<oembed><type>rich</type><version>1.0</version><author_name>npub10f96gqrsu4qpygfgvuvzce47aavjvql703egfde0l2hua8dzpszs67ej47</author_name><author_url>https://nostr.ae/npub10f96gqrsu4qpygfgvuvzce47aavjvql703egfde0l2hua8dzpszs67ej47</author_url><provider_name>njump</provider_name><provider_url>https://nostr.ae</provider_url><html>📅 Original date posted:2015-05-10&#xA;📝 Original message:Le 08/05/2015 22:33, Mark Friedenbach a écrit :&#xA;&#xA;&gt;   * For each block, the miner is allowed to select a different difficulty&#xA;&gt; (nBits) within a certain range, e.g. +/- 25% of the expected difficulty,&#xA;&gt; and this miner-selected difficulty is used for the proof of work check. In&#xA;&gt; addition to adjusting the hashcash target, selecting a different difficulty&#xA;&gt; also raises or lowers the maximum block size for that block by a function&#xA;&gt; of the difference in difficulty. So increasing the difficulty of the block&#xA;&gt; by an additional 25% raises the block limit for that block from 100% of the&#xA;&gt; current limit to 125%, and lowering the difficulty by 10% would also lower&#xA;&gt; the maximum block size for that block from 100% to 90% of the current&#xA;&gt; limit. For simplicity I will assume a linear identity transform as the&#xA;&gt; function, but a quadratic or other function with compounding marginal cost&#xA;&gt; may be preferred.&#xA;&gt; &#xA;&#xA;Sorry but I fail to see how a linear identity transform between block&#xA;size and difficulty would work.&#xA;&#xA;The miner&#39;s reward for finding a block is the sum of subsidy and fees:&#xA;&#xA; R = S + F&#xA;&#xA;The probability that the miner will find a block over a time interval is&#xA;inversely proportional to the difficulty D:&#xA;&#xA; P = K / D&#xA;&#xA;where K is a constant that depends on the miner&#39;s hashrate. The expected&#xA;reward of the miner is:&#xA;&#xA; E = P * R&#xA;&#xA;Consider that the miner chooses a new difficulty:&#xA;&#xA; D&#39; = D(1 + x).&#xA;&#xA;With a linear identity transform between block size and difficulty, the&#xA;miner will be allowed to collect fees from a block of size: S&#39;=S(1+x)&#xA;&#xA;In the best case, collected will be proportional to block size:&#xA;&#xA; F&#39; = F(1+x)&#xA;&#xA;Thus we get:&#xA;&#xA; E&#39; = P&#39; * R&#39; = K/(D(1+x)) * (S + F(1+x))&#xA;&#xA; E&#39; = E - x/(1+x) * S * K / D&#xA;&#xA;So with this linear identity transform, increasing block size never&#xA;increases the miners gain. As long as the subsidy exists, the best&#xA;strategy for miners is to reduce block size (i.e. to choose x&lt;0).</html></oembed>