Factorio Forums

vanatteveldt wrote:Well, if I can just ask

Assuming only depleted oil sources and 1 pumpjack per beacon, and assuming level 3 modules of the appropriate type in all relevant buildings, how many pumpjacks and refinery+processing infrastructure do I need per MW of generated energy (on top of energy consumption of the infrastructure)

1 net MW (post-burn, or hot water) = 
~ 0.6211 pumpjack
~ 0.0311 Oil Refinery
~ 0.1568 Chemical Plants

bobucles wrote:So how many pumps does a speed 3 beacon have to cover to be worth it?

XKnight wrote:Small addition to that post:

There was discussion on reddit about rocket fuel, so...

- assembler I/II/III + beacons: number of effect receivers - 8, beacons around - 8
- chemical plant + beacons: number of effect receivers - 8, beacons around - 8
- oil refinery + beacons: number of effect receivers - 4, beacons around - 8
- pumpjack + beacons: number of effect receivers - 1, beacons around - 8

{productivity3, speed3, efficiency3, speed3 in beacons, efficiency3 in beacons}
Energy net profit per solid fuel (I) - {0,0,0,0,0} 10710. kJ
Energy net profit per solid fuel (II) - {2,0,0,8,0} 12044.7 kJ
Energy net profit per solid fuel (III) - {4,0,0,8,0} 14365.6 kJ
Energy net profit per petrolium - {2,0,0,1,6} 8136.38 kJ
Energy net profit per light oil - {2,0,0,1,6} 16272.8 kJ
Energy net profit per light oil (cracking) - {2,0,0,1,6} 5972.44 kJ
Energy net profit per heavy oil - {2,0,0,1,6} 8136.38 kJ
Energy net profit per heavy oil (cracking) - {2,0,0,1,6} 14404. kJ
Energy net profit per crude oil - {2,0,0,1,6} 14546.6 kJ
Energy net profit per crude oil (advanced processing) - {2,0,0,1,6} 15483.3 kJ
Energy net profit per depletted pump. {0,2,0,8,0} One cycle - 788.328 kJ; Power - 4729.97 kW

dragontamer5788 wrote: Interesting. I'll have to double-check your math to fully trust it however. I'm kinda surprised that Efficiency3 actually helps with refineries, but it sorta makes sense. Each beacon saves 210kW per refinery (and with 4 refineries that's 840kW saved per beacon, with 480kW used. So net profit).

It should be noted that "compressing" to Rocket Fuel leads to a net profit with PM3 in Assembly Machines 3 (+40% productivity beats out the -25MJ loss per cycle). Do you think you can run that fancy program to figure out the best combination in AM3?

I'm betting PM3x4 + max speed beacons is the most efficient rocket fuel setup. But it can't hurt to be sure.

DaveMcW wrote:

bobucles wrote:So how many pumps does a speed 3 beacon have to cover to be worth it?

At 2 pumps, you pay 50% of the profit to run the beacon.
At 1 pump, you pay 100% of the profit to run the beacon.

If you use XKnight's productivity + efficiency beacon scheme farther down the line, you could even come out ahead with 1 pump per beacon.

B= beacons (2 x Speed3)
E= net power gain (MW)

      Usable Power gain     Pump power usage       Beacon power usage
E = 1*(1+0.5*B)             - 0.09*(1+0.7*B)          - 0.48*B
E = 0.91 - 0.043*B

E = net MW unmodded (assume non-negative)
P = MW produced/converted unmodded (non-negative)
C = MW consumed unmodded (non-negative)
Em = net MW modded w/ 1 x productivity.3 

E = P-C
E+C=P (for substitution later)

        Production   Usage     Productivity bonus
Em = 0.85*P        - 1.8*C       + 0.1*P
Em = 0.95*P - 1.8*C
Em = 0.95*(E+C) - 1.8*C
Em = 0.95*E - 0.85*C

Given that both variables are positive, you will always get a worse power per second production using productivity.

