One $40m windstorm, four reinsurers
analyseSpiral — the fixed point solved by iteration, one round per circuit
Loss that occurred
$40.00m
Own loss returned
$0
nothing came back
Circuits to settle
4
converged
Net finally retained
$40.00m
equals what occurred
| Participant | Originated | Arriving | Ceded on | Retained | Own loss back |
|---|---|---|---|---|---|
| Aurora Re | — | $4.80m | 0% | $4.80m | — |
| Castellan 2488 | — | $12.00m | 40% | $7.20m | — |
| Meridian Re | $40.00m | $40.00m | 60% | $16.00m | — |
| Northpoint Re | — | $24.00m | 50% | $12.00m | — |
Gross through the market
$80.80m
2.02× the loss that occurred
Cycles in the graph
0
Why it matters
Nothing returns to its origin here, and the loss settles in 4 passes. Note that the market's gross is already 2.02× the loss with no cycle at all — a loss travelling a four-party chain is counted by every party it passes. Inflation alone is not the tell. Close the loop.
Why iteration rather than matrix inversion
Arriving loss is a fixed point, not a sum: what a participant cedes on depends on what arrived. The iteration IS the Neumann series — each round is one more circuit of the spiral — so a programme that never settles is reported as converged rather than given a number. An inversion returns a figure for the divergent case too, and that figure is wrong in a way nobody can see.
Stated limit
Be precise about what the loop costs. It is not the gross figure — a chain inflates that on its own. It is that a participant now carries its own loss back, and that the loss takes 4 circuits to settle rather than 4. Every participant's figures stay correct and the market's net never moves; what the spiral destroys is any one participant's ability to know what it is finally carrying.