Extreme (infinite) mining with beacons

[vanatteveldt]

We had a very interesting discussion on extreme mining a while ago (viewtopic.php?f=8&t=2569), where the challenge was to fit as many miners on a field as possible.

Of course, real extreme mining only starts when you use modules. Now, adding a mix of prod and speed to every mining drill is pretty dull, but with beacons it becomes more fun. Problem is of course, you can never mine the central square of a beacon, so you'd have to move after the mines are depleted.

Now let's consider a field of infinite ores. What configuration of mines, modules, and beacons gets you the most ore per second per tile of ore?

My first attempt is to use interleaved columns of beacons and miners, with a 1-tile row between the miners for a pole and belts:

single pair

On a 45x43 patch of 100% (angels) infitite ore with all finite tech researched (i.e. creative mode default, set magic wand to 1500 resources to recreate patch), this yields 26.4k/min (=11 blue belts) for 126 miners, or 200k ore/min/miner, or 13.6 ore/min/resource tile.

screenshot

production

Is there a better mix of beacons and modules if the goal is to maximize ore yield per minute per tile, disregarding all else? For the 'challenge', you are free to use any sized resource patch, 'score' is production/min divided by the size of the patch.

[DaveMcW]

The obvious optimization is to use speed modules in mining drills.

But this challenge is not so interesting, because Mining productivity tech make it trivial to get all the ore you want.

[Jap2.0]

The annoying thing is that beacons are 3x3, so there is always some ore that you can't mine if you use beacons :\

[vanatteveldt]

The obvious optimization is to use speed modules in mining drills.

But this challenge is not so interesting, because Mining productivity tech make it trivial to get all the ore you want.

Do speed modules + speed modules give more output/sec than prod + speed? I thought that because of the multiplication there would be a point where they catch up, but maybe that point is never reached...

Of course after researching an infinite amount of mining prod, you get an infinite amount of resources. However, before you get there you still need to mine

[Jap2.0]

Wait, you can put productivity modules in mining drills?
Wow.

[leitk]

You can put prod modules in miners, but the prod reasearch mostly cancels the benefit when combined with speed modules. This is because normally prod multiples output but subtracts from speed. Speed beacons add the speed back (and then some) so you end up multiplying speed and prod.
Prod research just adds to prod modules, so in effect it reduces the huge bonus of prod+speed to some extent. I'm not sure exactly what level of mining prod research, but there is a point where speed modules in miners will make more sense than prod modules.

[eradicator]

Do speed modules + speed modules give more output/sec than prod + speed? I thought that because of the multiplication there would be a point where they catch up, but maybe that point is never reached...

I wrote a script to calculate the breakpoint ages ago. For a 3-slot machine configurations with mixed modules can only be faster with at least 10 beacons/machine. This does not account for productivity research however which would quickly push the breakpoint beyond 12 beacons (the maximum you can have per machine in vanilla).

PythonScript

Code: Select all

max_beacons = 12
#max_slots   = 2

prod  = {'speed' : -0.15, 'prod' : 0.1}
speed = {'speed' : 0.5}


##print(max_modules)
print('Actual Speed of machines with module boots:')
for max_slots in [2,3,4]:
    print('\nWith {:2d} machine slots:'.format(max_slots))
    for b in range(0,max_beacons+1):
        print('\n With {:2d} beacons:'.format(b))
        for s in range(0,max_slots+1):
        
        
            max_modules = b + max_slots # every beacon counts as one slot
            
            prodmods  = s
            prodfact  = 1 + prodmods * prod['prod']
            speedmods = max_modules - s
            speedfact = 1 + speedmods * speed['speed'] + prodmods * prod['speed']
            endspeed = prodfact * speedfact
            #print(prodmods+'prods+',speedmods+'speeds=', prodfact * speedfact)

            print('  {:2d} prods + {:2d} speeds -> {:.2f}'.format(prodmods,speedmods,endspeed))
        

Code: Select all

 With  9 beacons:
   0 prods + 12 speeds -> 7.00 <-fastest
   1 prods + 11 speeds -> 6.99
   2 prods + 10 speeds -> 6.84
   3 prods +  9 speeds -> 6.57

 With 10 beacons:
   0 prods + 13 speeds -> 7.50
   1 prods + 12 speeds -> 7.54 <-fastest
   2 prods + 11 speeds -> 7.44
   3 prods + 10 speeds -> 7.21

[Frightning]

Wait, you can put productivity modules in mining drills?
Wow.

