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j"19HxigV4QyBv3tHpQVcUEQyq1pzZVdoAutMe+Perica [Today at 6:11 AM] I was thinking along the line of mining when subsidy diminishes and fees become a dominant part of the mining reward. I'll share some of my thoughts in the thread. Maybe someone will find it interesting. 43 replies Perica [11 hours ago] Let's start with Craig's equation: https://metanet-icu.slack.com/archives/C5131HKFX/p1558016131036800 CSW This is how mining rewards are supposed to be/work Y=MX+B Y is the reward M is the fee X is the unit time B is the subsidy Thread in #generalMay 16thView message Perica [11 hours ago] Later in the same thread, Craig provided some graphs where block reward is represented on Y axes and time on X axes and that is the reason for choosing names of variables like he did. Let's, here, use F for the instantaneous amount of fee (accumulated fee at a moment) and t for time. >F = Mt + B Perica [11 hours ago] Next, assume B=0, meaning let's look at dynamics when the subsidy is zero and the block reward consists, only, of fees. M is a fee in unit time. It changes in time and is chaotic. But we can calculate an average value of M in a block, or average M in some other period of time. So, let M be constant in our model (or look at it as an average value). Perica [11 hours ago] Now the formula becomes: >F = Mt F is the amount of accumulated fee in a period and it will become a reward at some point when a block is mined. On the following graph M represents the slope of the line. Perica [11 hours ago] On this graph, we see the accumulation of fee in time, from the moment when one block is found, but before the next is mined. gf1.jpg Perica [11 hours ago] Let's now switch to mining part. First, look at the aggregated hashpower of network or alternatively you can see it as an example of a world where there is only one miner. Total power of the network, measured in hashes/sec is H, and in a period of time t it produces W amount of work >W = Ht Perica [11 hours ago] So here is the amount of work "produced" from the moment when one block is mined but before the next one. The graph is almost the same, and the slope of the line is H. gf2.jpg Perica [11 hours ago] The main cost for a miner is electric power and it is reasonable to assume that the cost C per unit time, is proportional to his haspower H deployed at that moment. Amount of money he spends in finding a solution to a block is S = Ct and is proportional to W by the same constant. This means we can look at miners instantaneous deployed hashrate H as an "instantaneous" cost and at his produced work while mining (W) as a total cost. That means the above graph represents miners cost in the unit of hash, too. Perica [11 hours ago] Now, since we established our context, let's look at a few examples. Perica [11 hours ago] First, slope M is greater than slope kH (k is constant which "translates" cost of hash to BSV, and is fixed - we don't acount for exchange rate fluctuations) This means that the amount of money flowing in the "system" (fees) is greater than the amount which is leaving (cost for hash). Perica [11 hours ago] Miners are more than happy. They are earning. gf3.jpg Perica [11 hours ago] Next, if M = kH slopes overlap and the amount of fee collected is the same as the cost of mining. Miners keep mining (we suppose that the cost of mining includes their (minimal) margin, too). Perica [11 hours ago] The third case is when M < kH, when there is more hashpower than "economic activity" represented by fees. This is where things are becoming interesting. Perica [11 hours ago] If miners use all hashpower their cost is greater than income and they are losing money. gf4.jpg Perica [11 hours ago] If you analyze it you'll see that a loss for a miner is proportional to his hashrate. For example miner with 33% of hashrate will take 33% of the overall loss, on average. Similar to when mining is profitable: the 33% miner will take on average 33% of the profit. Perica [11 hours ago] Since they are mining with loss when M < kH miners need to turn some machines off and cut total hashpower deployed. And they need to cut it to a point where M=kH and keep mining at that rate. Perica [11 hours ago] That is the dynamic of an ideal system. When fees are not enough, just cut the hashpower to the level to which "economic activity" represented by fees dictate. We can conclude that hashpower (tries to) mimic "economic activity" by lowering or raising its value. Perica [11 hours ago] Difficulty adjustment "averages out" hashpower deployed over the period of 2016 blocks. If hashpower mimics "economic activity" on chain (or at least tries to), we can conclude that difficulty indirectly "measures" the same "economic activity". It's acting similar to lowpass filter removing all the noise and at 2016 blocks it absorbs well all those "Black Fridays", "Chinese New Years" and similar impacts. Perica [11 hours ago] Let's look at the behaviour of a single miner in this environment. Perica [11 hours ago] As stated previously hashrate needs to adjust to a line where M=kH. What does it mean to a single miner? Perica [11 hours ago] Miner loss is proportional to his hashrate. If for an example, the rate at which fees are coming (M) is cut in half, miners need to cut overall hashrate in half, in order to mining overall become profitable. The ideal is for each miner to remove half of his hashrate. Everyone cuts in half, they are profitable, the size of one relative to others stays the same which