Productivity Module Math

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vanatteveldt
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Re: Productivity Module Math

Post by vanatteveldt »

Thanks for updating this :)

It might be worth making a marathon / expensive recipes column. Especially circuits and gears are more expensive, but that means that everything that uses these also gets more expensive.

Gears are now 1.25*2*4=10 r/s and green circuits 1.25*2*7=17.5, making them both pretty good targets for prod modules, although everything upstream of green circuits should also get a boost...

(steels is more expensive but also takes more time, so that shouldn't change things)

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Tev
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Re: Productivity Module Math

Post by Tev »

God this is so hard to find. Anyway - thanks for updating this!
It might be worth making a marathon / expensive recipes column.
+million

vanatteveldt
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Re: Productivity Module Math

Post by vanatteveldt »

Just for my own use some calculations for marathon. I simply took the "total raw" values as listed in the toolip and divide by time. Water is 0, all other fluids 0.1.

product = 1.25 * raw / time

cable = 1.25 * 1 / .5 = 2.5
gear = 1.25 * 4 / .5 = 10

green circuit = 1.25 * 7 / .5 = 17.5
red circuit = 1.25 * 24 / 6 = 5
blue circuit = 1.25 * 188.1 / 10 = 23.5

red science = 1.25 * 5 / 5 = 1.25
green science = 1.25 * 14.5 / 6 = 3.0
blue science = 1.25 * 137 / 12 = 14.3
military science = 1.25 * 104 / 10 = 13
prod science = 1.25 * 323.5 / 14 = 28.9
hitech science = 1.25 * 743 / 14 = 66.3 (!!)


My conclusion: First priority is hi-tech science. After that, it makes a lot of sense to use productivity on green and blue circuits, then prod, blue , and military science. It can also make sense to put prod modules in gear factories, e.g. in your "goods" (belts etc) line.

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Re: Productivity Module Math

Post by D0SBoots »

I think your math is off for the ROI calculations.

Resources/kW

You list the cost of 1kW of solar/accu power as 2.2 raw resources. As near as I can tell, you arrived at this number by adding the cost of one solar panel (67.5) and the cost of one accumulator (64) and dividing by 60 kW, the peak power of a solar panel. However, that's not how a pure solar setup works - the accumulators to solar ratio is 0.84:1, and you have to use the day-averaged power output of 42kW, not the peak power. That would come out to (67.5 + 64 * 0.84) / 42 ~= 2.887. However, we're still missing an important factor - the power poles needed to collect all the power being gathered! This varies based on the design, of course, but if we assume an efficient layout we can get as low as 1 pole per 5 panels (see viewtopic.php?f=208&t=1865 for examples). Medium power poles are 12 raw resources, so overall it comes to (67.5 + 64 * 0.84 + 12 * 0.2) / 42 ~= 2.944 per kW.

She needs more power, Cap'n!

In your ROI calculation, you accounted for the increased power due to the production module, but you are not accounting for the additional power draw of the additional factories that you need to balance the speed penalty. I believe the overall formula for your example should be as follows:

( 67.5 resources [productivity module] + 376 resources [assembling machine] + (210 kW [assembling machine] * 1.4 [power penalty] + 7 kW [power drain]) * 2.944 resources/kW) * 1/(1 - 0.15) [speed penalty] - 376 resources [we start with one assembling machine] - (210 kW + 7 kW) * 2.944 resources/kW [we start with power for one assembling machine] = 549.4 total additional investment

549.4 / 0.25 / 60 = 36.6 minutes to recoup investment

Note that there's a simpler way of looking at ROI that gives a different answer. Let's build one productivity module and enough extra power to cover it, and see how long it takes to break even:

67.5 resources [productivity module] + 210kW * 0.4 [power penalty] * 2.944 resources/kW = 314.8 additional resources
6.25 resources/sec [electronic circuit] * .04 [bonus productivity] * (1 - 0.15) [speed penalty] = 0.2125 bonus resources/sec

314.8 / 0.2125 / 60 = 24.7 minutes to break even. The difference is that you're building less extra infrastructure here, because you haven't overcome the speed penalty, so it takes less time to break even.

More Speed

You didn't show your calculations when you brought speed modules and beacons in to the mix, so I'm not going to even try to recreate what you did, I'm just going to work from scratch.

