AltF4 #47  Optimizing the Solution 17092021
Содержание

Optimizing the Solution
ilbJanissary  Why use a calculator?
 Concepts
 Crafting speed
 Recipe equations
 Recipe matrices
 Linear programming
 Why do we need a cost function?
 How do we define a cost function?
 Dual maximization problem
 Solving with simplex
 Matrix tab
 Result information
 Cost modifiers
 Simplex canonical tableau and solution
 Alternate algorithms
 TL;DR
 Contributing
After a threeweek hiatus, AltF4 is back, more motivated than ever! We start off with a bang as ilbJanissary goes into depth about how the calculation engine behind FactorioLab does its magic. It might seem complicated at the surface, but the process is broken down into granular parts that are illustrated with actual examples. No eyes will glaze over reading this piece (hopefully!).
Optimizing the Solution ilbJanissary
Hi! As an Angular developer and long time user of the Kirk McDonald Factorio Calculator, I decided to build my own webbased calculator for factory games called FactorioLab back in 2020. Kirk McDonald published an essay on his approach that helped me to get started, but it stops short of explaining the algorithm it uses to find a solution. After struggling through the math for some time and building two iterations of nonsimplex solvers in FactorioLab, I wanted to share with the community what I learned when implementing a true simplex algorithm solver. A TL;DR can be found at the very bottom of the article.
Why use a calculator?
Let’s start from the beginning. In factory games like Factorio, Dyson Sphere Program, and Satisfactory, the objective is to mine raw resources, use those resources to craft items following a recipe, and use those items to discover more advanced items and recipes. Early in these games, the recipes and items involved are simple and do not require extensive planning, but as more items and recipes are unlocked, the production chain can become incredibly complicated. Factory calculators like FactorioLab and the Kirk McDonald Factorio Calculator are intended to help determine how many raw resources and factories are required to sustainably produce items at a certain rate.
Concepts
There are two important concepts in factory games that are important to understanding the math behind calculations: items and recipes.
 Items are the products and ingredients in factory games. Some items only exist as intermediates, and cannot do anything on their own, such as most raw resources. Some items can also serve a purpose in the game, such as fuel, transport belts, furnaces, assemblers, power generation, power distribution, etc.
 Recipes are the ways in which players craft new, more advanced items in the game. Some items can be crafted by hand using a recipe, and other recipes can only be performed by specific factory buildings, like assemblers or furnaces. Recipes usually take a specific set of items as inputs, take a specific amount of time to perform, and produce a specific set of items as products. Recipes are often represented as a list of ingredients and an amount of time to craft to the left of an arrow, which points to the list of products on the right.
For example, in Factorio, a basic item is Iron ore. Iron ore is a raw resource, meaning it is produced via a recipe that takes no items as inputs:
This recipe can be crafted on an Iron ore resource patch, by hand or by a mining drill. Iron ore is primarily used as an ingredient to produce Iron plates, which use their own recipe:
Iron plates cannot be crafted by hand, and must be produced in a furnace. Recipes can also include many inputs and outputs, such as one of the more complicated recipes in Factorio, Advanced oil processing:
This recipe can only be crafted using an Oil refinery.
Crafting speed
The real crafting time of a recipe is also affected by the crafting speed of the machine used to craft the recipe. For example, the Electric mining drill has a crafting speed of 0.5
, which means when it is used to mine Iron ore using the recipe above, it actually takes two seconds per Iron ore. To simplify, we’ll generally treat the crafting speed as 1
.
Recipe equations
You may notice that if you substitute the arrow for an equals sign, a recipe can easily be interpreted as an equation. Since time itself is not an actual item, to make this equation meaningful, we need to divide both sides by the amount of time and thus treat the numbers in equation as rates instead of a simple number of items. Hence, the Iron ore recipe becomes:
In this case, the left side is zero since the recipe takes no inputs. Likewise, the Iron plate recipe becomes:
Up to a certain point, these equations are relatively simple to solve. Many items can only be produced by a single recipe, and thus it only requires simple algebra to determine the number of items and recipes involved to produce items at a certain rate. In FactorioLab, if the required products can only be produced by one recipe, these equations are solved directly as far down the production chain as possible. However, some items can be produced via multiple different recipes, making it more complex to determine which recipe, or often how many of each recipe, is optimal to produce the desired items while consuming less input.
Recipe matrices
The canonical problem in Factorio is Advanced oil processing and cracking, and will be used as the primary example of finding a solution. This problem usually involves at least five recipes and five items. The recipes are:
Crude oil (raw resource)
Crude oil production speed varies based on the resource on the map, but can be treated in a simplified form.
Water (raw resource)
Advanced oil processing
Heavy oil cracking
Light oil cracking
Note that other recipes may also be involved, such as Basic oil processing and Coal liquefaction, but these will be ignored at this time.
Dividing these recipes by their recipe time yields an equation based on rates, which is the form the calculator needs to calculate production requirements. Since the crafting speed of all the factories involved is 1
, we don’t need to make any adjustments to account for crafting speed, but that would be necessary for many factories, or any time modules or beacons are involved.
Crude oil (raw resource)
Water (raw resource)
Advanced oil processing
Heavy oil cracking
Light oil cracking
These recipes can be organized into a matrix. This matrix should have columns for recipes, and rows for items. This creates a different set of equations, one equation for each item. For example, for crude oil: 10  20 = # output
Having divided the item amounts by their recipe time, the full matrix looks like:
10  0  20  0  0  
0  1200  10  15  15  
0  0  5  20  0  
0  0  9  15  15  
0  0  11  0  10 
Next, we need to account for what we want to actually produce from our factory. To handle this, we can add a column to represent the final outputs of our factory. As an example, let’s say that we want to produce 5 Heavy oil/second and 100 Petroleum gas/second. In order to represent this idea in the matrix, we need to make these values negative, because we’re indicating that we’re taking that amount out of the system.
Out  

