Beaconed green circuit ratios/math
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Beaconed green circuit ratios/math
Green circuits production (especially non-beaconed) is very well covered in tutorials and blueprints. I thought I understood it and even designed my own.
This is what I know:
- Use 3 copper wire assemblers for every 2 GC assemblers, to get a proper ratio. Repeat enough times to saturate your output belts.
- Feed it 1 belt of Iron, 1.5 belts of copper and you will produce 2 belts of Green Circuits.
- Adding production modules to the assemblers and speed modules on beacons reduces iron/copper cost and improves productivity.
Sounds good right? Well, before doing a beaconed design of my own I went to review other people's designs to get layout ideas. To my shock I noticed that every single beaconed design uses a 1-1 wire/GC assembler ratio instead of the 3/2 I expected.
A few such designs can be attributed to sub-par approaches, but every single one... clearly points to a gap in my understanding.
Can someone please explain the gap in my understanding?
EDIT: THE ANSWER
Thanks the replies. Great "duuuuh" moment that removed my mental block. Here it is paraphrased in simple terms:
- The 4 Production modules reduce the Green Chip copper cable requirement by 40%. So, instead of 1.5 cable assemblers for each GC assembler, you need approx 1 (1.071 if you are nitpicky, but close enough).
- By removing that extra half copper cable assembler you also reduce the iron to copper belt ratio so that you now need one belt of each.
- From there it is just a matter of general efficiency since each GC costs less in resources due the productivity modules and in less time due to the speed modules in the beacons.
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In case someone is curious, what I am trying to create is a near perfect ratio, efficient, compact and tileable setup that produces 4 belts of GC per set.
And yes, I admit I am not very good about Factorio math (I am quite good at math in general but something about doing math for a game just makes my brain go on strike. Weird.)
This is what I know:
- Use 3 copper wire assemblers for every 2 GC assemblers, to get a proper ratio. Repeat enough times to saturate your output belts.
- Feed it 1 belt of Iron, 1.5 belts of copper and you will produce 2 belts of Green Circuits.
- Adding production modules to the assemblers and speed modules on beacons reduces iron/copper cost and improves productivity.
Sounds good right? Well, before doing a beaconed design of my own I went to review other people's designs to get layout ideas. To my shock I noticed that every single beaconed design uses a 1-1 wire/GC assembler ratio instead of the 3/2 I expected.
A few such designs can be attributed to sub-par approaches, but every single one... clearly points to a gap in my understanding.
Can someone please explain the gap in my understanding?
EDIT: THE ANSWER
Thanks the replies. Great "duuuuh" moment that removed my mental block. Here it is paraphrased in simple terms:
- The 4 Production modules reduce the Green Chip copper cable requirement by 40%. So, instead of 1.5 cable assemblers for each GC assembler, you need approx 1 (1.071 if you are nitpicky, but close enough).
- By removing that extra half copper cable assembler you also reduce the iron to copper belt ratio so that you now need one belt of each.
- From there it is just a matter of general efficiency since each GC costs less in resources due the productivity modules and in less time due to the speed modules in the beacons.
---
In case someone is curious, what I am trying to create is a near perfect ratio, efficient, compact and tileable setup that produces 4 belts of GC per set.
And yes, I admit I am not very good about Factorio math (I am quite good at math in general but something about doing math for a game just makes my brain go on strike. Weird.)
Last edited by zOldBulldog on Sat Oct 06, 2018 6:42 pm, edited 4 times in total.
Re: Beaconed green circuit ratios/math
It's because those designs use tier 3 prod modules in tier 3 assemblers which have 4 module slots.
That means, you get 40% more copper cables, which is very close to the actual 1.5 ratio for copper plates that you will need in the end.
Here is a calculator that lets you calculate ratios very easily. It's already set to green chips.
https://kirkmcdonald.github.io/calc.htm ... ircuit:f:1
That means, you get 40% more copper cables, which is very close to the actual 1.5 ratio for copper plates that you will need in the end.
Here is a calculator that lets you calculate ratios very easily. It's already set to green chips.
https://kirkmcdonald.github.io/calc.htm ... ircuit:f:1
Re: Beaconed green circuit ratios/math
40% extra bonus production for the four P3's means for every ten cycles you get 14 outputs, and copper cable produces 2 cable/output. Greenchip production needs three cable/cycle, 15 cable outputs per 10 of its own cycles, and with four P3's of its own a single A3 can deliver 14 of those all by its onesies.
So the OCD in us all is yelling "well, by God, I'm going to build that 15th cable assembler per 14 greenchip assemblers and distribute the output, then all the assemblers will be 100% busy all the time!", but OCD is like other pieces of our makeup that we listen to perhaps without thinking. Perhaps the best way to see it is to start from the 14 4×P3 cable A3's feeding 14 4×P3 greenchip A3's, the cable A3's output is short of the greenchip A3's demand by 1/15, so what to do? Compare your options, what's cheaper to build and more efficient to run? 1) one more cable assesmbler ... and the distribution network to route its output evenly among the 14 existing greenchip assemblers, or 2) another cable+greenchip pair just like the 14 you've got now? Both ways get the exact same boost in the results.
