Worked examples — Redundancy — cold standby, hot standby, active redundancy

3.6.32 · D3 · Physics › Spacecraft Structures & Systems Engineering › Redundancy — cold standby, hot standby, active redundancy

Yeh page parent topic ke redundancy formulas ko drill karti hai jab tak har case muscle memory na ban jaye. Hum koi naya theory introduce nahi karenge — hum reliability tools ko stress-test karenge har tarah ke inputs par jo ek problem throw kar sakti hai: normal numbers, zero time, infinite time, degenerate switches, word problems, aur ek exam twist.

Shuru karne se pehle, ek reminder un symbols ka jo humne parent note mein already earn kiye hain (yahan kuch naya nahi hai):

Recall Woh symbols jo humne already build kiye

• $\lambda$ ::: failure rate — ek single unit average mein kitne failures per hour deta hai. Bada $\lambda$ = jaldi fail hota hai.
• $R(t)$ ::: time $t$ par reliability — probability ki ek unit abhi bhi zinda hai time $t$ par. $1$ se start hoti hai (pakka zinda), $0$ ki taraf decay karti hai.
• $e^{-\lambda t}$ ::: ek unit ka survival curve — exponential kyun? Kyunki constant failure rate ka matlab hai ki agले ghante survive karne ka chance har ghante same hota hai, aur us constant survival ko kai ghanton par multiply karne se exponential milta hai. Yeh ek akela curve hai jisme "no memory" hoti hai.
• $\text{MTTF}={\int }_0^{\infty }R(t)dt$ ::: mean time to failure — survival curve ke neeche ka area. Zyada area = average mein zyada jeeta hai.

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Scenario matrix

Har redundancy problem jo tumhe milegi, in case classes mein se kisi ek mein aati hai. Neeche ke worked examples mein us cell ka label lagaya gaya hai jise woh cover karte hain, aur milkar woh har cell ko hit karte hain.

Cell	Case class	Kyun tricky hai
A	Normal cold vs hot MTTF	numbers plug karo, $2/\lambda$ vs $3/2\lambda$ compare karo
B	Ek specific finite time $t$ par reliability	$R(t)$ evaluate karo, MTTF se confuse mat ho
C	Zero-time input ($t=0$)	degenerate: har system $R=1$ padhta hai
D	Infinite-time limit ($t\to \infty$)	degenerate: har $R\to 0$
E	TMR (active, 2-of-3) crossover	TMR bade $t$ ke liye single se worse hoti hai
F	Imperfect switch (single point of failure)	switch reliability $R_s<1$ multiply ho jaati hai
G	Power-budget word problem	reliability gain vs watts cost
H	Exam twist — "TMR single ko kab beat karta hai?"	$R$ ke liye inequality solve karo

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Example A — Cold vs hot standby MTTF (normal numbers)

Forecast: Aage padhne se pehle ordering guess karo. Kaun sa redundancy scheme sabse lamba average life dega?

1. Single unit. ${\text{MTTF}}_{\text{single}}=\frac{1}{\lambda }=\frac{1}{2\times 10^{-5}}=50,000$ h. Yeh step kyun? Yeh hamara baseline hai — baaki sab kuch iske against measure hota hai.

2. Hot standby. Parent se, ${\text{MTTF}}_{\text{hot}}=\frac{3}{2\lambda }=1.5\times 50,000=75,000$ h. Yeh step kyun? Dono units saath age karte hain, isliye backup apni aadhi life sirf warm baithke burn kar deta hai — hum sirf aadhi lifetime gain karte hain.

3. Cold standby. ${\text{MTTF}}_{\text{cold}}=\frac{2}{\lambda }=2\times 50,000=100,000$ h. Yeh step kyun? Cold backup ka clock sirf switchover par shuru hota hai, isliye hum ek poori second lifetime gain karte hain.

