Reliability — MTTF, MTBF, exponential failure model: Exercises

3.6.31 · D4 · Physics › Spacecraft Structures & Systems Engineering › Reliability — MTTF, MTBF, exponential failure model

Yeh page parent topic ke liye ek self-test ladder hai. Pehle har problem ko pen aur paper se try karo, phir solution dekho. Difficulty L1 Recognition (kya tum tool ka naam bata sakte ho?) se shuru hokar L5 Mastery (kya tum apna approach khud bana sakte ho?) tak badhti hai.

Poore page mein hum sirf wahi tools use karte hain jo parent note mein bane hain:

Reliability $R (t) = e^{- λ t}$ — probability ki cheez time $t$ par abhi bhi zinda hai.
Failure rate $λ$ — failures per unit time (constant, useful-life region).
MTTF $= 1/ λ$ — pehli failure tak ka average time.
Time conversions ke liye hum $1 year = 365.25 \times 24 = 8766$ hours use karte hain.

Shuru karne se pehle, is page ke har calculation ko anchor karne ke liye ek picture.

Figure — Reliability — MTTF, MTBF, exponential failure model

Upar wala curve $R (t) = e^{- λ t}$ hai. Amber marker par dhyan do: $t = MTTF$ par survival probability sirf $e^{- 1} \approx 0.37$ hai, one-half nahi. Is shape ko apne dimag mein rakho — zyaadatar exercises bas "is curve se ek value padhna" ya "use invert karna" hain.

Level 1 — Recognition

Goal: sahi formula choose karo aur plug in karo. Units ke alawa koi trap nahi.

Exercise 1.1

Ek component ka constant failure rate $λ = 0.0002$ failures/hour hai. $t = 2000$ hours ke baad uski reliability kya hai?

Recall Solution 1.1

Kaunsa tool? Constant failure rate $\Rightarrow$ exponential model $R (t) = e^{- λ t}$ .

Yeh kyun, kuch aur kyun nahi? "Constant failure rate" exactly wahi assumption hai jo exponential curve produce karta hai (parent note derivation). Koi wear-out nahi, koi infant mortality nahi — bas bathtub curve ka flat middle.

$λ t = 0.0002 \times 2000 = 0.4$ $R = e^{- 0.4} = 0.6703$

Answer: $R \approx 0.670$ , yaani 2000 hours survive karne ka 67.0% chance.

Exercise 1.2

Wahi component: uska MTTF, hours mein, kya hai?

Recall Solution 1.2

Kaunsa tool? Exponential model ke liye, $MTTF = 1/ λ$ .

$MTTF = \frac{1}{0.0002} = 5000 hours$

Answer: 5000 hours. Note karo ki Ex 1.1 mein $t = 2000$ h MTTF se kaafi neeche hai, isliye $R$ comfortably $0.37$ se upar tha.

Level 2 — Application

Goal: multi-step plug-and-chug, jisme unit conversion aur formula ko invert karna bhi shamil hai.

Exercise 2.1

Ek star tracker ka $λ = 0.00003$ failures/hour hai. 3-year mission par uski reliability nikalo.

Recall Solution 2.1

Time convert karo: $t = 3 \times 8766 = 26, 298$ hours.

$λ t = 0.00003 \times 26, 298 = 0.78894$ $R = e^{- 0.78894} = 0.4544$

Answer: $R \approx 0.454$ , lagbhag 45.4%. Pehle se half se neeche — 3 saal ke liye ek akela star tracker accha bet nahi hai.

Exercise 2.2

Ek battery ko $t = 10, 000$ hours mein $R = 0.90$ tak pahunchna hai. Maximum $λ$ kitna allowed hai?

Recall Solution 2.2

Kaunsa tool? Humein $R$ aur $t$ pata hai, humein $λ$ chahiye — isliye humein exponential ko invert karna hoga. " $e$ ko kisi power tak uthane" ka inverse natural log $ln$ hai.

$R = e^{- λ t}$ se shuru karo, dono taraf $ln$ lo: $ln R = - λ t \Rightarrow λ = - \frac{l n R}{t}$

$λ = - \frac{l n 0.90}{10 , 000} = - \frac{- 0.105361}{10 , 000} = 1.0536 \times 1 0^{- 5}$

Answer: $λ \leq 1.05 \times 1 0^{- 5}$ failures/hour (equivalently $MTTF \geq 94, 912$ h).