10  0  20  0  0  0  
0  1200  10  15  15  0  
0  0  5  20  0  5  
0  0  9  15  15  0  
0  0  11  0  10  100 
Linear programming
A matrix of this nature is called a linear program in mathematics, and can be solved using linear programming techniques. The most common method for solving a linear program is the Simplex algorithm, and it is the method used by both FactorioLab and the Kirk McDonald Factorio Calculator. Since there are multiple possible solutions to this linear program, the Simplex algorithm takes the approach of maximizing an objective function, which is essentially another row in the matrix.
While we do not necessarily have anything we want to maximize in our linear program, we do want to minimize the amount of resources and factories required to produce our output items. So, instead of adding an objective we want to maximize, we can first add a “Cost” function to our linear program so that we can describe how we want to minimize the number of resources and factories in our solution.
Why do we need a cost function?
For simple factories where there is only a single recipe for each factory, we don’t even need a cost function. We can calculate how many of each ingredient we need, and keep going down the recipe chain, until we get a solution. So why does the linear program need a cost function?
Let’s look at the oil processing example in Factorio. In our example, we want to produce Petroleum gas and Heavy oil. Note that Advanced oil processing produces both of these products. Hence, a viable solution is to completely ignore the cracking recipes, and just produce all the oil products we need directly from Advanced oil processing. So, why do we ever use cracking in Factorio? There are two main reasons:
 The equation is not balanced, meaning in almost any factory the output oil products will not match the ratio of desired oil products, and eventually one of them will fill up all available storage and force production to halt completely. In most factories producing pure science products, the culprit will be Light oil. There is no way to just “burn off” excess in Factorio, and manually emptying storage tanks is tedious.
 It uses excess Crude oil. Crude oil is not as easy to obtain as water, so we usually want to minimize the amount that is required. By cracking the excess Light oil into Petroleum gas, we can both address the excess Light oil and reduce the amount of Crude oil required, at the expense of extra Water, which is easier to obtain.
A cost function quantifies how difficult resources are to obtain, and how much effort is involved in setting up factories. A solution that requires less resources is better than a solution that requires a lot of resources, and a solution that requires less factories is better than a solution that requires more factories. By putting specific numbers onto the cost of obtaining resources and setting up factories, we can directly quantify the cost of a solution, and the calculator can use that cost to determine which solution is optimal.
How do we define a cost function?
The cost function can be rather abstract and arbitrary as there is no specific cost of resources and factories in games like Factorio and Dyson Sphere Program. Instead, we have to rely on intuition of what is most difficult to obtain and set up in a factory. In FactorioLab, this comes down to a few specific variables that define the cost function:
 A single factory running a recipe costs 1.
 One unit of Water costs 100, since water is generally cheap to obtain.
 One unit of Crude oil costs 1000, since crude oil is generally produced in higher quantities than other mined resources.
 One unit of ‘mined’ resources costs 10,000. In Factorio, this applies to Coal, Uranium ore, Stone, Iron ore, and Copper ore. Generally, these are the resources we want to minimize most.
 One unit of ‘manually collected’ resources costs 100,000. In Factorio, this applies to Wood, for example. While Wood is generally not part of any important recipes in vanilla Factorio, we want to minimize any use of resources we cannot collect automatically.
We can represent this cost function as an additional row in our matrix. In order to accurately account for the cost of mined resources, we will simplify those recipes so that the cost is based on the number of items, rather than the number of factories mining the resource. Thus the Pumpjack and Offshore pump recipes are simplified into a recipe that has higher cost and only produces 1 item per second. Once we calculate an amount of Crude oil and Water for the solution, it is trivial to convert this production number back into a number of factories.
Out  

