Redundancy — TMR (triple modular redundancy), voting logic: Foundations

5.5.25 · D1 · Coding › Embedded Systems & Real-Time Software › Redundancy — TMR (triple modular redundancy), voting logic

TMR and voting logic ka parent note ek pile of shorthand pe rely karta hai. Uss shorthand ko abhi list karne ki bajaye (pehle mile bina woh sirf noise hoga), yeh page har symbol ko sirf tab introduce karta hai jab humein uski zaroorat ho, har ek ko scratch se build karte hue, ek aisi order mein jahan har idea apne pehle wale idea pe rest karta hai. Ek 12-saal ka bachcha jisne yeh kabhi nahi dekha, line one se follow kar sake.

1. Ek "module" aur letter $M$

Ise picture karo. Teen identical vending machines side by side khadi hain. Teeno pe same button dabaao; har ek ek can drop karta hai. Same machine, teen bodies.

Topic ko iske zaroorat kyun hai. TMR define hota hai teen modules se. "Kuch identical copies mein se ek copy" ka idea kiye bina, redundancy shabd ka koi matlab nahi.

Figure — Redundancy — TMR (triple modular redundancy), voting logic

Figure 1 — Ek input $x$ (yellow) teen identical modules $M_{1}, M_{2}, M_{3}$ (blue boxes) mein fan out karta hai, har ek apna output $y_{1}, y_{2}, y_{3}$ (green) produce karta hai. Takeaway: machine same hai; sirf body aur label number alag hai.

2. Input $x$ , function $f$ , aur output $y = f (x)$

Ise picture karo. Ek box jiske left mein ek slot hai ( $x$ andar jaata hai) aur right mein ek slot hai ( $y$ bahar aata hai). Box ka andar rule $f$ hai.

Kyunki har module same rule ko same input pe run karta hai, isliye hum har module ka output likhte hain: $y_{i} = M_{i} (x)$ Yahan $y_{i}$ hai "copy number $i$ ka output". Dhyaan se padho: $M_{i}$ machine hai (box), jabki $y_{i}$ woh number hai jo usse nikalta hai. Agar sabhi copies healthy hain, toh $y_{1} = y_{2} = y_{3}$ . Agar ek broken hai, uska $y$ disagree karta hai.

Topic ko iske zaroorat kyun hai. Voter $y_{1}, y_{2}, y_{3}$ compare karta hai. Baaki sab kuch inn teen outputs ka comparison hai, isliye pehle inhein name karna zaroori hai.

3. Probability $P$ , aur reliability $R$

Ise picture karo. Length $1$ ki ek bar. $R$ fraction ko green colour karo (works) aur baaki $1 - R$ ko red (fails). Dono pieces hamesha full bar tak add hote hain.

Figure 2 — Certainty bar ki total length $1$ hai. Green slice $R$ hai (module works); red slice $1 - R$ hai (module fails). Takeaway: work aur fail sirf do hi outcomes hain, isliye unki probabilities exactly $1$ tak add up honee chahiye.

Topic ko iske zaroorat kyun hai. TMR ka poora point reliability raise karna hai. Yeh prove karne ke liye ki yeh help karta hai, humein pehle aur baad $R$ measure karna hoga — isliye $R$ aur $1 - R$ poori derivation ki currency hain.

4. Kyun independent copies ke liye probabilities multiply hoti hain

Ise picture karo. Do coins toss karo. Heads-then-heads ka chance $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$ hai, kyunki $2 \times 2$ grid ke chaar equally-likely corners mein se sirf ek "both heads" hai. $\frac{1}{2}$ ki jagah $R$ rakh do aur tum paoge $R^{2}$ ; teen copies ke liye, $R^{3}$ .

Topic ko iske zaroorat kyun hai. Parent ka Step 1, " $P (all 3 work) = R^{3}$ ", tabhi sach hai jab failures independent hon. Yeh assumption real life mein fragile hai — jab yeh toot jaata hai tum paate ho Common-Cause Failures, jiske baare mein topic warn karta hai.

5. Ways count karna: $(2 3)$

Ise picture karo. Teen switches; exactly do ON karne hain. List karo: (1,2), (1,3), (2,3). Teen pictures, isliye number $3$ .

Topic ko iske zaroorat kyun hai. "Exactly 2 of 3 work" $3$ distinct arrangements mein ho sakta hai, har ek ki probability $R^{2} (1 - R)$ hai. Toh total $3 R^{2} (1 - R)$ hai — parent ka Step 2. Counting factor ke bina tum teen baar undercount karte.

