You have a family of 10 relays, each relay F1​(G1​,A10​) 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
F10​(G1​,A10​) : A family of 10 relays, each with 1 GB of bandwidth and 10 $ANYONE.
B10​: Boost factor for a family of 10.
M10​: The multiplier for having 10 times the amount of $ANYONE.
Given Relations
Equivalence of Family to Single Relay with Boost: F10​(G1​,A10​)=F1(G10​,A100​)×B10​ This means the family of 10, each with (G1, A10), boosted, is equivalent to one relay with 10 GB and 100 $ANYONE.
Relation of Individual to Group without Family Boost: 10×F1​(G1​,A10​)=F1​(G10​,A10​) 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 M10​:
We want to express M10​ in terms of known quantities. Start by considering the relationship between F1(G1​,A100​) and F1​(G1​,A10​) given M10​ : F1​(G1​,A100​)=M10​×F1​(G1​,A10​)M10​:F1​(G1​,A100​)=M10​×F1​(G1​,A10​)
This equation states that the reward for a relay with 100 $ANYONE is M10​ times the reward for a relay with 10 $ANYONE.
Use of F10​ in Calculating M10​:
Since F10​(G1​,A10​) must also be the sum of the rewards of each relay in the family, boosted by B10​, we can also say: F10​(G1​,A10​)=10×B10​×F1​(G1​,A10​)
Using the relationship F1​(G1​,A100​)=M10​×F1​(G1​,A10​) and substituting it into the equation derived from the family boost:
Equating the two expressions for F10​(G1​,A10​) : 10×B10​×F1​(G1​,A10​)=B10​×10×M10​×F1​(G1​,A10​)
Solving for M10​ : 10×B10​×F1​(G1​,A10​)=10×B10​×M10​×F1​(G1​,A10​) Cancel out common terms: M10​=1
Conclusion:
From the above, it appears that M10​=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.
Mn​ : Generalizations for all n∈[1,2,3,4,...]
Revised Definitions and General Assumptions
Let's define Mn​ 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:
Equivalence of Family to Single Relay with General Boost: Fn​(Gy​,Ar​)=F1​(Gyn​,Arn​)×Bn​ (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×n GB and r×n $ANYONE. Bn​ is the boost factor for a family of n.
Relation of Individual to Group without Family Boost: n×F1​(Gy​,Ak​)=F1​(Gyn​,Ak​) (Equation 2) This states that n individual relays, each with y GB and k $ANYONE, together provide the resources of one relay with y×n GB and k $ANYONE, without any additional boosts.
We can follow the exact same process as we did for M10​:
Since Fn​(Gy​,Ar​) must also be the sum of the rewards of each relay in the family, boosted by Bn​, we can also say: Fn​(Gy​,Ar​)=n×Bn​×F1​(Gy​,Ar​)
Using the relationship F1​(Gy​,Arn​)=Mn​×F1​(Gy​,Ar​) and substituting it into the equation derived from the family boost: F1​(Gyn​,Arn​)×Bn​=Bn​×n×F1​(Gy​,Arn​)=Bn​×n×Mn​×F1​(Gy​,Ar​)Fn​(Gy​,Ar​)=Bn​×n×Mn​×F1​(Gy​,Ar​)
Equating the two expressions for Fn​(Gy​,Ar​): n×Bn​×F1​(Gy​,Ar​)=Bn​×n×Mn​×F1​(Gy​,Ar​)
Solving for Cancel out common terms: Mn​=1
The conclusion that Mn​=1 for all positive integers n with any gigabyte relays y and minimum $ANYONE requirement r.