Reliability — MTTF, MTBF, exponential failure model: Question bank

3.6.31 · D5 · Physics › Spacecraft Structures & Systems Engineering › Reliability — MTTF, MTBF, exponential failure model

Yeh ek thinking gym hai, calculator drill nahi. Neeche har item ek question ::: answer reveal hai. Question padho, apna jawab ek reason ke saath zor se bolo, phir reveal karo. Agar tumhara reason galat tha — chahe yes/no sahi bhi tha — toh tum trap mein fas gaye — yahi toh point hai.

Prerequisites parent topic mein hain. Related traps Bathtub Curve, Series vs. Parallel System Reliability, Redundancy Design, aur Availability vs. Reliability mein chhupi hain.

True or false — justify

True/false batao, phir reason do. Sahi verdict lekin galat reason — fir bhi miss count hoga.

Ek component jiska MTTF = 1000 h hai, uske baare mein guarantee hai ki woh fail hone se pehle lagbhag 1000 h chalega

False — MTTF ek population ka average hai; sirf

R (1/ λ) = e^{- 1} \approx 37%

units itne lambe time tak survive karte hain, matlab 63% apne MTTF se pehle hi mar jaate hain.

Agar failure rate

λ

constant hai, toh ek component jo already 500 h survive kar chuka hai woh "worn in" hai aur ek fresh component ke comparison mein chalte rehne ki zyada probability rakhta hai

False — exponential model memoryless hota hai; ek survivor ka future statistically ek bilkul naye unit jaisa hi dikhta hai, kyunki

R (t + s) / R (s) = e^{- λ t} = R (t)

Failure rate

λ

ko double karne se fixed time

t

par reliability

R (t)

half ho jaati hai

False —

R = e^{- λ t}

mein

λ

ka relation exponential hai, linear nahi;

λ

double karne se

R

square ho jaata hai, kyunki

e^{- 2 λ t} = (e^{- λ t})^{2}

, jo halving se kaafi zyada harsh drop hai.

Exponential model ke liye, MTBF aur MTTF dono ki numerical value

1/ λ

hoti hai

True — jab repair time negligible ho toh dono

1/ λ

ke barabar hote hain; labels ka fark sirf intent mein hai (MTTF non-repairable ke liye, MTBF repairable ke liye), arithmetic mein nahi.

Kya exponential model mein kisi

t > 0

ke liye reliability

R (t)

exactly 1 ho sakti hai

False —

e^{- λ t} = 1

sirf

t = 0

par hota hai (ya trivial case

λ = 0

, matlab "kabhi fail nahi hoga"); kisi bhi real

λ > 0

aur

t > 0

ke liye,

R (t) < 1

strictly hota hai.

Ek redundant parallel unit add karne se system failure rate ek naye constant value par aa jaata hai

False — ek parallel system ka failure rate constant nahi hota; woh pehle near zero se shuru hota hai aur time ke saath badhta hai, isliye neat "

λ =

const" exponential form kho jaata hai — chahe har unit exponential ho.

MTBF mein repair time shamil hoti hai, isliye MTBF hamesha MTTF se bada hota hai

Strict definition

MTBF = MTTF + MTTR

mein True hai, lekin spacecraft ke liye hum usually MTTR ko zero set karte hain (orbit mein repair nahi ho sakti), jisse MTBF wapas MTTF ke barabar ho jaata hai.

Ek zyada MTTF ka matlab hamesha kisi given mission ke liye zyada reliability hota hai

Exponential model ke liye fixed

t

par True — zyada MTTF ka matlab chota

λ = 1/ MTTF

hai, isliye

e^{- λ t}

bada hota hai; dono monotonically saath saath chalte hain.

Spot the error

Har statement mein ek flawed step hai. Use identify karo.

"Do parallel units, har ek ka

λ = 0.001

, isliye system ka

λ = 0.002

hai."

Error yeh hai ki parallel redundancy ke liye

λ

add kiya gaya — yeh rule series ke liye hai; parallel reliability

R_{sy s} = 1 - (1 - e^{- λ t})^{2}

hoti hai, jo reliability ko improve karti hai, worsens nahi.

R (t) = e^{- λ t}

jahan

λ t

hours mein hai, toh

t = 1000

h par exponent 1000 hai."

Exponent

λ t

dimensionless hona chahiye;

λ

ki units 1/hours hoti hain, isliye failures/hour × hours cancel ho jaate hain — tum

λ

se multiply karna bhool gaye.

"Solar array ka MTTF 20000 h hai, aur mission 43830 h ka hai, isliye reliability

20000/43830 = 46%

hai."

Reliability kabhi bhi times ka ratio nahi hoti; use karo

R = e^{- λ t} = e^{- t / MTTF} = e^{- 2.19} \approx 11%

, jo naive 46% se kaafi neeche hai.

"Failure rate negative hai kyunki reliability decrease hoti hai, isliye

λ < 0

λ = - \frac{1}{R} \frac{d R}{d t}

; minus sign negative slope

d R / d t

ko cancel kar deta hai, aur

λ > 0

milta hai — ek rate hamesha positive hota hai.

f (t) = λ e^{- λ t}

reliability hai, isliye

t = 0

par reliability

λ

ke barabar hai."

f (t)

failure density (ek PDF) hai, reliability nahi; reliability

R (0) = 1

hoti hai, jabki

f (0) = λ

yeh describe karta hai ki shuruat mein probability mass kitni tezi se jaati hai.

"Kyunki MTTF

= \int_{0}^{\infty} t f (t) d t

diverge karta hai (t infinity tak jaata hai), isliye MTTF infinite hai."

Integrand

t λ e^{- λ t}

itni tezi se decay karta hai ki exponential linear growth ko maar deta hai; integral finite value

1/ λ

par converge karta hai.

