4th Economics 101

[mmartin798]

I am in the middle of building a spreadsheet to do all these calculations. So far using it, I have been able to get the same numbers as the examples, but I am only done with the agricultural part. The way it works best is to just start with how much food do the farmers need to produce. The number used is 300kg of grain equivalent and 150kg of meat or 15,000L of milk per person. In Sgt.'s example, 90,000kg of wheat and 45,000kg of meat.

Now let's start with the grain. Tech level E produces 1300kg per hectare. Ignore the 4.3 as that is just another way to look at the same number that is more confusing for our purposes. 90,000kg grain requires 69.23H to grow. Now at tech level E, each hectare take 80 hours annually, giving 5,538.4 hours. Divide this by 2000 man hours per year and we require 3 farmers to support the peoples grain requirements.

You do the same thing for the livestock, only you do need to figure that you keep a portion around for breeding increasing the total amount of meat on the hoof. Once you know how much meat you need, you figure out how much corn is needed and repeat the processes for people above. If calculating for corn, remember the crop multiplier, since cultivating corn generates twice the weight per hectare as wheat.

This process duplicated all the examples for ag, so I kind of trust it.

[ArmySGT.]

Originally Posted by mmartin798

I am in the middle of building a spreadsheet to do all these calculations. So far using it, I have been able to get the same numbers as the examples, but I am only done with the agricultural part. The way it works best is to just start with how much food do the farmers need to produce. The number used is 300kg of grain equivalent and 150kg of meat or 15,000L of milk per person. In Sgt.'s example, 90,000kg of wheat and 45,000kg of meat.

Now let's start with the grain. Tech level E produces 1300kg per hectare. Ignore the 4.3 as that is just another way to look at the same number that is more confusing for our purposes. 90,000kg grain requires 69.23H to grow. Now at tech level E, each hectare take 80 hours annually, giving 5,538.4 hours. Divide this by 2000 man hours per year and we require 3 farmers to support the peoples grain requirements.

You do the same thing for the livestock, only you do need to figure that you keep a portion around for breeding increasing the total amount of meat on the hoof. Once you know how much meat you need, you figure out how much corn is needed and repeat the processes for people above. If calculating for corn, remember the crop multiplier, since cultivating corn generates twice the weight per hectare as wheat.

This process duplicated all the examples for ag, so I kind of trust it.

Could you demonstrate in some examples? I am trying the process over completely with a Tech E homestead of 10 persons.... smaller more manageable calculations.

[ArmySGT.]

Farmstead…. Single family plus farm hands…. Tech level E (late steam) (Pg 208)…. 10 persons.
Domain equals 0.5km2 per person 10*0.5 = 5Km2
This family homestead comprises 5Km2 of land supporting 10 persons and livestock.

Labor Pool Labor pool: 47% under 15, 50% 15-64, 3% over 64%, 141, 150, 9 (pg 236)
10*0.47= 4.7, 10*0.50=5, 10*0.03= 0.3
5 adults, and five children.
Labor year = 2000 hours for each adult. 5*2000 = 10,000 labor hours. Tech level E modifier for this is 4. 10000*4=40000
The farmstead with implements and decent knowledge averages 40,000 man hours per year. (Pg 236)
Division of labor (Pg 237)……. Tech Level E…. Agriculture 11-25%, Mining/Manufacturing/Construction 25%, Distribution 20%, other Services 32%
5*0.25= 1.25, 5*0.25=1.25, 5*0.20=1, 5*0.32 = 1.6 = 5.1 Labors….. Problem…. One of the children is now farming.
One farmer, one builder, one delivery, one caretaker accounted for, fifth person? We will see if we need another adult in another job if the system will yield results.
Feeding one person equals 300Kg equivalent of wheat(Pg 237). 10x300kg equals 3000 kgs to feed the homestead. (Pg 237). People need 50 kilograms of meat and 150 liters of milk (dairy products) per year.
10*50=500Kg and 10*150=1500L …….. though this is not explained if this is in addition to the 300 Kg of wheat or substitutes for all or in part……. I can’t determine from the text.
So the Homestead needs 3000kgs of Wheat, 500 Kgs of Meat, and 1500L of milk. Per year to feed the inhabitants adequately, I think.
Confusing bit….. Agricultural Sector table (Pg 237)… Tech level E…. Time per Hectare for a Tech level E farmer is 80(25) 80 hours per hectare and a farmer does 25 hectares in one labor year.

