Infinity Fabric - mesh interconnects: Foundations — Hinglish deep dive

6.3.11 · D1 · Hardware › Interconnects, Buses & SoC › Infinity Fabric - mesh interconnects

Is page pe assume kiya gaya hai ki aap kuch nahi jaante. Har letter, symbol, aur buzzword jo parent note mein aata hai, woh yahan order mein build kiya gaya hai, har ek ke saath ek picture. Upar se neeche padhte jaiye.

1. The players: kaun kaun se boxes hain?

Road draw karne se pehle, humein jaanna hai ki un roads pe kaun chal raha hai.

Prerequisites ka building block: agar "core", "cache", aur "memory controller" words fuzzy lagte hain, toh 6.3.1-Bus-ArchitectureBasics mein wiring background ek achha companion hai — yeh page seedha usi pe build karta hai. (Hum har wiring word define karte hain jo hum use karte hain, jisme bus bhi shamil hai, section 2 mein neeche, toh aap yahan padhna jaari rakh sakte hain.)

Figure — Infinity Fabric - mesh interconnects

Figure dekho: teen squares (ek core, ek cache slice, ek memory controller) har ek ke paas ek chhota stub — woh stub plug-in point hai jahan baad mein ek road attach hogi. (Hum is plug-in point ko uska proper naam, port, section 3 mein denge.) In characters ko yaad rakho; neeche har topology sirf inhi squares ko rearrange karti hai.

2. Teen road shapes: bus, ring, mesh

Ab hum squares ko connect karte hain. Teen shapes hain jinhe parent note compare karta rehta hai.

3. Hop, router, aur port

"Fast" aur "slow" ke baare mein baat karne ke liye humein count karna hoga.

Figure ek single 5-port router dikhata hai. Chaar grey arrows neighbour routers ke compass directions hain; neeche jaata magenta arrow is router ke apne core ka local port hai. Parent note ka har mesh bas inhi tiles ko edge-to-edge wire karne se banta hai.

4. Coordinates aur Manhattan distance

Parent note $(0, 0)$ aur $(3, 2)$ jaise positions likhta hai aur hops compute karta hai. Yahan woh sab kya matlab rakhte hain.

Yeh formula kyun, aur straight-line distance kyun nahi? Grid pe aap buildings ke aarpaar diagonally nahi cut kar sakte — aapko streets ke along chalha padta hai, East/West phir North/South. Toh honest hop count horizontal aur vertical legs ko jodta hai. Isliye iska naam Manhattan ke block layout ke naam pe rakha gaya hai.

5. Symbol $N$ aur square root $N$

Common mistake Kya hoga agar

$N$ perfect square nahi hai? Neat formula $k = N$ tabhi whole number deta hai jab perfect squares hon ( $16, 25, 36, 64 \dots$ ). Real chips kabhi itne tidy nahi hote — ek "18-core" mesh square nahi ho sakta. Practice mein do cheezein hoti hain:

Rectangular meshes. Do side-lengths use karo, $k_{x}$ columns by $k_{y}$ rows, with $k_{x} \cdot k_{y} = N$ (e.g. $6 \times 3 = 18$ ). Averages cleanly generalise hote hain: column leg average $\approx k_{x} /3$ aur row leg $\approx k_{y} /3$ , toh avg hops $\approx (k_{x} + k_{y}) /3$ . Square case bas $k_{x} = k_{y} = N$ hai.
"Square" estimate round karna. Jab koi non-square $N$ ke liye $N$ quote karta hai toh unka matlab approximation hai; nearest rectangle $⌊ N ⌋ \times ⌈ N / ⌊ N ⌋⌉$ lo (floor $⌊ \cdot ⌋$ = round down, ceiling $⌈ \cdot ⌉$ = round up). Answer thoda hi shift hota hai.

Bottom line: $k = N$ ko clean square special case treat karo; messy core counts ke liye $k_{x}, k_{y}$ swap karo.

6. Averages, $E [\cdot]$ symbol, aur $k /3$ kahan se aata hai

Parent note $E [∣ x_{2} - x_{1} ∣]$ likhta hai. Ghabrao mat — yeh ek average hai.

