Factorio [early draft] Full guide to maximum (Pollution)-Efficiency with modules

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[German] Hallo. Dieser Guide ist in Englisch. Ich habe keine lust 90% davon ein 2.mal zu schreiben, also schreibe Ich nur die zusammenfassung im 5. Karpitel in Deutsch. Falls ihr verstehen wollt wie ich die zahlen und formeln errechnet habe, müsst ihr die anderen Karpitel lesen. Für die anwendung diesen wissens ist dies jedoch nicht nötig. Ich hoffe das Ihnen dieser Guide trotzdem helfen kann.

[English] A summary of everything is in chapter 5. It gives you the nessesary info for figuring out what to use, but it does not explain why. The other chapters are for the why.

If you just want a simple configuration without beacons or math, look at chapter 1.5

Chapter 3 starts explaining the math behind productivity modules (used together with beacons)

the next one goes on about how to do the math yourself

Chapter 3.3 Gives you all the info you need, as well as an example to how to calculate the most efficient setup (it depends on your production line. If this is too hard for you, just check chapter 1.5)

Chapter 4 is optional. If you have some other ideas for efficiency, look there first before you comment. I might have done the math on it already

|Chapter 1| Simple Start

Hello Steam.

This guide is for everyone who wants to use Modules for the porpoise of producing Items with as little pollution as possible. Maybe you want to save the trees, you want to play on a death world without a proper guide, or you are just a bit insane like me. This info might not be important to some players who don't care about the environment or try to prove the lack of spoon, but chapter 1 will still be useful for minimising electricity usage per item.

What is efficient?Everything in this guide is based on the following pieces of information. If you want to understand the Idea behind my approach, you should read the following axioms in descending order of importance.

1. A reduction of electricity consumption is a reduction in pollution. Unless you use productivity modules (or beacons), electricity efficiency is the same as pollution efficiency. The minimum consumption is at 20%. You generally want to operate at that level, because it makes the machine 5x as efficient

2. The second thing you should know, is that the stats displayed about electricity and pollution are over time, NOT per item. This means that if you can make 2 times the items without changing the pollution and electricity stats, you will have doubled the actual efficiency of your machine.

3. Productivity modules DO NOT make the machine they are being used for more efficient! Instead they make it so that all the machines that are needed to feed the last one do not need to work as much (Which might reduce electricity consumption of the entire system, but only if the difference of the power consumption between the last and the rest of the machines is great enough) In short: get electricity usage to 20% and then try to get the speed up without using more electricity

Productivity module can help you with more expensive, late game stuff, but it might not. Look at chapter 3 for more info

Simple MathEfficiency = Speed% / Electric usage%

Base Efficiency = 100%/100% = 1

Minimum power consumption Efficiency = 100%/20% = 5

Simple Modules, Simple Efficiency, Simple Math

First modules FirstIf you want to minimise pollution, you will want to research Efficiency MK1 first.

2 of them will reduce electric consumption by 60%, making things 2.5 times more efficient.

Adding the 3rd module will push it to the max with a total reduction of 80%, resulting in an efficiency factor of 5!

However, you want to start with giving every electric miner 2 modules, until you have produced enough to give every miner a 3rd one

The reason you want to start with the electric miners is because they have a base pollution of 10!

Making them more efficient will reduce pollution the most.

And the reason you want to give them 2 modules while you don't have enough models for everyone yet, is because the 3rd module will not be completely effective, because it will only reduce the consumption by 20% instead of 30%.

When you are done with your electric miners (which should have replaced all your burner miners at this point) you can start applying the modules to your assemblers mk2 next.

Obviously, you want to start using solar next, but that is something this guide will not cover. (The boilers make 3 times the pollution of unimproved electric miners, And they don't accept modules, so you CAN'T make it more efficient)

Simple late game efficiencyThe best you can do with most machines is just the 80% power reduction. Other than that, you'll want to upgrade your assemblers to Mk3 and use eff1 modules whenever they will do the job, eff2 when you only have 2 module slots

The 4th slot of the assembler will allow a 50% more efficient Module configuration ON TOP of the 5 times efficiency of the efficiency module. However, you will need level 3 speed and efficiency modules

Use 3 Eff3 with 1 Speed3 module. This will increase the speed by 50% while still reducing the power drain to 20%.

