1. So I'm a big fan of strategy games, and to some extent, you can solve math problems to let you know what decision to make.
I tried to tackle solving a problem and I'm just going around in circles on what to focus on, cos I'm ignorant when it comes to math.

The problem:

Output is determined by population size for the most part. For every person you have, they increase base cost of a technology by 1%.
For every planet you have, the base cost is increased by 10%. The mean size planet is 15, so what I did was simply average across planet and pop cost,
to estimate that for every person, on average it increases the cost by 10+15 = 25
25/15 = 1.66% recurring
However, the first 10 people you have, do not impact cost.

So, cost is the function C = BaseCost* [((P-10)*0.0166) + 1]
Note: Base cost ranges all the way from 360 to over 5000.
Mean cost is 2527, median is 1800.

So a base cost of 5000 and 100 people, would cost 149.4% more, or 5000*2.494 = 12, 470

To utilise the population, you must utilise science production in one of three categories.
Each of the categories has a building, which will produce 4 per month in its own category, and one point each in the two other categories.

So a physics research building will produce 4 physics, 1 society, and 1 engineering per month.

Each of these buildings has an upkeep of 2.5 energy. An energy building can be constructed which will produce 5 energy.
So for every 2 lab buildings, 1 energy building on average must be constructed to cover maintenance cost.

S = science building
E = energy building

E>= 0.5 S

Time to research something is T = C/M

With T in months
Where M = monthly production in a given category

Additional constraints: There is a third resource that cannot be ignored, which is required to construct buildings in the first place,
a mineral building will generate 5 minerals of its own resource, and has an upkeep of 2.5 energy.

B = mineral building

E>= 0.5 B

For every pop, there is a secondary upkeep of food. Each farm will cover 5 pops, and these farms also cost 2.5 energy upkeep.

E>= 0.5 F
F = 0.2 P

Putting these constraints into practice, I've determined that for a population P, the amount of people assigned to science
will be Number of science buildings (S) = 0.2 P
With 0.2 P dedicated to farming
0.4 P dedicated to energy (will cover costs and give a surplus 50 energy)
0.2P dedicated to minerals

Optimisation: Reduce T to its minimum amount by finding optimal P

Base cost of things may vary greatly, but the problem is knowing when you've reached the maximum P such that
the percentage increase to base cost can no longer be covered by simply constructing more science buildings.

I'll just approach this as for every time you increase P, you increase C by 0.0167.
Therefore, optimal P is where you can no longer generate an additional 1.67% output.

So, at what output can an additional lab no longer generate more than 1.67% output.
Lab output is 2 avg. per category
So 2 / 0.0167 = 119.76

When you are producing around 120 output from a population P, additional pops will no longer be beneficial.
120/2 = 60 labs
Every fifth person is working it, therefore 60* 5 = 300

Optimal P is therefore 300?

Any advice/tips would be greatly appreciated. It sucks I'm not better at math, because I like min-maxing and optimisation and they're mathematical problems   2.

3. Originally Posted by Curiosity So I'm a big fan of strategy games, and to some extent, you can solve math problems to let you know what decision to make.
I tried to tackle solving a problem and I'm just going around in circles on what to focus on, cos I'm ignorant when it comes to math.

The problem:

Output is determined by population size for the most part. For every person you have, they increase base cost of a technology by 1%.
For every planet you have, the base cost is increased by 10%. The mean size planet is 15, so what I did was simply average across planet and pop cost,
to estimate that for every person, on average it increases the cost by 10+15 = 25
25/15 = 1.66% recurring
However, the first 10 people you have, do not impact cost.

So, cost is the function C = BaseCost* [((P-10)*0.0166) + 1]
Note: Base cost ranges all the way from 360 to over 5000.
Mean cost is 2527, median is 1800.

So a base cost of 5000 and 100 people, would cost 149.4% more, or 5000*2.494 = 12, 470

To utilise the population, you must utilise science production in one of three categories.
Each of the categories has a building, which will produce 4 per month in its own category, and one point each in the two other categories.

So a physics research building will produce 4 physics, 1 society, and 1 engineering per month.

Each of these buildings has an upkeep of 2.5 energy. An energy building can be constructed which will produce 5 energy.
So for every 2 lab buildings, 1 energy building on average must be constructed to cover maintenance cost.

S = science building
E = energy building

E>= 0.5 S

Time to research something is T = C/M

With T in months
Where M = monthly production in a given category

Additional constraints: There is a third resource that cannot be ignored, which is required to construct buildings in the first place,
a mineral building will generate 5 minerals of its own resource, and has an upkeep of 2.5 energy.

B = mineral building

E>= 0.5 B

For every pop, there is a secondary upkeep of food. Each farm will cover 5 pops, and these farms also cost 2.5 energy upkeep.

E>= 0.5 F
F = 0.2 P

Putting these constraints into practice, I've determined that for a population P, the amount of people assigned to science
will be Number of science buildings (S) = 0.2 P
With 0.2 P dedicated to farming
0.4 P dedicated to energy (will cover costs and give a surplus 50 energy)
0.2P dedicated to minerals

Optimisation: Reduce T to its minimum amount by finding optimal P

Base cost of things may vary greatly, but the problem is knowing when you've reached the maximum P such that
the percentage increase to base cost can no longer be covered by simply constructing more science buildings.

I'll just approach this as for every time you increase P, you increase C by 0.0167.
Therefore, optimal P is where you can no longer generate an additional 1.67% output.

So, at what output can an additional lab no longer generate more than 1.67% output.
Lab output is 2 avg. per category
So 2 / 0.0167 = 119.76

When you are producing around 120 output from a population P, additional pops will no longer be beneficial.
120/2 = 60 labs
Every fifth person is working it, therefore 60* 5 = 300

Optimal P is therefore 300?

Any advice/tips would be greatly appreciated. It sucks I'm not better at math, because I like min-maxing and optimisation and they're mathematical problems a great strategy game: sid meiyer civilisation
what makes a strategy game great?
-rejection to some extend of a luck factor (throwing dice,getting the right card from a stack,...)
-confronting a player with multiple options (fight war,invest in defence, economics,infrastructure,...)
it looks like you're playing with the idea to design your own game...boardgame?computergame?
to some extend you gonna need some math:
qualifiers vs quantifiers
quantifiers are the numbers, subject to math
qualifiers is what this numbers represent
you cannot add numbers representing different qualifiers:
so 10% basecost+15 (planet size) is nonsense
there is one exception to the rule (lol):
2 figs + 1/2 coconut = 4 politicians  4. Game already exists, I'm not designing one fortunately XD  5. Any ideas on my estimate of N = 300?

I think I seriously screwed up and assumed that when the additional output of increasing N by 1
that the output would be average of 2 across all 3 categories, because that presumes that the additional person
isn't producing energy or minerals.  Bookmarks
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