What is the adequate fee for a reveal operation?How does Tezos manage its mempool?What elliptic curves are used by Tezos?what this 2 empty [] mean?What is the difference between implicit vs originated accounts?How does tezos-client compute a transaction fee?Why a reveal operation?How to manage rewards for small delegators?how to transaction from a KT1 address to another KT1 address?Can I add metadata to a transfer operation?When setting up a contract, what do the fields mean?
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What is the adequate fee for a reveal operation?
How does Tezos manage its mempool?What elliptic curves are used by Tezos?what this 2 empty [] mean?What is the difference between implicit vs originated accounts?How does tezos-client compute a transaction fee?Why a reveal operation?How to manage rewards for small delegators?how to transaction from a KT1 address to another KT1 address?Can I add metadata to a transfer operation?When setting up a contract, what do the fields mean?
When transfering funds to a non-revealed Tezos account, one must pay a "reveal fee". What is the optimal value to use for this fee? Or is this a fixed amount? What would be the adequate fee for a reveal operation?
account transfer reveal fee
asked 9 hours ago
Luiz MilfontLuiz Milfont
1025
add a comment |
When transfering funds to a non-revealed Tezos account, one must pay a "reveal fee". What is the optimal value to use for this fee? Or is this a fixed amount? What would be the adequate fee for a reveal operation?
account transfer reveal fee
asked 9 hours ago
Luiz MilfontLuiz Milfont
1025
add a comment |
When transfering funds to a non-revealed Tezos account, one must pay a "reveal fee". What is the optimal value to use for this fee? Or is this a fixed amount? What would be the adequate fee for a reveal operation?
account transfer reveal fee
asked 9 hours ago
Luiz MilfontLuiz Milfont
1025
When transfering funds to a non-revealed Tezos account, one must pay a "reveal fee". What is the optimal value to use for this fee? Or is this a fixed amount? What would be the adequate fee for a reveal operation?
account transfer reveal fee
account transfer reveal fee
asked 9 hours ago
Luiz MilfontLuiz Milfont
1025
asked 9 hours ago
Luiz MilfontLuiz Milfont
1025
asked 9 hours ago
Luiz MilfontLuiz Milfont
1025
asked 9 hours ago
Luiz MilfontLuiz Milfont
1025
asked 9 hours ago
Luiz MilfontLuiz Milfont
1025
1025
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
The question as stated confuses different things.
A "reveal" operation is an operation that writes on the chain the public key associated with a public key hash for an implicit account. The fee associated with this operation is up for the sender to decide, though most bakers implement default minimum fees (in this case it is currently 1267µꜩ).
The creation of a new account by sending tokens to an implicit address is not a "reveal" operation as the public key needs not be known in this operation. However, it does trigger the creation of an implicit account on the chain. Storage cost has to be paid, and it is not a part of the transaction fee. It is deducted directly from the account making the transaction, though it is subject to a cap specified by the sender. The default value for the storage cost incurred is indeed 0.257 XTZ.
answered 7 hours ago
Arthur BArthur B
2,376517
1
Thank you both. My intention is, indeed, specify a fee to cover the costs of a creation of a new account when sending tokens to it.
– Luiz Milfont
7 hours ago
add a comment |
The fee for adding a new account to the global state (--burn-cap with tezos-client) is 0.257 XTZ.
The origination burn is calculated as Tez_repr.(cost_per_byte *? (Int64.of_int origination_size)).
cost_per_byte is 0.001 XTZ. It is 1000 mutez and 1,000,000 mutez == 1 tez.
origination_size is 257.
Due to that the resulting cost for an origination of a completely new account is 257,000 mutez or 0.257 tez.
This is defined in proto_alpha/lib_protocol/fee_storage.ml, proto_alpha/lib_protocol/constants_repr.ml, and proto_alpha/lib_protocol/qty_repr.ml
answered 8 hours ago
adrianadrian
5917
1
That's incorrect, the origination burn is not part of the transaction fee.
– Arthur B
7 hours ago
Yes, you are correct. I have edited the above answer to clarify that the fee that I describe there is for adding a new account to the state.
– adrian
7 hours ago
I suggest keeping the terminology separate: "Fee" means paid to the baker. "Burn" means destroyed.
– Tom
4 hours ago
add a comment |
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2 Answers
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2 Answers
2
active
oldest
votes
active
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active
oldest
votes
The question as stated confuses different things.
A "reveal" operation is an operation that writes on the chain the public key associated with a public key hash for an implicit account. The fee associated with this operation is up for the sender to decide, though most bakers implement default minimum fees (in this case it is currently 1267µꜩ).