$E [∣ x_{2} - x_{1} ∣] = \frac{k ^{2} - 1}{3 k}$ (column part) derive karna, step by step. Hum sirf $x$ coordinate dekhte hain. Dono $x_{1}$ aur $x_{2}$ columns $0, 1, \dots, k - 1$ se uniformly pick kiye jaate hain, toh har ek ke liye $k$ choices hain — kul $k^{2}$ ordered pairs, har ek equally likely. Average hai sabhi gaps ka total divided by $k^{2}$ :

E [∣ x_{2} - x_{1} ∣] = \frac{1}{k ^{2}} S a = 0 \sum k - 1 b = 0 \sum k - 1 ∣ a - b ∣ .

Ab hum actually sum $S$ compute karte hain, sirf quote nahi karte. Pairs ko unke gap $g = ∣ a - b ∣$ ke hisaab se group karo:

gap $g = 0$ : pairs jaise $(0, 0), (1, 1), \dots$ — inke $k$ hain, contributing $0$ .
gap $g = 1$ : pairs $(0, 1), (1, 0), (1, 2), (2, 1), \dots$ — $k - 1$ adjacent positions mein se har ek ke liye 2 orderings hain, toh $2 (k - 1)$ pairs, har ek contributing $1$ .
gap $g$ generally: $k - g$ starting positions hain aur $2$ orderings, toh $2 (k - g)$ pairs, har ek contributing $g$ .

Inhe jodte hain:

S = g = 1 \sum k - 1 g \cdot 2 (k - g) = 2 g = 1 \sum k - 1 (k g - g^{2}) .

Dono schoolbook sums use karo $\sum_{g = 1}^{k - 1} g = \frac{( k - 1 ) k}{2}$ aur $\sum_{g = 1}^{k - 1} g^{2} = \frac{( k - 1 ) k ( 2 k - 1 )}{6}$ :

S = 2 [k \cdot \frac{( k - 1 ) k}{2} - \frac{( k - 1 ) k ( 2 k - 1 )}{6}] = \frac{( k - 1 ) k ( k + 1 )}{3} = \frac{k ( k ^{2} - 1 )}{3} .

$k^{2}$ se divide karo:

E [∣ x_{2} - x_{1} ∣] = \frac{S}{k ^{2}} = \frac{k ( k ^{2} - 1 ) /3}{k ^{2}} = \frac{k ^{2} - 1}{3 k} \approx \frac{k}{3} .

Bade $k$ ke liye $- 1$ negligible hai, tidy $\frac{k}{3}$ deta hai jo parent quote karta hai. Useful $\frac{1}{3}$ seedha $\sum g^{2}$ step se nikla.

Row part identical hai (bas $x$ ki jagah $y$ swap karo), aur ek aur $\frac{k}{3}$ milta hai. Kyunki Manhattan distance dono legs jodta hai:

avg hops = \frac{k}{3} + \frac{k}{3} = \frac{2 k}{3} = \frac{2 N}{3} .

Recall Ring ka

$\frac{1}{4} N$ kahan se aata hai? (Full derivation) $N$ nodes ki ring pe, ek fixed source se target ek direction mein $n$ nodes door baitha hai, lekin aap hamesha shorter arc lete ho, toh true distance $min (n, N - n)$ hai. Sabhi $N$ equally-likely targets pe average karo: $avg = \frac{1}{N} \sum_{n = 0}^{N - 1} min (n, N - n) .$ Sum ko halfway point pe split karo ( $N$ even lo). $n = 0 \dots \frac{N}{2}$ ke liye shorter arc hi $n$ hai; $n = \frac{N}{2} \dots N - 1$ ke liye yeh neeche mirror karta hai. $0$ se $\frac{N}{2}$ tak har value essentially do baar appear hoti hai, toh $\sum_{n = 0}^{N - 1} min (n, N - n) = 2 \sum_{n = 1}^{N /2 - 1} n + \frac{N}{2} = 2 \cdot \frac{( N /2 - 1 ) ( N /2 )}{2} + \frac{N}{2} = \frac{N ^{2}}{4} .$ (Middle sum $\sum_{n = 1}^{m} n = \frac{m ( m + 1 )}{2}$ with $m = \frac{N}{2} - 1$ kaam karta hai; leftover $\frac{N}{2}$ single un-mirrored halfway node hai.) $N$ se divide karo: $avg hops (ring) = \frac{1}{N} \cdot \frac{N ^{2}}{4} = \frac{N}{4} .$ Toh $\frac{1}{4}$ genuinely sum $\sum min (n, N - n) = N^{2} /4$ divided by $N$ hai — distances evenly $0$ se $N /2$ tak spread hoti hain, aur us spread ka midpoint $N /4$ hai.