Pre 0.15, it used to be pretty normal to do so in mega bases both to reduce the rate at which you needed to construct new mines, and to increase amount mined from each deposit, but this came at the cost of it taking longer and much more energy, hence you needed more simultaneous mines active and the correspondingly much larger amount of power (and train logistics) to handle that.

[Jap2.0]

Do speed modules + speed modules give more output/sec than prod + speed? I thought that because of the multiplication there would be a point where they catch up, but maybe that point is never reached...

I wrote a script to calculate the breakpoint ages ago. For a 3-slot machine configurations with mixed modules can only be faster with at least 10 beacons/machine. This does not account for productivity research however which would quickly push the breakpoint beyond 12 beacons (the maximum you can have per machine in vanilla).

PythonScript

Code: Select all

max_beacons = 12
#max_slots   = 2

prod  = {'speed' : -0.15, 'prod' : 0.1}
speed = {'speed' : 0.5}


##print(max_modules)
print('Actual Speed of machines with module boots:')
for max_slots in [2,3,4]:
    print('\nWith {:2d} machine slots:'.format(max_slots))
    for b in range(0,max_beacons+1):
        print('\n With {:2d} beacons:'.format(b))
        for s in range(0,max_slots+1):
        
        
            max_modules = b + max_slots # every beacon counts as one slot
            
            prodmods  = s
            prodfact  = 1 + prodmods * prod['prod']
            speedmods = max_modules - s
            speedfact = 1 + speedmods * speed['speed'] + prodmods * prod['speed']
            endspeed = prodfact * speedfact
            #print(prodmods+'prods+',speedmods+'speeds=', prodfact * speedfact)

            print('  {:2d} prods + {:2d} speeds -> {:.2f}'.format(prodmods,speedmods,endspeed))
        

Code: Select all

 With  9 beacons:
   0 prods + 12 speeds -> 7.00 <-fastest
   1 prods + 11 speeds -> 6.99
   2 prods + 10 speeds -> 6.84
   3 prods +  9 speeds -> 6.57

 With 10 beacons:
   0 prods + 13 speeds -> 7.50
   1 prods + 12 speeds -> 7.54 <-fastest
   2 prods + 11 speeds -> 7.44
   3 prods + 10 speeds -> 7.21

Okay - how much mining productivity research does it take to change it so that all speed is more efficient? Also, shouldn't it be 21 and 23 modules that you can fit? (3 in the drill, 2 in each beacon?)

Wait, you can put productivity modules in mining drills?
Wow.

Pre 0.15, it used to be pretty normal to do so in mega bases both to reduce the rate at which you needed to construct new mines, and to increase amount mined from each deposit, but this came at the cost of it taking longer and much more energy, hence you needed more simultaneous mines active and the correspondingly much larger amount of power (and train logistics) to handle that.

Okay cool.

[Frightning]

Do speed modules + speed modules give more output/sec than prod + speed? I thought that because of the multiplication there would be a point where they catch up, but maybe that point is never reached...

I wrote a script to calculate the breakpoint ages ago. For a 3-slot machine configurations with mixed modules can only be faster with at least 10 beacons/machine. This does not account for productivity research however which would quickly push the breakpoint beyond 12 beacons (the maximum you can have per machine in vanilla).

PythonScript

Code: Select all

max_beacons = 12
#max_slots   = 2

prod  = {'speed' : -0.15, 'prod' : 0.1}
speed = {'speed' : 0.5}


##print(max_modules)
print('Actual Speed of machines with module boots:')
for max_slots in [2,3,4]:
    print('\nWith {:2d} machine slots:'.format(max_slots))
    for b in range(0,max_beacons+1):
        print('\n With {:2d} beacons:'.format(b))
        for s in range(0,max_slots+1):
        
        
            max_modules = b + max_slots # every beacon counts as one slot
            
            prodmods  = s
            prodfact  = 1 + prodmods * prod['prod']
            speedmods = max_modules - s
            speedfact = 1 + speedmods * speed['speed'] + prodmods * prod['speed']
            endspeed = prodfact * speedfact
            #print(prodmods+'prods+',speedmods+'speeds=', prodfact * speedfact)

            print('  {:2d} prods + {:2d} speeds -> {:.2f}'.format(prodmods,speedmods,endspeed))
        