The creation of a new account by sending tokens to an implicit address is not a "reveal" operation as the public key needs not be known in this operation. However, it does trigger the creation of an implicit account on the chain. Storage cost has to be paid, and it is not a part of the transaction fee. It is deducted directly from the account making the transaction, though it is subject to a cap specified by the sender. The default value for the storage cost incurred is indeed 0.257 XTZ.
answered 7 hours ago
Arthur BArthur B
2,376517
1
Thank you both. My intention is, indeed, specify a fee to cover the costs of a creation of a new account when sending tokens to it.
– Luiz Milfont
7 hours ago
add a comment |
The question as stated confuses different things.
A "reveal" operation is an operation that writes on the chain the public key associated with a public key hash for an implicit account. The fee associated with this operation is up for the sender to decide, though most bakers implement default minimum fees (in this case it is currently 1267µꜩ).
The creation of a new account by sending tokens to an implicit address is not a "reveal" operation as the public key needs not be known in this operation. However, it does trigger the creation of an implicit account on the chain. Storage cost has to be paid, and it is not a part of the transaction fee. It is deducted directly from the account making the transaction, though it is subject to a cap specified by the sender. The default value for the storage cost incurred is indeed 0.257 XTZ.
answered 7 hours ago
Arthur BArthur B
2,376517
1
Thank you both. My intention is, indeed, specify a fee to cover the costs of a creation of a new account when sending tokens to it.
– Luiz Milfont
7 hours ago
add a comment |
The question as stated confuses different things.
A "reveal" operation is an operation that writes on the chain the public key associated with a public key hash for an implicit account. The fee associated with this operation is up for the sender to decide, though most bakers implement default minimum fees (in this case it is currently 1267µꜩ).
The creation of a new account by sending tokens to an implicit address is not a "reveal" operation as the public key needs not be known in this operation. However, it does trigger the creation of an implicit account on the chain. Storage cost has to be paid, and it is not a part of the transaction fee. It is deducted directly from the account making the transaction, though it is subject to a cap specified by the sender. The default value for the storage cost incurred is indeed 0.257 XTZ.
answered 7 hours ago
Arthur BArthur B
2,376517
The question as stated confuses different things.
A "reveal" operation is an operation that writes on the chain the public key associated with a public key hash for an implicit account. The fee associated with this operation is up for the sender to decide, though most bakers implement default minimum fees (in this case it is currently 1267µꜩ).
The creation of a new account by sending tokens to an implicit address is not a "reveal" operation as the public key needs not be known in this operation. However, it does trigger the creation of an implicit account on the chain. Storage cost has to be paid, and it is not a part of the transaction fee. It is deducted directly from the account making the transaction, though it is subject to a cap specified by the sender. The default value for the storage cost incurred is indeed 0.257 XTZ.
answered 7 hours ago
Arthur BArthur B
2,376517
edited 7 hours ago
answered 7 hours ago
Arthur BArthur B
2,376517
answered 7 hours ago
Arthur BArthur B
2,376517
answered 7 hours ago
Arthur BArthur B
2,376517
2,376517
1
Thank you both. My intention is, indeed, specify a fee to cover the costs of a creation of a new account when sending tokens to it.
– Luiz Milfont
7 hours ago
add a comment |
1
Thank you both. My intention is, indeed, specify a fee to cover the costs of a creation of a new account when sending tokens to it.
– Luiz Milfont
7 hours ago
1
1
Thank you both. My intention is, indeed, specify a fee to cover the costs of a creation of a new account when sending tokens to it.
– Luiz Milfont
7 hours ago
Thank you both. My intention is, indeed, specify a fee to cover the costs of a creation of a new account when sending tokens to it.
– Luiz Milfont
7 hours ago
add a comment |
The fee for adding a new account to the global state (--burn-cap with tezos-client) is 0.257 XTZ.
The origination burn is calculated as Tez_repr.(cost_per_byte *? (Int64.of_int origination_size)).
cost_per_byte is 0.001 XTZ. It is 1000 mutez and 1,000,000 mutez == 1 tez.
origination_size is 257.
Due to that the resulting cost for an origination of a completely new account is 257,000 mutez or 0.257 tez.
This is defined in proto_alpha/lib_protocol/fee_storage.ml, proto_alpha/lib_protocol/constants_repr.ml, and proto_alpha/lib_protocol/qty_repr.ml
answered 8 hours ago
adrianadrian
5917
1
That's incorrect, the origination burn is not part of the transaction fee.
