Theoretical delay before first approval

Question

I'm reading the white-paper about the tangle. In pages 8 to 10, the author is calculating the average time for first approval of a tip on a stationary tangle.

The first result is that under low load:

The first approval happens on an average timescale of order λ ⁻¹

where λ is the rate of arrival of new transactions. This result is clear to me.

But I must be missing something (maybe obvious) when the author present the result for a tangle under heavy load (page 10):

Let us now consider the high load regime, the case where L₀ is large. As mentioned above, one may assume that the Poisson flows of approvals to different tips are independent and have an approximate rate of 2λ/L₀. Therefore, the expected time for a transaction to receive its first approval is around L₀/(2λ) ≈ 1.45h (1)

Where:

• L₀ is the number of tips (large under heavy load).
• h is the average hidden time of a transaction (a transaction is hidden during the execution of the pow)

I don't see from where comes this factor: 1.45

I looked into reference "Sheldon M. Ross (2012) Introduction to Probability Models. 10th ed." (page 312) to have better understanding of the Poisson Process, but it didn't help me to understand why we obtain this result.

Answer 1

1/ln2 from equation (3) in page 8. (in version 1.2 of the whitepaper)

Answer 2

r/(r+λh) is a mean value ( < 1 ) r is tips before time [t-h] r+λh is a slightly higher number adding all tips from time interval [t-h,t]

since the node does not know λh tips are no tips any more at time index t (assuming λh is always stationary value), we have a total probability of choosing a tip of r/(r+λh) ( < 1 )

Total mean number of chosen tips is then 2*r/(r+λh) - you always select 2 tips. Why this is equal to 1 in a stationary regime I also do not fully understand. I assume that the r number of tips and the λh number of tips should be equal (half/half for instance) and then a new addition of one transaction would not change it by much at that very moment - so resulting in 1 quotient. Anyone can confirm this interpretation?

What remains unclear after comments issued in other posts: Why is the time for 1st approval L0/(2λ) ≈ 1.45h (or lets say ≈ h as in version 1.3 of paper)?

Paper states that during a period of h a transaction cannot be approved.

This is because, by our assumption, during the first h units of time a transaction cannot be approved, and after that the Poisson flow of approvals to it has rate approximately 2λ/L0.

So it should it total be the sum of h + L0/(2λ) ≈ 2h

I think white paper needs an update on this. Moreover the link to the equation L0/(2λ) ≈ 1.45h (1) is wrong in v1.3

Answer 3

Expected increment of # of tips at time t is: 1 - 2*r/(r+λh) .....eqn(1) where "1" refers to new tip created by the tx and "2*r/(r+λh)" refers to expected # of "erased" tips. If the two "erased" tips were already approved (usually for small L(t)), then "2*r/(r+λh)" = 0, thus eqn(1) becomes positive or increasing.
If the two "erased" tips were not previously approved (usually for large L(t)), then "2*r/(r+λh)" = 2, thus eqn(1) becomes negative or decreasing.
If one "erased" tip was already approved and one not previously approved (for L(t) becomes stationary or L0), then "2*r/(r+λh)" = 1, thus eqn(1) becomes zero, neither increasing nor decreasing. Thus L(t)=L0, eqn(1) = 0; 1-2*r/(r+λh) = 0; therefore 2*r/(r+λh) = 1.