B thoda kam hai ( $1.02 \times 1 0^{- 7} < 1.28 \times 1 0^{- 7}$ ), isliye pure probability par redundant solar array jeet jaata hai. Answer: baseline $P_{Top} \approx 6.38 \times 1 0^{- 6}$ ; Option A $\approx 1.28 \times 1 0^{- 7}$ ; Option B $\approx 1.02 \times 1 0^{- 7}$ . Dono "half" target ko lagbhag $25 \times$ se crush karte hain, redundant solar array (B) probability par marginally better hai. Practice mein final choice phir mass, cost, aur shared-hardware (common cause) risk par aati hai, na ki is choti probability gap par.

Common mistake L5 Trap — Redundancy add karne ke baad common cause bhool jaana

Galat path: "Maine teesra redundant string add kiya, toh meri AND branch ab $q^{3}$ hai — problem solved, risk negligible hai." Kyun sahi lagta hai: Multiplication real hai aur independent failure probability ko sach mein slash karta hai. Reality: Redundant copies almost hamesha kuch share karti hain — ek bus, ek regulator, ek software load, ek manufacturing lot. Jaise Ex 5.1 ne dikhaya, ek shared regulator $\sim 1 0^{- 3}$ par dominate kar sakta hai chahe kitne bhi strings add karo. AND math CCF term par floor karta hai, $q^{k}$ par nahi. Fix: Har redundancy addition ke baad poochho "yeh copies kya share karti hain?" aur ek explicit CCF basic event OR karke add karo. Reliability Block Diagrams aur Monte Carlo Reliability Simulation se cross-check karo taaki woh hidden couplings expose ho jayein jo tree structure akela miss kar sakta hai.

Recall Jaane se pehle ek-line self-check

AND probabilities multiply karta hai (redundancy, risk shrinks); OR $1 - \prod (1 - P)$ use karta hai (single points, risk grows); aur koi bhi shared component apne OR'd basic event ki tarah add hona chahiye.

Ek line mein test karo:

Gate for "any one fails"

OR gate (curved-shield shape), output

= 1 - \prod (1 - P_{i})

Gate for "all must fail"

AND gate (flat-bottomed dome shape), output

= \prod P_{i}

Kyun shared regulator ignore karna dangerous hai

yeh ek common cause failure hai jo branch dominate kar sakta hai, independence-based AND math (

82 \times

Ex 5.1 mein) ko bahut zyada optimistic bana deta hai