Worked examples — Reliability — MTTF, MTBF, exponential failure model

3.6.31 · D3 · Physics › Spacecraft Structures & Systems Engineering › Reliability — MTTF, MTBF, exponential failure model

Yeh page ek firing range hai. Hum yahan har tarah ke questions line up karte hain jo exponential reliability model aapse pooch sakta hai — har sign, har extreme value, degenerate cases, ek word problem, aur ek exam twist — aur har ek ko ek complete worked solution se solve karte hain. Agar aap neeche diye gaye saalon ko kar sako, toh is topic ka koi bhi reliability question aapko surprise nahi kar sakta.

Shuru karne se pehle, parent note se ek reminder: haari poori duniya yahan ek picture par tiki hai — ek curve $R(t)$ jo height $1$ se shuru hoti hai (launch par definitely alive) aur $0$ ki taraf slide karti hai (eventually definitely dead). Number $\lambda$ (lambda) failure rate hai: har surviving unit per unit time mein kitni failures hoti hain. Neeche sab kuch bas uss ek curve ko alag-alag points par padhna hai.

Recall Woh chaar formulas jinpar hum lean karenge

Exponential survival curve ::: $R(t)=e^{-\lambda t}$ Mean time to failure ::: $\text{MTTF}=1/\lambda$ Exactly MTTF par reliability ::: $R(1/\lambda )=e^{-1}\approx 0.368$ Parallel mein do units (kaam karta hai agar koi ek bhi kaam kare) ::: $R_{\text{sys}}=1-(1-R_1)(1-R_2)$

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Scenario matrix

Har reliability problem inhi case classes mein se ek hoti hai. Right-hand column us worked example ka naam batata hai jo us cell ko cover karta hai.

#	Case class	Kya khaas hai	Covered by
A	Forward: $\lambda ,t$ diya → $R$ nikalo	seedha plug in karo	Ex 1
B	Inverse: $R,t$ diya → $\lambda$ (ya $t$) nikalo	$e$ ko "undo" karne ke liye $\ln$ use karna padega	Ex 2
C	Degenerate $t=0$	curve apni starting height par	Ex 3
D	Limit $t\to \infty$	curve ki floor value	Ex 3
E	Degenerate $\lambda =0$	ek component jo kabhi fail nahi karta	Ex 4
F	$t=\text{MTTF}$ landmark	"$e^{-1}$" wala surprise	Ex 5
G	Parallel redundancy	do curves combine hoti hain	Ex 6
H	Series stacking	rates add hoti hain	Ex 7
I	Word problem (real mission)	English → math mein translate karo	Ex 8
J	Exam twist (unit trap / partial mission)	hidden conversion ya piecewise time	Ex 9, Ex 10

Hum inhe order mein cover karte hain.

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Worked examples

Forecast: Andaza lagao — kya $R$ $1$ ke kareeb hoga ya $0$ ke? Exponent $\lambda t$ thoda chhota sa hai, toh kuch $0.5$ se upar expect karo.

1. Exponent $\lambda t$ compute karo. $\lambda t=0.0002\times 3000=0.6$ Yeh step kyun? Poori curve sirf ek dimensionless number $\lambda t$ (failures/hour × hours = failures, ek pure number) par depend karti hai. $e$ ko touch karne se pehle hume ise build karna hoga.

2. Ise $R(t)=e^{-\lambda t}$ mein daalo. $R(3000)=e^{-0.6}=0.549$ Yeh step kyun? Yeh literally $t=3000$ par curve ki height padhna hai — figure s01 dekho, exponent axis par $x=0.6$ wala dot.

Answer: $R\approx 0.549$, yaani survive karne ka lagbhag 54.9% chance hai.

Verify: $\ln (0.549)=-0.600$ ✓ hamaara exponent recover karta hai. Units: exponent dimensionless hai ✓. Aur $0.549$ kisi bhi probability ki tarah $0$ aur $1$ ke beech hai ✓.

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Forecast: Lambe time mein high reliability → hum ek bahut chhota $\lambda$ expect karte hain.

1. Unknown ke saath equation likho. $0.99=e^{-\lambda \times 5000}$ Yeh step kyun? Ab $\lambda$ ek exponent ke andar band hai. Ise free karne ke liye hume ek aisa tool chahiye jo $e^{(\cdot )}$ ko undo kare.

2. Natural logarithm apply karo. Natural log $\ln$ exactly yeh poochhta hai "kis power par $e$ raise karna hoga?" — yeh exponential ka inverse hai. Isliye hum $\ln$ choose karte hain, koi aur function nahi. $\ln (0.99)=-\lambda \times 5000$ $-0.01005=-\lambda \times 5000$

3. $\lambda$ ke liye solve karo. $\lambda =\frac{0.01005}{5000}=2.010\times 10^{-6}\text{failures/hour}$

Answer: $\lambda \le 2.01\times 10^{-6}$ failures/hour (equivalently MTTF $\ge 497,500$ hours).

Verify: Wापस plug karo: $e^{-2.010\times 10^{-6}\times 5000}=e^{-0.01005}=0.990$ ✓. Chhota $\lambda$ jaise forecast kiya tha ✓.