Level 3 — Analysis

Goal: sochna ki ek number ka kya matlab hai, aur scenarios compare karna.

Exercise 3.1

Ek gyroscope ka $MTTF = 20, 000$ hours hai. Ek colleague claim karta hai "toh yeh basically 15,000 hours tak pahunchne ki guarantee hai." $R (15, 000)$ compute karke is claim ko evaluate karo.

Recall Solution 3.1

Pehle $λ$ recover karo: $λ = 1/ MTTF = 1/20, 000 = 5 \times 1 0^{- 5}$ /h.

$λ t = 5 \times 1 0^{- 5} \times 15, 000 = 0.75$ $R = e^{- 0.75} = 0.4724$

Interpretation: sirf 47.2% hi 15,000 h tak survive karte hain. Claim galat hai — yeh essentially ek coin flip hai. MTTF ek average hai, aur exponential distribution ki ek lambi tail hoti hai, isliye parts ka ek bada fraction mean se kaafi pehle mar jaata hai.

Exercise 3.2

Numerically dikhao ki kisi bhi exponential component ke liye, exactly utna hi fraction apne MTTF tak survive karta hai, chahe $λ$ kuch bhi ho. Woh fraction compute karo.

Recall Solution 3.2

Humne kya kiya: $t = MTTF = 1/ λ$ par $R$ evaluate kiya.

$R (MTTF) = e^{- λ \cdot (1/ λ)} = e^{- 1} = 0.3679$

$λ$ cancel ho jaata hai — answer ek universal constant hai. Kisi bhi exponential-model population ka sirf 36.8% hi apne MTTF tak survive karta hai; 63.2% already fail ho chuke hote hain. Yahi page ke top wali figure mein amber marker hai.

Level 4 — Synthesis

Goal: kai parts ki reliability combine karna — series aur parallel architectures. Dekho Series vs. Parallel System Reliability aur Redundancy Design.

Exercise 4.1

Mission par series (sab kuch kaam karna chahiye) mein teen subsystems ki reliabilities $R_{1} = 0.95$ , $R_{2} = 0.90$ , $R_{3} = 0.98$ hain. System reliability nikalo.

Recall Solution 4.1

Kaunsa tool? Series = ek chain; yeh tabhi survive karta hai jab har link survive kare. Independent parts ke liye, survivals ka "AND" probabilities ka product hota hai.

$R_{sys} = R_{1} R_{2} R_{3} = 0.95 \times 0.90 \times 0.98 = 0.8379$

Answer: 83.8%. Note karo yeh sabse kamzor single part (0.90) se kam hai — jaise jaise links badhate ho, chains sirf aur less reliable hoti jaati hain.

Exercise 4.2

Do identical units parallel mein (system kaam karta hai agar kam se kam ek kaam kare), har ek $λ = 0.00008$ /h ke saath, $t = 8000$ h par. System reliability nikalo.

Recall Solution 4.2

Single-unit reliability: $R_{1} = e^{- 0.00008 \times 8000} = e^{- 0.64} = 0.5273$

Parallel logic: system tabhi fail hota hai jab dono fail ho jaayein. Ek unit ke fail hone ki probability $1 - R_{1}$ hai; dono fail hona (independent) $(1 - R_{1})^{2}$ hai. $R_{sys} = 1 - (1 - R_{1})^{2} = 1 - (0.4727)^{2} = 1 - 0.2234 = 0.7766$

Answer: 77.7% — ek unit ke 52.7% se kaafi upar. Redundancy survival dilata hai.

Exercise 4.3

Ek subsystem ko mission par $R \geq 0.99$ chahiye jahan ek single unit sirf $R_{1} = 0.70$ deta hai. Parallel mein kitne identical units required hain?