Verify: Ratio ${\text{MTTF}}_{\text{cold}}:{\text{MTTF}}_{\text{hot}}:\text{single}=2:1.5:1$. Cold kaagaz par jeetta hai — parent ke key comparison $\frac{2}{\lambda }>\frac{3}{2\lambda }$ se match karta hai. Units poore mein hours hain. ✓ (Yeh cell A hai.)

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Example B — Ek finite mission time par reliability

Forecast: MTTF ne kaha tha cold best tha. Ek specific short time par, kya cold abhi bhi better lagega, aur kitna?

1. Single unit. $\lambda t=2\times 10^{-5}\times 10^4=0.2$, isliye $R=e^{-0.2}=0.8187$. Yeh step kyun? Hume pehle single-unit survival chahiye kyunki dono redundant formulas isi se bante hain.

2. Hot standby. $R_{\text{hot}}=2e^{-\lambda t}-e^{-2\lambda t}=2(0.8187)-e^{-0.4}=1.6374-0.6703=0.9671$. Yeh step kyun? Hot standby tabhi fail hota hai jab dono fail ho jaayein; yeh formula $1-(1-e^{-\lambda t}{)}^2$ hai.

3. Cold standby. $R_{\text{cold}}=e^{-\lambda t}(1+\lambda t)=0.8187(1+0.2)=0.8187\times 1.2=0.9824$. Yeh step kyun? Cold pehle do Poisson terms use karta hai kyunki yeh doosri failure tak survive karta hai, aur backup wait karte waqt age nahi hua.

Verify: Is time par order hai $R_{\text{cold}}=0.9824>R_{\text{hot}}=0.9671>R_{\text{single}}=0.8187$ — cold yahan bhi best hai, Example A se consistent. Sab $[0,1]$ mein probabilities hain. ✓

Neeche ki figure teeno survival curves plot karti hai. Notice karo ki sab left edge par $R=1$ se launch karte hain (Example C), grey single-unit curve sabse tez sagging karti hai, aur red dashed line $t=10,000$ h par teeno colored dots exactly unhi numbers par baithe hain jo humne abhi compute kiye — blue (cold) sabse upar, orange (hot) thoda neeche, grey (single) kaafi neeche.

(Cell B.)

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Example C — Zero-time degenerate case

Forecast: Compute karne se pehle — jis pal turn on karo usi instant mein kuch bhi fail hone ki probability kya hai?

1. Single: $e^{-\lambda \cdot 0}=e^0=1$.
2. Hot: $2e^0-e^0=2-1=1$.
3. Cold: $e^0(1+0)=1\cdot 1=1$. Yeh steps kyun? $t=0$ par kisi cheez ko fail hone ka time nahi mila. Ek sahi reliability curve $1$ se start honi chahiye (certain survival). Koi bhi formula jo $t=0$ par kuch aur deta woh galat hota.

Verify: Teeno $1$ equal hain. Yeh sanity check hai jo har reliability formula ko pass karna chahiye — agar tumhara $1$ se start nahi hota, to tumne ise galat copy kiya hai. ✓ (Cell C.)

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Example D — Infinite-time limit

Forecast: Infinite time dene par, kya koi bhi redundancy scheme hamesha ke liye survive karta hai?

1. Single: $\underset{t\to \infty }{\lim }{e}^{-\lambda t}=0$. Yeh step kyun? Hum simplest curve se shuru karte hain kyunki baaki teeno sab $e^{-\lambda t}$ se bane hain — agar yeh marta hai, to woh sab isse inherit karte hain.

2. Hot: $2e^{-\lambda t}-e^{-2\lambda t}\to 2\cdot 0-0=0$. Yeh step kyun? Dono terms mein decaying exponential hai, isliye har term independently zero par khich jaati hai — zeros ka sum zero hai.

3. Cold: $e^{-\lambda t}(1+\lambda t)\to 0$ (exponential linear $\lambda t$ ko beat karta hai). Yeh step kyun? Yeh woh term hai jo tumhe fool kar sakti hai — $(1+\lambda t)$ factor barhta hai, lekin $e^{-\lambda t}$ kisi bhi polynomial ke barhne se zyada tez shrink karta hai, isliye product phir bhi mar jaata hai. Humein sure hone ke liye sabse tez-barhne wala factor check karna hoga.

