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2024.01.17 14:37 MarioVX Manifold Production Delay & Ramp Up Time Analysis
IntroductionWhen distributing a stream of input items to an array of processing buildings, Ficsit employees typically choose between two major design principles for their distribution belt network: manifolds and balancers. Manifolds are widely appreciated for their compactness, simplicity and extensibility.It is well known that this comes at the (in most cases acceptable) cost of some delay in production behind the whole manifold, as the initially unbalanced distribution relies on the successive machines' internal buffers becoming filled and causing preceding belts to back up, causing the re-distribution of flow to the machines deeper in the manifold. Thus it takes some time for the production of the array as a whole to ramp up to full capacity. But as the sparse responses to this post I stumbled across a few days ago show, it remains so far largely uninvestigated and unknown how long this delay really is, depending on the setup - even approximately. The purpose of the following analysis is to change that. u/Cris-Formage , consider this an extensive response to your question, and u/Gorlough, a generalization to your correct answer for the specific example discussed. MethodGoalFor any given manifold, we would like to calculate two quantities of interest:
As usually in mathematical modeling, we need to make some difficult trade-offs between precision and universality. I want this analysis to be as universal as possible, so I have decided to ignore belt delays. These depend not only on the MK level of the belt, but also the exact lengths of belt segments and spaces between the buildings. If belt speeds are eventually changed or new MKs are introduced, the analysis would become outdated. Instead, we only consider the following:
Normalization It turns out there is quite a bit of redundance in the above specification, which can be eliminated by normalization as a pre-processing step. This translates a wide range of manifolds with different recipe speeds and buffer sizes to a small set of canonical standard cases, and hence the results directly transferable: We divide c, f & bs by c. This fixes c=1. It follows from f=n*c that f=n, hence f can be omitted as a parameter as well. Finally, instead of bs, we define b := bs/c. Since bs is in items and c in items per time, this quantity is a time - namely the buffer time of the individual processing buildings. That is, how many seconds or minutes of its own input consumption rate it would require to burn through its own filled buffer stack. Example: We make a manifold for smelters smelting copper ore into copper ingots. The smelters consume 30 copper ore/min, this is c. Copper ore stacks up to 100, this is bs. Suppose our total in-flow into the manifold is 180 copper ore /min. Then we have n = 180/30 = 6, and b = 100/(30/min) = 3.(3) min = 200 sec. This normalization thus reduces the number of relevant quantitative input parameters from 4 to 2. n and b are sufficient specification... except for one thing, and that's independent of the items, buildings and recipes involved: Topology As it turns out, there are two topologically distinct ways to construct a manifold:
top-3 manifold. connectivity of the last splitter depends on parity of n, in this example even. Both topologies qualify are manifolds by the usual understanding as they adhere to translational symmetry, making them easy to build, extensible and relatively compact. The at first glance obvious pros & cons are that top-2 is even more compact as it doesn't connect to the splitter outputs on the opposite side of the processing buildings, meanwhile top-3 uses only half as many splitters to connect the same number of machines which saves some system performance and counts up slower to the engine's object limit (splitters consist of multiple objects so this shouldn't be underestimated). But while all of these may be convincing arguments for one or the other in their own right, in this analysis we are only concerned with their behavior during the ramp-up process. Algorithmic Computation With all relevant quantitative and structural input parameters in place, it's time to actually perform the computation which will yield us the ramp-up time and later the production delay. The following lends itself to automation via a script, which is how I got the results I will present later. But for small n, it is quite simple to do these with pen and paper, which is useful for verification purposes and quite instructive to make sure one understands the computational process. The core idea is to essentially simulate the whole ramp-up process until the maximum output rate is reached. For this, we need to track the following quantities across time:
