How can I optimize the following production simulator?

Question

There are two types of mines: iron ore and coal.Plus there is a steel factory.You start with 500 iron and 500 coal.Initially the production level of everything is zero.Levels can be upgraded as below:

Iron:60*1,5^(level-1) Iron + 15*1,5^(level-1) Coal
Coal:48*1,6^(level-1) Iron + 24*1,6^(level-1) Coal
Steel:225*1,5^(level-1) Iron + 75*1,5^(level-1) Coal

For example, by spending 60 Iron and 15 Coal, you can upgrade your Iron Mine to Level 1.

Also, you can instantly exchange/trade iron, coal, or steel according to this formula: 1 steel = 1,2 coal = 2 iron

Productions are:
Iron=30*level*(1,1^ level)
Coal=20*level*(1,1^ level)
Steel=10*level*(1,1^level)

Which sequence of upgrades shall I follow to reach Iron Level 35 quickest?

Please note,you do not have to put iron or coal into the factory to produce steel. You just "upgrade" your mines or factory once ... and from that point on, they keep producing the amount referring to their production level indefinetely.

The main point of the question is the fact that efficiency of the mines and the factory decrease as you keep upgrading them.

So initially you start with upgrading your iron and coal mines and spend your resources on them. At that level steel just requires too much of an investment and is not worth it. You cannot use steel to upgrade anything, all upgrades are done through iron and coal. Steel can only be traded for iron or coal.

But after a while, spending your resources to upgrade the steel factory becomes profitable because the efficiencies in the iron and coal mines have dropped too much. It makes sense to produce steel and trade it for iron or coal.

But how do you optimize this? Which sequence of upgrades should be followed to reach Iron Mine Level 35 quickest?

Oh, and by the way, the production formulas I gave above describe production per unit time. It does not matter whether it is per hour, per day, or per month ...

Answer 1

Since iron and coal can be converted to steel, let us work in units of steel and call the units $. Let ni, nc, ns be levels of iron, coal, steel, respectively. Initially we have "in the bank" an amount
B = 500/2 + 500/1,2 = $666,667
Production rates in $:
Iron= $15*ni*(1,1^ ni)
Coal= $16,6667*nc*(1,1^ nc)
Steel= $10*ns*(1,1^ns)
Upgrade costs from level n-1 to n in $
Iron: $42,5*1,5^(ni-1)
Coal: $44*1,6^(nc-1)
Steel: $175*1,5^(ns-1)
Consider P/U = increase in production rate/upgrade cost:
Iron: P/U=0,3529*(1+0.1*ni)
*0,7333^(ni-1)
Coal: P/U=0,3788*(1+0.1*nc)
*0,6875^(nc-1)
Steel: P/U=0,057*(1+0.1*ns)
*0,7333^(ns-1)
which for upgrades to level 1 are
Iron: P/U=0,3882
Coal: P/U=0,4167
Steel: P/U=0,0627
Thus we should immediately upgrade coal to level 1 at a cost of $44 leaving B = $622,667.
Now P/U for upgrading coal to level 2 is
Coal: P/U = 0,3125
Comparing with the previous P/Us, we see that the next immediate upgrade should be iron to level 1 at a cost of $42,5 leaving B = $580,167
Next compare the P/U for upgrading each of the three to the next level and upgrade the one with the highest P/U and deduct the up grade cost from B. Eventually B will be <0 and you will have to produce from all three at their current levels for time t in order to generate enough $ for the next upgrade. The one to upgrade is selected from the highest current P/U and since it's upgrade cost is known, then the production time t can be calculated. This selection process is repeated until iron reaches level 35. Near the end it may be better to upgrade iron faster in order to get it to level 35 faster. It appears to be necessary to write a computer program to do the calculation since there are a lot of levels. Good luck!