4. TMR: $3R^2-2R^3$ jisme $R=e^{-\lambda t}\to 0$ hai, $0$ deta hai. Yeh step kyun? TMR ki reliability $R$ mein ek polynomial hai jisme koi constant term nahi hai, isliye jab underlying unit reliability $R\to 0$ hoti hai to poora voter output bhi zero par gir jaata hai.

Verify: Har system $0$ ki taraf tend karta hai. Redundancy time kharidti hai, kabhi immortality nahi — yeh decay ko slow karti hai, ise rok nahi sakti. Slower-decaying curve ke neeche ka area exactly woh MTTF gain hai jo humne Example A mein compute kiya. ✓ (Cell D.)

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Example E — TMR aur uska crossover point

Forecast: TMR hamesha single unit ko beat karta hai... ya karta hai? Guess karo jab har unit already unreliable ho ($R=0.4$) to kya hoga.

1. High-quality units, $R=0.9$. $R_{\text{TMR}}=3(0.81)-2(0.729)=2.43-1.458=0.972$. Yeh step kyun? Jab units acchi hoti hain, "2-of-3" rare single failure ko mask kar deta hai — TMR ($0.972$) single ($0.9$) ko beat karta hai.

2. Poor units, $R=0.4$. $R_{\text{TMR}}=3(0.16)-2(0.064)=0.48-0.128=0.352$. Yeh step kyun? Jab units kharab hoti hain, do ka simultaneously kaam karna ek ke kaam karne se mushkil hota hai — TMR ($0.352$) single ($0.4$) se worse hai!

Verify: Crossover exactly $R=0.5$ par hai (Example H dekho). Iske upar TMR help karta hai; neeche TMR hurt karta hai. Isliye $0.9$ help karta hai, $0.4$ hurt karta hai — match karta hai. ✓

Neeche ki figure grey diagonal (single unit, output = input) ko green TMR curve ke against draw karti hai. Jahan green curve diagonal ke upar hai (green shaded, $R>0.5$) TMR help karta hai; jahan yeh neeche dip karti hai (red shaded, $R<0.5$) TMR hurt karta hai. $R=0.9$ par blue dot green zone mein land karta hai, $R=0.4$ par orange dot red zone mein, aur dono curves exactly red dot $R=0.5$ par cross karti hain.

(Cell E.)

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Example F — Imperfect switch (single point of failure)

Forecast: Parent ne warn kiya tha ki switch khud ek single point of failure hai. Guess karo: kya ek $0.98$ switch redundancy benefit ko wipe out kar deta hai?

1. Ideal redundant pair. $R_{\text{pair}}=1-(0.05{)}^2=1-0.0025=0.9975$. Yeh step kyun? Yeh "do computers, perfect switch" fantasy hai — woh number jo log quote karte hain.

2. Switch se multiply karo. Poori chain tabhi kaam karti hai jab switch bhi kaam kare, isliye multiply karo: $R_{\text{real}}=R_{\text{pair}}\times R_s=0.9975\times 0.98=0.97755$. Yeh step kyun? Switch redundant block ke saath series mein hai — ek series weak link poore system ko cap kar deta hai (yeh ek reliability block diagram hai jisme ek series element hai).

3. Single unit se sanity check. Single unit $=0.95$. Isliye real redundancy $0.9776>0.95$ — abhi bhi ek gain hai, lekin $0.9975$ fantasy se kaafi kam.

Verify: $0.95<0.9776<0.9975$: redundancy abhi bhi help karta hai lekin switch zyaatar gain throw away kar deta hai. Isliye real spacecraft fault detection and isolation plus cross-strapping use karte hain taaki switch khud reliable ho. ✓ (Cell F.)