For the total processed items, we need the previously calculated durations of all segments individually, and in each segment the sum of the consumption rates over all buildings. The total processed items are then a piecewise defined linear function of time. If a queried time lies in segment k, sum up the product of total consumption rate and duration of all segments up to k-1, then add for the k-th segment the product of total consumption rate with just the time difference between the queried time and the last transition point. For the production delay, we simply compare this production curve to that of a hypothetical load balancer - the linear function n * t. Beyond the last segment of the ramp-up process, the curves are parallel and thus have constant difference. This difference is the terminal production delay. But especially for comparing different manifolds, all the intermediary delays can be interesting too. If this sounded a little technical or vague, you're invited to the following example. If it was already clear to you, skip ahead to the next section. We're picking up the old example of a copper core manifold that translated to b=200sec, n=6. Suppose we connect it in top-3. b_0 = 0, 0, 0, 0, 0, 0 i_0 = 2, 2, 2/3, 2/3, 1/3, 1/3 c_0 = 1, 1, 2/3, 2/3, 1/3, 1/3 n_0 = 1, 1, 0, 0, 0, 0 t_0 = (200 - 0)/1 = 200 b_1 = 200, 200, 0, 0, 0, 0 i_1 = 1, 1, 4/3, 4/3, 2/3, 2/3 c_1 = 1, 1, 1, 1, 2/3, 2/3 n_1 = 0, 0, 1/3, 1/3, 0, 0 t_1 = (200 - 0)/(1/3) = 600 b_2 = 200, 200, 200, 200, 0, 0 i_2 = 1, 1, 1, 1, 1, 1 ; terminal state T = 200 + 600 = 800 PD(t): 0 =< t =< 200: 4 * t 200 =< t =< 800: 800 + (5 + 1/3) * (t - 200) 800 =< t: 4000 + 6 * (t - 800) = -800 + 6 * t TPD = -800So it will take this manifold 800 seconds or 13 minutes and 20 seconds - plus the neglected belt delay times - to reach its maximum output rate from a cold start. By then, it will have accumulated a terminal production delay of 800 seconds worth of base consumption rate in items compared to a balancer that had cold started at the same time. To re-convert this into an actual item count, we can multiply with said consumption rate: 800 seconds * 0.5 items/second = 400 items of Copper Ore that it lags behind. If we instead want to convert this delay into a time rather than item delay for the whole manifold, we instead divide by n: 800 seconds / 6 = 133.33 seconds, or 2 minutes 13.33 seconds that the manifold as a whole is behind in production compared to a balancer (plus neglected belt delays). ResultsSo, let's see what we got! There are some findings here that are surprisingly simple and seemed obvious to me in hindsight, nevertheless I didn't anticipate them beforehand, so I didn't want to take them away beforehand either. Then some other findings are just surprising, but not simple. Let's go through all of it:Contribution of Buffer Time This is a huge one. As complicated as the ramp-up time works out to be, it turns out that the buffer time is a multiplier that can be cleanly factored out to allow even more normalization! I.e.: T(n,b,top) = b * T(n,1,top) This translates to the accumulated production function as a stretching in x-direction. The transition points' times are multiplied by b and so are the production amounts at these points. As such, the TPD is multiplied by b as well. This means that henceforth, the buffer can be ignored. We understand the following time values as multiples of the buffer time, and production quantities as buffer time worth of individual consumption rate in items. But why is the total ramp-up time proportional to buffer time? Well, the very first segment's time is proportional to it: T_0 = (b-0)/x = b * 1/x, and the subsequent segments are proportional if the preceding segments time and hence buffer fill states are proportional: T_n+1 = (b - b_n,b)/x = (b - b * b_n,1)/x = b * (1 - b_n,1)/x. It follows by induction that the total time is proportional too. Terminal Production Delay It turns out there is an easy shortcut to the TPD of a manifold: Think about where the items are going that have entered the manifold but not exited it through processing. Since our belts have no capacity, they must all be hung up in building buffers. So we only need to imagine the buffer fill states in the terminal segment (which has 100% production) and sum them up.
