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Author Topic: Mandatory transaction fees??  (Read 1200 times)
digimag
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June 21, 2011, 08:13:23 AM
 #1

Hello,

Sometimes I get a warning from my bitcoin client :



And this, even for tiny amount from 0.01 to 1.00 BTC !

Sometimes 0.20 BTC works, but then I put the same amount and send it again and it doesn't work anymore.

Why is the bitcoin client so hungry for transaction fees? What is it supposed to mean?

You already get 500 $ for one generated block, why is the software bothering with those transaction fees?

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Grinder
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June 21, 2011, 08:40:14 AM
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It'll be interesting to see what happens when the reward is halved. Either people will have to pay about 400x more fees than they do today, or the security of the network will be significantly reduced because it will no longer be profitable enough for many of the miners.
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June 21, 2011, 09:27:09 AM
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I guess https://en.bitcoin.it/wiki/Transaction_Fee needs an update to reflect the lastest client changes https://github.com/bitcoin/bitcoin/commit/6de1326ba4a35ab781107a8b56f74affaed87cba.

@digimac: Which bitcoin client version are you using? You are probably not using the latest (0.3.23) version, are you?
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June 21, 2011, 09:29:49 AM
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Since 0.3.23 fees are 0.0005

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June 21, 2011, 09:37:43 AM
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0.3.21-beta.

I'll update.

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DamienBlack
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June 21, 2011, 09:40:31 AM
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When the block reward is halved, it won't really be that big a deal. We've already seen difficulty rises that are >80%. That is effectively the same thing.

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digimag
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June 21, 2011, 09:43:15 AM
 #7

Since 0.3.23 fees are 0.0005

I just updated to the latest software.

It still asks me for a fee.

A small fee, 0.0005, sure, but it's still very fucking annoying.

A few months ago we could send several bitcoins for free, what the fuck happened?

How can a 1 BTC transaction be “over the size limit”? Please enlighten me.

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digimag
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June 21, 2011, 09:46:26 AM
 #8

I'll try to rebuild the OS X package so it doesn't ask for a fee systematically like a beggar.

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June 21, 2011, 09:54:22 AM
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I'll try to rebuild the OS X package so it doesn't ask for a fee systematically like a beggar.

Sheesh! just use the command line client. It has none of this bullshit.
blueling
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June 21, 2011, 10:04:01 AM
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How can a 1 BTC transaction be “over the size limit”? Please enlighten me.

Size is measured in bytes, not BTC. You can use http://blockexplorer.com/ to inspect completed blocks. If you combine many BTC amouts that you received independently in a transaction the size of the tansaction will be larger than when you use only a single source address.

(please correct me if I am wrong..)
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June 21, 2011, 10:19:06 AM
 #11

Sheesh! just use the command line client. It has none of this bullshit.

As mining becomes harder transations sent without transaction fees will be unlikely included in one of the next blocks. I think it is best to think of transaction fees as a bid offer towards the miners to include your transaction. It is up to you to pay a fee and up to them to confirm your transaction in a block. There in no fixed transaction fee and since finding a block is a random process (i.e. a mining pool will find blocks in a frequency based their share of the total computation power of the network) sometimes you might be included earlier .. sometimes later (depending on the policy of the mining pool that found the block). The higher your bid for inclusion in a block the more likely it is going to happen with the next block.
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