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Author: Admin | 2025-04-28
Every transaction.The Amount of FeeAs for how much the fee will be, let's say you pay a fee of ρ/Byte where ρ is around 10 Satoshi. This is equal to what you mean by Would it be .0001 x 20 = .002 in fees?if you mean ".0001 BTC/kB" by ".0001" because .0001 BTC/kB * 100'000'000 Satoshi/BTC / (1000 B/kB) = 10 Satoshi/BYou can inform yourself about fee recommendations in articles like this one.For simplicity, let's assume you use P2PKH with compressed keys (which is standard), have a UTXO large enough for the entire transaction but not of just the right size, so you need a change address (which is almost always the case).Under these assumptions, you have to use 148 Byte per input and 34 Byte. In addition to this, the transaction header needs 10 more Bytes.We can now write down a function φ : ℕ×ℕ×ℝ → Satoshi which tells us the transaction fee provided the number of inputs and outputs as well as ρ. This function is: φ(ni, no, ρ) := (10 + ni*148 + no*34) * ρ * SatoshiLet γ : ℕ×ℝ → Satoshi be the function which tells us the fee for a transaction with 1 input, 1 change address, and and n addresses you want to send money to. This function is defined as: γ(n, ρ) := (192 + n*34) * ρ * SatoshiConcrete NumbersIf you pay the 20 people individually, you will spend 20*γ(1, ρ) = 4'520*ρ*Satoshi = 0.0452*ρ*mBTC as the fee. On the other hand, if you pay them all at once, you only spend γ(20, ρ) = 872*ρ*Satoshi = 0.00872*ρ*mBTC. If we assume ρ=10, that's 0.452 mBTC or 0.0872 mBTC respectively.The difference is that you would spend (0.452 mBTC)/(0.0872 mBTC) = 5.18 times as much of a fee if you made individual transactions compared to if you made one cumulative transaction.If you pay the 200 people individually, you will spend 200*γ(1, ρ) = 45'200*ρ*Satoshi = 0.452*ρ*mBTC as the fee. On the other hand, if you pay them all at once, you only spend γ(200, ρ) = 6'992*ρ*Satoshi = 0.06992*ρ*mBTC. If we assume ρ=10,
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