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Author Topic: Parano1d ① Proof-native Layer 1 ordered by PoW  (Read 2877 times)
max2000irc
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Today at 08:54:21 AM
 #161

Just one question , how much is the min fee ? For example is someone decide to make micro txs just to take more slots and speed up the halving, how many coins he will need.
GofyGofy
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Today at 08:59:06 AM
 #162

Just one question , how much is the min fee ? For example is someone decide to make micro txs just to take more slots and speed up the halving, how many coins he will need.

You will need more than 12,5 million live utxo  Grin
max2000irc
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Today at 09:04:37 AM
 #163

Just one question , how much is the min fee ? For example is someone decide to make micro txs just to take more slots and speed up the halving, how many coins he will need.

You will need more than 12,5 million live utxo  Grin

That, I understood but I am trying to figure out whats the cost for those 12.5 mil utxo .
aria.devcode
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Today at 10:16:50 AM
 #164

Your blocks table shows TEMPLATE 0.4s. Just to understand what's measured: is that the full HistoryStep proof generated after the parent block arrives, or template assembly with a pipelined/precomputed proof? And do you fully validate the parent block before proving on top of it, or start from the unvalidated header?"
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Today at 10:28:04 AM
 #165

Just one question , how much is the min fee ? For example is someone decide to make micro txs just to take more slots and speed up the halving, how many coins he will need.

You will need more than 12,5 million live utxo  Grin

That, I understood but I am trying to figure out whats the cost for those 12.5 mil utxo .

A theoretical look at how much activity would be needed to reach the first NOID halving

If we use a purely theoretical model starting from an empty State, where every transaction is:

Code:
1 input -> 2 outputs

then each transaction increases Live State by exactly +1 UTXO.

The minimum value of a single output is:

Code:
1 μNOID = 0.000001 NOID

The initial State has:

Code:
2^24 = 16,777,216 slots

The first halving is triggered when sustained State occupancy reaches 75%, so the threshold is:

Code:
16,777,216 × 75%
= 12,582,912 live UTXOs

So in this simplified +1 UTXO per transaction model, it would take approximately:

Code:
12,582,912 transactions

to reach the first halving threshold.

If every newly created UTXO contained only the minimum possible amount of 1 μNOID, then the total value held in those UTXOs would be only:

Code:
12,582,912 μNOID
= 12.582912 NOID

The interesting part is the transaction fees.

For a 1-input / 2-output transaction:

Code:
Base fee:       5,000 μNOID
1 input:          100 μNOID
2 outputs:      1,400 μNOID

Below 50% State occupancy, the growth fee is 2,500 μNOID, giving:

Code:
9,000 μNOID
= 0.009 NOID per transaction

From 50% to 75% State occupancy, the growth fee rises to 5,000 μNOID, giving:

Code:
11,500 μNOID
= 0.0115 NOID per transaction

So, hypothetically starting from zero UTXOs:

Code:
0% -> 50%

8,388,608 transactions
× 0.009 NOID

= 75,497.472 NOID

Then:

Code:
50% -> 75%

4,194,304 transactions
× 0.0115 NOID

= 48,234.496 NOID

Total theoretical minimum consensus fees:

Code:
≈ 123,731.968 NOID

Summary

Code:
First halving threshold:
12,582,912 live UTXOs

Minimum value per output:
0.000001 NOID

Minimum total value in those UTXOs:
12.582912 NOID

Theoretical minimum consensus fees:
≈ 123,731.968 NOID

This is only a theoretical model.

In reality, the network already contains UTXOs, some UTXOs are spent and reused, block rewards also contribute to State, and the dynamic mempool relay floor can make actual fees higher.

The main point is that filling the State does not necessarily require a huge amount of NOID to be locked in UTXOs. The much larger cost comes from creating and maintaining enough new State through on-chain activity.
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