title: Where the boundary should fall
summary: >-
Which side of a die boundary a thing belongs on — the traffic classes ranked,
why a fabric spanning SLRs was rejected on measurement, and the three hard
constraints that leave one shape standing.
tags:
- architecture
- physical
Where the boundary should fall
Every die boundary in the machine has to be crossed by something. The question this page answers is which thing, and it has an answer rather than a preference, because the traffic classes in a machine of this shape are very unequal. A boundary should land on the cheapest one.
Roughly, in descending order of bandwidth:
- inside a compute unit;
- compute unit to memory agent;
- between compute units;
- control traffic;
- host to control plane.
Class 1 is not a candidate at all — a datapath built on a cascade cannot be cut by a boundary, because cascades do not propagate across one. That is a correctness rule, not a cost: device-facts.
A fabric cannot be stretched across a boundary
That leaves the question of whether the mesh can span SLRs, and the answer came back on measurement rather than on argument. A mesh spanning several dice was implemented, and its worst path was almost entirely route delay with essentially no logic in it at all.
That reading is the general lesson, and it is worth stating in the form that transfers: a path that is nearly all route delay at zero or near-zero logic levels is a placement failure, stated as plainly as a tool can state it. Pipelining it adds latency and moves nothing, because the signal is not passing through logic — it is travelling. There was no version of that design that pipelining would have rescued.
It was not the wire count that killed it. A full-width fabric link is a small fraction of one boundary's cross-die wires. It was that a fabric whose whole premise is locality stops having any once it spans dice.
What replaced it is the arrangement in ship: one mesh per SLR, each with its own memory channel, joined edge to edge by an explicit registered link. The instance that was built that way, the alternative it beat, and the measured worst path of each are in projects/kohakutpu/ship.
The three constraints that leave one shape standing
All three are hard rather than preferential, and together they admit essentially one arrangement:
- a datapath on a cascade cannot cross a boundary — so a compute unit is a unit of placement;
- a memory channel cannot cross a boundary — so a mesh that wants its own memory without crossing must sit on the die that memory is wired to;
- every crossing signal is flop to flop, one cycle plus pipelining — so anything spanning a boundary needs a protocol that tolerates latency and never sends a combinational answer back.
The third is why the interlink is credit-based. A ready travelling back across
a boundary is exactly the combinational crossing the whole arrangement exists to
avoid, so the receiver is unconditionally ready and credit reserved the space
before the beat was sent.
Assume more pipeline stages than the delay suggests
For wide buses at the frequencies this kind of machine targets, vendor guidance asks for several stages, and its own worst case needs more than that. On the reference part the crossing delay alone consumes something close to a quarter of the clock period before any fabric routing to reach the transmit register or leave the receive one — the measured breakdown is in projects/kohakutpu/ship.
So the stage count is a parameter to set generously and leave to the tool to place, not a number to derive. See floorplan for why those registers are the one thing in the design that is deliberately left unpinned.
Which side a thing belongs on
Given the above, the placement question for any new block reduces to three checks:
| Ask | If yes |
|---|---|
| Does it contain a cascade — carry chain, arithmetic cascade, memory cascade? | It is one unit of placement. It goes wholly on one die, and which one is a floorplan decision made at design time |
| Does it own or attach to a memory interface? | It is pinned to that interface's die. The interface cannot move |
| Does it need an answer back within a cycle? | Both ends go on the same die. If they cannot, the protocol between them is wrong, not the floorplan |
A block that fails all three checks is free to sit anywhere, and should be written against parameters rather than assuming a position — which is the same advice a compute unit gets from ship.