将您的超额收益委托给独奏面包师的最佳频率是多少？

Question

将您的超额收益委托给独奏面包师的最佳频率是多少？

- 8
- 208 2019-02-05
想象一下,我是一个面包师,面包卷为1辊(10k XTZ),考虑到当前每个周期我应该赚取6.5 XTZ(0.065％)的主动辊数,粗略估算一下,让我们假设这是静态的,不变,也就是说,每个周期我将准确获得0.065％.

现在,随着奖励的开始支付,它们将不足以让我再赚一卷,因此我的烘焙权仍然相当于一卷,因此我的奖励将保持在每个周期6.5 XTZ.由于我的奖励太小,除非我将6.5委派给另一位面包师,否则我似乎无法从复利中受益.

但是,委派行为需要进行交易,才能将6.5 XTZ从我的烘烤tz1帐户转移到KT1帐户,这会产生交易费.看来最低交易费用为0.001272mtz,所以我们假设这是我们必须支付的.

如果我们假设从代表团获得的任何奖励都收取10％的费用(按照我自己的面包师收取的6.5％的固定回报),我的6.5 XTZ奖励将获得(6.5 - .001272) * .00065 * (1 - 0.10) = 0.00380175588 xtz.假设面包师要承担交易费用.

因此,从.00380> .001272开始,每个周期之后都应该转移多余的XTZ.即使面包师转嫁交易费也应该值得.

现在,我们知道烘烤并不是那么线性,某些周期我会或多或少地烘烤,所以让我们假设一个糟糕的情况,我只认可一个区块以获得2 XTZ奖励.现在在这种情况下,如果我立即转移多余的款项,我的预期奖励为(2 - .001272) * .00065 * (1 - 0.10) = 0.001169,根据我的理解,这低于交易费用那么您最好等到下一个周期的回报,然后再将资金转移到您的KT1帐户.

我的计算正确吗?是否只需检查您的预期收益是否大于交易费以确保您获得最佳回报,还是我错过了任何东西,那么简单吗?

Imagine I am a solo baker with 1 roll (10k XTZ), as a rough estimate given the current number of active rolls I should earn around 6.5 XTZ (0.065%) per cycle, let's assume for a moment that is static and not variable, i.e I will receive exactly 0.065% every cycle.

Now as the rewards start being paid out they will not be enough to earn me another roll, therefore my baking rights will still correspond to 1 roll and so my rewards will remain at 6.5 XTZ per cycle. Since my rewards are too small it looks like I do not benefit from compounding unless I delegate that 6.5 to another baker.

However the act of delegating requires a transaction to move the 6.5 XTZ from my baking tz1 account to a KT1 account which incurs a transaction fee. It seems like the minimum transaction fee is 0.001272mtz so let's assume that is what we must pay.

If we assume a 10% fee for any rewards from delegation (along with a static 6.5% return as per my own baker), my 6.5 XTZ reward will earn (6.5 - .001272) * .00065 * (1 - 0.10) = 0.00380175588 xtz. let's assume the baker absorbs the transaction fee.

So it looks like it makes sense to transfer the excess XTZ after every cycle since .00380 > .001272. Even if the baker passes on the transaction fee it should be worthwhile to do.

Now we know that baking is not as linear as that, some cycles I will bake more or less, so let's assume a bad scenario where I only endorse a single block for 2 XTZ reward. Now in this scenario if I transfer the excess straight away my expected reward is (2 - .001272) * .00065 * (1 - 0.10) = 0.001169 which is less than the transaction fee so as I understand it you are better off waiting until the next cycles rewards before moving the money to you KT1 account.

Are my calculations correct? Is it as simple as checking whether your expected rewards are greater than the transaction fee to ensure you are receiving the optimal return or is there anything I have missed?

baking solo-baking delegating

- 您的第二个陈述是不正确的:您可以通过成为一名独奏者而受益,而无需委托他人
  
  your second statement is incorrect: you benefit from compounding as a solo-baker without needing to delegate to someone else
  - 0
  - 2019-03-02
  - Ezy
- @Ezy如果您的报酬还不成问题,那就不要了.例如,如果我只有10k xtz.我要花一年多的时间才能积累足够的奖励,才能再次获得名次
  
  @Ezy Not if your rewards don't make up another roll. For example if all I have is 10k xtz. It would take more than a year for me to accumulate enough rewards to get another roll at which point I would start compounding
  - 0
  - 2019-03-02
  - xtzbaker
- 我编辑了您的问题,以澄清您的第二句话
  
  i edited your question to clarify your second statement
  - 1
  - 2019-03-02
  - Ezy
- 另一个似乎不太可能的情况是,面包师会接受6.5XTZ委托:如果您不能从单独烘焙中获得额外的保费,那么假设的面包师也是如此,因此,假设您将设法从另一位面包师那里获得复利收入只是意味着其他面包师将不得不给您奖励,而无需自己额外增加一卷(换句话说,其他面包师将用自己的扑克向您付款).注意:我实际上将此贴发布为我的答案.
  
  another premice that seems unlikely is that a baker would accept a 6.5XTZ delegation: if you cannot earn an extra premium from solo-baking, the same goes for the hypothetical baker, so assuming that you will manage to earn your compounding income from another baker just means that that other baker will have to give you rewards without an additional roll of his own (in other words that other baker would be paying you out of his own poket). NB: I posted this as my answer actually.
  - 0
  - 2019-03-02
  - Ezy
- 如果我已经将可接受的数量委托给面包师,那么大多数人将接受6.5充值,而每个周期都没有任何问题
  