This results in 150% / 20% = 7.5 times the efficiency compared to a module-less assembler

In summary

Module configuration Speed Power Efficiency Nothing 100% 100% 1 Enough Eff mods 100% 20% 5 3 Eff3 + 1 Speed3 150% 20% 7.5

Ensure that you have your power consumption to 20%, if you want a bit more efficiency, do the 3 Efficiency3 with 1 Speed3 Configuration. This is Very late game and not worth it for almost everyone, but it is slightly more efficient if you only care about pollution

|Chapter 2| 12 Beacons Of Pollution-reduction

Beacons require a relative high amount of electricity. Because of this, every notion of efficiency goes out the window when they are introduced. However, because beacons themselves do not produce pollution, we can still use them for out porpoises. Just remember that this chapter is NOT about reducing your electric bill. You might want to use nuclear power plants from here on out.

Assemblers and other machines of the same size can be affected by 12 beacons. Because every beacon has 2 slots, but only 50% effective, it's like adding 12 slots per machine. Beacons have only an effect on machines that accept modules. They are an expansion, not an enabler for your modules.

It's also important to note that beacons do not accept productivity modules. We do not care about that for now, but it does make the calculation simpler, because we don't have to worry about more that 4 productivity modules per machine.

Now, what is the best combination of Efficiency and speed modules in our 12 beacons?

Assuming that our machine is already at 20% power consumption (and 150% speed) we only need to maximise speed without increasing electricity usage. The ratio for that is 5 speed3 to 7 eff3.

5x50% Speed = +250% Speed

5x70% power = +350% power

7x(-50%) power = -350% power

This gives us +250% speed at no downsides (other than the power cost of the beacons) Our beacons will need 10 speed3 and 14 Eff3 modules in order to give us that speed. It does not matter which beacon has what module, not whether or not we mix the modules within a beacon. Just have 10|14 modules in the beacons, as well as the 1|3 modules in the assembler.

This will give us a total of (150% + 250%) / 20 % = 20 Times base efficiency

|Chapter 3| Productivity And A Whole Lot Of Math

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Now that we have extensively talked about speed and efficiency modules, as well as beacons, you might ask about the 3rd kind of module that also appears to be the most popular of them all. The Productivity module.

The math behind this module is a bit more complicated. It's still easy (Compared to what I will do further down the guide), but I will need to explain the meaning of the new stats. In this guide I will only talk about Productivity3, because the 3rd level of modules is the most effective and I really need to simplify as much as possible. This chapter is meant for the late game, so I assume that you can use lvl 3 modules for everything.

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-15% Speed | This is something you should be mindful of! Because of how the game calculates the modifiers to the machines base stats, any configuration that ends up with below 100% of something will have a big impact. Bigger if it goes lower

+80% Energy consumption | Self explanatory. This obviously decreases your efficiency if not taken care of. do note that a jump from 20% to 100% decreases the efficiency by a factor of 5, while a jump from 100% to 180% decreases it by only 1.8

+10% Productivity | This stat is the main feature of this module. It adds 10% to a second bar in the machine per produced item. If that bar reaches 100%, it makes another item without consuming any products! It is vital if you want to maximise output per item. For that you just need to use as many prod3 modules as possible. (Not all products accept prod. modules!). For our porpoises, it's only useful, because less production could save pollution and electricity, but it might cost too much in the device being used in to make it not worth it. Because of that, I'll do a lot of math to figure out if the usage is worth it or not

+10% Pollution | This is the biggest downside for us. The multiplier is applied after the power reduction. This changes our formula like this:

Efficiency = Speed% / (Electric usage% x Pollution multiplier)

Base = 100%/(100% x 1) = 1

minimum electricity and the added pollution of 1 prod3 module = 100%/(20% x 1.1) = 100%/22% =4.545454...This alone shows that it's not (pollution) efficient to use Productivity modules on miners/Pump-jacks since they don't depend on any other thing that pollutes. But just to show you maximum efficiency (with 12 beacons)

#Prod3 Module configuration Pollution Speed Efficiency Productivity bonus 0 10+6 20% 400% 20 +0% 1 10+4.5+0.5 20%(22%) 320% 16 +10% 2 10.5+3.5 20%(24%) 245% 12.25 +20% 3 10.5+2.5 20%(26%) 180% 9 +30% 4 10.5+1.5 20%(28%) 115% 5.75 +40% [Efficiency = (speed*Productivity) / (power*pollution)]

Because Pollution is always equal to the productivity gain, they cancel each other out and the formula gets simplified back to

[Efficiency = speed / power]

Less-than-simple Math

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#Modules 0 1 2 3 4 Efficiency Factor (with productivity) 20 16 12.25 9 5.75

The above table sums up the efficiency of the machine you use the modules on. The problem is that it does not take the machines before the last one into account. Because it also reduces how much power is needed in them, the total efficiency is unknown until you apply it to the entire system. How many Prod3 modules you should use depends on the ratio of the power usage of the previous machines and the last machine . But first, I have to remind you of some counter intuitive math.