1  0  20  0  0  0  
0  1  10  15  15  0  
0  0  5  20  0  5  
0  0  9  15  15  0  
0  0  11  0  10  100  
Cost  1000  100  1  1  1  0 
Dual maximization problem
Now, the issue with this linear program is that we want to minimize the cost of our solution, while the Simplex algorithm is designed to maximize an objective. Fortunately, there is a simple way to convert a minimization problem into a maximization problem that can be solved using the Simplex algorithm. This is accomplished by converting the minimization problem into its “dual maximization problem.” Once a minimization problem is converted into its dual maximization problem, it is feasible to solve the dual problem using the Simplex algorithm. The solution to the maximization of the dual problem is the same as the solution to the minimization of the original problem, so we can accept this as our solution.
The dual problem is simply a transpose of the original problem, or swapping the rows and columns. This results in:
Cost  

1  0  0  0  0  1000  
0  1  0  0  0  100  
20  10  5  9  11  1  
0  15  20  15  0  1  
0  15  0  15  10  1  
Out  0  0  5  0  100  0 
In order to handle potential surplus values, linear programming generally adds ‘surplus’ or ‘slack’ variables to the program. For instance, for a factory to produce only Heavy oil, there is no solution that does not involve some surplus amount of Light oil and Petroleum gas. These are represented by additional columns in the matrix where each column has a ‘1’ value in the row matching its item.
+  +  +  +  Cost  

