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Topic: Why is 550 (MiB) chosen for minimum storage size for prune mode? (Read 203 times)

newbie
Activity: 19
Merit: 7
Thank you @BitCryptex for the great starting point in Github issues. Just reviewed all pruning related stuff at 0.11 Changelog & later versions.

Say, I had just 1 sMerit & couldn't send more than that to you! But I appreciate your answer much more! Smiley
legendary
Activity: 1876
Merit: 3132
Would you please explaining what the logic behind 550 is? Is there any kind of difficulty level relating to 51% attack or any other stuff like that?

Basically, pruned nodes should be able to withstand chain reorganizations lasting at least 2 days. Take a look at the following quote for more detailed explanation.

The minimum allowed is 550MB. Note that this is in addition to whatever is required for the block index and UTXO databases. The minimum was chosen so that Bitcoin Core will be able to maintain at least 288 blocks on disk (two days worth of blocks at 10 minutes per block). In rare instances it is possible that the amount of space used will exceed the pruning target in order to keep the required last 288 blocks on disk.
newbie
Activity: 19
Merit: 7
As you know Bitcoind can run in pruned mode, thus not advertising blocks to other nodes. Even though it has some drawbacks, saving you storage space isn't a benefit you can ignore.

My question is about the magic 550(MiB) default/minimum number. Why they've selected that limit? (& Why not less or more?)

I know it implicitly indicates blocks will be kept for the last (about) 3.8 days (91 hours) (one MiB because of one block every 10 minutes). But where that number come from? I know there is also a dust limit equal to 546 Satoshis.[182*3 ~ 550 here! Even though it is unlikely to be related to this case]

Would you please explaining what the logic behind 550 is? Is there any kind of difficulty level relating to 51% attack or any other stuff like that?
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