Infinite resource (e.g., oil) calculations

[jcranmer]

I was mathing out the requirements for the 8 hour speed run, and I needed to know how much oil I could expect from oil fields. Since I couldn't find detailed technical information on how infinite resources work, I decided to experiment myself and then run the numbers. Here's the results for those who have the same questions.

Infinite resources internally have an amount value (like finite resources); this value is compared to a Lua-specified "normal" value to compute the yield: an amount equal to the normal value is 100% yield. There is also a minimum amount beyond which the number cannot fall. Every mining cycle (calculated as for regular resource mining drills), the amount value falls by 1 unit, and the output is multiplied by the yield fraction to output the number of resources.

In the following calculations, let s be the starting yield, measured as a fraction (so 100% yield means s = 1); n the normal Lua yield amount (for oil, this is 15000 as of 0.13), m the minimum yield fraction (for oil m = 1500 / 15000 = 0.1), u the resource output per cycle at 100% yield (1 for oil), and t be a parameter indicating the number of mining cycles elapsed.

It then holds that the internal amount value at time t is a(t) = sn - t. The time it takes this field to deplete is then ṫ = sn - mn. (I'll give a table later for several values). The resource yield at any given time t is y(t) = u·a(t) / n = u·(s - t / n). We can then solve for the amount of resources yielded per depletion:

Code: Select all

Sum[y(t) for t = 0 .. ṫ] = Sum[u·(s - t/n) for t = 0 .. sn - mn]
= u·Sum[s - t/n for t = 0 .. sn - mn]
= u·(s·(sn - mn + 1) - Sum[t/n for t = 0 .. sn - mn])
= u·(s·(sn - mn + 1) - 1/n · Sum[t for t = 0 .. sn - mn])
= u·(s·(sn - mn + 1) - 1/n · (sn - mn)(sn - mn + 1) / 2
= u·(sn - mn + 1)(s - (s - m) / 2)
= ½u·(sn - mn + 1)(2s - s + m)
= ½u·(sn - mn + 1)(s + m)

Depletion is quadratic in the starting yield, so doubling the yield quadruples the output. For oil, the resulting formula is (s · 7500 - 749.5) · (s + 0.1). Example values for oil are:

Code: Select all

Yield      Total  Time (hours)
  30%      600.2  0.83
  35%      844.0  1.04
  40%     1125.2  1.25
  45%     1444.0  1.46
  50%     1800.3  1.67
  55%     2194.1  1.88
  60%     2625.3  2.08
  65%     3094.1  2.29
  70%     3600.4  2.50
  75%     4144.2  2.71
  80%     4725.4  2.92
  90%     6000.5  3.33
 100%     7425.6  3.75
 125%    11644.4  4.79
 150%    16800.8  5.83
 175%    22894.7  6.88
 200%    29926.1  7.92
 250%    46801.3 10.00
 350%    91801.8 14.17
 400%   119927.0 16.25
 500%   187427.5 20.42
 600%   269928.0 24.58
 700%   367428.5 28.75
 800%   479929.0 32.92
 900%   607429.5 37.08
1000%   749930.0 41.25

For estimation purposes, a rocket requires 41,505 petroleum gas to launch (required techs (and no more!), rocket silo, 72 rocket parts, 1 satellite, 4 productivity 3 modules for the rocket silo (that's why you only need 72 parts, not 100)). That comes out to about 103,763 oil without cracking or advanced oil processing (i.e., gas is 40% of oil as opposed to 55%/75%/90%, depending on your mixture of processes), which is about 12 oil fields at 60% starting yield for 6 hours. Given that the average oil yield is about 45% in the starting field today at default settings, the 8-hour achievement run is virtually impossible on default settings for oil (12 oil fields for 8 hours at 45% yields 41,029 gas at a 90% conversion ratio--the max you get without productivity).

[Frightning]

I was mathing out the requirements for the 8 hour speed run, and I needed to know how much oil I could expect from oil fields. Since I couldn't find detailed technical information on how infinite resources work, I decided to experiment myself and then run the numbers. Here's the results for those who have the same questions.

Infinite resources internally have an amount value (like finite resources); this value is compared to a Lua-specified "normal" value to compute the yield: an amount equal to the normal value is 100% yield. There is also a minimum amount beyond which the number cannot fall. Every mining cycle (calculated as for regular resource mining drills), the amount value falls by 1 unit, and the output is multiplied by the yield fraction to output the number of resources.