1  0  0  0  0  1  0  0  0  0  1000  
0  1  0  0  0  0  1  0  0  0  100  
20  10  5  9  11  0  0  1  0  0  1  
0  15  20  15  0  0  0  0  1  0  1  
0  15  0  15  10  0  0  0  0  1  1  
Out  0  0  5  0  100  0  0  0  0  0  0 
Now, we finally have a canonical matrix that can be solved using the Simplex algorithm.
Solving with simplex
The end goal of the Simplex algorithm is to arrive at a matrix where the objective function (in this case, the last row of the matrix) contains no negative numbers. In our initial matrix, note that our objective row has two columns with negative values, representing the two items we want to produce. In order to reach our solution, we simply need to apply ‘pivot’ operations until our objective row no longer has any negative values.
Pivot 1
A Simplex pivot operation involves four steps.
 Choose a column to pivot. This must be a negative column, and is usually chosen as the largest negative number. In our matrix, this would be the Petroleum gas column.
 Choose a row to pivot. Check each row where the chosen column has a value greater than zero, and calculate the ratio of the cost column in that row to the value of the chosen column in that row. The pivot row is the row with the lowest ratio. In our matrix, we can check each row:
 : column is 0, ignore this row.
 : column is 0, ignore this row.
 : column is 11, cost column is 1, so the ratio is 1/11.
 : column is 0, ignore this row.
 : column is 10, cost column is 1, so the ratio is 1/10.
 The lowest ratio is the row at 1/11, so we will pivot this row.
 Divide the pivot row by the reciprocal of the pivot element. In our pivot row, this yields 20/11, 10/11, 5/11, 9/11, 1, 0, 0, 1/11, 0, 0, 1/11
 Add multiples of the pivot row to other rows to change the pivot column to 0. The final result is:
+  +  +  +  +  Cost  

1  0  0  0  0  1  0  0  0  0  1000 
0  1  0  0  0  0  1  0  0  0  100 
1 9/11  10/11  5/11  9/11  1  0  0  1/11  0  0  1/11 
0  15  20  15  0  0  0  0  1  0  1 
18 2/11  5 10/11  4 6/11  23 2/11  0  0  0  10/11  0  1  1/11 
181 + 9/11  90 10/11  40 5/11  81 9/11  0  0  0  9 1/11  0  0  9 1/11 
Now, let’s continue with the next pivot.
Pivot 2
Column 1 (181 9/11), Row 5 (Ratio 1/200)
+  +  +  +  +  Cost  

0  13/40  1/4  1 + 11/40  0  1  0  1/20  0  11/200  9999 + 199/200 
0  1  0  0  0  0  1  0  0  0  100 
0  1 1/2  0  1 1/2  1  0  0  0  0  1/10  1/10 
0  15  20  15  0  0  0  0  1  0  1 
1  13/40  1/4  1 11/40  0  0  0  1/20  0  11/200  1/200 
0  150  5  150  0  0  0  0  0  10  10 
Pivot 3
Column 2 (150), Row 2 (Ratio 100)
+  +  +  +  +  Cost  

0  0  1/4  1 + 11/40  0  1  13/40  1/20  0  11/200  9967 + 99/200 
0  1  0  0  0  0  1  0  0  0  100 
0  0  0  1 1/2  1  0  1 1/2  0  0  1/10  150 + 1/10 
0  0  20  15  0  0  15  0  1  0  1501 
1  0  1/4  1 11/40  0  0  13/40  1/20  0  11/200  32 + 101/200 
0  0  5  150  0  0  150  0  0  10  15010 
Pivot 4
Column 4 (150), Row 4 (Ratio 133 2/5)
+  +  +  +  +  Cost  

0  0  1 19/20  0  0  1  1 3/5  1/20  17/200  11/200  9839 91/100 
0  1  0  0  0  0  1  0  0  0  100 
0  0  2  0  1  0  3  0  1/10  1/10  300 + 1/5 
0  0  1 1/3  1  0  0  1  0  1/15  0  100 1/15 
1  0  1 19/20  0  0  0  1 3/5  1/20  17/200  11/200  160 9/100 
0  0  205  0  0  0  300  0  10  10  30020 
Pivot 5
Column 3 (205), Row 1 (Ratio 5046 7/65)
+  +  +  +  +  Cost  

