# How do I calculate the number of people who have ever lived in a place over a given span of time assuming a fixed population?

500 years ago, the mountain dwelling Carac clan was cursed with undeath: whenever they or any of their descendants die, their corpses rise as undead creatures, or if burned, as shadowy specters. The Carac quickly adopted a custom of casting their deceased into a chasm. This does not prevent the occurrence of undeath, but nothing has ever crawled back out of the hole so it's always been considered a wise solution.

I would like to know approximately how many undead creatures are in this chasm. I know there are resources online for how to estimate the length of a generation etc., but I'm a complete blockhead at this sort of thing and can't figure out for the life of me how to calculate the number of people who have ever died in the Carac clan.

In the mountain valley the Carac call home, let's say the clan has had a stable population of 1000 for the last 500 years. Not a realistic assumption, but I'm only trying to figure out how many undead might plausibly occupy the chasm, not an exact figure. I'm sure there are other figures you will need from me to answer this question: please feel free to make any other demographic assumptions that you wish for an iron age barbaric mountain folk, or ask in comments and I'll try to respond promptly.

Thank you!

• Remember that in an iron-age society, half of your undead will be infants who died in their first couple of years. Oct 14, 2018 at 6:52
• @MikeScott I completely forgot to consider that. What a grisly detail! Will have to create a special stat block for swarm of infant undead I guess? Oct 14, 2018 at 14:36
• yanno, if dismemberment was an option, that would totally keep them from climbing out. Oct 15, 2018 at 1:00

What’s the average life expectancy? A fixed population makes the maths very easy (ignoring edge effects where people are alive at the beginning or end of the period, which are negligible over a long enough interval). The number of people who have lived is:

Population x length of period / life expectancy

So for your 1,000 population over 500 years, if their life expectancy is 50 years then there have been 10,000 people alive during that period (1000 x 500 / 50).

• Perfect, thank you! I'm a little confused why life expectancy matters, though. If every couple has enough children to replace themselves at age 20, why does it matter at what age they will eventually die thereafter? Maybe I'm asking too much, because my original question did not ask for an explanation of the formula, but if it's easy to explain and you have a moment I would appreciate it! Oct 14, 2018 at 14:43
• @PinkSweetener Looking at the extremes may help you to see it. If your people lived for 500 years, then the initial 1,000 people would die but no one else. But if they only lived for one year then 1,000 would die every year and there would be 500,000 deaths in total. Oct 14, 2018 at 14:51
• @PinkSweetener: That's a good question, and the answer is surprisingly deep. The reason that life expectancy can appear in the formula is that you've specified a constant population, which implies a certain relationship between life expectancy, how many children the average couple has, and how long it takes them to have that many. So Mike Scott's formula doesn't represent causation -- increased life expectancy doesn't cause a drop in the number of people who've ever lived -- but a calculation using an already-known outcome (constant population). Oct 14, 2018 at 17:15
• @PinkSweetener: More specifically, you can work it out this way: if population is constant, then population = birth rate * life expectancy, birth rate = death rate, and the result you want is death rate * length of period. Therefore, the result you want is (population / life expectancy) * length of period. Oct 14, 2018 at 17:17
• I would assume that life expectancy in iron age is far less than 50 years. I can't show you proofs immediately but at this time life expectancy was ~25-30 years which could double number of undeads.