means one earn the same percentage of block reward as before. It is fair, too. :slightly_smiling_face: Perica [11 hours ago] But mining is competition. What are the options? Should a miner lower his hashrate? Can he outsmart others? Can bigger ones "throw out" small ones? Perica [11 hours ago] When M=kH, if a miner lowers his hashrate, the "hole" will be filled with hashrate of others so he will earn less. If he "pushes pedal" and adds hashrate, mining becomes unprofitable and not just for him but for all. Can he force others to lower their hashrate and earn more than his size assumes? Can he play poker and bluff? Is there a flow and bigger grows more by not lowering? Perica [11 hours ago] Let's look at an example. Miner A has 33.33 EH/s and the other(s) B have 66.67 EH/s. 100 eH/s was profitable "line" and all are mining with full capacity, but at some point in time "profitability" is cut in half and 50 eH is the line of profitability. Ideally, all cut in half: miner A mines at 16.66 EH/s and B with 33.33 eH/s of power. A will get every third block on average, the same as if all hash is utilized (as earlier). Perica [11 hours ago] What if B cuts in half and A keeps mining at "full steam"? Then A is mining at 33.33 EH/s and B is mining at 33.33 EH/s. It is 66.66 eH/s combined and 50 EH is profitable. The equivalent of 16.66 EH/s is loss in money terms. Miner A mines half of blocks and B mines the other half, on average. Perica [11 hours ago] Does this make that B cuts more in order to be profitable (33.33 - 16.66 = 16.66) which leads that 2/3 of blocks are solved by A and 1/3 by B? If others (B) cut that means A "outsmarted" them, since having the 1/3 of the overall hashpower, now he mines with the power of 2/3 and mining is now profitable. Perica [11 hours ago] Do others (B) need to cut in this "hash war" or A will need to cut first? Perica [11 hours ago] Well if you think the "size" of a miner is represented by his overall hashrate and "size" means moneywise, you can see A needs to give up first. The overall loss is equivalent of 16.66 EH/s per block as stated before, so the A is losing the equivalent of 8.33 EH/s per block, which is the same amount the others (B) are loosing. They suffer the same absolute loss, but the others (B) are as twice as big as A, so they can sustain the loss twice longer, theoretically. Perica [11 hours ago] A similar situation is if others (B) don't cut and stay at 66.67 EH/s when 50 EH/s is the line of profitability, and A cuts from 33.33 EH/s to 16.66, then the loss for them will be greater not just in absolute terms, but in relative, too. Their loss will be 66.67/16.66 = 4 times bigger but they are twice as big as A, so they cannot sustain it in the long term. This means there is no flaw in the system and there is no "solution" where biggest ones get ever bigger on the cost of smaller ones. Perica [11 hours ago] This was an idealised system where each miner knows total hashpower of the network, knows what amount of hashpower is deployed at each moment. As Craig said miner can only estimate it with a certain probability after an hour or two. Perica [11 hours ago] Here I looked at an idealized model in an interval between two blocks are mined and assumed M is constant in time while it is actually chaotic. I did it to get a "feel" of the process. The miner needs to look at fees among a longer period of time, among many blocks. If there is a "fat" block which you mine in full steam and you don't solve it you may accumulate a relatively big loss which you need to recover among the longer period of "thin" blocks. The miner needs to look at longer interval containing many blocks in order to estimate others hashrate, anticipate missing hashrate, spot oportunities... CSW [11 hours ago] Not even intervals Poisson time CSW [11 hours ago] Some small CSW [11 hours ago] Some large CSW [10 hours ago] But, you are getting the right idea Perica [10 hours ago] Yes, it gets complicated (edited) Perica [10 hours ago] :slightly_smiling_face: Perica [10 hours ago] I was just going to write that you provided earlier those kind of graphs which illustrate the chaotic nature of fees and are spanned among many blocks (edited) Perica [10 hours ago] I didn't do my homework and look into Stackelberg game and links provided with it (shame on me, but I'll do it very soon), but when I add to this analysis, the chaotic nature of fees in time, fluctuation of exchange rate, the fact that miner can estimate, but only to a certain degree, things like his relative hashrate, overall hashrate, size of others, behaviour (bluffing) of others, etc I can try to imagine why notes of some Russian guy who solved "dolichobrachistochrone differential game of Rufus Isaacs", while the lady was swimming in the lake, were needed, and why "Chaos and Nonlinear Forecastability in Economics and Finance Pty Ltd" was formed. :smiley: Perica [10 hours ago] I was trying to see where is the point when mining starts. Does miner wait for an accumulation of fee to some certain value, before he starts? And the conclusion is no he doesn't. He mines based on his estimation of "fee rate" over time and adapts according to new inputs. He mines all the time - intensity is what fluctuates. macsga [4 hours ago] https://nicolewhite.github.io/2015/05/23/understanding-waiting-times.html CSW [3 hours ago] Same slope, but the distance between then is variable. Some tall, some short. CSW [3 hours ago] Try adding a time variable and now see text/markdownUTF-8Mining Revenues with no subsidy