6. Disjoint cases add karna, aur full TMR reliability

Abhi hamare paas do pieces hain: $R^{3}$ (teeno kaam karte hain) aur $3 R^{2} (1 - R)$ (exactly do kaam karte hain). Ek TMR system healthy hai jab kam se kam do modules kaam karein, yaani "teeno kaam karte hain" ya "exactly do kaam karte hain". "Or" combine karne ke liye humein probability ka ek aur rule chahiye.

"Teeno kaam karte hain" aur "exactly do kaam karte hain" disjoint hain: tum same trial mein teeno ko working aur exactly ek ko failing nahi rakh sakte — survivors ki count ya toh $3$ hai ya $2$ , kabhi dono nahi. Toh hum unhe add kar sakte hain.

Sanity check. $R = 0.9$ ke saath: $R_{TMR} = 0.81 \times (3 - 1.8) = 0.81 \times 1.2 = 0.972$ — ek single module ke $0.9$ se better. Yeh exactly woh formula hai jo parent note use karta hai; ab tum jaante ho har term kahan se aaya.

7. Failure rate $λ$ aur curve $e^{- λ t}$

Definition Number

$e$ aur $e^{- λ t}$ ka matlab == $e$ == $\approx 2.718$ ek fixed mathematical constant hai, jaise $π$ . Yeh woh special base hai jiske liye continuous, steady decay sabse clean hai: yeh woh number hai jo tum tab approach karte ho jab ek tiny growth "ek saath" compound karo, $e = (1 + \frac{1}{n})^{n}$ jaise $n$ bahut bada hota jaata hai. Exponentiation $e^{k}$ ka matlab hai " $e$ ko $k$ baar khud se multiply karo" jab $k$ ek whole number ho, aur yeh smoothly kisi bhi $k$ ke liye extend hota hai. Ek negative exponent ise ek fraction mein flip karta hai: $e^{- k} = 1/ e^{k}$ , isliye jaise $k$ badhta hai value $0$ ki taraf shrink hoti hai par kabhi reach nahi karti. $k = λ t$ plug karo aur milta hai == $e^{- λ t}$ ==: ek curve jo $t = 0$ pe $1$ equal karta hai (kyunki $e^{0} = 1$ : ek brand-new module certainly working hai) aur time $t$ badhne ke saath $0$ ki taraf slide karta jaata hai.

Ise picture karo. Ek slide jo full height se start hoti hai aur gently neeche curve hoti jaati hai, kabhi floor ko touch nahi karti. Steep slide = bada $λ$ .

Figure 3 — Reliability over time, $R (t) = e^{- λ t}$ , ek slow-failing module ke liye (blue, small $λ$ ) aur ek fast-failing ke liye (red, large $λ$ ). Dono $R = 1$ (yellow dot) se start hote hain aur $0$ ki taraf curve karte hain bina use touch kiye. Takeaway: bada $λ$ = steep drop = chhoti useful life.

Yahi tool kyun, straight line kyun nahi? Ek straight line eventually zero se neechey cross kar jaati (probability ke liye impossible) aur assume karti ki cheezein ek fixed clock time pe fail hoti hain. $e^{- λ t}$ ki jagah kehta hai "har equal time slice mein, survivors ka same fraction fail hota hai" — random hardware death ka natural model. Yahi wajah hai ki parent likhta hai $R (t) = e^{- λ t}$ .

8. Majority gate aur voter output $V$

Ab tak $M_{i}$ matlab tha poori machine aur $y_{i}$ matlab tha uska output number. Voter circuit ke baare mein cleanly baat karne ke liye, hum sabse simple case mein zoom karte hain jahan har output sirf ek bit hai — sirf $0$ ya $1$ . Inn teen bits ko == $b_{1}, b_{2}, b_{3}$ == kaho, jahan $b_{i}$ module $i$ ka one-bit output hai (toh $b_{i} = y_{i}$ jab output already ek single bit hai).

Voter ko ek sentence ki tarah padho. $V = b_{1} b_{2} + b_{2} b_{3} + b_{1} b_{3}$ " $V$ $1$ hai agar bits 1-and-2 dono $1$ pe agree karein, OR bits 2-and-3 karein, OR bits 1-and-3 karein." Koi bhi pair $1$ pe agree kare toh $V = 1$ force ho jaata hai — yeh exactly "teen mein se kam se kam do" hai.