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Forecast: Launch par cheez kaam karti hai; hamesha ke liye wait karo toh definitely dead. Toh $1$ aur $0$ expect karo.

1. $t=0$ par: $R(0)=e^{-\lambda \cdot 0}=e^0=1$. Yeh step kyun? $t=0$ "start of mission" ka degenerate input hai. Curve zaroor height $1$ se guzarni chahiye — koi system on hone se pehle fail nahi ho sakta.

2. Jab $t\to \infty$: kyunki $\lambda >0$ hai, exponent $-\lambda t\to -\infty$, toh $e^{-\lambda t}\to 0$. Yeh step kyun? Yeh curve ka floor hai. Yeh batata hai ki exponential components kabhi immortal nahi hote — itna time do, failure certain hai.

Answer: $R(0)=1$ aur $R(\infty )=0$. (Figure s01 dekho: curve top-left corner ko touch karti hai aur right side mein bottom axis ke saath hugs karti hai.)

Verify: $e^0=1$ ✓; $e^{-\lambda t}$ decreasing aur neeche $0$ se bounded ✓.

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Forecast: Bilkul failure rate nahi → ise kabhi fail nahi hona chahiye.

1. $\lambda =0$ substitute karo: $R(t)=e^{-0\cdot t}=e^0=1\text{for all }t$ Yeh step kyun? Decay ke driver ko zero karna curve ko flat kar deta hai — height $1$ par ek horizontal line.

2. MTTF check karo: $\text{MTTF}=1/\lambda =1/0\to \infty$. Yeh step kyun? Infinite expected life honest answer hai: ek component jo kabhi fail nahi hota uska koi finite average failure time nahi hota. Yeh model ki boundary hai — physically kisi cheez ka $\lambda =0$ nahi hota, lekin yeh ideal limit hai.

Answer: $R(t)\equiv 1$, $\text{MTTF}=\infty$. Idealized never-failing part.

Verify: Jab $\lambda \to 0^{+}$, kisi bhi fixed $t$ ke liye $R(t)=e^{-\lambda t}\to 1$ ✓, aur $1/\lambda \to \infty$ ✓.

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Forecast: Bahut log $0.5$ guess karte hain ("yeh average hai, toh 50/50"). Savdhan raho.

1. Yaad karo $t=\text{MTTF}$ ka matlab $t=1/\lambda$ hai. Toh exponent hai $\lambda t=\lambda \cdot \frac{1}{\lambda }=1$ Yeh step kyun? MTTF par exponent hamesha exactly $1$ ke barabar hota hai, chahe $\lambda$ kuch bhi ho. Isliye yeh point ek universal landmark hai.

2. Evaluate karo: $R=e^{-1}=0.368$

Answer: Sirf 36.8% apne MTTF tak survive karte hain — 50% nahi. Kyunki curve bent (convex) hai, zyaadatar failures jaldi ho jaati hain. Yeh "MTTF = safe lifetime" wali intuition ko khatam kar deta hai.

Verify: $e^{-1}=0.3679$ ✓. Parent note ke Mistake 1 ko confirm karta hai.

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Forecast: Survive karne ke do chances → system ek single unit se better hona chahiye.

1. Pehle single-unit reliability nikalo. $R_1=e^{-0.00008\times 10000}=e^{-0.8}=0.449$ Yeh step kyun? Redundancy math individual survival probabilities se build hoti hai, isliye pehle ek unit ka chance compute karte hain.

2. System tab hi fail hota hai jab DON'T fail karti hain. Har unit $1-R_1=0.551$ probability se fail hoti hai. Kyunki units independent hain, dono ke fail hone ka chance product hai: $P(\text{both fail})=(0.551{)}^2=0.3036$ Yeh step kyun? Independence hume multiply karne deti hai — yahi multiplication poori wajah hai ki redundancy help karta hai (figure s02 dekho: woh chhota purple overlap jahan dono curves dead hain).

3. System survive karta hai agar not-both-fail ho. $R_{\text{sys}}=1-0.3036=0.696$

Answer: $R_{\text{sys}}\approx 0.696$ (69.6%), single unit ke 44.9% se kaafi zyada. Series vs. Parallel System Reliability aur Redundancy Design se compare karo.

Verify: $1-(1-e^{-0.8}{)}^2=1-(0.5507{)}^2=0.6968$ ✓, aur $0.696>0.449$ (redundancy hamesha help karta hai) ✓.

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Forecast: Series mein zyaada parts → kisi bhi single part se worse → kisi bhi best unit se chhota MTTF.

1. Series ke liye survival curves multiply karo: $R_{\text{sys}}(t)=e^{-{\lambda }_{1}t}{e}^{-{\lambda }_{2}t}{e}^{-{\lambda }_{3}t}=e^{-({\lambda }_1+{\lambda }_2+{\lambda }_3)t}$ Yeh step kyun? Sab parts survive karne chahiye, isliye multiply karo — aur exponentials ko multiply karne se unke exponents add hote hain. Isliye precisely series mein failure rates add hoti hain.

