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Topic: Transaction Fee Clarification (Read 145 times)

legendary
Activity: 1904
Merit: 1563
August 15, 2020, 03:25:45 AM
#12
Thanks everyone for the clarification especially with regards to the difference between MB size and the amount of sats/(v)byte and what is the correlation between them.

Anyhow, locking this thread now.
copper member
Activity: 2562
Merit: 2510
Spear the bees
August 15, 2020, 02:32:27 AM
#11
..or more (not less). meaning the size is the sum of transactions paying anything higher than or equal to the value on the left (eg >=2).
My mistake. Mixing up the system. Roll Eyes

technically any fee value that shows up to 1 MB has the potential of being included in the next block. the problem is if more transactions pop up paying more which is not possible to predict.
Like I said previously in the thread, using the peaks is a decent predictor for the future fees. Usually, though, I like to just do 10-20% more than the fee at the time, or even 1-3MB size fee values if I'm not in a hurry.

Have to balance the free market Wink
legendary
Activity: 3472
Merit: 10611
August 15, 2020, 02:26:36 AM
#10
The value on the right of the fee is the size, and it represents the total volume of transactions in the mempool that have a fee equal to the value on the left, or less.

..or more (not less). meaning the size is the sum of transactions paying anything higher than or equal to the value on the left (eg >=2).

technically any fee value that shows up to 1 MB has the potential of being included in the next block. the problem is if more transactions pop up paying more which is not possible to predict.
copper member
Activity: 2562
Merit: 2510
Spear the bees
August 15, 2020, 01:55:00 AM
#9
So it simply means that the total MB size is accumulating overtime. In this example, 2+ sat/(v)byte up to the highest transaction fee is included on the 0.481 MB making it possible for these transactions to be included in the next few blocks due to uncertainty that there will be more tx with a higher transaction fee.
The value on the right of the fee is the size, and it represents the total volume of transactions in the mempool that have a fee equal to the value on the left, or less.

That means that if you have a size of N MB, if you pay the corresponding fee for your transaction you can expect it to be confirmed in 10N minutes... + incoming transactions, of course Wink
sr. member
Activity: 910
Merit: 351
August 15, 2020, 01:47:14 AM
#8
You can also use mempool.space for easier calculation. They showed the lowest fee in each next block by default so that should solve your problem. Still, that doesn't guarantee that there won't be any spike since that's not a factor that you can predict easily.
legendary
Activity: 1904
Merit: 1563
August 15, 2020, 01:18:26 AM
#7
you should keep in mind that these numbers are "total" sizes not the size at that fee level (hence the + sign).
for example when on 2+ shows 0.481 that means there are 0.481 MB (mega virtual byte) of transactions paying more than 2 (2, 3, 4,...) and with a quick look at next level you can see that only 0.020 MB tx is paying 2 sat/vbyte here and the rest is paying higher.
So it simply means that the total MB size is accumulating overtime. In this example, 2+ sat/(v)byte up to the highest transaction fee is included on the 0.481 MB making it possible for these transactions to be included in the next few blocks due to uncertainty that there will be more tx with a higher transaction fee.

Right?
legendary
Activity: 3472
Merit: 10611
August 14, 2020, 11:31:15 PM
#6
you should keep in mind that these numbers are "total" sizes not the size at that fee level (hence the + sign).
for example when on 2+ shows 0.481 that means there are 0.481 MB (mega virtual byte) of transactions paying more than 2 (2, 3, 4,...) and with a quick look at next level you can see that only 0.020 MB tx is paying 2 sat/vbyte here and the rest is paying higher.
as a result it would take a smaller amount of transaction paying higher fee for any tx @2 to become low priority compared to if the sizes were like this:
2+: 0.481
3+: 0.010

For Segwit transactions, its 4MB per block. 
the chart is based on virtual bytes and it is including the SegWit discount as is explained at the bottom of the page. so the total size that is picked up by a miner each time is 1 MB.
sr. member
Activity: 1554
Merit: 413
August 14, 2020, 11:31:14 PM
#5
......

So to simplify, we can easily determine the lowest transaction fee possible when the mempool size is close to 1 MB right? In this case, 1-2+ sats/vb or 2-3+ sats/vb is the recommended tx fee isn't it?
For Segwit transactions, its 4MB per block. 
legendary
Activity: 2310
Merit: 4085
Farewell o_e_l_e_o
August 14, 2020, 11:24:38 PM
#4
Let's start off with Bitcoin block size. One bitcoin block has its size ~ 4MB [1]

From Johoe's site, you can check at the bottom (a third) graph that shows you size of waiting transaction (total and at specific fee-rate range). As of writing, the total size in mempool is ~ 1.83 MB that is lower than 4MB. So you can expect your bitcoin transaction will be confirmed at next one block, even with the fee at 1 satoshi/ (v)byte.

If you are in a hurry and do neither want to wait and see your luck, you can use 2 satoshi/ (v)byte for (instantly) transactions. You can do a substraction with total size (all fee rate) - size of fee rate at 1 sat/(v)byte = size of waiting transactions with fee rate > 1 sat/(v)byte. From the result, you can imagine the prob for your transaction will be confirmed.

It is only estimate because of fast changes in mempool!

[1] https://en.bitcoin.it/wiki/Block_size_limit_controversy
[2] https://cointelegraph.com/explained/bitcoin-block-size-explained


I moved my coin at 1 sat/(v)byte less than 30 mins ago and now it has 3 confirmations.  Cheesy
legendary
Activity: 3290
Merit: 16489
Thick-Skinned Gang Leader and Golden Feather 2021
August 14, 2020, 11:20:48 PM
#3
the minimum transaction fee that we can use in order for our transaction to be mined in the few blocks
You can never be sure: the estimate is based on past blocks and transactions, while the reality is based on future blocks. In this case it would probably be okay, but if you're unlucky, and the next block takes much longer to be found, fees will spiral up. The same happens when more new transactions are created.
In that case the minimum fee will quickly be above your fee, and it can stay high for a long time.

In the rare case that I want to be very sure my transaction gets included in the next block, I pay a much higher fee than recommended (for instance when ordering dinner). In any other case, I go as low as possible.
You can of course go low, and use RBF in case fees go up unexpectedly.
copper member
Activity: 2562
Merit: 2510
Spear the bees
August 14, 2020, 11:15:56 PM
#2
Sort of. That's the mempool as of right now, but it doesn't mean that if you broadcast a transaction, there won't be transactions with much higher priority than yours.

I like to estimate it by looking at the previous block's peak: see the highest point right before the sharp drop? Whatever the blocks are saying there is the estimated block size for the next one. Each MB corresponds to 10 minutes, basically.

Recommended resource: https://mempool.observer
legendary
Activity: 1904
Merit: 1563
August 14, 2020, 11:11:02 PM
#1
Greetings!

Just want some clarification regarding transaction fees based on the mempool created by Johoe.



I just want to clarify if the minimum transaction fee that we can use in order for our transaction to be mined in the few blocks is at least 2 sats/vb? Right?

So to simplify, we can easily determine the lowest transaction fee possible when the mempool size is close to 1 MB right? In this case, 1-2+ sats/vb or 2-3+ sats/vb is the recommended tx fee isn't it?

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