  If I already have an acceptable amount delegated to the baker most of them will accept the 6.5 top up without any problems each cycle
  - 0
  - 2019-03-02
  - xtzbaker
- 它不会改变结论.您的充值不会给面包师带来任何额外的报酬,因此您的额外收入将来自稀释该面包师的其他代表以及面包师自己(取决于他如何计算分配).另外,您现在添加了另一种假设,即您已经拥有另一位被面包师锁定的资金来源.那是几卷?这很重要,因为如果您想平分收益,将资金分散到多个烘焙操作中是次优的
  
  It does not change the conclusion. Your top-up earns no marginal extra reward to the baker, so your extra income would come out of diluting other delegators to this baker as well as the baker himself depending on how he calculates the distribution. In addition you are now adding another hypothetical that you already have another source of capital that is locked with a baker. How many rolls is that ? this is important cause it is suboptimal to split your capital accross several baking operations if you want to smooth your rewards
  - 0
  - 2019-03-02
  - Ezy

3 个回答

投票数

- 5
- 2019-02-05
委派"多余的XTZ"完全有道理.只是整卷为您烤.整卷之上的所有未授权的XTZ(考虑费用)都是浪费的奖励.

计算对我来说很好.

您只需要关注"下一滚动目标"(之前7个循环).这样您就可以将自己委托给自己.

It makes totally sense, to delegate the "excess XTZ". Just full rolls bake for you. Every undelegated XTZ (considering the fee) above a full roll are wasted rewards.

The calculation is looking good to me.

You just need to have an eye on the "Next Roll Goal" (7 cycles before). So you can delegate back to yourself.
- 0
- 2019-03-02
似乎不太可能的前提是,面包师会接受6.5XTZ委派.

如果您无法通过单独烘焙从6.5XTZ获得额外的保费,那么假设的面包师也是如此.

因此,假设您将设法从另一位面包师那里获得您的复利收入,则意味着该另一位面包师将不得不给您奖励,而无需额外支付自己的费用(换句话说,另一位面包师将向您支付其收入)自己的口袋).

因此,尽管您的数学是正确的,但前提是要找到一个慷慨的面包师从自己的口袋里掏钱给您,这在我看来是不现实的.

A premise that seems unlikely is that a baker would accept a 6.5XTZ delegation.

If you cannot earn an extra premium from these 6.5XTZ by solo-baking, the same goes for the hypothetical baker.

So assuming that you will manage to earn your compounding income from another baker just means that that other baker will have to give you rewards without an additional roll of his own (in other words that other baker would be paying you out of his own pocket).

Therefore although your math is correct the premise is conditional on find a generous baker to pay you out from his own pocket which is unrealistic in my opinion.
- 原文
- Ezy
- 是的,绝对必须确保面包师的支出结构能够为您带来期望的回报.大多数面包师(尤其是较小的面包师)讨厌小型代表团,因为每个周期都有很多人付出,而您的代表团很少会帮助他们全面发展.
  
  yes, definitely have to make sure the payout structure of the baker will give you the rewards you expect. Most bakers(especially smaller) hate small delegations as a lot pay out every cycle and your delegation rarely will help them get to a full roll.
  - 0
  - 2019-03-02
  - Frank
- 当前有许多面包师正在接受这种行为.当然,这不是给定的,并且因面包师而异,但这绝对不是不可能.
  
  There are a number of bakers currently operating that accept this kind of behaviour. Granted it is not a given, and varies by baker but it is definitely not unlikely.
  - 0
  - 2019-05-31
  - xtzbaker

xtzbaker · Accepted Answer · 2019-05-31

0

2019-05-31

通常,最佳行为是在每个周期之后转移过多的tezo,只要委托的预期收益超过交易费用,并在委托的预期收益不超过交易费用时将多余的tezo留在您的帐户中

但是,在某些情况下,这不是最佳行为,这在很大程度上是一种边缘情况,并且差异并不是真正的实质性问题,而是出于完整性考虑问题进行了描述,因为问题在于最佳解决方案.

如果您有过多的XTZ,那么如果委派太多的XTZ不能支付转账费用,并且您不希望在当前周期内收到任何奖励,那么最佳选择是直接转账.希望下面的示例可以说明原因.

周期0中的多余Tezos:2 XTZ
在周期0中委派的预期奖励:(2 - .001272) * .00065 * (1 - 0.10) = 0.001169
在第0周期中自己烘焙的预期奖励:0 XTZ
从周期1的委托中获得的预期奖励:(2 - .001272) * .00065 * (1 - 0.10) = 0.001169 // The compounding is negligible at these levels so not including it for simplicity
在第1周期中从自己的烘焙中获得的预期奖励:10 XTZ//让我们假设这是物质金额
如果在第0周期和第2周期中进行了委托,则可以从周期2获得委托的预期奖励:((10 - .001272) + (2 - .001272)) * .00065 * (1 - 0.10) = 0.007018 // The compounding is negligible at these levels so not including it for simplicity
如果在第2周期中进行了委派,则会从第2周期的委派中获得预期的奖励:(12 - .001272) * .00065 * (1 - 0.10) = 0.007019

在周期0 + 2中委派的超额XTZ期望总值(10 - .001272) + (2 - 0.001272) + 0.001169 + 0.001169 + 0.007018 = 12.006812
等待直到第2周期委派过量的XTZ的过量XTZ总期望值:(12 - 0.001272) + 0.007019 = 12.005747

立即委派会导致另外的.001065 XTZ.