How fast would a colonizing population grow on virgin soil?

Question

I have a story where the setup is that, ~600 years prior to the story's beginning, there was an apocalypse of a magical nature that rendered most of the world uninhabitable. A few survivors from various places managed to escape to the one inhabitable landmass, an island roughly the size of Greenland. The island has a latitude, climate and environment similar to New Zealand. It had no prior inhabitants. The survivors are coming from a world that was approaching an Industrial Revolution, but their technological prowess will probably be set back because of the disaster.

So, say the starting population was ~10,000 people, after which there were no newcomers because, well, everyone else is dead. At least some of the surviving groups would've brought animals and plants, since they knew they would essentially have to restart civilization. What would the population of this island be 600 years later?

Answer 1

Your question is asking for a calculation of the biotic potential. The biotic potential combines growth factors, i.e. reproduction, and survival factors to reach an estimate of the unconstrained growth of an organism. Countering that is environmental resistance, such as food availability. In principle a population will grow until it reaches the environmental carrying capacity. Wikipedia

There are different models available. The Malthusian model assumes exponential growth. A probably better model, especially given your time span, would be the logistic growth model which explicitly caters for the carrying capacity.

The carrying capacity varies over time, as agricultural methods improve, and there are also models which cater for this.

An estimate of the upper bound of the carrying capacity of your island could be the biomass of New Zealand. This would assume that the entire biomass is made up of humans and their foodstuffs, which could be vat-grown produce. Not as nice to live in as New Zealand. Isaac Asimov wrote a short story about the Earth being in this state (actually, he wrote two).

To calculate the reproductive potential, you need to consider the number of humans surviving to reproductive age, the number of years they reproduce and the average number of children per birth. These statistics are readily available, as is the average reproductive rate for various societies.

Calculator for various growth models

Answer 2

Impossible to say for sure, due to many factors

How well picked are the colonists? Do they have the skills to rapidly re-establish modernity? Do the seeds/animals they have suit the environment? How much of the land is good arable land? How much can have, e.g. sheep grazed on it? Do they rapidly establish a single polity, or several friendly polities, or engage in repeated, destructive wars, etc.

However, we can give a fair estimate by comparison to other countries

Given that length of time, and with modern agricultural technology, and medical science, they should be able to achieve levels similar to modern nations. If New Zealand had the same population density as Europe, it'd hit 31.3 million people; if it was as populous as England that would boost to 75 million. These numbers may seem high but both require population growth of less than 1.5%/year, substantially less than current growth levels in Nigeria or Ethiopia (~2.5%/year) that growth level would punt the population to an eye-watering 2.7 billion, and a population density of 101,000/km2 which is surely not realistic.

Answer 3

I agree with the logarithmic model, but wanted to explore this topic a bit more. Specifically, I wanted to try applying something like a sales funnel.

Edit: 2-23-2022

Wanted to amend this answer to correct a very bad oversimplification. It's easy to concentrate on women, as the only producer of children. You might be tempted to assume, in this example, that for your 10,000 people, the best chances of survival would be if 9,999 of them are women + 1 man. But, on further study, the best ratio is 1 man per woman (or very close to it), for the following reasons

Genetic Diversity

If you have a 1:1 ratio of men to women, then --

• In the first generation you have 5,000 sufficiently different lineages.
• In the second generation, and carrying forward you have 2,500 families that are not at high risk of defects because they are siblings or first cousins.

If the ratio is adjusted as little as 1:2 (a man, a wife, and an ex-wife) --

• 1st generation, 2,500 sufficiently different lines to avoid inbreeding problems
• 2nd generation, only 625 unique families far enough apart.

In a 1:$\infty$, everyone in the first generation of children in a sibling.

How important is genetic diversity? This study attempts to put some numbers to the situation. They found, roughly a 4-fold increase in negative traits (such as expression of proteins showing a susceptibility to tuburculosis infection) being expressed among the most inbred populations.

This study, has generously attempted to develop criteria we can plug in to the survivability ($P_s$) and fertility ($P_f$) model below.

Whatever your base survivability ($P_s$) is, reduce it by roughly double your inbreeding factor (F) -- 25% for siblings, 3% to 6% for first cousins, negligible for more distant. So, a $P_s$ of 0.999 drops to about 0.949 for a society that's just a twinge inbred... which doesn't seem like much, but that has a pretty dramatic effect.

The fertility rate ($P_f$) drops by about four times the inbreeding factor (F). So, our base $P_f$ of 0.84 would drop to 0.72 for the same barely inbred society of F = 0.03 (first cousins). A very dramatic effect.

Disease

This might also be easy to overlook, but 1:1 pairing is a firewall against contracted disease. With anything like a 1:$\infty$ one male for everyone relationship, every member of the community has everyone's illnesses, including many fatal ones that might wipe out the population.

Back to statistics

Let's say our population of survivors $n$ can be broken into demographic buckets by gender $n_w$, $n_m$ and age $n_{w,20}$, $n_{w,21}$, ...