100% * 1.1 = 110% BUT 110% * 0.9 = 99%

100% * 0.9 = 90% BUT 90% * 1.1 = 99%

CORRECT : 100% * 1.1 = 110% THEN 110% / 1.1 = 100%

It does make a difference whether you calculate with / [1+Prod.Multiplier] or * [1-Prod.Multiplier]

The correct way is to use the former. This results in the following formulas

E=(10*Eb)/(10+n)+Ea/m

where Ea= base pollution of the machine/s you are adding to the production line. Eb= improved pollution of Machines already in the production line. n=amount of Production modules per machine you are adding.(0-4) m= efficiency factor according to the table above.

In order to find out which configuration is the most efficient depending on the ratio of Ea to Eb, we simplify the formula by substituting it with X=Eb/Ea, which allows us to look at the case Ea=1; Eb =x, which turns our formula into:

fn(x)=(10x)/(10+n)+1/m Which gives us the following formulas depending on n:

f0(x)=x+0.05

f1(x)=10x/11+1/16

f2(x)=10x/12+1/12.25

f3(x)=10x/13+1/9

f4(x)= 10x/14+1/Trivial Math (no seriously)by analysing the formulas (y=mx+b), we know that all formulas are rising (positive x), the higher the n, the lower the rate (m), and the higher the n, the higher the starting point (b).

This tells us that the most efficient function is the one with the lowest n, until it crosses over with the next n, which is the most efficient until IT crosses over with the next one. So we only need to get the crossing over values, which we can do the easiest by simply calculating Fn(x)=Fn+1(x)

This gives us the following values

#Modules Efficient above x= Precise Value 0 0 0 1 0.138 11/80 2 0.253 99/392 3 0.46 338/735 4 1.143 1183/1035 Here's what that looks like if you draw it

Reminder: X= (Pollution of the current production line)/(Base pollution of the Machines you want to add to it)

But wait! There's moreup until now we have assumed that Electricity consumption = Pollution production. This is true as long as you use only 1 type of machine. Production lines however, use different machines. You have miners to furnaces to assemblers. Each type has different pollution values. Thanks to us calculating according to pollution and not power usage, we can still use the above values for our further math. However, we have to use pollution values to make the following make sense.

This is how: Base Pollution = Base Pollution of device * amount of devices used

For 4 Electric miners: 40=10*4

We use that base value and modify it according to our module configuration. Remember that if you go down this deep into efficiency, you use 12 beacons (the max for most devices) and lvl 3 modules. We spare no expense here.

One final complicationOur calculations are for machines that have 4 slots for modules and accept a max of 12 beacons. This only applies to Assemblers Mk3. We will need to use different tables and configurations for the other devices like assemblers 2 (which you should upgrade away from) Electric miners and furnaces. For the sake of completeness, I will quickly add the other devices depending on their module (and max beacon) count.

Math Cheat Sheet + Example

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Machines with 4|3|2 Module slots and 12 beacons (size 2x2 to 4x4)

#Mods Configuration Alpha Configuration Alpha Configuration Alpha 0 10+6 20 9.5+5.5 18.75 9+5 17.5 1 10+4.5+0.5 16 9.5+4.5 15.5 9+4 14.25 2 10.5+3.5 12.25 9.5+3+0.5 11.5 9+3 11 3 10.5+2.5 9 10+2 7.75 4 10.5+1.5 5.75

The Rocket silo and Oil Refinery (with 20|16 beacons)

#Mods Configuration Alpha Configuration Alpha 0 15+9 27.5 12+7 22.5 1 15+8 24.25 12+6 19.25 2 15+7 21 12+5 16 3 15+6 17.75 12+4 12.75 4 15+5 14.5

The formula of fn(x)=(10x)/(10+n)+1/m still applies btw. After doing fn(x)=fn+1(x) on all of those new values, we get the following tables:

