Topic: MOORE: Mining Bond Beating the Moore's Law (Collecting Shareholder Data) - page 9. (Read 19165 times)

Update

The price of IPO is set at 0.480.50 BTC/s.

The price of bulk purchase is 0.45 BTC/s(6.6710.00% discount), but each trade has to involve larger than 500 shares(225 BTC). Please PM me for trades, and find witnesses/make records on replies if necessary.

The total amount of our initial release will be less than or equal to 20,000 shares.
It depends on how the bulk purchase goes on.

Thanks to everyone who's interested.

I wrote a little script to generate the data, then using Excel to plot them.
It's the standard and most painless way for me.

Thanks for the graphs. I like graphs!  What program did you use to produce them?

The price is now finally determined at 0.48BTC/s. The amount is to be announced yet.

The following are two pictures showing that if the difficulty increases according to Moore's Law, how our bonds will perform compared to normal bonds.

Normal Mining Bonds are assumed to have a price of 0.3BTC/share, plotted in blue.
Moore Bonds are assumed to have a price of 0.48BTC/share, plotted in red.

X-axis is the number of weeks, Y-axis is the expected total return rate from coupons.

The difficulty is initially set at 1,733,208, and increases by 0.89% each week.
The date when block reward reduces to 25 is set at 30 weeks later.

The first picture assumes that the difficulty always smoothly increases:



The second picture assumes that when the block rewards turns to 25, the difficulty also turns to a half, as some miners will close their operations:

I'm looking forward to the IPO.

In the post above he/she valued them at 0.48 BTC per initial MH/s, about 60% above the typical 0.30 BTC that 1 MH/s bonds sell atm.

Another interesting plot would be with difficulty that increases according to Moore's law - your bonds then should show a straight line while constant bonds (while initially cheaper) will rapidly decline at some point.

How many shares are you going to sell? How much does each share cost? At least I would like to know an estimated price :-)

Estimated Returns vs Normal Mining Bonds

Normal Mining Bonds are assumed to have a price of 0.3BTC/share, plotted in blue.
Moore Bonds are assumed to have a price of 0.48BTC/share, plotted in red.

X-axis is the number of weeks, Y-axis is the expected total return rate from coupons.

The difficulty is set at 1,733,208.
The date when block reward reduces to 25 is set at 30 weeks later.

The first picture assumes that the difficulty does not change:



The second picture assumes that when the block rewards turns to 25, the difficulty also turns to a half, as some miners will close their operations:

I'm now actively working on it, besides the planning of the initial portfolio, and the coordinating jobs of miners and mining operations.
Expect a detailed estimation soon.

see what I mean by confusing!


Well, I have to disagree with you there. There's no reason a bond has to be faithful to the real mining process, just faithful to the promised dividend. I guess what I'd really like is a bond that pays the same amount per dividend for a given difficulty period - or as in MOORE, an exponentially ramping amount.

Anyway, this is total offtopic, and your current system will serve just as well.

Fixed. Thanks for informing.


I think a more faithful connection to real mining process is more important than easier coupon calculation. So I may just stick to my new calculation formula.


Glad to know this. Thank you very much.

1.089 != 1 + 0.89%...

1 + 0.89% is 1.0089!

Meni does this thing with 336 block "macroblocks" - there might still be an issue though with being able to game the system a bit by buying in on weeks where it's likely that they contain 1 more macroblock than other weeks, since on average one macroblock is a bit longer than 1 day.

The correct (though a bit complicated) way to do it (if you want constant time window payouts) is as friedcat already outlined: Calculate for each second how much 1 hash would be worth and sum it up for 1 week or whatever your payout frequency is.

Looking good. Just a couple of points:
1. I find

a bit confusing.


is a bit simpler and easier to parse (nitpicky I know, but still...)

2. You're still basing payment on a chronological basis which I'm sure makes ROI easier to calculate, but mean you have to use a complex function to calculate f(i). Basing the coupon period on the greatest common denominator of the block count to halving the bitcoin reward (210000) and between difficulty changes (2016) mean that you'd calculate the dividend on 336 blocks at a time. You can still pay weekly if you want, although the weekly payment will vary depending on how many sets of 336 blocks were in that time period. But the divdend calculation becomes a lot simpler. 

Good luck. I'll purchase some of these with my next lot of bond dividends.

Perfect!

Just a recommendation: Plot this (with assumed constant difficulty) for 1 year and show people the nice exponential curve!

Revision of Coupon Payments

Definitions
Financial Week: Starting from 16:00:00(GMT Time) each Tuesday, ending at 16:00:00(GMT Time) the succeeding Tuesday.

Coupon Unit: The quantity of Bitcoins paid each share each financial week.