I've built the robot distribution network to efficiently balance an exact 14:15 pod, and it was a royal pain in the ass—satisfying and gratifying to achieve and it was fun to watch it run, enjoyable hours all around, but less UPS-friendly than just leaving well enough alone. I haven't tried to distribute the extra with belts while keeping the max-beaconed thing working, that's adding yet _more_ inserter swings to the cable path, and when you're going for UPS you start counting inserter swings, especially when belts are involved.
You can get an _apparent_ 3% boost by using 2×P3,2×S3 in the cable assemblers, (9.625×1.2)/(8×1.4)==1.03125, but that like so many apparently efficient builds is just shoving costs off into a corner you're not paying attention to, suddenly your single greatest cable consumer needs 16⅔% more copper, which means that many more miners and smelters and that much more transport overhead. Let your OCD contemplate that tradeoff for a while. "Premature optimization is the root of all evil"? I call them "immature optimizations" when I catch myself in the act.
So the OCD in us all is yelling "well, by God, I'm going to build that 15th cable assembler per 14 greenchip assemblers and distribute the output, then all the assemblers will be 100% busy all the time!", but OCD is like other pieces of our makeup that we listen to perhaps without thinking. Perhaps the best way to see it is to start from the 14 4×P3 cable A3's feeding 14 4×P3 greenchip A3's, the cable A3's output is short of the greenchip A3's demand by 1/15, so what to do? Compare your options, what's cheaper to build and more efficient to run? 1) one more cable assesmbler ... and the distribution network to route its output evenly among the 14 existing greenchip assemblers, or 2) another cable+greenchip pair just like the 14 you've got now? Both ways get the exact same boost in the results.
I've built the robot distribution network to efficiently balance an exact 14:15 pod, and it was a royal pain in the ass—satisfying and gratifying to achieve and it was fun to watch it run, enjoyable hours all around, but less UPS-friendly than just leaving well enough alone. I haven't tried to distribute the extra with belts while keeping the max-beaconed thing working, that's adding yet _more_ inserter swings to the cable path, and when you're going for UPS you start counting inserter swings, especially when belts are involved.
You can get an _apparent_ 3% boost by using 2×P3,2×S3 in the cable assemblers, (9.625×1.2)/(8×1.4)==1.03125, but that like so many apparently efficient builds is just shoving costs off into a corner you're not paying attention to, suddenly your single greatest cable consumer needs 16⅔% more copper, which means that many more miners and smelters and that much more transport overhead. Let your OCD contemplate that tradeoff for a while. "Premature optimization is the root of all evil"? I call them "immature optimizations" when I catch myself in the act.
Last edited by quyxkh on Sat Oct 06, 2018 12:00 pm, edited 1 time in total.
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Re: Beaconed green circuit ratios/math
Beacons and modules alter the ratios as DerGraue says.
Suppose you're using prod 3 modules and need an output of 100 GCs/sec. How many GCs do you actually need to pay for? 71.42857143 GCs/s because you get 28.57142857 free GCs/s. Using prod 3 modules in the cable machines too? You need 71.42857143*3 = 214.2857143 Cables/sec of which you will need to pay for 153.0612245/sec because 61.2244898/sec are free.
Machines Needed = (Desired Output per Second / Items per Craft / Productivity Percentage) * (Recipe Craft Time / Machine Speed)
Items Payed For = (1/Productivity Percentage) * Desired Output per Second
Note that productivity percentage would be in decimal format, not percentage format. 1.4 not 140%
Suppose you're using prod 3 modules and need an output of 100 GCs/sec. How many GCs do you actually need to pay for? 71.42857143 GCs/s because you get 28.57142857 free GCs/s. Using prod 3 modules in the cable machines too? You need 71.42857143*3 = 214.2857143 Cables/sec of which you will need to pay for 153.0612245/sec because 61.2244898/sec are free.
Machines Needed = (Desired Output per Second / Items per Craft / Productivity Percentage) * (Recipe Craft Time / Machine Speed)
Items Payed For = (1/Productivity Percentage) * Desired Output per Second
Note that productivity percentage would be in decimal format, not percentage format. 1.4 not 140%
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Re: Beaconed green circuit ratios/math
quyxkh wrote: ↑Sat Oct 06, 2018 11:46 amYou can get an _apparent_ 3% boost by using 2×P3,2×S3 in the cable assemblers, (9.625×1.2)/(8×1.4)==1.03125, but that like so many apparently efficient builds is just shoving costs off into a corner you're not paying attention to, suddenly your single greatest cable consumer needs 16⅔% more copper, which means that many more miners and smelters and that much more transport overhead.
