Continuing from my previous post, I have attempted to determine when the effective R = 1 for this population (herd immunity). Let me know if you find something questionable.

The primary assumption made along with the two previously calculated R-values is that “R” describes not only an individual’s propensity to spread the virus, but also describes their propensity to get the virus. On some level, that makes sense. To spread the virus to many people one needs to interact with many people. It’s not a stretch to believe that the same activity that allows one to infect many others also makes one more likely to get the virus. This assumption takes that idea to the extreme, but, it is illustrative.

Assume:

1. Population R0 = 3

2. 10% of the population has an R = 24

3. 90% of the population has an R = 2/3

Method: I use weights to allocate 1% of the population at a time to each group. The weights depend on the ratio of individuals with R = 24 to those with R = 2/3. This ratio will change as those in the group with R = 24 will get the virus at a higher rate. Thus the weights are constantly changing.

Initially, 100% of the population is susceptible:

10% weight: 0.1*24 = 2.4

90% weight: 0.9*2/3 = 0.6

The proportion associated with the 10% = 2.4/(2.4 + 0.6) = 0.8

- For the first 1% of the population infected, 0.8% are the “24’s” and 0.2% are the “2/3’s”

- So, the “24’s” drop from 10% to 9.2% and the 2/3’s” drop to 89.8%

- Re becomes (0.092*24 + 0.898*2/3) = 2.84

- Note that for the homogeneous population where R = 3 for everyone, after 1% are infected, Re = 0.99*3.0 = 2.97

Now, 99% of the population is susceptible

10% weight: 0.092*24 = 2.208

90% weight: 0.898*2/3 = 0.059866…

The proportion associated with the 10% = 2.208/(2.208 + 0.059866..) = 0.787

…

I summarize this process in the table below which was done on a spreadsheet

As a final adjustment, I averaged the proportion associated with 10% so that it doesn’t always take the value at the beginning of the “1% allocation” process and re-calculated.