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 $r$.