## Manual Calculations for Items/Second and Assembler Ratios

Calculate optimal ratios for feeding recipes, search through the research-tree, specialized tools to view game-information.
AidanNJ
Manual Inserter
Posts: 1
Joined: Sat Nov 26, 2016 4:15 am

### Manual Calculations for Items/Second and Assembler Ratios

First off I'd like to say hello! I'm new to forms and their format, but hopefully over time I will find I quite enjoy being on the forums!

...But back to the topic.

I see a lot of calculators, but not any raw calculations. For those who would rather solve their problems in a different way, I have some of that covered. Having raw calculations opens a lot of room for flexibility, as is shown in one of the examples.

Now I myself have not gotten very far in factorio yet and haven't messed around with enough to say with complete certainty that all of these are correct, but I would be open to any suggestions on things I missed or messed up on or to clarify and the like. This is from a text folder that I have checked for grammar mistakes and copy-pasted.

*Note that anything that involves assemblers will assume that all assemblers are the same level.

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Getting the actual production speed of an assembler
C - Crafting speed of assembler N - Normal production speed P - Actual production speed

N/C=P

For example...

I have tier two assemblers which have a crafting speed of 0.75; this is C. I want to craft red science, which is crafted in
5 seconds; this is N. 5/0.75=6.666...(2/3) repeating; this is P. This means that the actual production speed for red science in
a tier two assembler is 6+2/3.
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Getting items/second from production speed

It is simply the reciprocal of the production speed divided by the items created.
P - Production speed M - Equivalent i/s of the given production speed A - Items created per instant

1/(P/A)=M

For example...

I want to craft copper cables in my inventory. It takes 0.5 seconds to craft; this is P. Two items are produced, so divide
by two; this is A. The reciprocal of 0.25 is 4; this is M. This means that the recipe outputs 4 i/s.
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Assemblers required for given output (in items/second)

I - Goal i/s output M - i/s of individual assemblers R - Assemblers needed to sustain this i/s

I/M=R

For example...

I have twenty labs. I want to produce enough red science to sustain one science being consumed in thirty seconds per lab.
So thirty seconds per lab divided over twenty labs (seconds divided by labs equals the reciprocal of the i/s required to sustain the amount of labs
consuming one science every x seconds) is 30s/20=1.5, reciprocal is 0.666...(2/3) i/s (red science/second); this is I. I am using tier two
assemblers, so it's 0.15 i/s per machine; this is M. 2/3 /0.15=4.444...(4/9) tier two assemblers, rounded up to five; this
is the answer, R.

Note*
One iron gear assembler can provide for ten red science assemblers, so I might as well have ten assemblers.
Given the amount of labs, the amount of assemblers, and their production speed we can find how much science consumption the ten assemblers can take!
10*0.15=1.5
1/1.5=0.666...(2/3)
2/3*20=13+2/3
So ten assemblers can support up to twenty labs consuming one science every 13.666 seconds!
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Since I mentioned it, why not give the equation for required i/s to keep a lab running indefinitely?

C - How fast one science is consumed in seconds (if it consumes multiple, divide the consumption time by the amount consumed)
L - The amount of labs I - Science in i/s needed to sustain research

1/(C/L)=I

Notice it's the same as the production speed to items/second equation—That's because science consumption falls under production speed.
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Assemblers required to maintain output for given amount of receiving assemblers (for intermediate steps)

Time for a bunch of (totally not made up for the purpose of this equation) definitions.

In the relationship of electronic circuits and copper cables:

Product - The end item that is created at the end of a production line. In this instance, they're electronic circuits.

Step - Assembler(s) creating a single part for a product of a single production line. This would be both the copper wires and electronic circuit.

Intermediate step - The non-final step to creating a product; this is also the set of assemblers that we are trying to find the amount of.
This would be the copper wires.

Receiving step - The next step after the corresponding intermediate step. This can be the final or not the final step, meaning that the receiving step
isn't always the product. This would be the electronic circuits.

Receiver(s) - Individual assemblers on the receiving step.

Needed - The amount of intermediate items required to craft one set of items for the receiver. This would be that three copper cables are
needed to create one electronic circuit.

Created - The amount of intermediate items created per instance. This would be that two copper cables are created per iteration.

Now we can get to the equation!

A - Quantity of receiving step assemblers B - i/s of receiver C - Quantity of items created by receivers per instance
D - i/s of individual intermediate step assemblers E - Quantity of items created by individual intermediate step assemblers per instance
F - Needed quantity of intermediate items per receiver G - Quantity of intermediate assemblers
daum_equation_1480134756787.png (2.74 KiB) Viewed 2155 times
For example...

*Keep in mind the ratio of electronic circuit assemblers to copper cable assemblers is 2:3.
Before we get to the main part, let's get something else out of the way first.
Let's say we want electronic circuits to be produced at 13.33 i/s. Each tier two assembler produces these at a rate of 1.5 i/s. 13.33/1.5=8.8866, rounded
up to nine.

All right, so now we have our quantity of receiving step assemblers; this is A.
As previously mentioned, tier two electronic circuit assemblers run at 1.5 i/s; this is B.
One electronic circuit is produced per instance; this is C.
Copper cables are created at a rate of 3 i/s per assembler: this is D.
Two copper cables are created per instance; this is E.
Three copper cables are needed to craft an electronic circuit; this is F.

9((1.5/1)/(3/2))(3/2)=G
9(1.5/1.5)*1.5=G
9*1.5=G
13.5=G
So this is our answer! Lets check it.
9/13.5=0.666...=2/3
Yep, that's a ratio of 2:3!
So we need 9 electronic circuit assemblers and 14 copper cable assemblers to sustain an item output of 13.33 i/s, rounded up.
Original Post Do Not Steal!

faloun
Burner Inserter
Posts: 14
Joined: Thu Dec 18, 2014 4:59 pm

### Re: Manual Calculations for Items/Second and Assembler Ratios

Thanks for putting all the math down in one place, person from almost a year ago!

I think it's much harder to use the automatic calculators and google sheets for ratios until you understand the math behind it.