Recall Solution 4.3

$n$ units ke liye parallel formula set up karo: $R_{sys} = 1 - (1 - R_{1})^{n} \geq 0.99$ $(1 - 0.70)^{n} \leq 0.01 \Rightarrow 0.3 0^{n} \leq 0.01$

Logs lekar $n$ solve karo (log woh tool hai jo $n$ ko exponent se neeche kheenchta hai): $n ln 0.30 \leq ln 0.01 \Rightarrow n \geq \frac{l n 0.01}{l n 0.30} = \frac{- 4.60517}{- 1.20397} = 3.825$

(Inequality flip hoti hai kyunki $ln 0.30 < 0$ .) $n$ ek integer hona chahiye, isliye upar round karo.

Answer: $n = 4$ units. Check: $1 - 0.3 0^{4} = 1 - 0.0081 = 0.9919 \geq 0.99$ . ✓ (Teen units sirf $1 - 0.3 0^{3} = 0.973$ dete hain, kaafi nahi.)

Level 5 — Mastery

Goal: koi template directly fit nahi hota; tumhe pieces khud assemble karne hain.

Exercise 5.1

Ek spacecraft power chain hai: ek solar array ( $λ_{A} = 3 \times 1 0^{- 5}$ /h) series mein ek parallel pair of batteries ke saath (har ek $λ_{B} = 1 \times 1 0^{- 4}$ /h). Poori chain ki reliability 2-year mission par nikalo.

Recall Solution 5.1

Step 1 — time convert karo. $t = 2 \times 8766 = 17, 532$ h.

Step 2 — array reliability. $R_{A} = e^{- 3 \times 1 0^{- 5} \times 17, 532} = e^{- 0.52596} = 0.5910$

Step 3 — ek battery. $R_{B} = e^{- 1 \times 1 0^{- 4} \times 17, 532} = e^{- 1.7532} = 0.1732$

Step 4 — battery pair parallel mein (survive karta hai agar kam se kam ek battery zinda rahe): $R_{batt} = 1 - (1 - R_{B})^{2} = 1 - (0.8268)^{2} = 1 - 0.6836 = 0.3164$

Step 5 — array series mein battery block ke saath (dono blocks kaam karne chahiye): $R_{sys} = R_{A} \times R_{batt} = 0.5910 \times 0.3164 = 0.1870$

Answer: 18.7% — poor. Yahi woh number hai jo ek engineer ko aur battery redundancy ya higher-grade array add karne par majboor karta hai. Dekho Mission Design Constraints.

Exercise 5.2

Ek single exponential component ke liye, kaunsa mission duration $t$ (MTTF ke fraction ke roop mein) exactly $R = 0.99$ deta hai? Answer ko MTTF ke multiple ke roop mein express karo.

Recall Solution 5.2

MTTF ke terms mein set up karo. $λ = 1/ MTTF$ likho, toh $λ t = t / MTTF$ . Maano $x = t / MTTF$ . $R = e^{- x} = 0.99 \Rightarrow x = - ln 0.99 = 0.010050$

Answer: $t \approx 0.01005 MTTF$ . 99% reliability hit karne ke liye tum MTTF ka sirf lagbhag 1% hi use kar sakte ho. Yeh L3 warning ko quantify karta hai: high reliability ke liye mean life se kaafi neeche operate karna padta hai.

Exercise 5.3

Ek repairable subsystem ka $MTTF = 800$ h aur $MTTR = 4$ h hai. Nikalo (a) uska MTBF aur (b) uski steady-state availability $A = \frac{MTTF}{MTTF + MTTR}$ . Dekho Availability vs. Reliability.

Recall Solution 5.3

(a) MTBF. Ek repairable system ke liye, $MTBF = MTTF + MTTR = 800 + 4 = 804$ h.

(b) Availability. $A = \frac{800}{800 + 4} = \frac{800}{804} = 0.99502$

Answer: MTBF $= 804$ h aur availability $\approx 99.50%$ — system time ka lagbhag $99.5%$ "up" rehta hai. Note karo ki reliability (kya yeh bina ruke survive karta hai?) aur availability (time ka kitna fraction yeh usable hai?) alag sawal hain; availability high reh sakti hai chahe repairs frequent bhi hon agar woh short hain.

Related deep-dives: Bathtub Curve · Weibull Distribution · Poisson Process · Fault Tree Analysis · Safety-Critical Systems · Probability and Statistics Fundamentals.