Leech Principle

Bond’s social value calculation algorithm help determine a user's social value based on their activity, influence, and position within the network, rather than just their adoption time. This means that even those who join later can achieve a high social value by being active and influential within the network. Additionally, we've implemented a decay factor (Penality). If the system detects a decline in a user's staking situation or social activity, the quantity of their rewards will correspondingly diminish. In essence, Bond's "Leech Principle" is a just and reasonable reward mechanism that benefits both the project side and all project users. It serves to motivate value creators to consistently contribute to the project, preventing those who haven't genuinely participated from reaping excessive rewards, thus enhancing the project's long-term viability and fairness.

Here we have:

$$\begin{gathered} \mathrm{Penalty}=1-(P_{\mathrm{Team}}+P_{\mathrm{Token}}+P_{\mathrm{Social}}) \\ P_{\text{Team}}=\begin{cases}0.15&\text{ if }\Delta\text{Team-Value}<0\\0&\text{ otherwise}\end{cases} \\ P_{\mathrm{Token}}=\begin{cases}0.6&\mathrm{~if~}\Delta\text{Token-Stake}<0\\0&\mathrm{~otherwise}\end{cases} \\ P_{\mathsf{Social}}=\begin{cases}0.25&\mathrm{~if~}\Delta\text{Social-Value}<0\\0&\mathrm{~otherwise}\end{cases} \end{gathered}$$

We can start by calculating the reward for each cycle from the beginning and summing them up.

Now let's define some parameters in our senario

Initial reward: ​ $R=100$

Fixed reward rate: $f=25​$

Time intervals: $6$ ​

Decay rates: We have three: $d1=15​, d2=25​, d3=40​$

In the 1st cycle:

$Reward=100×0.25×0.851​$

$Balance=100+100×0.25×0.851​$

In the 2nd cycle:

$Reward=(100+100×0.25×0.851)×0.25×0.852​$

$Balance=100+100×0.25×0.851+(100+100×0.25×0.851)×0.25×0.852​$

And so on, for the $t-th$ cycle:

$Reward=(Balancefromthepreviouscycle)×0.25×0.85t​$

Thus, for time t​, the total amount is the original principal plus the reward from each cycle.

We can formulate this as a recursive relation:

$$y(t)=100+\sum_{i=1}^t(100+y(i-1))\times0.25\times0.85^i$$

However, this remains a recursive relation, not a closed-form solution. To obtain the balance at any given $t$, we would need to iterate through this process.

Considering different decay rates, we can write a formula for each rate:

No decay (0% decay): This means that the reward rate for every cycle remains 0.25, with no decay.

$$y_1(t)=100+\sum_{i=1}^t(100+y_1(i-1))\times0.25$$

Note: The reward is 25% of the previous cycle in each cycle and there's no decay. Hence, this series will not converge as it adds a value of 25% each time, rather than approaching a fixed value.

15% decay: The reward rate starts at 0.25, but with each cycle, it is multiplied by 0.85.

$$y_2(t)=100+\sum_{i=1}^t(100+y_2(i-1))\times0.25\times0.85^i$$

25% decay: The reward rate starts at 0.25, but with each cycle, it is multiplied by 0.75.

$$y_3(t)=100+\sum_{i=1}^t(100+y_3(i-1))\times0.25\times0.75^i$$

40% decay: The reward rate starts at 0.25, but with each cycle, it is multiplied by 0.60.

$$y_4(t)=100+\sum_{i=1}^t(100+y_4(i-1))\times0.25\times0.60^i$$

Note: These four formulae describe the change in account balance over time t, based on different decay rates. $y2$, ​,​and $y4$​ do not converge to a fixed point, but the slopes of their curves gradually decrease. This means that the rewards received by users with decay will become progressively smaller over time."

Now we have:

Week	Reward1 (No Decay)	Reward2 (15% Decay)	Reward3 (25% Decay)	Reward4 (40% Decay)
1	150.000	142.500	137.500	130.000
2	212.500	186.302	170.898	150.700
3	290.625	230.258	199.470	164.238
4	388.281	273.357	223.158	172.799
5	510.352	314.772	242.330	178.102
6	662.939	353.880	257.562	181.346
7	853.674	390.256	269.494	183.315
8	1092.093	423.653	278.742	184.505
9	1390.116	453.975	285.851	185.222
10	1762.645	481.241	291.283	185.653

To accentuate the curves with decay rates, we can employ subplots to plot individual curves for each decay rate, or we can use a logarithmic scale for the y-axis.

Here, we present two methods:

Method 1: Subplots

Plot a distinct subplot for each decay rate.

Method 2: Logarithmic Scale

Utilising a logarithmic scale can better illustrate the growth trend of the data, especially when some data grows much faster than others.

Both methods have their merits. The subplot method allows for a more detailed examination of each curve, while the logarithmic scale approach provides a holistic view, aiding in comparing different curves.

Generalisation

Let's generalise these parameters:

Alright, let's express these parameters in a more generalised form:

• $P$​: Pledged amount (originally the 100 principal)

• $r$​: Reward rate (originally 0.25)

• $d$: Decay rate as a percentage (for instance, 15% would be 0.15)

• $t$: Time period

Note: The decay factor is $1−d$​. For instance, with a 15% decay rate, the decay factor becomes 0.85.

Hence, our recursive formula becomes:

$$y(t)=y(t-1)+(P+y(t-1))\times r\times(1-d)^t$$

Where the initial condition is $y(0)=P$.​

Do bear in mind: this is a recursive formula, describing the balance at each time period t​. To compute the balance at any specific t​, one would have to start from t=0​ and apply this formula recursively until the desired t​ is reached.

Hence, our recursive formula becomes:The increment can be represented as:

$$\text{Increment}=(P+y(t-1))\times r\times(1-d)^t$$

Summing this increment from t=1 to t, we get:

$$y(t)=P+\sum_{i=1}^t(P+y(i-1))\times r\times(1-d)^i$$

For $d=0$​，

$$y_{d=0}(t)=P+\sum_{i=1}^t(P+y_{d=0}(i-1))\times r$$

That's when there's no decay.

For $0<d<1​$,

$$y_d(t)=P+\sum_{i=1}^{\cdot}(P+y_d(i-1))\times r\times(1-d)^i\quad;$$

Therefore, the difference between these two at time t​ is:

So the difference between them when t​ is:

$$\begin{aligned} &\Delta y(t) =y_{d=0}(t)-y_d(t) \\ &\Delta y(t) =\sum_{i=1}^t(P+y_{d=0}(i-1)-P-y_d(i-1))\times r+\sum_{i=1}^t(P+y_d(i-1))\times \\ &r\times d^i \end{aligned}$$

Finally, let's draw the curve the this difference:

Summary

Without reward decay: This takes the form of a geometric series, and as time progresses, it will persistently grow, meaning it won't converge to a finite value. In essence, it tends towards infinity.

With reward decay: The reward rate gradually diminishes, but doesn't become nil. Under these circumstances, each additional period will contribute some added value to the account, yet these increments will dwindle over time.

Whilst the magnitude of growth diminishes over time, given that new values are introduced in every period, $y(t)$​ doesn't converge to a set figure either. However, the rate of growth will progressively decrease, suggesting that the slope of the function will approach zero, yet the overall value will persist in its ascent, tending towards infinity.