Payment Time
We assume T(0) is the IPO starting time, T(i) is the time of i-th coupon payment, and it should satisfy:

16:00:00(GMT Time) Tuesday of the i-th financial week <= T(i) < 16:00:00(GMT Time) Wednesday of the i-th financial week

The substraction on T(i) is defined in seconds, which means that (T(i)-T(i-1)) represents "how many seconds elapse between T(i) and T(i-1)".

Amount of Payments
The coupon unit of the i-th financial week is:

10^6 * (1.0089)^(i-1) * f(i) / 2^32

in which f(i) is calculated as follows:
  f(i) = (t(1)-t(0))*B(0)/D(0) + ... + (t(n)-t(n-1))*B(n-1)/D(n-1)
     where:
        the number of the changes of difficulty and block reward between T(i-1) and T(i) is (n-1).
        t(0) equals to T(i-1).
        t(j) (when 01) means the time of the j-th change of either the difficulty or the block reward during T(i-1) and T(i).
        t(n) equals to T(i).
        B(j) means the last block reward before t(j+1).
        D(j) means the last difficulty number before t(j+1).
        The substraction on t(j) is also defined in seconds.

This matches real mining process most accurately. Thanks for figuring out for me. I think I fell into the pitfall of assuming that the difficulty change will always be mild so that I chose the over-simplified model. You are right. Extreme cases should be taken into account.


Thanks for your criticism. I won't "feel free". I will do my best to revise my plan. I'm sorry for the informality and flaws of the OP, and will make a more detailed appendix to it.

Not at all, as this means you pay out more on rounds where difficulty is on the rise (average time between blocks is less than 600 seconds) and less if it is going down (slower than 1 block every 10 minutes).

Just pay the expected 100% PPS rate:

Sum of ( (Hash rate * Time in seconds of difficulty 1 * block reward 1) / (difficulty 1 * 2^32) + (Hash rate * Time in seconds of difficulty 2 * block reward 2) / (difficulty 2 * 2^32) + ... )

The time in seconds either starts at the beginning of the week or at the last block of the old difficulty and ends at the last second of the week or at the last block of the current difficulty.

Example:

Hashrate of the network explodes, difficulties are 1, 4 and 16, 2 difficulty changes after 3 days, 0:00 (timestamp of last block with difficulty 1) and 2 days later 0:00 (timestamp of last block with difficulty 4)and another time on the last second of the week, hashrate of the bond = 1 MH/s:

( ( 1*10^6 H/s * 259200 s * 50 BTC) / (1 * 2^32) ) + ( ( 1*10^6 H/s * 172800 s * 50 BTC) / (4 * 2^32) ) + ( ( 1*10^6 H/s * 172800 s * 50 BTC) / (16 * 2^32) ) = 3 646.12788 BTC (1 MH/s was a lot in the beginning of a bitcoin block chain!)

With your old calculation (R*(86400*7*B*10^6)/(D*2^32)):
( 1 * 86400 * 7 * 50 * 10^6 ) / ( 1 * 2^32 ) = 7 040.79866 BTC

or the updated one (R*(N*600*B*10^6)/(D*2^32)):
( 1 * 6048 (= 3 times 2016 blocks) * 600 * 50 * 10^6 ) / ( 1 * 2^32 ) = 42 244.792 BTC

Yes, the examples are highly exaggerated, but you still see that you clearly will pay out too much (in extreme cases far too much!) if difficulty rises. As you increase your hash rate per your calculations as well, you'll increase that effect too. On the other hand, if difficulty goes down, you underpay or get closer to the real PPS values.

Feel free to pay out on your flawed simplified model, but if a little bit of math is already too difficult to do, it makes me wonder if you're up to the task of handling a mining operation that has to double it's size consistently every 1.5 years...

Yes, they only make sense when we pre-make an assumption on future difficulty evolution.


It seems you are concerned with my dividends calculation formula, and some other people too.
Would it be better to change it to:
R*(N*600*B*10^6)/(D*2^32)
in which R, B and D keep the original meaning, and N means the number of actual blocks between the time of two payments?

Thank you very much.

Update

MOORE is already listed on the IPO page of GLBSE.

The number of shares for selling, the IPO price, and the price for bulk purchase are to be determined.

"How long until it will have solved more virtual hashes per BTC invested than a cheaper static 1 MH/s contract" is calculable. Not perfectly, as you have a weird payout scheme/calculation, but it should be close.
"How long until your investment is gained as dividends" not, as it depends on difficulty
"How long until you will have gained 10x the dividends of a static bond" is not calculable as well, as this also depends on difficulty.

An ideal case for buying this bond here would be a strong difficulty increase in the beginning and then, after you passed the hashes per BTC invested of 1 MH/s bonds, a decline or at least slow increase in difficulty again. Also you should take into account when buying this on which weekdays difficulty increases can be expected, unless the payout calculation gets closer to the real expected values.