Now, concentrating only on the part of the population pipeline that can give birth $n_w$ :

• Pf : percentage of the population that is able to have children is 0.84 (84%) -- you can dive deeper by age, if you like. Reference.
• Pb : percentage of pregnancies that will end in live birth is 0.1857 (18.75%) -- with 75% of pregnancies ending in miscarriage before week 5, and other ~25% ending in miscarriage after week 5. Source
• The youngest pregnancy ever recorded is at 5 years of age, and the oldest at 74 years of age; although most agree that, physically, this is most likely between the ages of 13 and 40.

Each "turn of the crank" in time, there are new people created in the $n_{m,0}$ and $n_{w,0}$ buckets at the front of our population funnel. The number of people created $n_{i,0}$ is equal to the ${\sum_{i=13}^{40}}Pf \times Pb \times n_{w,i}$

Half this number is new baby boys $n_{m,0}$ and half is new baby girls $n_{w,0}$.

Also, each "turn of the rank" every demographic group moves a level down the age part of the funnel. The chance of surviving the year Ps varies by age group. According to Unicef $P_{s,0}$ is 0.995 (99.5%) and $P_{s,1..5}$ is a sad 0.9 (90%) And then, according to actuarial tables I've looked at, survivability improves to $P_s$ 0.999 (99.9%).

The survivability for new mothers is a little lower. Using a source from memory of the mortality statistics for hospitalized mothers when Ignaz Semmelweis was trying to convince doctors to wash their hands between studying cadavers and working with living patients, the survivability of new mothers $P_s$ was 0.7 (70%). So, $P_{s_{w,13..40}}$ is $[n_{w,i} - (n_{w,i} P_f P_b 0.7)] P_s$

Putting that all together, you get a funnel like this :

$\begin{array}{aaaa} & a & 0 & 1-5 & ... & 13 - 40 & ... \\ & P_s & 0.995 & 0.9 & 0.999 & 0.999 & ... \\ & P_f & 0 & 0 & 0.16 & ... \\ t = 0 & n_{w,a} & 0 & 0 & 0 & 10,000 & ... \\ 1 & & 787 & 0 & 0 & 9,527 & ... \end{array}$

Iterating this for five years gives me a population of 14,086. Figuring out the logarithmic exponent is $\ln{{n_5} \over {n_0}} = \ln{1.4} = 0.336 \div 5 = 0.067$ .. or about doubling $e^{0.067t} = e^1$ every ten years

Which is a really long and hard way of re-deriving the rule of thumb that "a population roughly doubles every ten years".

But we learned some new things from this long slog :

• the rule of thumb assumes only biological (not social) limits. If half the eligible women choose to opt-out, or wait until they can afford a child, or wait until they find the right man, this number of years "per doubling" can very easily dramatically increase
• in fact, the rule of thumb assumption that ALL of the available women are doing nothing with their time but trying to have babies seems a bit extreme
• also, the assumptions on generally surviving in the environment : having enough food, dealing with disease, are so optimistic that a model that doesn't include those risks at all produces very similar results.
• we've also assumed that the men don't have an impact. As a social limit, if women decide to only have children with a partner, then the man's fertility (Pf = 0.6 or 0.8, depending on source) affects the fertility of matched couples $P_{f, total} = 0.6 \times 0.84 \approx 0.48$
• also deaths among the men (whatever $P_{s,m}$ is), would result in widows, or just not enough men for every woman to have one. $P_{f, total(2)} = P_{f, total} \times P_{s,m}$

Making this Algebraic

Working with a spreadsheet with 100 columns and 600 rows isn't fun.

To simplify the math, we can see that if $P_s$ doesn't change, then the compounded $P_s$ over $t$ iterations is $(P_{s,m})^t$ or $0.999^t$for men, and also for women not having children.

Note: $t$ is being used incorrectly as both "9 months" and "1 year". This mistake will effect the outcome, but I can think of a few ways to correct it later, and I think it only makes things murkier right now.

• $P_{s,base} = 0.999^t$
• $P_{s,w,13..40} = [(0.16 + (0.84 \times 0.81) + (0.84 \times 0.18 \times 0.7))\times 0.999]^t = (0.946 \times 0.999)^t = 0.945^t$

Against the number of children being born :

• $n_{i,0} = {\sum_{i=13}^{40}(n_{w,i}) P_f P_b} $

• where $n_{w,i} = n_{w,i-1} P_{s,w,i-1} = n_w (0.945)^t$

• so, $n_{i,0} = {\sum_{i=13}^{40}(0.945^t) P_f P_b}$

We can integrate this :

• $\int P_f P_b (0.945^t) = P_f P_b \int(0.945^t)$
• $\int (0.945^t) = { {(0.945^t)} \over {\ln{0.945} } } + C = { {(0.945^t)} \over {-0.0565 } } + C $
• $\int P_f P_b (0.945^t) = P_f P_b { {(0.945^t)} \over {-0.0565 } } + C$
• C, working it out by solving at t=0 is 28,302

This algorithm seems produce errors < 4% for the 4 t-values that I've manually calculated.