– Arthur B
7 hours ago
Yes, you are correct. I have edited the above answer to clarify that the fee that I describe there is for adding a new account to the state.
– adrian
7 hours ago
I suggest keeping the terminology separate: "Fee" means paid to the baker. "Burn" means destroyed.
– Tom
4 hours ago
add a comment |
The fee for adding a new account to the global state (--burn-cap with tezos-client) is 0.257 XTZ.
The origination burn is calculated as Tez_repr.(cost_per_byte *? (Int64.of_int origination_size)).
cost_per_byte is 0.001 XTZ. It is 1000 mutez and 1,000,000 mutez == 1 tez.
origination_size is 257.
Due to that the resulting cost for an origination of a completely new account is 257,000 mutez or 0.257 tez.
This is defined in proto_alpha/lib_protocol/fee_storage.ml, proto_alpha/lib_protocol/constants_repr.ml, and proto_alpha/lib_protocol/qty_repr.ml
answered 8 hours ago
adrianadrian
5917
1
That's incorrect, the origination burn is not part of the transaction fee.
– Arthur B
7 hours ago
Yes, you are correct. I have edited the above answer to clarify that the fee that I describe there is for adding a new account to the state.
– adrian
7 hours ago
I suggest keeping the terminology separate: "Fee" means paid to the baker. "Burn" means destroyed.
– Tom
4 hours ago
add a comment |
The fee for adding a new account to the global state (--burn-cap with tezos-client) is 0.257 XTZ.
The origination burn is calculated as Tez_repr.(cost_per_byte *? (Int64.of_int origination_size)).
cost_per_byte is 0.001 XTZ. It is 1000 mutez and 1,000,000 mutez == 1 tez.
origination_size is 257.
Due to that the resulting cost for an origination of a completely new account is 257,000 mutez or 0.257 tez.
This is defined in proto_alpha/lib_protocol/fee_storage.ml, proto_alpha/lib_protocol/constants_repr.ml, and proto_alpha/lib_protocol/qty_repr.ml
answered 8 hours ago
adrianadrian
5917
The fee for adding a new account to the global state (--burn-cap with tezos-client) is 0.257 XTZ.
The origination burn is calculated as Tez_repr.(cost_per_byte *? (Int64.of_int origination_size)).
cost_per_byte is 0.001 XTZ. It is 1000 mutez and 1,000,000 mutez == 1 tez.
origination_size is 257.
Due to that the resulting cost for an origination of a completely new account is 257,000 mutez or 0.257 tez.
This is defined in proto_alpha/lib_protocol/fee_storage.ml, proto_alpha/lib_protocol/constants_repr.ml, and proto_alpha/lib_protocol/qty_repr.ml
answered 8 hours ago
adrianadrian
5917
edited 7 hours ago
answered 8 hours ago
adrianadrian
5917
answered 8 hours ago
adrianadrian
5917
answered 8 hours ago
adrianadrian
5917
5917
1
That's incorrect, the origination burn is not part of the transaction fee.
– Arthur B
7 hours ago
Yes, you are correct. I have edited the above answer to clarify that the fee that I describe there is for adding a new account to the state.
– adrian
7 hours ago
I suggest keeping the terminology separate: "Fee" means paid to the baker. "Burn" means destroyed.
– Tom
4 hours ago
add a comment |
1
That's incorrect, the origination burn is not part of the transaction fee.
– Arthur B
7 hours ago
Yes, you are correct. I have edited the above answer to clarify that the fee that I describe there is for adding a new account to the state.
– adrian
7 hours ago
I suggest keeping the terminology separate: "Fee" means paid to the baker. "Burn" means destroyed.
– Tom
4 hours ago
1
1
That's incorrect, the origination burn is not part of the transaction fee.
– Arthur B
7 hours ago
That's incorrect, the origination burn is not part of the transaction fee.
– Arthur B
7 hours ago
Yes, you are correct. I have edited the above answer to clarify that the fee that I describe there is for adding a new account to the state.
– adrian
7 hours ago
Yes, you are correct. I have edited the above answer to clarify that the fee that I describe there is for adding a new account to the state.
– adrian
7 hours ago
I suggest keeping the terminology separate: "Fee" means paid to the baker. "Burn" means destroyed.
– Tom
4 hours ago
I suggest keeping the terminology separate: "Fee" means paid to the baker. "Burn" means destroyed.
– Tom
4 hours ago
add a comment |
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