You have a family of 10 relays, each relay F1β(G1β,A10β) 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
F10β(G1β,A10β) : A family of 10 relays, each with 1 GB of bandwidth and 10 $ANYONE.
B10β: Boost factor for a family of 10.
M10β: The multiplier for having 10 times the amount of $ANYONE.
Given Relations
Equivalence of Family to Single Relay with Boost: F10β(G1β,A10β)=F1(G10β,A100β)ΓB10β 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: 10ΓF1β(G1β,A10β)=F1β(G10β,A10β) 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 M10β:
We want to express M10β in terms of known quantities. Start by considering the relationship between F1(G1β,A100β) and F1β(G1β,A10β) given M10β : F1β(G1β,A100β)=M10βΓF1β(G1β,A10β)M10β:F1β(G1β,A100β)=M10βΓF1β(G1β,A10β)
This equation states that the reward for a relay with 100 $ANYONE is M10β times the reward for a relay with 10 $ANYONE.
Use of F10β in Calculating M10β:
Since F10β(G1β,A10β) must also be the sum of the rewards of each relay in the family, boosted by B10β, we can also say: F10β(G1β,A10β)=10ΓB10βΓF1β(G1β,A10β)
Using the relationship F1β(G1β,A100β)=M10βΓF1β(G1β,A10β) and substituting it into the equation derived from the family boost:
Equating the two expressions for F10β(G1β,A10β) : 10ΓB10βΓF1β(G1β,A10β)=B10βΓ10ΓM10βΓF1β(G1β,A10β)
Solving for M10β : 10ΓB10βΓF1β(G1β,A10β)=10ΓB10βΓM10βΓF1β(G1β,A10β) Cancel out common terms: M10β=1
Conclusion:
From the above, it appears that M10β=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.
Mnβ : Generalizations for all nβ[1,2,3,4,...]
Revised Definitions and General Assumptions
Let's define Mnβ 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: Fnβ(Gyβ,Arβ)=F1β(Gynβ,Arnβ)ΓBnβ (Equation 1) Here, a family of n relays, each with y GB and r $ANYONE, when boosted, this is equivalent to one relay with yΓn GB and rΓn $ANYONE. Bnβ is the boost factor for a family of n.
Relation of Individual to Group without Family Boost: nΓF1β(Gyβ,Akβ)=F1β(Gynβ,Akβ) (Equation 2) This states that n individual relays, each with y GB and k $ANYONE, together provide the resources of one relay with yΓn GB and k $ANYONE, without any additional boosts.
We can follow the exact same process as we did for M10β:
Since Fnβ(Gyβ,Arβ) must also be the sum of the rewards of each relay in the family, boosted by Bnβ, we can also say: Fnβ(Gyβ,Arβ)=nΓBnβΓF1β(Gyβ,Arβ)
Using the relationship F1β(Gyβ,Arnβ)=MnβΓF1β(Gyβ,Arβ) and substituting it into the equation derived from the family boost: F1β(Gynβ,Arnβ)ΓBnβ=BnβΓnΓF1β(Gyβ,Arnβ)=BnβΓnΓMnβΓF1β(Gyβ,Arβ)Fnβ(Gyβ,Arβ)=BnβΓnΓMnβΓF1β(Gyβ,Arβ)
Equating the two expressions for Fnβ(Gyβ,Arβ): nΓBnβΓF1β(Gyβ,Arβ)=BnβΓnΓMnβΓF1β(Gyβ,Arβ)
Solving for Cancel out common terms: Mnβ=1
The conclusion that Mnβ=1 for all positive integers n with any gigabyte relays y and minimum $ANYONE requirement r.