How much CPU power to mine 1 coin a day? Monero Stack Exchange

This reaction will attempt to response the question at numerous levels, spil goes after:

  • Section I explains the basics of how to calculate this yourself.
  • Section II provides updated calculations based on Section I.
  • Section III provides an abstracted way to determine this based only on the total network hashrate and the total number of coins te circulation.

I. Explaining how to determine this on your own:

  • Monero has an average block time of Two minutes, meaning that there are approximately 720 blocks vanaf day.
  • Your expected prize is proportional to your share of the network hashrate. The current hashrate is

23.Trio MH/s (http://moneroblocks.informatie/), or 23300000 H/s.

  • Monero’s block prize is decreasing slightly every block. Right now it is a little less than 11 XMR, but I’m going to just use 11.0 te my calculations.
  • To find out the precies prize at any given time, you could go to a block explorer (http://moneroblocks.informatie/) and click on the most latest block. It will tell you the prize for that block.
  • Disclaimer: The prize switches based on how many transactions are included te a block. For example, block 1124279 wasgoed empty and had a prize of Ten.826400000000, while block 1124278 had Four transactions and a higher prize of 11.078600000000. Nonetheless, you can get a rough idea using this method.
  • Okay, putting it all together, where your share of the hashrate is “n”:

    Since the question wasgoed for a prize of 1 XMR vanaf day, using the numbers above:

    and solving for n:

    which is pretty close to @villabacho’s reaction.

    I just want to get a general idea.

    Generally speaking, the formula to determine the hashrate needed to mine 1 XMR vanaf day is:

    What CPU / GPU would be required to solo mine 1 coin a day?

    The most energy efficient GPU that I know of for mining XMR is the GTX 750 Ti, which gets approximately 250 H/s for a little more than $100/GPU. 12 of thesis GPUs would give you 3000 H/s, or a little more than 1 XMR vanaf day at the current mining and prize levels.

    II. 20 June 2018 Update to Section 1:

    • Hashrate:

    7.22 XMR/block (including fees)

    Using our formula n = (Network Hashrate) / (720 * Avg Block Prize) :

    n = 16812 H/s, or 16.81 kH/s to mine 1 XMR vanaf day.

    You would need approximately 67 GTX 750 Ti’s at 250 H/s each, OR approximately 28 RX 470’s at 600 H/s each, OR approximately 22 RX 480’s at 750 H/s each.

    III. 20 June 2018 addition abstracting calculation further:

    A more general formula can be developed that calculates the Average Block Prize used above from the total coins te circulation.

    The base block prize is calculated by

    where M = 2^64 and A is the current amount of XMR te circulation (te terms of atomic units, where 1 XMR = 10^12 atomic units). Wij can reduce this for ease of use to be spil goes after:

    where a = A * 10^(-12) and represents XMR “coins” te circulation spil wij traditionally think of them.

    From this formula, it should be clear that the base prize for each block is progressively decreasing. However, wij can consider it harshly onveranderlijk ter the brief term for our purposes here, spil overheen a 720 block period (one day) the prize drops

    0.01 XMR at today’s rate. Therefore, wij can substitute this Prize value spil an approximation of the Avg Block Prize overheen a relatively brief period with an error of much less than 1%. Combining our two formulas:

    and reducing gives

    According to at the current block height of 1336863,

    • The network hashrate is 87.Four MH/s or 87400 kH/s
    • There are 14658448 total coins te circulation

    Substituting thesis values into our formula

    and noting that n is ter kH/s, wij would need 16.796 kH/s of hashing power to mine 1 XMR today.

    Note: This calculation does not include fees, so you should actually mine slightly more than 1 XMR today with that hashrate. But you will have to pay pool fees, account for variance, etc, so it’s a fairly good approximation still.

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