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The formal condition

What load band makes a system bistable, and why does amplification set it rather than headroom?

This formula is short, and it is the whole subsection. Let:

  • λ = the base arrival rate (useful demand),
  • C = capacity,
  • k = the work amplification factor in the degraded state: the number of units of work the system performs per unit of useful demand once retries fire, caches miss and trip.

The healthy state is stable if λ < C. The broken state is stable if λ × k > C (the system is still overloaded in the degraded state, so it stays degraded). Both are true at the same time when

the base arrival rate (useful demand)
capacity
units of work per unit of useful demand once retries fire, caches miss and timeouts trip
Data table
Utilization above which a broken state can sustain itself, by degraded amplification
Degraded amplification kVulnerable above this utilization
1100%
1.1189.9%
1.2381.6%
1.3474.8%
1.4569%
1.5664%
1.6859.7%
1.7955.9%
1.952.6%
2.0149.7%
2.1347.1%
2.2444.7%
2.3542.6%
2.4640.6%
2.5838.8%
2.6937.2%
2.835.7%
2.9134.3%
3.0233.1%
3.1431.9%
3.2530.8%
3.3629.7%
3.4828.8%
3.5927.9%
3.727%
3.8126.2%
3.9225.5%
4.0424.8%
4.1524.1%
4.2623.5%
4.3822.9%
4.4922.3%
4.621.7%
4.7121.2%
4.8320.7%
4.9420.3%
5.0519.8%
5.1619.4%
5.2819%
5.3918.6%
5.518.2%
5.6117.8%
5.7217.5%
5.8417.1%
5.9516.8%
6.0616.5%
6.1716.2%
6.2915.9%
6.415.6%
6.5115.4%
6.6315.1%
6.7414.8%
6.8514.6%
6.9614.4%
7.0814.1%
7.1913.9%
7.313.7%
7.4113.5%
7.5313.3%
7.6413.1%
7.7512.9%
7.8612.7%
7.9712.5%
8.0912.4%
8.212.2%
8.3112%
8.4311.9%
8.5411.7%
8.6511.6%
8.7611.4%
8.8811.3%
8.9911.1%
9.111%
9.2110.9%
9.3210.7%
9.4410.6%
9.5510.5%
9.6610.3%
9.7810.2%
9.8910.1%
1010%

That band is the region. In it the system is fine, and would be equally content being broken.

Two stable states at one loadC = 2,000/s · k = 4

So the quantity that governs your exposure is not . Headroom sets where the band ends (at C). k sets where it starts (at C/k). And k is the parameter nobody measures.