If there are 600 mines at Truk on turn 1 how long will they last and when will they need to be...

[Tanaka]

....replaced. In other words how long will mines last before you need to replace the bulk of them???

[Bradley7735]

Mines located in friendly hex's decay at 1% per day. So, that comes to about 50% every 60 days (a little less, but I like to think in terms of months).

If you start with 600 mines on Dec 7th, then around Feb 7th you'll have about 300. On April 7th you'll have about 150 mines. etc etc.

By the way, I REALLY like the method for mine decay. I always thought that Mine were WAAAAAYYYYY to effective in UV. Now you can't just mine the crap out of the slot. They'll decay fast enough to be historic.

bc

[von Murrin]

Formula for a friendly base is: Current Mines - Current Mines(.01) = Remaining Mines

Remaining Mines becomes Current Mines for the next turn and so on. So: 600, 594, 588, 582, 576, 570, 564, 558, 552, 546, 540, 534, 528, 522, 516, 510, 504, etc.

The question remains. Do mine fractions round up, down or are they kept? That's what I want to know.

[tsimmonds]

Depends on how low you are willing to let the number of mines fall. Since you lose 1% a day it is easy to figure out that:

after 18 days you are down to 500
after 40 days you are down to 400
after 69 days you are down to 300
after 110 days you are down to 200
after 179 days you are down to 100
after 338 days you are down to 20
the last 20 last almost forever.

Results will vary depending on how WitP handles rounding.

[Bradley7735]

The way to check if the calculation rounds down or up is to place exactly 50 mines in a firendly port. Then check back in a day and see if any decayed.

50 *.01=.5 If the .5 rounds up, then no mines decay below 50 in a friendly port.

bc

[von Murrin]

Exactly. ROUNDING!!![:D]

This I want to know.[&:]

[von Murrin]

ORIGINAL: Bradley7735

The way to check if the calculation rounds down or up is to place exactly 50 mines in a firendly port. Then check back in a day and see if any decayed.

50 *.01=.5 If the .5 rounds up, then no mines decay below 50 in a friendly port.

bc

Care to try? I'm at work.[:D]

[Bradley7735]

Maybe I should have said 49 mines. Of course .5 would round up to 1. So, I don't think you'd decay below 49 mines if there is normal rounding on the calculation.

It's also possible that a minimum of one mine decays each day.

[Mr.Frag]

Pretty simple to tell, stick 100000 mines in a port and watch. You'll know quick enough if it rounds up or down.

[Bradley7735]

I wish I could try. I'm at work too. In fact, I'd probably get in trouble if my boss see's me on the forum.

[von Murrin]

Yes. This must be answered; it's been annoying me for a week now.

[von Murrin]

ORIGINAL: Mr.Frag

Pretty simple to tell, stick 100000 mines in a port and watch. You'll know quick enough if it rounds up or down.

Ok. I'll go play with the editor when I get home.[:@][:D]

[ShakyJake]

Just to back up what Bradley and the others have said, there are also additional tools you can use to determine how many mines there will roughly be in the hex after decay.

You could also use the original equation for exponential growth and decay:

X=X0*e^(-r*n), where:

X is current value,
X0 is initial value,
r is the rate (calculated to be 0.0100503358 for 1% decay),
and n is the number of days you're looking at.

Also, there is a much more friendly equation one might use, being:

X=X0*(1-.01)^n,

with the same variables as above.

There might be rounding cutoff error at play that'll make any results you get from the equations a little different than what the game gets, but this should be pretty handy for determining roughly how many mines you can expect after a given period, or how long it would take those mines to decay to a certain point.

[Mr.Frag]

ORIGINAL: ShakyJake

Just to back up what Bradley and the others have said, there are also additional tools you can use to determine how many mines there will roughly be in the hex after decay.

You could also use the original equation for exponential growth and decay:

X=X0*e^(-r*n), where:

X is current value,
X0 is initial value,
r is the rate (calculated to be 0.0100503358 for 1% decay),
and n is the number of days you're looking at.

Also, there is a much more friendly equation one might use, being:

X=X0*(1-.01)^n,

with the same variables as above.

There might be rounding cutoff error at play that'll make any results you get from the equations a little different than what the game gets, but this should be pretty handy for determining roughly how many mines you can expect after a given period, or how long it would take those mines to decay to a certain point.

Oh! The pain! The horror!

I'm back in math class!

Seriously ... great for spreadsheets.

Now, give an agreggate sum version of that so I can handle repairs of factories increasing production output over time at the expence of supplies [:D]

[Charles2222]

Does the game keep track of dead or dud mines? I ask this because mines don't vanish even if they aren't lethal anymore. Naturally any minesweeper would have to sweep them all up, not just the live ones, which means, that the game should keep track of dead mines. If the game keeps track of dead mines it would take minesweepers longer to clear the water of mines, so, if one were to mine the living daylights out of an area and all were to go dead, there still would be some advantage beyond the scale of their lethality. BTW, anyone know if aerial recon can spot a mine or two or occasion (especially if at low altitude)?

[ShakyJake]

Oh! The pain! The horror!

I'm back in math class!

