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Does that mean someone can denial-of-service all of bitcoin just by moving a balance between a couple wallets repeatedly? Can one person (or group)'s activity
by mynewwork 13y ago
Does that mean someone can denial-of-service all of bitcoin just by moving a balance between a couple wallets repeatedly? Can one person (or group)'s activity crowd out other people so their transactions can't be added to the blockchain? What prevents that from happening?
25,000/hour is a shockingly small number to me. I feel like I must be misunderstanding something, or else someone from 4chan would have already DOS'd bitcoin for fun.
- knowitall 13y agoThere already is a transaction fee - miners will ignore your transaction if you pay too little or nothing.
- teraflop 13y agoTransactions normally include a small fee, to incentivize miners and disincentivize abuse. You don't have to pay the fee, but if there are more transactions being broadcast than the blockchain can accommodate, the fee-paying ones get priority. (It's a little more complicated than that: when you generate a transaction, you choose whatever fee you want, and they get prioritized based on a combination of the fee, the transaction size in KB, and how long it's been since those coins were last moved.) Current transaction fees are typically in the range of 0.1-0.5mBTC, so in order to continuously flood out everyone else, you'd need to spend more than 2.5BTC/hour in fees.
- gibybo 13y agoIt is definitely a small number. Visa is capable of handling 80 million+ per hour, for example. However, transactions are prioritized by transaction fee so you would have to pay a fee higher than all other transactions to block them out. Currently this would cost you something like $100k per day, but if you started to do this you would rapidly increase the cost of transactions, so the cost to perform this attack would grow much more expensive.