Except later in text farmers are farming 46 hectares for 1840 hours… baffled… is 25 hectares a limit that is then ignored? Is this meant to mean that at Tech Level E a farmer can work 25 hectares in 80 hours?
Next is yield….. Tech level E is 1300kg per hectare and supports 4.3 persons on that hectare (302.3kgs per person). Why does the persons per hectare matter? It isn’t used in any of the following calculations.
3000Kg/1300kg = 2.3 Hectares needed for just wheat. Domain equals 5KM2 or 500 hectares….. so space is covered.
3000kgs/1300kg = 2.3 hectares…. 2.3 = 160hours ( our farmer has it easy!)
On this table is featured the Base Cost and the Input cost….. and no reason why you need to calculate this.
Live stock and Aquaculture… (Pg 238)
You have an average yield per hectare…. Then a modifier….. to me ….. these contradict each other..
I can use the average and get one sum or the modifier and get another sum…..
So is it 180 liter per hectare with an animal using 2-10 hectares for pasture; or is the modifier of whatever the tech level produces….. Tech Level E is 1300Kg per hectare… 1/12 of this is 108kg or liter of milk with an animal using 2-10 hectares….. I am assuming (I know!) the 2 hectares is a milk goat and 10 hectares is a milk cow.
So which one?
Then, there is the corn equivalent ( corn modifier is 2) at Tech level E this would be 2600Kg of corn per hectare. …… so this is the corn needed to feed animals in grain to have them produce…… Does this come out of the farmers output or does there need to be another farmer..... See where the farmers production limit has produced something confusing?
So… Um milk cows. Do I have one or two?
The homestead needs 1500L of milk for a year…. This could mean 1500/180=8.3 hectares…. Or 13.8 hectares… Does the modifier 1/12 affect the manhours? Is it 80 (Tech E) divided by 1/12? Or is it still 80 hours per hectare? I think it would still be the full time requirement… so. 664 hours for 8.3 hectares or 1104 hours for 13.8 hectares. (farmer is a little busier (824 hours or 1264 hours)
Meat. The Homestead requires 500kgs of meat per year to feed the 10 persons.
Beef = 25kg average or 1/88 the equivalent of wheat per hectare.. 1300*(1/88) or 1300*0.011 or 14.3kg per hectare…… again which is it? 500kgs / 25kg per hectare equals 20 hectares of pastured beef or 500/14.3kg per hectare equals 34.9 hectares. 80hrs*20= 1600 and 80*34.9=2792 (our farmers is at either at 2424 labor hours or 4056 hours)

But, wait the animals need grain… milk needs 1kg of corn per liter produced and beef needs 13 kgs of corn per kilo of meat. So the homestead must produce 1500kg of corn for milk cows and 6500 kgs of corn for beef cattle. At tech level E…. 1300kg per hectare with a x2 modifier for corn. Or 2600kgs per hectare. 1500 + 6500 = 8000 kgs of corn required. 8000/2600= 3.07 hectares or 245.6 labor hours.
( the farmer is now at 2669.6 or 4301.6 labor hours.)

So we need 1.33 farmers or 2.15 farmers….. to feed the Home stead.

That would mean that this Homestead would have 2 farmer, one house wife, one ranch hand fixing everything, and one ranch hand in Delivery.

[mmartin798]

Sure. I just realized I quoted some wrong number for the meat anyway, so I will correct that here.

I will assume no exceptional attributes, so this is where we start:

Tech Level (TL): E
Population: 10
Volatile: 50%
Extrovert: 50%
Compassion: 50%
Discipline: 50%
Curious: 50%
Organization: 50%

The available labor pool is 2/3 of 50% of 10, or 3 people. At 2000 man-hours per year, we have 6000 hours available.

To feed 10 people, we need 3000kg of wheat and 500kg of meat.

At TL E, each hectare yields 1300kg of wheat per year and takes 80 hours per hectare to work and harvest. So we need 3000/1300, or 1.36 hectares of land to provide the people's wheat. This require 108.8 hours, so less than one farmer so far.

