jschoubben
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Jochen asked whether the order made sense. It did not -- it followed when things happened to be decided, which after consolidation is fictional anyway since record 5 alone folds decisions taken across a week. Concretely wrong before: the domain statement sat at 8, after five engineering rules; the constitution was scattered across 5, 12 and 17; the tiers landed at 15, 16, 21 and 22 with process records in between. Now it walks: what the mesh is (1-3), its tiers from the bottom up (4-8), what runs on them and how it gets there (9-10), how it is built (11-16), how it is checked (17-18), how we work (19-23). Two things made this safe rather than free. It is a permutation, not a compaction, so the renames go through temporary names -- otherwise two files want one slot and one is lost. And the reference rewrite is a single simultaneous pass, because almost every number moved into a slot another number was vacating; replacing one at a time would have cascaded and pointed things at the wrong record while still resolving. Verified: 284 [ADR NNNN](path) links across the repository, all with matching text and target. The ordering principle is now stated in 19 rather than left implicit -- the repository already said "the numbering is the flow" about its folders, and there was no reason for the records to be the exception.
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status, initiated, touches, became
| status | initiated | touches | became | |||
|---|---|---|---|---|---|---|
| active | 2026-08-24 |
|
010 — What the lab's inner loop actually costs
What is being investigated
The lifecycle design closes on an open question that is measurable rather than arguable:
What a snapshot costs. Whole-scenario snapshots of several machines are the operation the inner loop repeats most, so their cost sets the loop's speed. If restoring is slow, the loop is slow, and everything above is theory.
Measured on a workstation, 2026-08-24. Numbers in measurements.md.
Why it matters
ADR 0016 makes the bootstrap scenario the inner development loop for tiers 0 and 1 — the argument being that raising a node from nothing stops being the least-exercised path and becomes the most-exercised one. That argument is only true if raising and resetting are cheap. A loop that costs minutes is a loop people avoid, and the least-exercised path stays least-exercised.
Status
Measured, and the finding is a blocker rather than a data point.
A snapshot on the current host is a full copy of the machine's disk. 1.6 GB and ten seconds for one small virtual machine at best — and over two minutes when observed a second time. Cost scales with the number of machines and the size of their disks, not with what changed.
With copy-on-write it is 0.13 seconds and costs the delta. Verified, not assumed. The projected four-machine reset cycle falls from roughly ninety seconds to roughly fifteen, of which almost all is a boot that cannot be avoided.
The cause is not virtual machines and not incus. It is that the host offers incus exactly one
storage driver, dir, which has no copy-on-write and therefore no cheap snapshot. The kernel
supports btrfs; the userspace tool that would let incus use it is simply not installed.
So the question "is the lab's inner loop fast enough" currently answers itself the wrong way, for a reason that is one declared package away from being fixed — and declaring packages is something the mesh already does.
Open questions
| Question | Why it matters |
|---|---|
| Answered. The inner-loop argument holds with copy-on-write and did not without it. | |
| How does the lab install its own prerequisites on a clean machine? | The lab needs a virtualisation daemon, copy-on-write tooling and a pool before it can do anything — and it cannot depend on the mesh for them, since it is where the mesh is built. |
| Why was the second snapshot more than twelve times slower than the first? | If snapshot cost is unpredictable rather than merely high, that is worse — a loop with a variable multi-minute step is one nobody trusts. |
| Does a scenario snapshot need the machines stopped? | Stateless snapshots of a running virtual machine capture the disk but not memory. Whether a mesh restored that way is coherent is not established. |
| What is the cost at scenario scale — four machines rather than one? | Only single-machine numbers were taken. If the operation is serial, four machines is four times the wait. |