Seriously ... great for spreadsheets.

Now, give an agreggate sum version of that so I can handle repairs of factories increasing production output over time at the expence of supplies

Heya Frag, Engineering student here, so it seems I practically breathe mathematics. [:D] I haven't bought the game yet so I don't know exactly how the supply costs work, but I seem to remember you saying that the facilities started off damaged and would repair 1% each day? Is the cost of repair each day for a facility directly proportional to the amount of damage it currently has? And 1% is indeed the amount repaired per day? If so, then that aggregate sum you're asking for is a snap to calculate.

Let:
C = the supply cost per day for each 1% of damage to the facility,
X represent the %damage (and also number of days to repair).

Your equation for the total cost to upgrade/repair a facility would be:

Total Cost = 0.5*C*(X²+X)

Again, since I don't have the game I don't know if this is exactly how it works. Either way, tell me if this helps you, or more particulars if you would like me to give you something you can use.

[AmiralLaurent]

ORIGINAL: Charles_22

Does the game keep track of dead or dud mines? I ask this because mines don't vanish even if they aren't lethal anymore. Naturally any minesweeper would have to sweep them all up, not just the live ones, which means, that the game should keep track of dead mines. If the game keeps track of dead mines it would take minesweepers longer to clear the water of mines, so, if one were to mine the living daylights out of an area and all were to go dead, there still would be some advantage beyond the scale of their lethality. BTW, anyone know if aerial recon can spot a mine or two or occasion (especially if at low altitude)?

I think disappearing mines in minefields are not no more fonctionning but are lost by sea causes and either sink or are carried away by currents aftert heir moorings break, then being washed on beaches or exploding on reefs. If you want to be really realistic, a small part of those mines should appear as 'hostile' for both sides in the same hexs and the hexs around, but in reality the probability to hit a drifiting mine was rather small (even if tens of ships did during the war and the years after, but the probability is still small) so I think this is not required in the game. Minesweepers were occasionnaly sweeping these mines but the real danger were minefields. Moored mines floating under the sea are far more dangerous than those floating over the waves.
On the other hand, the fact that minefields need to be maintained by DM and ML is historical and is well simulated in the game, as a minefield will disappear without a regular visit by those ships.

As for air recon watching mines, it happens during the war but all sightings were probably drifting mines or errors. Planes flying very low (for ASW or ASR) report mines from times to times... at least in Europe, where the numebr of mines dropped is far more important than in the Pacific before the 1945 minelaying mine by B-29s.

[Mr.Frag]

Heya Frag, Engineering student here, so it seems I practically breathe mathematics. [:D] I haven't bought the game yet so I don't know exactly how the supply costs work, but I seem to remember you saying that the facilities started off damaged and would repair 1% each day? Is the cost of repair each day for a facility directly proportional to the amount of damage it currently has? And 1% is indeed the amount repaired per day? If so, then that aggregate sum you're asking for is a snap to calculate.

Let:
C = the supply cost per day for each 1% of damage to the facility,
X represent the %damage (and also number of days to repair).

Your equation for the total cost to upgrade/repair a facility would be:

Total Cost = 0.5*C*(X²+X)

Again, since I don't have the game I don't know if this is exactly how it works. Either way, tell me if this helps you, or more particulars if you would like me to give you something you can use.

Basically looking at this:

Factory = X + Y

X = production output, Y = Damaged

Each turn assuming supply (which we have to assume or you just can't work the math), Y = Y-1, X=X+1 until Y=0.

I need to Sum of X based on X+Y over the duration of Y. The Cost of Y is easy enough to figure out.

Most of the production stuff pretty much falls into this type of catagory.

Where it gets complicated is these need to be expressed based on a Variable number of turns.

ie: X = 100, Y = 100, but whats the total for 30 days? ie: I need to look into the future a variable number of days to produce results and forcasts of when things will be needed. Something like x+(y-days)! ?

[ShakyJake]

A few more questions, then, just so that I'm sure I understand what you're saying. Do your variables X and Y sum to 100? Or are you saying that production starts at 100 + 100 additional which is damaged (such that when fully repaired both sum to 200?). Also, 1% is repaired daily? Additionally, do you know what the cost is in supplies for production points, and for repair? Are both of these directly proportional to to production points and damage percentage, respectively?

[Mr.Frag]

ORIGINAL: ShakyJake

A few more questions, then, just so that I'm sure I understand what you're saying. Do your variables X and Y sum to 100? Or are you saying that production starts at 100 + 100 additional which is damaged (such that when fully repaired both sum to 200?). Also, 1% is repaired daily? Additionally, do you know what the cost is in supplies for production points, and for repair? Are both of these directly proportional to to production points and damage percentage, respectively?

X can be any value (functional production). Y is an upgrade to the functional production (generally 100, but can be less). Each turn, Y has a point repaired and X increases by that point. (not a percent, an actual point so y-1)

I need the X over time view ... ie: whats X's output on day 11?

Basically if x = 5 and y = 5 on day 1, output = 5, day 2 output = 6, day 3 output = 7, etc ... I need the sum of the outputs taking the Y (repairs if any) into account.