By knowledge mentioned in EstimateBlockFee and BlockIntervalTime, one can derive the optimal ratio about the delayed time of block propagation. When miners assess zero revenue for orphan probability and aims to maximize the the expected value of the block fee, they come to maximize
or in case of exponential payment distribution:
With x being defined such x = B / MT = ND / M and k being defined k = M / DnA , this formula is translated to
which is maximized at x when
When k is large, the optimal x approaches zero. As mentioned in a long run the n or A may adapt with the x decided by the miners to have long term kx being 1 and then the expected value becomes
More precisely, the orphan does not necessarily mean zero revenue because either chain might lose or not. With knowledge mentioned in ScaleDebate, the overall impact is the reduced coin price and production efficiency and the miners are to maximize which, being decreasing beyond 1/k, is maximized at x when
When block reward R and network latency l being ratio of connection latency time to block time are considered, the maixmization is about the revenue per unit of time:
It shows that considering latency and R / T being constant for design, the latency ratio shall be as small as ideal which means the block time shall be long rather than short. Further, the optimal x which must be between 0 and 1/k happens when
Two calculation examples assuming zero latency and zero block reward.
Two calculation examples assuming 0.02 latency and 50 block reward.
One calculation example of mining monopoly where h is 1.
Note that for generic transaction fee distribution G, the goal is to maximize
and the optimal equation is
and similar stories like the exponential distribution assumption will be told.
Also note that the two maximization using orphan probability vs using efficiency loss agree on the first order by the Taylor series expansion at x = 0 when the miners are small meaning h being almost zero. Again, in a long run the n or A may adapt with the x decided by the miners to have long term kx being 1 and then the expected value becomes . By the point of view of the macro, it is the reduced price of the coin and a regular block fee of qVT. As time goes by, k increases because miners choose higher M for the sake of higher block fee without sacrifice the security as mentioned in AttackCost.
While one may argue that the block size B is not the "transmitted" block size thanks to the fact that miner may see many transactions already, in which only 6 bytes check sum is transmitted rather than the whole D.
It comes to some modeling of probability of knowing at time T of a transaction broadcast at time t during the time interval 0 to T, w(t). Then the average transmitted block size B will be
So the above reasoning remains but for a smaller effective D. For example of a simple modeling, the transaction information before timing t is of the size DnAt and is broadcast to the miner taking time T-t, and all transactions before t are known and all transactions after t are unknown to the miner, therefore
And k shall be defined as
Further, the above reasoning assumes that the CPU speed is infinite or the major bottleneck is internet transmission. Otherwise, a CPU speed C for validating the transactions can be introduced as below.
The x in the optimization is the ratio of overall delay time over T beside latency,
where B' be the transmitted block and B be the real block therefore
In terms of B, the optimal equation is
The better the infrastructure, the lower b. Beside the theoretical deduction, can be directly observed and and can be obtained by regression of block delay time against block size. After that, the optimal block size B, or equivalently the optimal x, is derived accordingly.
Let r be the block reward of unit time of a design, L be the latency in seconds so that , the above equation does hint about the optimal design about T. At every time interval T, a revenue of
is collected. Knowing that q is also time preference rate, the overall present value of all future revenue is
So the optimal T is about maximization of
whose optimal equation is
When qT is tiny, the left hand side is , this solves to
For example, assume h being zero and x being tiny and L being 10 second and q being 0.005 annually, T is 355168 seconds.