0  0  1  0  0  20/39  32/39  1/39  17/390  11/390  5046 7/65 
0  1  0  0  0  0  1  0  0  0  100 
0  0  0  0  1  1 1/39  1 14/39  2/39  1/78  17/390  10392 27/65 
0  0  0  1  0  80/117  11/117  4/117  1/117  22/585  6828 41/195 
1  0  0  0  0  1  0  0  0  0  1000 
0  0  0  0  0  105 5/39  131 31/39  5 10/39  1 5/78  4 17/78  1064472 1/13 
The final row in the matrix now has no negative values, so this is recognized as the solution. The solution can be extracted from the last row, where the first five columns represent surplus values (in this case, all zero), the next two values represent the input Crude oil and Water, the next three values represent the number of factories running Advanced oil processing, Heavy oil cracking, and Light oil cracking, and the last value represents the ‘cost’ of this solution.
Thus, our solution requires 105 5/39 , 131 31/39 , 5 10/39 , 1 5/78 , and 4 17/78 .
This solution can be compared to the result in FactorioLab.
Matrix tab
The Matrix tab of FactorioLab, a relatively new feature, allows you to peek under the hood of the simplex solver. For recipe chains that can be solved without simplex, this tab is mostly empty, but when simplex is used, the tab exposes many details and even allows you to customize the cost function used by the solver.
Result information
The first table exposes details on the current simplex solver result.
 The Simplex Result from the solver
 Skipped: The simplex solver was not used, either because it is disabled or because it was not required to solve the recipe chain.
 Failed: The simplex solver encountered an error, usually due to a recipe loop that cannot be solved (e.g. if iron plates require two iron gears and iron gears require two iron plates)
 Cancelled: The user cancelled the solver manually when prompted after running for at least five seconds.
 Solved: A valid solution was found.
 The Runtime of the simplex solver, in ms
 The total number of Pivots performed by the solver
 The Size of the canonical matrix
Cost modifiers
The second table allows the user to tweak the cost function used in the canonical matrix.
 Recipe is a factor that multiplies the cost of recipes with a defined cost, usually mining recipes representing raw inputs
 Factory is a factor that multiplies the cost of the number of factories themselves, which is usually only a secondary consideration
 Input defines the cost of items that cannot be produced by any recipe, such as Wood in Factorio
 Ignored defines the cost of items which have been explicitly ignored, which defaults to zero
Simplex canonical tableau and solution
The third table shows both the canonical matrix generated by the calculator and the solution row with the final cost. Note that this is displayed as the dual minimization problem, which is how the calculator minimizes cost using the simplex method. In each recipe row, the cost of the recipe is displayed and can also be overridden using the cost input field to the right. Values are represented as items/second in this table regardless of the user’s display rate selection, since this is how calculations are always performed internally.
Alternate algorithms
The Simplex algorithm using dual maximization has proven to be a very effective method of solving factory calculations. However, simplex is not the only algorithm to solve a linear program. There are other methods such as revised simplex, crisscross algorithm, ellipsoid method, Karmarkar’s algorithm, and the interiorpoint method. In fact, in just 2019 the running time was improved by a new matrix multiplication time algorithm. As linear programs grow large, especially in modded Factorio, the simplex algorithm used by FactorioLab can take a very long time to find a solution, especially when mods introduce ‘recycling’ recipes that introduce large numbers of loops. With further research, it is possible that a better algorithm for solving factory linear programs could be developed.
TL;DR
 Simple factories are easy to calculate by hand, when items are only produced by a single recipe.
 When items are produced by multiple different recipes, we need a way to figure out how much of each recipe to use.
 Since there are multiple potential solutions, we need to measure the cost of a solution to determine which is best.
 Simplex is the most commonly used algorithm (though not the fastest) to solve these kinds of problems, in both the Factorio and mathematics communities.

FactorioLab uses the Simplex algorithm to optimize the solution for complex factories and allows users to:
 Customize the cost function since it is subjective
 Use common mod data sets
 Run calculations and plan the factory from mobile devices or at work
 Calculate how much rocket fuel you can produce from a single refinery
Join the FactorioLab Discord to learn more!
Contributing
As always, we’re looking for people that want to contribute to AltF4, be it by submitting an article or by helping with translation. If you have something interesting in mind that you want to share with the community in a polished way, this is the place to do it. If you’re not too sure about it we’ll gladly help by discussing content ideas and structure questions. If that sounds like something that’s up your alley, join the Discord to get started!