<p>It`s a <a href="http://en.wikipedia.org/wiki/Geometric_progression" rel="noreferrer">geometric series</a> with the base 1/2. After half of the remaining coins are mined, rate is decreased until half of remaining coins since the drop are mined.</p>

<p>The rate is convergent, that is, it approaches a finite number and doesn't go on until infinity (like say, 1, 1/2, 1/3... 1/n would). This makes it so that there will never be more than certain amount of Bitcoins created.</p>

<p>As for why it is so dramatic, probably for the ease of calculating whether the block is correct. If you would have some complicated way of calculating the reward some of the coins might be lost due to rounding error (below 1 Satoshi), and different (non-official) clients might have issues with calculating the result effectively. Moreover, these sudden decreases in amount of coins generated probably will increase the worth of Bitcoins.</p>

CC BY-SA 3.0
2011-09-28T15:05:56.657Z
Well it is worth pointing out that these steps, no matter how dramatic, should be well-priced in before the event, and that "pricing in" can still be fairly smooth and gradual.
lemonginger
CC BY-SA 3.0
2011-09-28T16:59:34.167Z
Sorry but my eyes bleed, [1 + 1/2 + 1/3 + … + 1/n](http://math.stackexchange.com/questions/30907) is *not* convergent and approaches +∞.
Stéphane Gimenez
CC BY-SA 3.0
2011-09-28T17:49:07.307Z
@Stephane: "doesn't go on until infinity (like say, 1, 1/2, 1/3... 1/n would)" or to put it shortly: "1/2^n does not diverge, like 1/n would". I know 1/n diverges, I did state so, but maybe I worded it in a bit of a complicated way.
ThePiachu
CC BY-SA 3.0
2011-09-28T19:28:02.180Z
Why not a geometric series with a base of 9/10? It still converges, it is still easy to calculate, and it is less dramatic of a change.
JGWeissman
CC BY-SA 3.0
2011-09-28T19:58:06.807Z
@ThePiachu: Ok, sorry, I read it the wrong way.
Stéphane Gimenez
CC BY-SA 3.0
2011-09-28T11:52:55.227Z
2011-09-28T11:52:55.227Z
ThePiachu
<p>The mining rewards are scheduled to drop from 50btc to 25btc. The next drop again will be from 25 to 12.5btc. In each of the steps into a new "era" the mining reward is reduced by 50%. </p>

<p>This seems like a very dramatic pay cut at each juncture, Is there a reason why the amount isn't being decreased more gradually by smaller and more frequent amounts?</p>

<mining-profitability><money-supply><price-volatility><reward-schedule>
Why are the mining rewards set to reduce in such dramatic steps?
Andrew Vit
# Why are the mining rewards set to reduce in such dramatic steps?

(Asked by: username_0 on 2011-09-28)

The mining rewards are scheduled to drop from 50btc to 25btc. The next drop again will be from 25 to 12.5btc. In each of the steps into a new "era" the mining reward is reduced by 50%.

This seems like a very dramatic pay cut at each juncture, Is there a reason why the amount isn't being decreased more gradually by smaller and more frequent amounts?

**Tags:** mining-profitability, money-supply, price-volatility, reward-schedule


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## 1 answers

### Answer by username_1 on 2011-09-28. Score: 6 | Accepted answer

It`s a [geometric series](http://en.wikipedia.org/wiki/Geometric_progression) with the base 1/2. After half of the remaining coins are mined, rate is decreased until half of remaining coins since the drop are mined.

The rate is convergent, that is, it approaches a finite number and doesn't go on until infinity (like say, 1, 1/2, 1/3... 1/n would). This makes it so that there will never be more than certain amount of Bitcoins created.

As for why it is so dramatic, probably for the ease of calculating whether the block is correct. If you would have some complicated way of calculating the reward some of the coins might be lost due to rounding error (below 1 Satoshi), and different (non-official) clients might have issues with calculating the result effectively. Moreover, these sudden decreases in amount of coins generated probably will increase the worth of Bitcoins.

> **Comments:**
> **username_2**: Well it is worth pointing out that these steps, no matter how dramatic, should be well-priced in before the event, and that "pricing in" can still be fairly smooth and gradual.  
> **username_3**: Sorry but my eyes bleed, [1 + 1/2 + 1/3 + … + 1/n](http://math.stackexchange.com/questions/30907) is *not* convergent and approaches +∞.  
> **username_1**: @Stephane: "doesn't go on until infinity (like say, 1, 1/2, 1/3... 1/n would)" or to put it shortly: "1/2^n does not diverge, like 1/n would". I know 1/n diverges, I did state so, but maybe I worded it in a bit of a complicated way.  
> **username_4**: Why not a geometric series with a base of 9/10? It still converges, it is still easy to calculate, and it is less dramatic of a change.  
> **username_3**: @username_1: Ok, sorry, I read it the wrong way.  

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