Figure 4 — Majority voter: teen AND gates (yellow), har ek module ke pair ke liye, ek single OR gate (green) ko feed karte hain jo $V$ output karta hai. Takeaway: $V = 1$ jab bhi koi ek pair $1$ pe agree kare, yaani jab bhi teen inputs mein se kam se kam do $1$ hon.

Topic ko iske zaroorat kyun hai. Yeh woh digital voter hai jo $V$ produce karta hai. Continuous (analog) outputs ke liye same "odd one out ignore karo" kaam median karta hai — aage define kiya gaya hai.

9. Median vs average, symbolically define kiya gaya

Median kyun, average kyun nahi? Average $\frac{a + b + c}{3}$ kisi bhi wild value ki taraf drag ho jaata hai; median simply ise ignore karta hai.

$average (100, 100, 1000) = \frac{1200}{3} = 400$ — broken $1000$ result pollute kar deta hai.
$median (100, 100, 1000) = 100$ — outlier throw away ho jaata hai. ✅

Yahi ek difference hai kyun analog voters median use karte hain: yeh "2-out-of-3 majority" ka continuous cousin hai.

Prerequisite map

Neeche ka diagram ek dependency graph hai: ise top-to-bottom padho, har box is page ka ek idea hai, aur har arrow ka matlab hai "tail pe idea pehle samajhna zaroori hai head pe idea se pehle". Koi bhi arrows ka path follow karo aur tum exactly usi order mein chal rahe ho jisme humne cheezein build ki; har path eventually "TMR topic" box mein funnel ho jaata hai.

Aage feed hota hai: Fault Tolerance Fundamentals, Watchdog Timers, Redundancy vs. Diversity, aur Byzantine Fault Tolerance ke deeper failure models mein.

Equipment checklist

Khud test karo — reveal karne se pehle answer do.

M_{2}

mein subscript ka kya matlab hai?

Yeh copy label karta hai — teen identical modules mein se "module number 2".

f (x)

ka kya matlab hai, aur kya nahi?

"Rule

f

ko input

x

pe apply karo"; yeh

f

multiplied by

x

bilkul nahi hai.

M_{i}

aur

y_{i}

mein kya fark hai?

M_{i}

machine hai (box);

y_{i} = M_{i} (x)

woh number hai jo woh machine output karta hai.

Agar reliability

R

hai, toh failure ka chance kya hai?

1 - R

, kyunki working aur failing milke poori probability bar ki length

1

fill karte hain.

Hum

P (all 3 work) = R^{3}

kyun likh sakte hain?

Kyunki modules independently fail hote hain, isliye unki working probabilities multiply hoti hain.

Do probabilities ADD karne ki permission kab hai?

Jab events disjoint (mutually exclusive) hon — woh ek saath dono nahi ho sakte.

(2 3) = 3

kyun hai?

Teen distinct ways hain yeh choose karne ke liye ki teen modules mein se kaunse do survivors hain.

Full TMR reliability formula kya hai?

R_{TMR} = R^{3} + 3 R^{2} (1 - R) = R^{2} (3 - 2 R)

Kin values of

R

ke liye TMR single module ko beat karta hai?

Sirf jab

R > \frac{1}{2}

;

R = \frac{1}{2}

pe tie hota hai, aur

R < \frac{1}{2}

ke liye TMR worse hai.

e

kya hai, aur

e^{- λ t}

time ke saath kya karta hai?

e \approx 2.718

ek fixed constant hai;

e^{- λ t}

1

se start hota hai aur smoothly

0

ki taraf decay karta hai, bade

λ

ke liye tez.

Voter

V = b_{1} b_{2} + b_{2} b_{3} + b_{1} b_{3}

mein

b_{i}

\cdot

aur

+

kya hain?

b_{i}

modules ke one-bit outputs hain;

\cdot

(juxtaposition) logical AND hai;

+

logical OR hai — arithmetic nahi.

median (a, b, c)

kaise compute karte hain aur average se better kyun hai?

Teeno ko sort karo aur middle lo (ya

max (min (a, b), min (b, c), min (a, c))

); yeh ek outlier ignore karta hai jabki average uski taraf drag ho jaata hai.

Recall Self-check: ek sentence mein core idea state karo.

Same kaam teen identical copies pe chalao aur us answer par trust karo jis pe unme se kam se kam do agree karein, taaki ek faulty copy mask ho jaaye.