為什麼 1000 sat/kb 的 minRelayTxFee 不等於 250 sat/kw？

天真地，我們期望 1000 sat/kb 的 minRelayTxFee 對應於 250 sat/kw。但是，根據這個問題（ACINQ）顯然不是這樣，它也引用了這個問題（c-lightning）。

  /** 
    * why 253 and not 250 since feerate-per-kw is feerate-per-kb / 250 and the minimum relay fee rate is 1000 satoshi/Kb ? 
    * 
    * because bitcoin core uses neither the actual tx size in bytes or the tx weight to check fees, but a "virtual size" 
    * which is (3 * weight) / 4 ... 
    * so we want : 
    * fee > 1000 * virtual size 
    * feerate-per-kw * weight > 1000 * (3 * weight / 4) 
    * feerate_per-kw > 250 + 3000 / (4 * weight) 
    * with a conservative minimum weight of 400, we get a minimum feerate_per-kw of 253 
    * 
    * see https://github.com/ElementsProject/lightning/pull/1251 
    **/ 
   val MinimumFeeratePerKw = 253

我認為 vbytes 等於重量單位除以 4 並四捨五入，所以我不明白從何(3 * weight) / 4而來。

我認為 vbytes 等於重量單位除以 4

它是 [1]，但四捨五入到下一個整數，所以實現bitcoind是：

int64_t GetVirtualTransactionSize(int64_t nWeight, int64_t nSigOpCost, unsigned int bytes_per_sigop)
{
   return (std::max(nWeight, nSigOpCost * bytes_per_sigop) + WITNESS_SCALE_FACTOR - 1) / WITNESS_SCALE_FACTOR;
}

（哪裡WITNESS_SCALE_FACTOR = 4）。

此外，bitcoind不僅vbyte使用者界面使用 s，記憶體池邏輯也使用 s，DEFAULT_MIN_RELAY_TX_FEE常量設置在vbytes 中：

static const unsigned int DEFAULT_MIN_RELAY_TX_FEE = 1000;

為了檢查它的標準性，bitcoind將計算您的交易規模向上取整（如果不是 的倍數4），並將您的交易費率記錄為向下取整。

然後它會將其與最小1000中繼250費率4（

使用程式碼可能會更清楚，這是一個說明此行為的 Python 函式：

>>> def bitcoind_fun(tx_weight, feerate_perkw):
...     fees = tx_weight * feerate_perkw // 1000
...     print("Your transaction will pay {} sats of fees, is {} WU large (a feerate of {}sat/Kw)".format(fees, tx_weight, feerate_perkw))
...     tx_vbytes_for_bitcoind = (tx_weight + 3) // 4
...     tx_feerate_vbytes_for_bitcoind = fees * 1000 // tx_vbytes_for_bitcoind
...     print("bitcoind reads your transaction as paying {} sats of fees for a transaction of {} vbytes, so a {}sat/perKvb feerate".format(fees, tx_vbytes_for_bitcoind, tx_feerate_vbytes_for_bitcoind))

如果您使用 a tx_weightwhich is a multiple of4將通過，否則它將不會：

>>> bitcoind_fun(1600, 250)
Your transaction will pay 400 sats of fees, is 1600 WU large (a feerate of 250sat/Kw)
bitcoind reads your transaction as paying 400 sats of fees for a transaction of 400 vbytes, so a 1000sat/perKvb feerate
>>> bitcoind_fun(1601, 250)
Your transaction will pay 400 sats of fees, is 1601 WU large (a feerate of 250sat/Kw)
bitcoind reads your transaction as paying 400 sats of fees for a transaction of 401 vbytes, so a 997sat/perKvb feerate
>>> bitcoind_fun(1602, 250)
Your transaction will pay 400 sats of fees, is 1602 WU large (a feerate of 250sat/Kw)
bitcoind reads your transaction as paying 400 sats of fees for a transaction of 401 vbytes, so a 997sat/perKvb feerate

所以我們使用 253 作為最安全的地板：

>>> bitcoind_fun(1600, 253)
Your transaction will pay 404 sats of fees, is 1600 WU large (a feerate of 253sat/Kw)
bitcoind reads your transaction as paying 404 sats of fees for a transaction of 400 vbytes, so a 1010sat/perKvb feerate
>>> bitcoind_fun(1601, 253)
Your transaction will pay 405 sats of fees, is 1601 WU large (a feerate of 253sat/Kw)
bitcoind reads your transaction as paying 405 sats of fees for a transaction of 401 vbytes, so a 1009sat/perKvb feerate
>>> bitcoind_fun(1602, 253)
Your transaction will pay 405 sats of fees, is 1602 WU large (a feerate of 253sat/Kw)
bitcoind reads your transaction as paying 405 sats of fees for a transaction of 401 vbytes, so a 1009sat/perKvb feerate

順便說一句，您輸入的程式碼中的註釋是錯誤的：不是(3 * weight) / 4 但是(3 + weight) / 4。

---

[1] 排除人為增加虛擬大小以進一步防止 DOS 的 sig OPs 限制。

PS：同樣，請參閱<https://github.com/bitcoin/bitcoin/issues/13283>。

引用自：https://bitcoin.stackexchange.com/questions/95926