SgtSpike
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June 20, 2011, 10:51:22 PM 

The equation is quite simple... Mining is profitable, for you, as long as Price/Difficult is more than:
(Your electric cost per kwh) / (41,908 * (Your MH/s per Watt))
For me, that number comes to about 0.000001590757697776 The current Price/Difficulty ratio is 0.000015412202315
So, I am mining at about 9.7 times my cost of electricity... and it would take a price drop down to $1.37 to make me unprofitable at the current difficulty level.







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bcpokey


June 20, 2011, 10:59:45 PM 

What is 41,908?
The units of your ratio are ($ / kWh) / (Mh/s / W) = $ / kWh * W*s / MH = $ / MH, so it's a little confusing as difficulty isn't really a measurement of MH (at least not directly).




SgtSpike
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June 20, 2011, 11:09:06 PM 

41908 is the estimated bitcoins generated by 1 mhash/s at 1 difficulty for 1000 hours, according to numbers output by the bitcoin calculator on the web. Then this number is manipulated by how many mhash/s you can generate for 1 watt of power, and by how much your power costs per kwh (hence the 1000 hours in the aforementioned number). I don't know, it's rather confusing looking at it myself, but the answer makes sense. Prove it wrong, please.




[Coins!]


June 21, 2011, 04:09:42 AM 

It makes sense for the current difficulty, right? That is 887k.
What will the profitability be if the difficulty is 1.4million? 3 million? 7 million?




Fjordbit


June 21, 2011, 04:26:09 AM 

It makes sense for the current difficulty, right? That is 887k.
What will the profitability be if the difficulty is 1.4million? 3 million? 7 million?
My understanding is that this works for all difficulties, because you need to divide difficulty into price. So if the price drops or the difficulty rises, then the ration falls and you get closer to the watt comparison. My issue with this is that it does not consider opportunity cost of the hardware. So for example, if you could sell your hardware and buy bitcoins with it, would you end up with more btc?




Moussekateer


June 21, 2011, 04:43:49 AM 

It makes sense for the current difficulty, right? That is 887k.
What will the profitability be if the difficulty is 1.4million? 3 million? 7 million?
The difficulty is included in Price/Difficulty ratio is 0.000015412202315 A 40% difficulty increase to 1241800 would be 0.000015412202315 / 1.4 = 0.000011008715939

17BbBd3HqbSXPBTUipRyaAMaQ2NLnyoNPf



bcpokey


June 21, 2011, 05:29:39 AM 

41908 is the estimated bitcoins generated by 1 mhash/s at 1 difficulty for 1000 hours, according to numbers output by the bitcoin calculator on the web. Then this number is manipulated by how many mhash/s you can generate for 1 watt of power, and by how much your power costs per kwh (hence the 1000 hours in the aforementioned number). I don't know, it's rather confusing looking at it myself, but the answer makes sense. Prove it wrong, please. I'm confused as to why you chose 1000 hours? I mean it's a nice round number but how to does it relate to profitability exactly? I'm not trying to prove you wrong, I'm just trying to work out where the numbers are coming from.




Moussekateer


June 21, 2011, 06:42:24 AM 

41908 is the estimated bitcoins generated by 1 mhash/s at 1 difficulty for 1000 hours, according to numbers output by the bitcoin calculator on the web. Then this number is manipulated by how many mhash/s you can generate for 1 watt of power, and by how much your power costs per kwh (hence the 1000 hours in the aforementioned number). I don't know, it's rather confusing looking at it myself, but the answer makes sense. Prove it wrong, please. I'm confused as to why you chose 1000 hours? I mean it's a nice round number but how to does it relate to profitability exactly? I'm not trying to prove you wrong, I'm just trying to work out where the numbers are coming from. 1000 hours to cancel out the factor of 1000 in kWH in the formula.

17BbBd3HqbSXPBTUipRyaAMaQ2NLnyoNPf



Sukrim
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June 21, 2011, 11:49:58 AM 

I'm confused as to why you chose 1000 hours? I mean it's a nice round number but how to does it relate to profitability exactly? I'm not trying to prove you wrong, I'm just trying to work out where the numbers are coming from.
He produces MH/s per Watthour but electricity costs are paid per kiloWatthour For this kind of calculations you have a few factors: Internal (can be influenced by you):  cost of electricity
 efficiency of hardware (MH/s per Watt)
 initial cost of hardware
External (cannot really influenced by you):  exchange rate at the time you want to cash out
 difficulty
The external ones can be expressed as a single factor, initial cost is (too) often neglected. The OP also chose to keep the exchange rate from the equation to have something to compare with more easily (is it higher/lower). It might be more useful though to have also a more meaningful number as ratio (like the charts in my sig: USD per 100MH/s) as it is quite easy to determine if mining is profitable vs. electricity costs BUT it is harder to see how much and also how much the gains would be (Example: I pay 1 Cent/day for my whole cluster and gain 3 times as much! Still the absolute gains are not really breathtaking...)

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nebiki


June 21, 2011, 02:15:57 PM 

i use this: cost := watt*24*elecost/100000; // electricity cost in cents money := btccost*50*24*60*60/(difficulty*Power(2,32) / (hash*1024*1024) ); // where hash = hashrate in Mhash/s and btccost = price of 1 btc in usd. profit := moneycost; that's your daily profit. furthermore, i use this to calculate the average time for each retarget in days: procedure getdays(); var bt, bs: double; // where bt = time to solve 1 block and bs = number of blocks solved begin bt := 600; bs := 0; days := 0; while (true) do begin bt := bt/(1+0.01*increase); // increase = the daily increase in total network hashing power in % bs := bs + 24*60*60/bt; if bs > 2016 then break; days := days+1; increase := increase*0.96; // i assume increase is going down by 4% per day. you can leave it out if you don't believe in this. end; end; then i use this number of average days for my difficulty formula: difficulty := difficulty*Power((1+0.01*increase),days); with these calculations i can simulate my profit over a period of time. although i can only assume a constant bitcoin value, because nobody can predict the future, right? :/ edit: i know it's not too precise, because i "round" everything to days, but i don't think it makes that much of a difference. not sure, though.

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Sukrim
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June 21, 2011, 03:25:49 PM 

increase := increase*0.96; // i assume increase is going down by 4% per day. you can leave it out if you don't believe in this.
I strongly disagree, based on previous data over the past months: http://bitcoin.sipa.be/growth10k.png45% increase per day is a reasonable assumption, no need to make it fall. With this method you could calculate how many Bitcoins you approx. would get after some time and also exactly how much money you would have spent until then. Then you can calculate how much Bitcoins must be worth at least to have a certain profit/0 profit.

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SgtSpike
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June 21, 2011, 03:56:25 PM 

Well, certainly there are other factors to consider, such as opportunity cost. But this is meant to be as simple as possible, to where you can just find the price/difficulty ratio that is the break even point for your setup specifically, and remember it. If you want to get all complicated about days and difficulty increases and profitability over time and opportunity costs and depreciation and heck, even pool fees, use the spreadsheet I created. It includes all of that stuff. http://forum.bitcoin.org/?topic=7531.0 . This is just meant to be a simple metric that gets "pretty close" to the real deal as far as a pointintime calculation goes.




