When we try to send monero to someone using
transfer [integrated address] 0.0016
how can we calculate the size where the priority is normal and ring size is default(may be 7)
So I need an algorithm to calculate the transaction's actual size.
Please give me an example with the algorithm
For the example you've given, where there is only one wallet destination, the transaction will have two outputs (one for the recipient, one sent back to you with your change).
A simple approximation is that the transaction size will be 6176 bytes * [output count] for the range proofs, plus for the ring signatures: [input count] * 64 bytes * [ring size].
So if you spend a single input (which contains enough funds), with a ring size of 7, this formula gives you an approximate transaction size of 6176*2+1*64*7 = 12800 bytes. In reality, it'd probably be closer to 13200 bytes, because of the simplification we've made.
When you spend funds, you probably won't be able to tell whether you're spending funds out of a single input you own, or whether the source of the funds is spanned over multiple smaller inputs you own. As you can see from the formula, in some cases, the transaction could easily be many KB larger than usual if you have to spend together lots of small inputs in order to have enough funds in total for the transaction. The answer also varies a little according to the random indices of the decoy inputs that you select during the transaction.
Therefore the simplification this answer presents should be good enough, given that you won't have the information available in advance to make a more accurate prediction of the transaction size.
Also note that Monero fees are charged on KB boundaries. Therefore a 13KB transaction will cost the same in fees as a 13.99KB transaction, but a 14.01KB transaction will cost more than a 13.99KB transaction.
It is expected that later in 2018, a new bulletproof scheme will replace Monero's existing range proofs. This will significantly lower the size of transactions.