2. Rates add karo: ${\lambda }_{\text{sys}}=0.0001+0.00005+0.00002=0.00017\text{/hr}$

3. Series MTTF: $\text{MTTF}=\frac{1}{{\lambda }_{\text{sys}}}=\frac{1}{0.00017}=5882\text{hours}$

Answer: $\approx 5882$ hours — best single part se chhota ($1/0.00002=50,000$ h), exactly jaise forecast kiya tha.

Verify: $1/0.00017=5882.35$ ✓; aur $5882<50000$ ✓. Ex 6 ke parallel case se contrast karo (yeh Series vs. Parallel System Reliability split hai).

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Forecast: 60,000 hours mein 12 failures ek modest rate hai; 1-year ka kaam chhota hai. Shayad pass kar le.

1. Test data se $\lambda$ estimate karo. Failure rate hai failures per hour: $\lambda =\frac{12}{60000}=0.0002\text{/hr}$ Yeh step kyun? Real data hume empirically $\lambda$ deta hai — failures count karo, total operating time se divide karo. Yeh rare events count karne ke Poisson Process view se jodta hai.

2. 1 year ko hours mein convert karo. $t=365.25\times 24=8766\text{hours}$ Yeh step kyun? Rate per hour hai, isliye time bhi hours mein hona chahiye — exponentiate karne se pehle units match mandatory hai.

3. Reliability compute karo. $\lambda t=0.0002\times 8766=1.753$ $R=e^{-1.753}=0.173$

Answer: $R\approx 0.173$ (17.3%) — 90% requirement fail ho gayi, buri tarah. Aapko redundancy (Ex 6) ya bahut better part chahiye hoga. Yeh directly Mission Design Constraints mein feed hota hai.

Verify: $12/60000=0.0002$ ✓; $e^{-1.7532}=0.1732$ ✓; $0.173<0.90$ toh requirement meet nahi hui ✓.

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Forecast: 6 months 15 years ka sirf ek chhota sa slice hai → $R$ ko $1$ ke kareeb expect karo.

1. MTTF se $\lambda$ convert karo, EK unit system rakhte hue. Years mein kaam karo: $\lambda =\frac{1}{\text{MTTF}}=\frac{1}{15}=0.06667\text{/year}$ Yeh step kyun? Trap months aur years ko mix karna hai. Sab kuch years mein convert karo aur mismatch gayab ho jaata hai.

2. Mission time usi unit mein express karo. Chhe months $=0.5$ year. $\lambda t=0.06667\times 0.5=0.03333$

3. Reliability: $R=e^{-0.03333}=0.9672$

Answer: $R\approx 0.967$ (96.7%). Unit conversion hi poora exam tha.

Verify: $e^{-0.03333}=0.96722$ ✓, aur forecast ke anusaar $1$ ke kareeb ✓. (Cross-check: $0.5/15=0.0333$ — same exponent chahe aap years use karo ya hours mein convert karo, kyunki ratio $t/\text{MTTF}$ unit-free hai ✓.)

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Forecast: Exponential "memoryless" hone ke liye famous hai. Guess: pichle 4000 hours matter nahi karte.

1. Conditional-survival ratio likho. $10,000$ h tak survive karne ka chance given ki $4000$ h tak pahuncha: $P(T>10000\mid T>4000)=\frac{R(10000)}{R(4000)}=\frac{e^{-\lambda \cdot 10000}}{e^{-\lambda \cdot 4000}}$ Yeh step kyun? Conditional probability = (poora survive karo)/(ab tak survive kiya). Hum dono curve heights divide karte hain.

2. Simplify karo — exponents subtract hote hain. $=e^{-\lambda (10000-4000)}=e^{-\lambda \cdot 6000}=e^{-0.00005\times 6000}=e^{-0.3}$ Yeh step kyun? Exponentials ko divide karne se exponents subtract hote hain, sirf extra 6000 hours bachte hain. Starting 4000 cancel ho jaata hai — yahi ek line mein memoryless property hai.

3. Evaluate karo: $R=e^{-0.3}=0.741$

Answer: $74.1\%$ — exactly wahi jo ek brand-new unit ka 6000 hours mein chance hoga. Exponential model ki koi memory nahi hoti age ki (yeh limitation sirf Weibull Distribution / Bathtub Curve wear-out phase se cure hoti hai).

Verify: $R(10000)/R(4000)=e^{-0.5}/e^{-0.2}=e^{-0.3}=0.7408$ ✓, aur yeh ek fresh $e^{-\lambda \cdot 6000}$ ke barabar hai ✓.

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Recall Self-test

$t=\text{MTTF}$ par $R$ kya hai? ::: $e^{-1}\approx 0.368$, $0.5$ nahi. Teen parts ki series — rates kaise combine hoti hain? ::: Add hoti hain: ${\lambda }_{\text{sys}}={\lambda }_1+{\lambda }_2+{\lambda }_3$. Do parallel units har ek $R=0.4$ — system $R$? ::: $1-(0.6{)}^2=0.64$. Parallel units ke liye $\lambda$ add kyun nahi kar sakte? ::: Parallel ko dono ka fail hona chahiye (failure probs ka product), rate sum nahi.