Machines with 4|3|2 Module slots and 12 beacons | Rocket silo | Oil Refinery

# 4 Mods 3 Mods 2 Mods Rocket Silo Oil Refinery 0 0 0 0 0 0 0 0 0 0 0 1 0.138 11/80 0.123 286/2325 0.1434 286/1995 0.0536 26/485 0.0825 26/315 2 0.253 99/392 0.296 1056/3565 0.2738 26/95 0.0842 286/3395 0.139 39/280 3 0.46 338/735 0.656 468/713 0.136 338/2485 0.2485 169/680 4 1.143 1183/1035 0.23 2366/10295 Remember: X=pollution current / base-pollution new

And: Pollution = [Pollution current*10/(10+n)]+Base-pollution new / Alpha

ExampleLets say we want to make 150 Gears per minute. We'd need 9 electric miners feeding 8 electric furnaces, feeding 1 Assembler Mk3

We start off with 0 pollution and the miners

Pollution Current = 0 ; Base-pollution new = 9 * 10 = 90 ; x= 0/90 = 0

The miners have 3 slots and 12 beacons, so we look into the 2nd set of values. Because 0<=x<0.123 , we go with 0 Productivity modules, which gives us an alpha of 18.75 . We do 90/18.75 and get 4.8

we have 4.8 Pollution and do the furnaces now

Pollution current = 4.8 ; Base-pollution new = 8 * 1 = 8 ; x = 4.8/8 = 0.6

The furnace has 2 slots, so we look at the 3rd set of values. Because x > 0.2738 , we use 2 Productivity modules. This gives us an alpha of 11 . 4.8*10/12 + 8/11 = 4.727272...

Yes, we just decreased our total pollution. we now have 4.7272727... pollution and add our 1 assembler

Pollution current = 4.727272... Base pollution new = 2*1 = 2 ; x= 4.727272... / 2 = 2.36363636...

The assembler Mk3 has 4 slots, so we use the 1st set of values and get WAY over the 4 module threshold. So n=4 and Alpha = 5.75

4.727272...*10/14 + 2/5.75 ~ 3.72445

This means that for our 150 Gears/min, we need to produce 3.7245 pollution per min. It also tells us which modules to use in the machines and the beacons. It does NOT tell us how many machines we need of each type. Despite having the base amount of machines, we don't know how many of them will be needed afterwards (we will need less than originally) I could do the math on it, but I can't really be bothered to (yet).

But it is interesting that you go from using 9 miners, 8 Furnaces and 1 assembler

to 2(1.4)Miners, 3(2.2)Furnaces and 1(0.7)Assemblers. While the beacons take away space, having to use less machines saves a bit of it. But in this case, you still need more space than you would without beacons

|Chapter 4| No Beacons And Other Nonsense Ideas

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Beacon-less efficiencyThis chapter is for discussing other option than chapter 3 that are more efficient than chapter 1. This is quite a tall order, because we'd have to fight an alpha of 7.5 with only productivity modules. The logic from chapter 1 will need to be thrown out the window, because we can't always get power usage to 20%. Sometimes, especially with more Prod. Modules, it will be better to use speed modules over efficiency. I will try out all configurations, but only show you the most efficient ones

4 slots | 3 slots | 2 slots. Size does not matter this time

#Mods Config Alpha Config Alpha Config Alpha 0 3+1 7.5 3 5 2 5 1 2+1 4.25 2 17/16 1 17/26 2 1+1 3/7 1 4/11 7/26 3 1 21/82 11/68 4 2/21

# 4 Slots 3 Slots 2 Slots 0 0 0 0 0 0 0 1 0.388 33/85 8.153 693/85 14.62 1243/85 2 Nope Never 23.87 4059/170 28.84 3432/119 3 26.236 18733/714 53.54 468/713 4 120 3601/30

Just do the math like before if you want to figure out what configuration to use. Ignore the part where it says "Nope Never", it turns out that using 2 Prod modules in a machine that uses 4, is never more efficient than using 1 or 3.

Using infinite speedUsing a lot of speed modules decreases the amount of machines necessary, but it does so at the cost of efficiency. The math for it is quite simple:

Base speed + N*50 = new speed | Base power usage + N*70 = New power usage

If we turn N towards infinity (not equal, because that violates mathematical rules. Though it's not like I'm a mathematician ), we can divide by N.