Code: Select all

 With  9 beacons:
   0 prods + 12 speeds -> 7.00 <-fastest
   1 prods + 11 speeds -> 6.99
   2 prods + 10 speeds -> 6.84
   3 prods +  9 speeds -> 6.57

 With 10 beacons:
   0 prods + 13 speeds -> 7.50
   1 prods + 12 speeds -> 7.54 <-fastest
   2 prods + 11 speeds -> 7.44
   3 prods + 10 speeds -> 7.21

Yea, I actually took advantage of this fact in my random world challenge because I was stuck with a single pumpjack, so I 12x beaconed it, and used prod inside instead of speed to max rate on it. Very much a corner case use for prod in pumpjacks.

[eradicator]

Yea, I actually took advantage of this fact in my random world challenge because I was stuck with a single pumpjack, so I 12x beaconed it, and used prod inside instead of speed to max rate on it. Very much a corner case use for prod in pumpjacks.

You probably did it "wrong" though. For 3-slot-machines putting 3 prod modules is never fastest. Though in most cases the gain is less than 2% overall anyway. I just checked for productivity research. If you have just a single level (+2%) researched, then any more than one productivity module makes the pumpjack slower again.

Code: Select all

 With 12 beacons (+0%):
   0 prods + 15 speeds -> 8.50
   1 prods + 14 speeds -> 8.63
   2 prods + 13 speeds -> 8.64 <-fastest
   3 prods + 12 speeds -> 8.52

Code: Select all

 With 12 beacons (+2%):
   0 prods + 15 speeds -> 8.67
   1 prods + 14 speeds -> 8.79 <-fastest
   2 prods + 13 speeds -> 8.78
   3 prods + 12 speeds -> 8.65

 With 12 beacons (+22%):
   0 prods + 15 speeds -> 10.37 <-fastest
   1 prods + 14 speeds -> 10.36
   2 prods + 13 speeds -> 10.22
   3 prods + 12 speeds -> 9.96

ps: forgot to mention that all those numbers are based on Tier3 modules.

[vanatteveldt]

Thanks all for the replies. Very helpful for me to realize that prod modules in miners don't make sense, even at relatively low levels of mining prod research and high # of beacons.

So, the challenge is now: what configuration of miners, beacons , and belts/bots gives the highest extraction rate (ore/second) on a field?

[Frightning]

Yea, I actually took advantage of this fact in my random world challenge because I was stuck with a single pumpjack, so I 12x beaconed it, and used prod inside instead of speed to max rate on it. Very much a corner case use for prod in pumpjacks.

You probably did it "wrong" though. For 3-slot-machines putting 3 prod modules is never fastest. Though in most cases the gain is less than 2% overall anyway. I just checked for productivity research. If you have just a single level (+2%) researched, then any more than one productivity module makes the pumpjack slower again.

Code: Select all

 With 12 beacons (+0%):
   0 prods + 15 speeds -> 8.50
   1 prods + 14 speeds -> 8.63
   2 prods + 13 speeds -> 8.64 <-fastest
   3 prods + 12 speeds -> 8.52

Code: Select all

 With 12 beacons (+2%):
   0 prods + 15 speeds -> 8.67
   1 prods + 14 speeds -> 8.79 <-fastest
   2 prods + 13 speeds -> 8.78
   3 prods + 12 speeds -> 8.65

 With 12 beacons (+22%):
   0 prods + 15 speeds -> 10.37 <-fastest
   1 prods + 14 speeds -> 10.36
   2 prods + 13 speeds -> 10.22
   3 prods + 12 speeds -> 9.96

ps: forgot to mention that all those numbers are based on Tier3 modules.

Pumpjacks were 2-slot machines in 0.14 days, and there was no productivity research back then either.

[Frightning]

Thanks all for the replies. Very helpful for me to realize that prod modules in miners don't make sense, even at relatively low levels of mining prod research and high # of beacons.

So, the challenge is now: what configuration of miners, beacons , and belts/bots gives the highest extraction rate (ore/second) on a field?

I would imagine the usual alternating rows setup with a coverage pattern of Roboports (and maybe some extras near the perimeter to provide enough charging ports).