Looking to gain insight on the matter of the **inactivity penalty quotient**.

While the Serenity Design Rationale document states that

With the current parametrization, if blocks stop being finalized, validators lose 1% of their deposits after 2.6 days, 10% after 8.4 days, and 50% after 21 days. This means for example that if 50% of validators drop offline, blocks will start finalizing again after 21 days.

If we examine the current parametrization we have

```
penalties[index] += Gwei(effective_balance * finality_delay // INACTIVITY_PENALTY_QUOTIENT)
```

with `INACTIVITY_PENALTY_QUOTIENT`

= `2**25`

(`33,554,432`

).

Now, if we build and run a quick python script

```
balance = 100.0
for i in range(4,4726):
balance -= (i * balance) / 2**25
print(str(i) + "\t" + str(balance))
```

our results are, for an initial balance of `100`

```
4 99.99998807907104
5 99.99997317791163
6 99.99995529652298
[snip]
4723 71.71416019772546
4724 71.70406383570766
4725 71.69396675817036
```

adjusting the exponent of `2`

in the above script, with `2**23.94128`

we get

```
[snip]
567 99.00533232662947
568 99.00184121683876
569 98.99834408404766
[snip]
1840 90.01886280962862
1841 90.00857450307387
1842 89.99828178458043
[snip]
4723 50.026236935763414
4724 50.01156578060706
4725 49.996895823295965
```

Which are *closer* to the statements in the rationale, namely, losing 1% at 2.6 days (568 epochs), 10% after 8.2 days (1841 epochs), and 50% after 21 days (4724 epochs).

So the **short question** is why **2**25** was chosen as the Inactivity Penalty Quotient instead of **2**24**. And the **long question** is about the methodology used to compute the coefficient .

For the latter, we tried an analytical approach and end up with the following equation to compute the exponent x:

Suppose we are looking for a value x such that if we apply B_i = B_{i-1}(1-\frac{i+4}{2^{x}}) we obtain that B_{4725} \approx 0.5B_0. In other words we want the balance B to be halved after 21 days, (4,725 epochs)

B\prod_{n=4}^{4725} (1-\frac{i}{2^{x}}) = 0.5B

Simplifying the product,

\prod_{n=4}^{4725} (1-\frac{i}{2^{x}}) = 1 - [\frac{4725 * 4726}{2}-6]\frac{1}{2^x}+...\approx1 - [\frac{4725 * 4726}{2}-6]\frac{1}{2^x}=1-\frac{11165169}{2^x}

Equaling to 0.5 and solving for x we have x=24.412. We can attribute the difference between this result and the ran experiment (23.94128) to the dismissed terms of the product.