Base speed/N + 50 = new speed/N | Base power usage/N + 70 = New power usage/N

Base speed and Power usage are divided by basically infinity. This means we can play physicist and say they are 0 (they are not, they are just very close to it) doing so simplifies the calculation to:

Speed/Power usage = 50/70 * N/N = 50/70 * 1 ~= 0.7143

This means that the more Speed modules you use, the more it moves to an Alpha value of 5/7. That is the most efficient you can do. My configurations are more efficient, because they have higher Alpha values, or not enough slots to get towards 5/7 with speed alone.

Using this many speed modules would only make sense if your goal is to save space

Minimising Resource InputIf you want to get the most items out of your ores, just maximise Productivity modules and the research of mining productivity. It's that easy

If you get +110% Productivity or more, you might be able to partially produce 1 Product without consuming any items. Demolishing and rebuilding the machine would reset that while still giving you your 1 free production. this would give you an infinite amount of items (given enough electricity)(Bear in mind that in vanilla, only miners can reach that much productivity)

Using productivity modules without even doing what little math I require you to doThe ideal amount of Productivity modules is always equal to the item you want to produce. I might make a list ... But at the moment I just can't be bothered to

|Chapter 5| All The Results In One

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EnglishHow to calculate which option is the best:

X=current pollution / added base-pollution * new machines

look at the highest N that has a threshold less than X

use the configuration of that N. Remember N and Alpha

New pollution = 10*Current pollution/(10+N)+base pollution * amount of new machines / AlphaRepeat until you have added every machine of the production line. Start from the ores/Oil and a pollution value of 0

If you hit a fork (2 or more inputs are necessary) Simply calculate the polution of each side, add them, and then continue with this new value. How the pollution was generated does not matter to the machines further up the production chain

No beacon N Config Alpha Config Alpha Config Alpha 0 3+1 7.5 3 5 2 5 1 2+1 4.25 2 17/16 1 17/26 2 1+1 3/7 1 4/11 7/26 3 1 21/82 11/68 4 2/21

N 4 Slots 3 Slots 2 Slots 0 0 0 0 0 0 0 1 0.388 33/85 8.153 693/85 14.62 1243/85 2 Nope Never 23.87 4059/170 28.84 3432/119 3 26.236 18733/714 53.54 468/713 4 120 3601/30

Max Beacon4|3|2 Slots

N Configuration Alpha Configuration Alpha Configuration Alpha 0 10+6 20 9.5+5.5 18.75 9+5 17.5 1 10+4.5+0.5 16 9.5+4.5 15.5 9+4 14.25 2 10.5+3.5 12.25 9.5+3+0.5 11.5 9+3 11 3 10.5+2.5 9 10+2 7.75 4 10.5+1.5 5.75

The Rocket silo and Oil Refinery

N Configuration Alpha Configuration Alpha 0 15+9 27.5 12+7 22.5 1 15+8 24.25 12+6 19.25 2 15+7 21 12+5 16 3 15+6 17.75 12+4 12.75 4 15+5 14.5

Machines with 4|3|2 Module slots and 12 beacons | Rocket silo | Oil Refinery

X=Bis jetzt errechnete verschmutzung / Grundwert der verschmutzung einer Maschiene * anzahl der neuen maschienen

Guckt in der richtigen Tabelle nach dehn größten N, mit einen Schwellwert von weniger als X

Benutze die configuration für dieses N. Merke dir das zugehörige N und Alpha

Neu errechnete verschmutzung = 10*Alter wert/(10+N)+Grundwert der verschmutzung * anzahl neuer Maschienen / AlphaWiederholt dies, biss Sie alle Maschienen der Produktionskette habt. Fängt mit den Erzen und einem Schmutz-wert von 0 an.

Falls sie eine vergabelung haben (mindestens 2 Eingaben sind nöting für diesen schritt), müssen Sie nur die verschmutzungen von dehn Zweigen errechnen. Das produkt dieser kann wie normal für die nächsten rechnungen verwended werden, da es dehn maschienen weiter oberhalb der Produktionskette egal ist, wie die verschmutzung entschtanden ist.

(Ich habe grad keine lust die rechtschreib Fehler zu korriegieren)

Source: https://steamcommunity.com/sharedfiles/filedetails/?id=2485119176

[early draft] Full guide to maximum (Pollution)-Efficiency with modules