"Parallel redundancy reliability 100% kar deti hai, kyunki backup hone par fail nahi ho sakta."

Dono units abhi bhi fail ho sakte hain; parallel reliability

1 - (1 - R_{1})^{2} < 1

hoti hai — redundancy reliability badhati hai lekin certainty kabhi nahi aati.

Why questions

Reason explain karo, sirf fact nahi.

Exponential model bathtub curve ke flat middle mein kyun appear karta hai lekin ends mein nahi?

Sirf middle "useful life" region mein roughly constant

λ

hota hai; infant mortality (falling

λ

) aur wear-out (rising

λ

) ke liye Weibull Distribution chahiye kyunki constant rate bend nahi kar sakta.

R (t) = e^{- λ t}

radioactive decay se identical kyun lagta hai?

Dono

d R / d t = - λ R

follow karte hain — ek quantity jiska loss rate us cheez ke kitne bacha hai uske proportional hota hai — aur yeh ek differential equation physical context ki parwah kiye bina exponential solution force karta hai.

Engineers missions ko MTTF se kaafi neeche run karne ke liye kyun design karte hain, MTTF tak nahi?

t = MTTF

par reliability sirf

e^{- 1} \approx 37%

hoti hai; 90% survival tak pahunchne ke liye

t \approx 0.1/ λ

chahiye, yani MTTF ka lagbhag ek-dasva hissa.

Ek target reliability se

λ

solve karte waqt logarithm kyun aata hai?

Unknown ek exponential ke andar hai (

R = e^{- λ t}

); natural log exactly woh inverse operation hai jo

e^{(\cdot)}

ko "unwrap" karke

- λ t

expose karta hai.

Failure rate ko

- \frac{1}{R} \frac{d R}{d t}

kyun define kiya jaata hai, simply

- \frac{d R}{d t}

kyun nahi?

R

se divide karna surviving population ke against normalize karta hai — yeh poochha jaata hai ki "abhi bhi zinda hain unme se kitne fraction per unit time fail ho rahe hain," jo ki ek engineer jo parts replace kar raha hai uske liye actually matter karta hai.

Ek doosra parallel unit add karna tab bhi kyun help karta hai jab har unit fail hone ki zyada likely ho survive hone se?

System death ke liye dono ka fail hona zaroori hai, aur do sub-one probabilities multiply karne se chota number milta hai:

(0.63)^{2} \approx 0.40 < 0.63

, isliye joint failure single failure se rarer hai.

Hum series system ki overall reliability sirf uske worst single component se kyun describe nahi kar sakte?

Ek series system fail ho jaata hai agar koi bhi link fail ho jaaye, isliye reliabilities multiply hoti hain:

R_{sy s} = \prod R_{i}

, jo hamesha sabse chhote

R_{i}

\leq

hoti hai — har weak part poori chain ko aur neeche kheenchta hai. Dekho Series vs. Parallel System Reliability.

Edge cases

Model ko uski limits tak push karo aur dekho kya bachta hai.

Jab

λ = 0

ho toh

R (t)

kya hoga?

R (t) = e^{0} = 1

sabhi

t

ke liye — ek hypothetical component jo kabhi fail nahi hota; MTTF

= 1/ λ

infinite ho jaata hai, jo "immortal" ke saath consistent hai.

Kisi bhi

λ > 0

ke liye jab

t \to \infty

toh reliability ka kya hota hai?

R (t) \to 0

— infinite time dene par har real component aakhirkar fail ho jaata hai, isliye long-term survival probability zero par drain ho jaati hai chahe

λ

kitna bhi chota ho.

Exactly

t = MTTF

par reliability kya hoti hai?

R (1/ λ) = e^{- 1} \approx 0.368

— kisi bhi exponential system ke liye ek fixed universal number, specific

λ

se independent.

Agar MTTR (repair time) negligible nahi hai, toh kya MTBF abhi bhi

1/ λ

ke barabar hai?

Nahi — tab

MTBF = MTTF + MTTR = 1/ λ + MTTR

, aur uptime aur cycle-time ka fark Availability vs. Reliability ke liye matter karta hai.

Ek parallel system ke liye

t = 0

par reliability kya hai?

R_{sy s} (0) = 1 - (1 - 1)^{n} = 1

— har unit shuru mein working hota hai, isliye system bilkul ek single unit ki tarah launch par perfectly reliable start karta hai.

MTTF ka kya matlab hai ek truly non-repairable, single-use item jaise pyrotechnic bolt ke liye?

Yeh ek batch par average time-to-failure hai; koi "between failures" nahi hota kyunki item ek baar fire hota hai aur khatam ho jaata hai, isliye MTBF meaningless hai aur sirf MTTF apply hota hai.

Agar do parallel units independent nahi hain (ek shared power bus dono ko maar sakta hai), toh kya

R_{sy s} = 1 - (1 - R_{1})^{2}

abhi bhi valid hai?

Nahi — woh formula independence assume karta hai; ek common-cause failure dono ko correlate kar deta hai, isliye real

R_{sy s}

aur neeche hoti hai, aur yahi exactly woh situation hai jiske liye Fault Tree Analysis bana hai.

Recall Is page ko close karne se pehle fast self-check

Ek-line answers; reveal se pehle khud force karo. $t =$ MTTF par survivor fraction? ::: Lagbhag 37% ( $e^{- 1}$ ). Series reliability rule? ::: Multiply karo: $R_{sy s} = \prod R_{i}$ . Parallel reliability rule? ::: $R_{sy s} = 1 - \prod (1 - R_{i})$ . Exponential $λ$ constant hai ya time-varying? ::: Constant — yahi toh poora assumption hai. Kaunsi property "already survived" ko irrelevant banati hai? ::: Memorylessness.