In the following calculations, let s be the starting yield, measured as a fraction (so 100% yield means s = 1); n the normal Lua yield amount (for oil, this is 15000 as of 0.13), m the minimum yield fraction (for oil m = 1500 / 15000 = 0.1), u the resource output per cycle at 100% yield (1 for oil), and t be a parameter indicating the number of mining cycles elapsed.

It then holds that the internal amount value at time t is a(t) = sn - t. The time it takes this field to deplete is then ṫ = sn - mn. (I'll give a table later for several values). The resource yield at any given time t is y(t) = u·a(t) / n = u·(s - t / n). We can then solve for the amount of resources yielded per depletion:

Code: Select all

Sum[y(t) for t = 0 .. ṫ] = Sum[u·(s - t/n) for t = 0 .. sn - mn]
= u·Sum[s - t/n for t = 0 .. sn - mn]
= u·(s·(sn - mn + 1) - Sum[t/n for t = 0 .. sn - mn])
= u·(s·(sn - mn + 1) - 1/n · Sum[t for t = 0 .. sn - mn])
= u·(s·(sn - mn + 1) - 1/n · (sn - mn)(sn - mn + 1) / 2
= u·(sn - mn + 1)(s - (s - m) / 2)
= ½u·(sn - mn + 1)(2s - s + m)
= ½u·(sn - mn + 1)(s + m)

Depletion is quadratic in the starting yield, so doubling the yield quadruples the output. For oil, the resulting formula is (s · 7500 - 749.5) · (s + 0.1). Example values for oil are:

Code: Select all

Yield      Total  Time (hours)
  30%      600.2  0.83
  35%      844.0  1.04
  40%     1125.2  1.25
  45%     1444.0  1.46
  50%     1800.3  1.67
  55%     2194.1  1.88
  60%     2625.3  2.08
  65%     3094.1  2.29
  70%     3600.4  2.50
  75%     4144.2  2.71
  80%     4725.4  2.92
  90%     6000.5  3.33
 100%     7425.6  3.75
 125%    11644.4  4.79
 150%    16800.8  5.83
 175%    22894.7  6.88
 200%    29926.1  7.92
 250%    46801.3 10.00
 350%    91801.8 14.17
 400%   119927.0 16.25
 500%   187427.5 20.42
 600%   269928.0 24.58
 700%   367428.5 28.75
 800%   479929.0 32.92
 900%   607429.5 37.08
1000%   749930.0 41.25

For estimation purposes, a rocket requires 41,505 petroleum gas to launch (required techs (and no more!), rocket silo, 72 rocket parts, 1 satellite, 4 productivity 3 modules for the rocket silo (that's why you only need 72 parts, not 100)). That comes out to about 103,763 oil without cracking or advanced oil processing (i.e., gas is 40% of oil as opposed to 55%/75%/90%, depending on your mixture of processes), which is about 12 oil fields at 60% starting yield for 6 hours. Given that the average oil yield is about 45% in the starting field today at default settings, the 8-hour achievement run is virtually impossible on default settings for oil (12 oil fields for 8 hours at 45% yields 41,029 gas at a 90% conversion ratio--the max you get without productivity).

You might find the thread I did on windfall oil useful, the additional income beyond windfall is a constant rate (before speed modifiers from modules are applied), so the only variable is how much windfall you get before it reaches depletion (I recently updated the numbers for 0.13 too; check last post I made in thread, has updated formulas too).
Thread: viewtopic.php?f=18&t=24748

[jcranmer]

You might find the thread I did on windfall oil useful, the additional income beyond windfall is a constant rate (before speed modifiers from modules are applied), so the only variable is how much windfall you get before it reaches depletion (I recently updated the numbers for 0.13 too; check last post I made in thread, has updated formulas too).
Thread: viewtopic.php?f=18&t=24748

Yeah, I did see that thread, and I indeed checked that my formulas get the same results you do after conversion (I'm counting total oil, not windfall oil, and the result is off by 0.1, so I'm counting one extra depleted cycle). More importantly, I wanted to get the time to depletion metrics factored in, as well as extend how resources work for other infinite ores. Give a fully general solution and then specialize it for oil, instead of trying to generalize it from oil's numbers. I also needed the time to verify that my understanding of infinite mining mechanics worked properly---I don't trust the wiki a lot here because it can be notoriously out of date (e.g., belt mechanics info pages still have the corner slowdown information listed in places).