Let's assume they keep chickens, since they take up less space than beef. We assume they only eat 10% of the flock each year. So to get the 500kg of meat, we need 500/115 hectare of land, or 4.35 hectares for their coop and "grazing" area. They grow corn to feed the chickens. The amount of corn is 2.5*500/2 (the corn multiplier) or 625 kg wheat equivalents. But since only 10% of the flock is eaten, we need to multiply that by 10, giving 6250kg. Again, TL E produces 1300 kg/hectare, we need 6250/1300, or 4.8 hectares to grow the corn on. We add the two areas together, 4.35+4.80, to give 9.15 hectares. At TL E, each hectare takes 80 hours, so we need 9.15*80 or 732 man-hours per year for the chickens and their feed.

So to feed the settlement, we need to supply 841 man-hours, which is still one farmer. We now add the inputs for one person in the ag sector at TL E, which is 2000. The total man-hours to feed this settlement is 2841.

[mmartin798]

The biggest problem is the use of different units all the time. The calculations really need to be calculated in hours, but they keep quoting labor years. It really messes things up.

[ArmySGT.]

Originally Posted by mmartin798

The biggest problem is the use of different units all the time. The calculations really need to be calculated in hours, but they keep quoting labor years. It really messes things up.

The livestock table uses hectares for pasture on some animals and square meters for other animals. another area they should have stayed with one unit.

[ArmySGT.]

Originally Posted by mmartin798

So we need 3000/1300, or 1.36 hectares of land to provide the people's wheat. This require 108.8 hours, so less than one farmer so far.

3000/1300=2.307 ........ Wouldn't that be 2.3 Hectares... 80*2.3= 184 labor hours?

[mmartin798]

Originally Posted by ArmySGT.

3000/1300=2.307 ........ Wouldn't that be 2.3 Hectares... 80*2.3= 184 labor hours?

You are quite correct. So that gives us... 916 man-hours and we will need 2000 man-hours of inputs since that is still one farmer.

[ArmySGT.]

Originally Posted by mmartin798

They grow corn to feed the chickens. The amount of corn is 2.5*500/2 (the corn multiplier) or 625 kg wheat equivalents. But since only 10% of the flock is eaten, we need to multiply that by 10, giving 6250kg. Again, TL E produces 1300 kg/hectare, we need 6250/1300, or 4.8 hectares to grow the corn on. We add the two areas together, 4.35+4.80, to give 9.15 hectares. At TL E, each hectare takes 80 hours, so we need 9.15*80 or 732 man-hours per year for the chickens and their feed.

Shouldn't this be 2.5*500*2=2500 Kg of corn? 1300kg per hectare with x2 for corn = 2600kg produced for one hectare of corn. 2500 kg of corn for 10% of the flock?

[mmartin798]

Originally Posted by ArmySGT.

Shouldn't this be 2.5*500*2=2500 Kg of corn? 1300kg per hectare with x2 for corn = 2600kg produced for one hectare of corn. 2500 kg of corn for 10% of the flock?

No, this is another area where the text is confusing. Let me rephrase things to make it clearer. Just like explosives have an RE value where C4 is 1, so too does crop yield have a Yield Effectiveness (YE) value where wheat is 1. Replace crop multiplier with YE and the calculations are the same. 1 of wheat = 0.5 of corn = 100 of cocaine.

So if I need x hectares for 50kg of wheat, I need 0.5x hectares for 50kg of corn and 100x hectares for 50kg of cocaine. I hope that is clearer.

[ArmySGT.]

Originally Posted by mmartin798

No, this is another area where the text is confusing. Let me rephrase things to make it clearer. Just like explosives have an RE value where C4 is 1, so too does crop yield have a Yield Effectiveness (YE) value where wheat is 1. Replace crop multiplier with YE and the calculations are the same. 1 of wheat = 0.5 of corn = 100 of cocaine.

Hmmm ok..... What do you make of the multiplier? 1/12 for milk or 1/20 for poultry. Where does this fit in?

[mmartin798]

Originally Posted by ArmySGT.

Hmmm ok..... What do you make of the multiplier? 1/12 for milk or 1/20 for poultry. Where does this fit in?

Since they are roughly linear to the hectare yield, I think it was an attempt to simplify calculations that just adds to the noise. It is never used in any of the examples.

I think I figured out what the multiplier is in the livestock. It is kind of stupid. It lets you express the land the livestock uses to graze in terms of wheat. It is just way too confusing to use when we can calculate it from the other numbers easier.

[ArmySGT.]