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Example G — Power-budget word problem

Forecast: Reliability kehti hai cold ka waise bhi zyada MTTF hai. Kya power budget agree karta hai ya conflict karta hai?

1. Hot-standby power draw check karo. Extra load $=12$ W. Available margin $=5$ W. Kyunki $12>5$, hot standby power budget violate karta hai. Yeh step kyun? Jo design power budget bust karta hai woh reliability se regardless infeasible hai — power ek hard constraint hai.

2. Cold-standby power draw check karo. Extra load $=0$ W (backup OFF). $0\le 5$: fit karta hai. Yeh step kyun? Cold ka zero standby power battery-tight spacecraft par uska headline advantage hai.

3. Timing requirement check karo. Cold ka $90$ s switchover $\le$ tolerable $90$ s gap: acceptable. Yeh step kyun? Cold ka ek hi downside (delay) yahan allowed limit ke andar hai.

Verify: Power ($0\le 5$ ✓), timing ($90\le 90$ ✓), aur cold ka pehle se Example A se zyada MTTF tha. Har constraint cold standby ki taraf point karta hai — yeh ek rare case hai jisme koi tradeoff nahi hai. ✓ (Cell G.)

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Example H — Exam twist: TMR crossover ke liye solve karo

Forecast: Example E ne hint kiya tha crossover $0.5$ hai. Ise prove karo aur baaki roots nikalo.

1. Equation set up karo. $3R^2-2R^3=R$. Yeh step kyun? "TMR single ko beat karta hai" ⇔ $3R^2-2R^3>R$; boundary equality hai.

2. Ek side par laao aur factor karo. $2R^3-3R^2+R=0\Rightarrow R(2R^2-3R+1)=0\Rightarrow R(2R-1)(R-1)=0$. Yeh step kyun? Factoring ek cubic ko teen simple roots mein turn kar deta hai — koi formula nahi chahiye.

3. Roots padho. $R=0$, $R=\frac{1}{2}$, $R=1$. Yeh step kyun? Ek product zero hota hai tabhi jab uska koi ek factor zero ho, isliye hum teeno factors $R$, $2R-1$, $R-1$ ko ek ek karke zero set karte hain — factoring ka poora point yahi hai, yeh seedha roots haath mein de deta hai.

4. Winning region decide karo. $R=0.9$ test karo (Example E): TMR $0.972>0.9$ ✓. $R=0.4$ test karo: TMR $0.352<0.4$ ✓. Isliye TMR single ko exactly tab beat karta hai jab $R>\frac{1}{2}$. Yeh step kyun? Roots sirf woh jagah mark karti hain jahan dono curves cross karti hain; kaun si side jeetti hai jaanne ke liye humein roots ke beech har interval mein ek point sample karna hoga — wahan difference ka sign poora interval settle kar deta hai.

Verify: Roots $\{0,\frac{1}{2},1\}$ cubic satisfy karte hain; sign tests confirm karte hain ki TMR $0.5$ ke upar help karta hai aur neeche hurt karta hai. Example E se consistent. ✓ (Cell H.)

Recall Quick self-test

$R=0.5$ par, kya TMR single unit ko beat karta hai? ::: Nahi — woh exactly equal hain; $R=0.5$ crossover hai. Cold standby ka MTTF hot se zyada kyun hai? ::: Cold backup OFF rahte waqt age nahi karta, isliye tum ek poori second lifetime gain karte ho sirf aadhi ki jagah. Ek "redundant" system jisme fragile switch hai — switch reliability block diagram mein kahan baitha hai? ::: Redundant block ke saath series mein, isliye yeh total reliability ko cap kar deta hai.

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Har idea kahan connect hota hai

Yeh bhi dekho: Failure Modes and Effects Analysis (FMEA) — kaun si failures ko cover karne ki zaroorat hai yeh jaanne ke liye, Single Event Upset (SEU) — radiation bit-flips ke liye jo switchovers trigger karte hain, aur Byzantine Fault Tolerance — jab koi faulty unit voter se jhooth bole.