2x productivity module, 2x speed module (+8% productivity) with electronic circuit:
6.25 resources/sec [electronic circuit] * .08 [bonus productivity] = 0.5 bonus resources/sec

( 67.5 resources * 4 [level 1 modules] + 376 resources [assembling machine] + (210 kW [assembling machine] * (1 + 0.4 * 2 + 0.5 * 2) [power penalty] + 7 kW [power drain]) * 2.944 resources/kW) * 1/(1 - 0.15 * 2 + 0.2 * 2) [speed modifier] - 376 resources [we start with one assembling machine] - (210 kW + 7 kW) * 2.944 resources/kW [we start with power for one assembling machine] = 1164.9 total additional investment
1164.9/0.5/60 = 38.8 minutes

3x productivity module, 1x speed module (+12% productivity): 2099.6 investment, 0.75 bonus, 46.7 minutes
1x productivity module, 3x speed module (+4% productivity): 681.4 investment, 0.25 bonus, 45.4 minutes

2x productivity module 2, 2x speed module 2 (+12% productivity):
( 689.25 resources * 4 [level 2 modules] + 376 resources [assembling machine] + (210 kW [assembling machine] * (1 + 0.6 * 2 + 0.6 * 2) [power penalty] + 7 kW [power drain]) * 2.944 resources/kW) * 1/(1 - 0.15 * 2 + 0.3 * 2) [speed modifier] - 376 resources [we start with one assembling machine] - (210 kW + 7 kW) * 2.944 resources/kW [we start with power for one assembling machine] = 3027.9 total additional investment
3027.9/0.75/60 = 67.3 minutes

3x productivity module 2, 1x speed module 2 (+18% productivity): 5168.2 investment, 1.125 bonus, 76.6 minutes
1x productivity module 2, 3x speed module 2 (+6% productivity): 1988.4 investment, .375 bonus, 88.4 minutes

Module 3's have a different formula now: 5x module 2 (689.25) + 5x advanced circuit (11) + 5x processing unit (72.85) = 3865.5

2x productivity module 3, 2x speed module 3 (+20% productivity):
( 3865.5 resources * 4 [level 3 modules] + 376 resources [assembling machine] + (210 kW [assembling machine] * (1 + 0.8 * 2 + 0.7 * 2) [power penalty] + 7 kW [power drain]) * 2.944 resources/kW) * 1/(1 - 0.15 * 2 + 0.5 * 2) [speed modifier] - 376 resources [we start with one assembling machine] - (210 kW + 7 kW) * 2.944 resources/kW [we start with power for one assembling machine] = 9768.4 total additional investment
9768.4/1.25/60 = 130.2 minutes

3x productivity module 3, 1x speed module 3 (+30% productivity): 16502.7 investment, 1.875 bonus, 146.7 minutes
1x productivity module 3, 3x speed module 3 (+10% productivity): 6759.5 investment, .625 bonus, 180.3 minutes

Beacons

Each beacon is 325 resources and 480kW of power. If we assume each beacon boosts 2 assemblers, then if each assembler is being boosted by X beacons, it means a total ratio of X/2 beacons per assembler. So our already-complicated formula evolves to:

( 3865.5 resources * 4 [level 3 modules] + 376 resources [assembling machine] + (325 resources [beacon] + 3865.5 resources * 2 [modules] + 480kW [beacon power] * 2.944 resources/kW) * (X / 2) [beacons per assembler] + (210 kW [assembling machine] * (1 + 0.8 * 4 + 0.7 * X) [power penalty] + 7 kW [power drain]) * 2.944 resources/kW) * 1/(1 - 0.15 * 4 + 0.5 * X) [speed modifier] - 376 resources [we start with one assembling machine] - (210 kW + 7 kW) * 2.944 resources/kW [we start with power for one assembling machine]

4x productivity module 3, 2x speed module 3 in beacon (+40% productivity): 25232.4 investment, 2.5 bonus, 168.2 minutes
4x productivity module 3, 4x speed module 3 in beacon (+40% productivity): 19549.3 investment, 2.5 bonus, 130.3 minutes
4x productivity module 3, 6x speed module 3 in beacon (+40% productivity): 16857.3 investment, 2.5 bonus, 112.4 minutes
... too lazy to do all the ones in-between ...
4x productivity module 3, 16x speed module 3 in beacon (+40% productivity): 12574.6 investment, 2.5 bonus, 83.8 minutes

Before I ran the numbers, I was sure that beacon costs and power drain must not have been accounted for properly in your calculations; it was the only way that more beacons being more efficient made any sense. However, even with the initial numbers being off, the later ones rapidly converged towards the same ones as in your original post, which is why I didn't bother calculating the middle values.

Thinking about it more, this actually makes sense: The label of "16x speed module 3" is deceptive, because it sounds like you're building a lot more beacons. What's actually going on is that you're building fewer assemblers/beacon - since the formula is keeping production rate constant at 1x the rate of an unmodified assembler 3, you never actually end up building more than one beacon, no matter how high X goes. But while # beacons approaches 1 as X -> infinity, the number of assemblers goes to 0, taking with it the more expensive part of the equation.

Coda

This stuff is complicated; it wouldn't surprise me if I've made mistakes too. Please double-check my math if you've got the time.

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Vladmirangel
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Re: Productivity Module Math

Post by Vladmirangel »

So, i dont really know how to apply any of those numbers into building my factory, however can i assume as a rule of thumb that its never a bad idea to place productivity module 1s on every possible slot in the whole factory?

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