This is almost always good advice. If you're going to cut costs by using productivity, then cut costs by reducing productivity. There is simply no good reason to try and mix productivity and speed in the same machine in a design from the outset.
There is only one good reason I can think of to do it, and that's in trying to squeeze everything you can out of an existing ad-hoc design.
For example:
You're rounding the corner from the late game to the end game.
You want to launch a few dozen rockets with your current build to get some techs that you want so you don't want to interrupt current production.
Your current refinery isn't putting out the amount of rocket fuel you need
You need to increase output it as a stopgap measure until you're ready to build a new refinery
There isn't space available there to add more solid fuel and rocket fuel machines.
IMO, balancing speed and productivity modules to get the output you need while saving on resources as much as possible is completely fine in this situation. It's just a stopgap measure though.
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Re: Beaconed green circuit ratios/math
Still not right.zOldBulldog wrote: ↑Sat Oct 06, 2018 10:50 am EDIT: THE ANSWER
Thanks the replies. Great "duuuuh" moment that removed my mental block. Here it is paraphrased in simple terms:
- The 4 Production modules reduce the Green Chip copper cable requirement by 40%. So, instead of 1.5 cable assemblers for each GC assembler, you need approx 1 (0.9 if you are nitpicky, but close enough).
The 4 prod modules in the Green Chip Factory reduce the copper cable requirement by 1-(1/1.4) or 28.57%. Assuming that both machines are using 4Xprod 3 modules and both are running at the same speed, you need 1.071 cable assemblers / Green Chip Assembler.
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Re: Beaconed green circuit ratios/math
Hmmm, close enough to answer the why 1:1 ratio of assemblers and belts, but I am afraid I don't follow your math.ColonelSandersLite wrote: ↑Sat Oct 06, 2018 6:22 pmStill not right.zOldBulldog wrote: ↑Sat Oct 06, 2018 10:50 am EDIT: THE ANSWER
Thanks the replies. Great "duuuuh" moment that removed my mental block. Here it is paraphrased in simple terms:
- The 4 Production modules reduce the Green Chip copper cable requirement by 40%. So, instead of 1.5 cable assemblers for each GC assembler, you need approx 1 (0.9 if you are nitpicky, but close enough).
The 4 prod modules in the Green Chip Factory reduce the copper cable requirement by 1-(1/1.4) or 28.57%. Assuming that both machines are using 4Xprod 3 modules and both are running at the same speed, you need 1.071 cable assemblers / Green Chip Assembler.
I calculated as follows:
- Non-beaconed GC assembler requires the output of 1.5 copper wire assemblers.
- 4 Prod modules reduce the required input by 40%.or is it 28.57? I was going by the 40% number people mentioned earlier.
- 1.5 x 0.60 (40% reduction) =0.9.
- Speed modules on the beacons seem to apply equally to all, so I assume they can be ignored in the calculations.
Oh.... But prod assemblers also slow down the output, in particular the output of the copper wire assembler. Is that where the 1,071 comes from?
I have to say, the interactions between the different bonuses give me a headache
EDIT:. I just had it explained. I updated the OP
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Re: Beaconed green circuit ratios/math
zOldBulldog wrote: ↑Sat Oct 06, 2018 6:39 pm - 4 Prod modules reduce the required input by 40%.or is it 28.57? I was going by the 40% number people mentioned earlier.
4 Prod 4 modules increase the *output* by 140%.
4 Prod 4 modules decrease the *cost of a fixed output* by 1-(1/140%) = 28.57% .
Does that make more sense?
Re: Beaconed green circuit ratios/math
No. The reason you need fewer cable assemblers is they're producing more cables per assembly cycle, not that you need fewer cables, not that greenchip assembly needs fewer cables, but that you're producing more cables. 40% more, to be exact. The two descriptions are not fungible, you're equivocating, treating two different descriptions as the same when they're not. Ten cycles of greenchip production requires ten cycles' worth of input, 10 iron plate and 30 copper cable, with or without prod modules.zOldBulldog wrote: ↑Sat Oct 06, 2018 6:39 pm - 4 Prod modules reduce the required input by 40%.or is it 28.57? I was going by the 40% number people mentioned earlier.
Re: Beaconed green circuit ratios/math
I never use any other ratios than 1:1.
Late game I often use 3:3 with 15 beecons to make one blue belt of gc.
Early game I use one asembler1 and one asembler2, for perfect ratio. This is cost efficient and is simple to upgrade the gc assembler to a asembler2 with two prod1 modules. This is also very cost efficient and only has a 5% idle time.
Late game I often use 3:3 with 15 beecons to make one blue belt of gc.
Early game I use one asembler1 and one asembler2, for perfect ratio. This is cost efficient and is simple to upgrade the gc assembler to a asembler2 with two prod1 modules. This is also very cost efficient and only has a 5% idle time.