Clarify why it's called INVERSE_SQRT_E_DROP_TIME (#261)
* (cleanup) Rename `SQRT_E_DROP_TIME` to `INVERSE_SQRT_E_DROP_TIME` to reflect `1/sqrt(e)` * (cleanup) Justify why it's called INVERSE_SQRT_E_DROP_TIME * (cleanup) Polish and simplify presentation * (bug fix) Fixed 2**34 value * (bug fix) Replace epochs with slots to make the calculations work
This commit is contained in:
parent
238c0e10fe
commit
70836ecd1c
|
@ -184,10 +184,10 @@ Unless otherwise indicated, code appearing in `this style` is to be interpreted
|
||||||
| `BASE_REWARD_QUOTIENT` | `2**11` (= 2,048) |
|
| `BASE_REWARD_QUOTIENT` | `2**11` (= 2,048) |
|
||||||
| `WHISTLEBLOWER_REWARD_QUOTIENT` | `2**9` (= 512) |
|
| `WHISTLEBLOWER_REWARD_QUOTIENT` | `2**9` (= 512) |
|
||||||
| `INCLUDER_REWARD_QUOTIENT` | `2**3` (= 8) |
|
| `INCLUDER_REWARD_QUOTIENT` | `2**3` (= 8) |
|
||||||
| `INACTIVITY_PENALTY_QUOTIENT` | `2**34` (= 131,072) |
|
| `INACTIVITY_PENALTY_QUOTIENT` | `2**34` (= 17,179,869,184) |
|
||||||
|
|
||||||
* The `BASE_REWARD_QUOTIENT` constant dictates the per-epoch reward. It corresponds to ~2.54% annual interest assuming 10 million participating ETH in every epoch.
|
* The `BASE_REWARD_QUOTIENT` parameter dictates the per-epoch reward. It corresponds to ~2.54% annual interest assuming 10 million participating ETH in every epoch.
|
||||||
* The `INACTIVITY_PENALTY_QUOTIENT` equals `SQRT_E_DROP_TIME**2` where `SQRT_E_DROP_TIME := 2**17 slots` (~9 days) is the amount of time it takes for the inactivity penalty to cut deposits of non-participating [validators](#dfn-validator) by ~39.4%. The portion lost by offline [validators](#dfn-validator) after `D` epochs is about `D*D/2/INACTIVITY_PENALTY_QUOTIENT`.
|
* The `INACTIVITY_PENALTY_QUOTIENT` equals `INVERSE_SQRT_E_DROP_TIME**2` where `INVERSE_SQRT_E_DROP_TIME := 2**17 slots` (~9 days) is the time it takes the inactivity penalty to reduce the balance of non-participating [validators](#dfn-validator) to about `1/sqrt(e) ~= 60.6%`. Indeed, the balance retained by offline [validators](#dfn-validator) after `n` slots is about `(1-1/INACTIVITY_PENALTY_QUOTIENT)**(n**2/2)` so after `INVERSE_SQRT_E_DROP_TIME` slots it is roughly `(1-1/INACTIVITY_PENALTY_QUOTIENT)**(INACTIVITY_PENALTY_QUOTIENT/2) ~= 1/sqrt(e)`.
|
||||||
|
|
||||||
### Status codes
|
### Status codes
|
||||||
|
|
||||||
|
|
Loading…
Reference in New Issue