# M Derivation ## Appendix 3: $$M\_N$$ Derivation Setup for an special case $$M\_{10}$$ You have a family of 10 relays, each relay $$F\_1(G\_1, A\_{10})$$ providing 1 GB of bandwidth and staked with 10 $ANYONE. You want to compare this to the family as a whole having the collective resources of 10 GB of bandwidth and staked with 100 $ANYONE, and examine how the rewards scale. Definitions and Equation * $$F\_{10}(G\_1, A\_{10})$$ : A family of 10 relays, each with 1 GB of bandwidth and 10 $ANYONE. * $$B\_{10}$$: Boost factor for a family of 10. * $$M\_{10}$$: The multiplier for having 10 times the amount of $ANYONE. Given Relations 1. Equivalence of Family to Single Relay with Boost: $$F\_{10}(G\_1, A\_{10}) = F1(G\_{10}, A\_{100}) \times B\_{10}$$ This means the family of 10, each with (G1, A10), boosted, is equivalent to one relay with 10 GB and 100 $ANYONE. 2. Relation of Individual to Group without Family Boost: $$10 \times F\_1(G\_1, A\_{10}) = F\_1(G\_{10}, A\_{10})$$ Ten individual relays each with 1 GB and 10 $ANYONE have the collective power of one relay with 10 GB and 10 $ANYONE without considering any boost. Calculate $$M\_{10}$$: We want to express $$M\_{10}$$ in terms of known quantities. Start by considering the relationship between $$F1(G\_1, A\_{100})$$ and $$F\_1(G\_1, A\_{10})$$ given $$M\_{10}$$ : $$F\_1(G\_1, A\_{100}) = M\_{10} \times F\_1(G\_1, A\_{10})$$$$M\_{10} : F\_1(G\_1, A\_{100}) = M\_{10} \times F\_1(G\_1, A\_{10})$$ This equation states that the reward for a relay with 100 $ANYONE is $$M\_{10}$$ times the reward for a relay with 10 $ANYONE. Use of $$F\_{10}$$ in Calculating $$M\_{10}$$: * Since $$F\_{10}(G\_1, A\_{10})$$ must also be the sum of the rewards of each relay in the family, boosted by $$B\_{10}$$, we can also say: $$F\_{10}(G\_1, A\_{10}) = 10 \times B\_{10} \times F\_1(G\_1, A\_{10})$$ * Using the relationship $$F\_1(G\_1, A\_{100}) = M\_{10} \times F\_1(G\_1, A\_{10})$$ and substituting it into the equation derived from the family boost: $$F\_1(G\_{10}, A\_{100}) \times B\_{10} = B\_{10} \times 10 \times F\_1(G\_1, A\_{100}) = B\_{10} \times 10 \times M\_{10} \times F\_1(G\_1, A\_{10})$ $F\_{10}(G\_1, A\_{10}) = B\_{10} \times 10 \times M\_{10} \times F\_1(G\_1, A\_{10})$$ * Equating the two expressions for $$F\_{10}(G\_1, A\_{10})$$ : $$10 \times B\_{10} \times F\_1(G\_1, A\_{10}) = B\_{10} \times 10 \times M\_{10} \times F\_1(G\_1, A\_{10})$$ * Solving for $$M\_{10}$$ : $$10 \times B\_{10} \times F\_1(G\_1, A\_{10}) = 10 \times B\_{10} \times M\_{10} \times F\_1(G\_1, A\_{10})$$ Cancel out common terms: $$M\_{10} = 1$$ Conclusion: From the above, it appears that $$M\_{10} = 1$$ under the assumptions used for the calculations. However, this result seems counterintuitive as it implies no scaling effect of rewards despite a tenfold increase in staking. However, if the assumptions are true, then token scaling efforts should not affect the rewards or else the assumptions cannot be considered true. $$M\_n$$ : Generalizations for all $$n \in {\[1,2,3,4,...]}$$$$M\_n$: Generalizations for all $n \in {\[1,2,3,4,...]}$$ Revised Definitions and General Assumptions Let's define $$M\_n$$ as the reward multiplier for a relay when the $ANYONE staked is increased $$n$$-fold, while bandwidth remains constant. This generalization will help establish a broader rule applicable to any positive integer $$n$$, $$y$$ gigabytes of bandwidth per relay, and a minimum $ANYONE stake of $$r$$ $ANYONE. Generalized Given Relations: 1. Equivalence of Family to Single Relay with General Boost: $$F\_n(G\_y, A\_{r}) = F\_1(G\_{yn}, A\_{r n}) \times B\_n$$ (Equation 1) Here, a family of $$n$$ relays, each with y GB and $$r$$ $ANYONE, when boosted, this is equivalent to one relay with $$y \times n$$ GB and $$r \times n$$ $ANYONE. $$B\_n$$ is the boost factor for a family of $$n$$. 2. Relation of Individual to Group without Family Boost: $$n \times F\_1(G\_y, A\_{k}) = F\_1(G\_{yn}, A\_{k})$$ (Equation 2) This states that $$n$$ individual relays, each with $$y$$ GB and $$k$$ $ANYONE, together provide the resources of one relay with $$y \times n$$ GB and $$k$$ $ANYONE, without any additional boosts. Definition of $$M\_n$$ Let’s Define $$M\_n$$ the following way: EQ3: $$F\_1(G\_y, A\_{rn}) = M\_{n} \times F\_1(G\_y, A\_{r})$$ (Equation 3) Derivation of $$M\_n$$: We can follow the exact same process as we did for $$M\_{10}$$: * Since $$F\_{n}(G\_y, A\_{r})$$ must also be the sum of the rewards of each relay in the family, boosted by $$B\_{n}$$, we can also say: $$F\_{n}(G\_y, A\_{r}) = n\times B\_{n} \times F\_1(G\_y, A\_{r})$$ * Using the relationship $$F\_1(G\_y, A\_{rn}) = M\_{n} \times F\_1(G\_y, A\_{r})$$ and substituting it into the equation derived from the family boost: $$F\_1(G\_{yn}, A\_{rn}) \times B\_{n} = B\_{n} \times n \times F\_1(G\_y, A\_{rn}) = B\_{n} \times n \times M\_{n} \times F\_1(G\_y, A\_{r}) F\_{n}(G\_y, A\_{r}) = B\_{n} \times n \times M\_{n} \times F\_1(G\_y, A\_{r})$$ * Equating the two expressions for $$F\_{n}(G\_y, A\_{r})$$: $$n\times B\_{n} \times F\_1(G\_y, A\_{r}) = B\_{n} \times n \times M\_{n} \times F\_1(G\_y, A\_{r})$$ * Solving for $$M\_{n}$: $n\times B\_{n} \times F\_1(G\_y, A\_{r}) = n \times B\_{n} \times M\_{n} \times F\_1(G\_y, A\_{r})$$ Cancel out common terms: $$M\_{n} = 1$$ The conclusion that $$M\_n = 1$$ for all positive integers $$n$$ with any gigabyte relays $$y$$ and minimum $ANYONE requirement $$r$$. --- # Agent Instructions: Querying This Documentation If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question. 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