Homestead #2
Tech D (Early Electric)
10 persons. 15-64=66% or 6.6 rounding up to 7 adults . Labor pool = 2/3 of 7= 4.62 rounding up to 5.

Tech level D Agriculture 11-25%, Mining / manufacturing/construction 25%, Distribution=20%, Other services 32%
5*0.25= 1.25, 5*0.20= 1, 5*0.32= 1.6
Five adults, One Farmer, One maker, One Trader, 2 Other.

Consume = 300kg wheat equivalents x 10 persons = 3000Kg

A Tech level D farmer produces 1300kg per hectare at 40 hours of work per hectare and can harvest 50 hectares per labor year (2000 labor hours).
3000kg/1300kg= 2.30 hectares of wheat produced for a labor hours of 92 hours.

The homestead requires 50Kg of meat and 150L of milk per person to adequately feed them.

50Kg*10= 500kg of meat and 1500L of milk.

The Homestead raises hogs for meats like hams and bacon. Hogs require 6kg of corn for each kilogram of pork at harvest. 6*500= 3000 kilos of corn….. only 10% of the herd can be harvested to keep the stock healthy though. 6*500 =3000/2*1300 = 1.2 hectares or 40*1.2= 48 labor hours * 10 (10% is consumed) 480 labor hours keeping hogs for meat.
2.3 hectares wheat and 1.2 hectares corn. 2.3+1.2= 3.5 hectares * 40 hours per hectare…. 140 labor hours plus 480 hours herding hogs for 620 labor hours per year farming the absolute minimum.

[ArmySGT.]

Extraction.... the Tech D homestead is producing natural gas.....

A natural gas well is tapped to produce fuel for the tractor, heating the home and out buildings and cooking meals. The gas well equipment, pumps, vacuums, refining, and pressure bottling is occupies one laborer (1 (laborer) x 5 (output) x 500 (material conversion factor)= 2500 and occupies 125,000M2 ((2500x100)/0.5=125000) This produces 2500kg of gas or 2500*35.3= 8825 cubic meters of natural gas.

8825 cubic meters of gas for one labor year.

To build the well took 1 (laborer)*6 (output) = 6 labor years (6*2000=12,000hrs) to get started and 1(laborer)*3(input)= 3 labor years (3*2000=6000 hours) annually in upkeep.

[mmartin798]

Originally Posted by ArmySGT.

Homestead #2
Tech D (Early Electric)
10 persons. 15-64=66% or 6.6 rounding up to 7 adults . Labor pool = 2/3 of 7= 4.62 rounding up to 5.

Tech level D Agriculture 11-25%, Mining / manufacturing/construction 25%, Distribution=20%, Other services 32%
5*0.25= 1.25, 5*0.20= 1, 5*0.32= 1.6
Five adults, One Farmer, One maker, One Trader, 2 Other.

Consume = 300kg wheat equivalents x 10 persons = 3000Kg

A Tech level D farmer produces 1300kg per hectare at 40 hours of work per hectare and can harvest 50 hectares per labor year (2000 labor hours).
3000kg/1300kg= 2.30 hectares of wheat produced for a labor hours of 92 hours.

The homestead requires 50Kg of meat and 150L of milk per person to adequately feed them.

50Kg*10= 500kg of meat and 1500L of milk.

The Homestead raises hogs for meats like hams and bacon. Hogs require 6kg of corn for each kilogram of pork at harvest. 6*500= 3000 kilos of corn….. only 10% of the herd can be harvested to keep the stock healthy though. 6*500 =3000/2*1300 = 1.2 hectares or 40*1.2= 48 labor hours * 10 (10% is consumed) 480 labor hours keeping hogs for meat.
2.3 hectares wheat and 1.2 hectares corn. 2.3+1.2= 3.5 hectares * 40 hours per hectare…. 140 labor hours plus 480 hours herding hogs for 620 labor hours per year farming the absolute minimum.

Your pork section is off. You are harvesting 500Kg of pork and leave an additional 4500Kg of pork alive. You need to feed them all and have land for them. I come up with 6*500*10=30000Kg of corn. That comes to 11.5 hectares, 462 man-hours. The land to graze the pork is 10*500/50 = 100 hectare, using 4000 man-hours. So the total land usage for just the pork is 111.5 hectare and requires 4462 man-hours a year. You need three people.