7. Bandwidth, cycles, aur clock

Parent ke worked examples bytes ko frequency se multiply karte hain. Yahan har piece hai.

Is arithmetic ke memory-side ke liye, 5.1.7-Memory-BandwidthCalculation dekho.

8. Coherence: copies kyun agree karni chahiye

9. Chiplets aur NUMA — "kahan" kyun matter karta hai

10. Real chips perfect grids nahi hote

11. Sab kuch topic ko kaise feed karta hai

Ise padhein: raw players tiles bante hain, tiles ek topology mein arrange hote hain, hum us topology ko hops aur coordinates se measure karte hain, ise $N$ aur $k = N$ (ya $k_{x}, k_{y}$ ) se size karte hain, $E$ se average karte hain, bandwidth se uski speed price karte hain, coherence se iske copies honest rakhte hain, aur finally accept karte hain ki real dies irregular grids use karte hain — yeh sab milke parent topic mein combine hote hain.

Equipment checklist

Right side cover karo; kya aap parent note par jaane se pehle har ek answer de sakte ho?

Node ek sentence mein kya hai?

Chip pe koi bhi box (core, cache slice, memory/I-O controller) jo messages send ya receive karta hai.

Bus jam kyun hoti hai lekin mesh nahi?

Bus ek shared wire hai toh ek time pe sirf ek node baat kar sakta hai; mesh kai parallel roads aur shortcuts deta hai.

Hop kya hai?

Node se uske immediate neighbour tak message ka ek step.

Kya "router" aur "switch" yahan different cheezein mean karte hain?

Nahi — is topic mein woh same component hain, tile pe 5-port traffic cop.

Intel mesh router ke kitne ports hote hain, aur woh kya hain?

5 — North, South, East, West, aur apne core/cache ka ek local port.

(x_{1}, y_{1})

(x_{2}, y_{2})

tak Manhattan distance likho.

d = ∣ x_{2} - x_{1} ∣ + ∣ y_{2} - y_{1} ∣

N

nodes ke square mesh mein side length

k

kya hai?

k = N

nodes per edge.

Agar

N

perfect square nahi hai toh?

Rectangular mesh

k_{x} \times k_{y}

use karo with

k_{x} k_{y} = N

; avg hops

\approx (k_{x} + k_{y}) /3

E [\cdot]

ka matlab kya hai?

Sabhi random cases mein average (typical) value.

Average column gap

(k^{2} - 1) / (3 k) \approx k /3

kyun hai?

Pairs ko gap

g

se group karne par

S = \sum_{g} 2 (k - g) g = k (k^{2} - 1) /3

milta hai;

k^{2}

se divide karne par

(k^{2} - 1) / (3 k)

milta hai.

Ring ka

N /4

kahan se aata hai?

\sum_{n} min (n, N - n) = N^{2} /4

;

N

se divide karne par

N /4

milta hai.

Kaun sa assumption

k /3

aur

N /4

valid banata hai, aur kya use todta hai?

Uniform random traffic (no favourites); hot-spot / non-uniform traffic ise tod deta hai.

Bandwidth formula do.

Bandwidth = bus width (bytes) × clock frequency (cycles per second); bidirectional ke liye double karo.

32 B fabric at 1.8 GHz ki one-direction bandwidth calculate karo.

32 \times 1.8 = 57.6

GB/s.

N \times N

mesh vs ring ki bisection width kya hai?

Mesh

= N

links, ring

= 2

links.

Directory vs broadcast coherence — kaun sa better scale karta hai aur kyun?

Directory, kyunki yeh sirf thode sharers ko message karta hai (

O (k_{s})

) instead of har core ko (

O (N)

xGMI ka kya matlab hai, aur IFIS?

xGMI = inter-Chip Global Memory Interconnect (die-to-die); IFIS = Infinity Fabric Inter-Socket (package-to-package).

Real meshes irregular kyun ho jaate hain?

Die ek rectangle hai jisme I/O aur memory controllers edges pe fixed hote hain, toh grid mein missing tiles aur extra links aa jaate hain die shape match karne ke liye.

Ready ho? parent topic pe wapas jaao — wahan ka har symbol ab tumhara hai.