Solar Heretic wrote: It's actually slightly worse than that. If the beacon is not shared with any other pumps, then each beacon actually decreases the net gain:

Solar Heretic wrote:Actually, the only difference I can see in our formulas is what our E is. In your case, you denote it as net energy. For me, I used it as gross energy gain, which is why I just replaced it with 1 MW for 0.1 oil/s. And the more I look at the calculation, the more I feel like you are wrong on your end conclusion. I feel like you may have forgot your hypothesis and got confused at the end. Either that, or the oil to usable energy math I did in my original post was off.

Calculations

modules = { productivity2, speed2, efficiency2, speed2 in beacons, efficiency2 in beacons};
Energy net profit per solid fuel (I) - {0,0,0,0,0} 10710. kJ
Energy net profit per solid fuel (II) - {0,0,2,0,0} 11130. kJ
Energy net profit per solid fuel (III) - {4,0,0,8,0} 12062.6 kJ <- still useless
Energy net profit per petrolium - {2,0,0,0,5} 6413.71 kJ
Energy net profit per light oil - {2,0,0,0,5} 12827.4 kJ
Energy net profit per light oil (cracking) - {0,0,2,0,0} 4219.81 kJ
Energy net profit per heavy oil - {2,0,0,0,5} 6413.71 kJ
Energy net profit per heavy oil (cracking) - {2,0,0,0,5} 10481.9 kJ
Energy net profit per crude oil - {2,0,0,0,5} 10216.7 kJ
Energy net profit per crude oil (advanced processing) - {2,0,0,0,5} 11101.3 kJ
Energy net profit per depletted pump. {0,2,0,0,0} One cycle - 986.377 kJ; Power - 1578.2 kW

XKnight wrote: My hypothesis? In this thread I didn't make any assumptions...
Probably we are talking about different initial conditions, in my case any modules may be used (that's why E=1.5483), and your case seems to be without modules at all.

E = base usable energy produced (0.1 oil/s, or 1 usable MW)
Em = modded usable energy produced, using speed3 only
B = number of beacons filled w/ Speed3

Em = E*(100% + direct mod boost + becaon mod boost * B) - 0.09*(100% + direct mod boost + beacon mod boost * B) - 0.48*B
Em = E*(1+1+0.5B) - 0.09*(1+1.4+0.7B) - 0.48B
Em = 2E + 0.5EB - 0.216 - 0.543B
Substitute E=1
Em = 2 + 0.5B - 0.216 - 0.543B
Em = 1.784 - 0.043B

B(0) = 1.784 MW
B(1) = 1.741 MW
B(2) = 1.698 MW
...


Alternatively, using calculus and proof by contradiction:
Em = 2E + 0.5EB - 0.216 - 0.543B
d.Em/d.B = 0.5E - 0.543
Second deriv(Em,B) = 0

Let's assume that my statement regarding that beacon were wasteful is wrong, then that means beacons either break even or produce a net of power...
d.Em/d.B >= 0
0.5E - 0.543 >= 0
Substitute E=1
0.5*1 - 0.543 >= 0
-0.043 >= 0
Because the last part is not true, this disproves that beacons either break even or produce a net power gain.

QED: Speed beacons are not efficient at any number.

Solar Heretic wrote: - Modules effects are completely additive regardless of the source. You're usage of the formula partly implies that the direct modules take affect first, then the beacons, resultuing in a partly multiplicative effect. The same applies to any combination of mods. (I tested in game to make sure of this)
- Using modded net energy is incorrect, as the growth of speed and "de-efficiency" is not equal (+50%/+70% respectively). They need to be split up to their base components (production and consumption), and evaluated from there.

Solar Heretic wrote:Ok, I found the point of difference in all our math: the beacon sharing.

Your formula is based on sharing the beacon effect among as many machines as possible, effectively distributing the beacon burden across all machines. My math was based on if beacons were not shared, meaning that each pump had to pay the beacon cost on their own. This was primarily to answer the 1 pump-all beacon scenario listed below (to simulate scenarios where none of the oil spots were within beacon reach of each other). I do concur that as soon as the beacon gets shared, the system actually start producing more net energy.

XKnight wrote: - assembler I/II/III + beacons: number of effect receivers - 8, beacons around - 8
- chemical plant + beacons: number of effect receivers - 8, beacons around - 8
- oil refinery + beacons: number of effect receivers - 4, beacons around - 8
- pumpjack + beacons: number of effect receivers - 1, beacons around - 8