It's just an example which shows that the absolute fee can be lower than the dust limit (which shouldn't be linked together in the first place)
What I'm telling is if the fee rate is @1sat/vB and the outputs aren't dust, the transaction is standard.
Just take note of the thing's I've previously listed.
usually it can still pay for the additional size that it will add to that transaction.
But since it's only based from the input's size, if both inputs are only 547sats, it may not be enough to pay 2sat/vB (needs computation).
But for 1 sat/vB, it's definitely more than enough.
Here are some examples (testnet):
- 9e3d0c4089ea141ca9f9870985c4bb0e37f05231628454381df92ce56fe1b7d1
Two 546 sats input (min) sending 752 sats to one output @ 1sat/vB. - 05ffb639cc25b0c792e9dd075f3e1af5f92e5984260942b9b14b68255327c6cb
Three 546 sats input (min) sending 1150 sats to one output @ 1sat/vB.
Thanks for granting my request to unlock the topic for me clarify my previous post.
Hope this can clear the confusion.