This means if you compare topologies based on the criterion of TPD alone, top-2 and top-3 are equal for even n, top-3 is only better for odd n. Transient Production Delays Perhaps you're not just interested in the terminal delays, as perhaps you already have use for a smaller quantity of produced items that can be obtained before a complete ramp-up of the manifold. So let's look at the ramp-up process output dynamically. As the TPD hints, it is quite important to distinguish by parity of n. The differences are more apparent for smaller n, so here are the production graphs for n=5 and n=6: https://preview.redd.it/26ve4ooxjzcc1.png?width=600&format=png&auto=webp&s=5017107fcc60aadd1d1f4a977cc65527597783d1 https://preview.redd.it/hapujooxjzcc1.png?width=600&format=png&auto=webp&s=9ab959af7d59459bf0f5e8d09793838587391ddf As we can see here, top-3 gets a head start on production. For even n, top-2 catches up to be tied in the terminal state by reaching its max production slightly sooner. Nevertheless, at any point in time, top-3 is ahead of or even with top-2 in terms of accumulated production. For odd n, top-3 is also always ahead or even with top-2, but as we know from the previous result maintains a genuine lead in the end. Ramp-up time dependence on n Finally, the last and most difficult piece of the puzzle. How does a growing number of attached buildings (and hence depth of the manifold, and multiplicity of the input stream) influence the ramp-up time of the manifold? Well, without further ado: linear plot of ramp-up times vs n for both topologies, for small n semi-log plot of ramp-up times vs n for both topologies, with logarithmic regression curves for top-2 and for either parity n with top-3, for larger n Pay attention to the logarithmic scaling of the x-axis in the second plot. The behavior for large n attunes to a logarithmic function, not a linear function as the scaled plot may suggest at first glance. The logarithmic regressions don't fit well for very small n. The values may be read off the first plot, but here is a little lookup table with the values to three decimal places for reference:
DiscussionEvaluation of Results, Practical AdviceIt is eye-catching how extremely much faster top-3 is for odd n than both for even n and top-2. Even a lot more machines can be ramped up in shorter time this way. The difference is so vast I initially suspected an error in my code, but manually re-calculating with pen & paper revealed these numbers to be correct and this extreme zig-zagging behavior to be genuine. This has an immediate practical application: When concerned with ramp-up time, overbuild to an odd number (possibly underclock) and connect in top-3. For even n, top-2 reaches maximum output rate slightly faster than top-3 - however keep in mind the previous result that nevertheless, top-3 is still ahead or even at all times in the number of items it has actually outputted. Intuitively, top-3 distributes the items "more evenly" than top-2. This gets buildings further down the manifold working sooner (and hence output up quicker), but it fills the buffers of earlier buildings slower (and hence reach full buffers later). So here the choice depends on how you value stableness versus earliness of the output (and the other considerations briefly hinted at in the introduction, not the topic of this analysis). Origin of the roughly logarithmic dependence Finally, one might be wondering, why the hell the ramp-up time depends roughly logarithmically on n? My best explanation goes like this: Consider a slightly simplified ramp-up process, where only the in-flow into the buildings at the first non-filled splitter (and before) is considered, and the rest - rather than already slightly filling successive buildings - simply vanishes. Let's assume top-2. Then the first building fills up (normalized buffer) in time 1/(n/2) = 2/n. After it is full, the second splitter receives only n-1 flow (because 1 flow goes and is consumed by the first, filled, building). Only (n-1)/2 goes into the second building, so the time needed to fill it in our simplified model is 1/((n-1)/2) = 2/(n-1). The next one will be 2/(n-2), then 2/(n-3), and so on, all the way down to 2/1. When we add these up, we have T = 2/1 + 2/2 + ... + 2/n = 2 * (1/1 + 1/2 + ... + 1/n). The sum in parentheses has a name, it's called the n-th Harmonic number. Famously the Harmonic numbers can be asymptotically approximated with the natural logarithm and the Euler-Mascheroni constant (about 0.577) as H_n ~ ln(n) + 0.577 for large n. For readers familiar with calculus, it may help to consider that the antiderivative of 1/x is ln(x) to make sense of this. If we plug this in for this simplified ramp-up process, we get T_n ~ 1.154 + 2 ln(n). A closer comparison of the simplified with the more accurate ramp-up process from our full model reveals that this simplified one must always be slower to ramp-up than the complete one, as we only let flow vanish and not create more. This means the times derived from the formula for the simplified process are a reliable upper bound for the times of the accurate process. This means the accurate process' ramp-up time can grow at most logarithmically with n. Closing ThoughtsThis was a surprisingly vast rabbit hole to delve in, but I'm happy with the clarity of the results. We finally got some quantitative estimates on by how much a manifold actually delays your production until it's ramped up to parity with a balancer that instead might have been more elaborate to plan and build and take away more space. This wasn't done before to this extent in the Satisfactory community as of my knowledge.Some aspects or doubts you want to discuss? Some part of the derivation you wanted to but couldn't quite follow along and want a more thorough explanation? Some specific values you want the time to be computed for? Other thoughts? Please comment! If you feel like these results are worth buying me a coffee for my time, you can. Thanks! Now, happy manifolding and back to work, for Ficsit! |
2024.01.14 09:19 Friendly-Employ-5595 How do definite integrals work?
2023.12.12 20:42 CheshireKat-_- [calc] How do I know when to take the derivative and when to take the antiderivative in a problem where you have to evaluate the integral by making the given substitution? In the first problem, I took the derivative but in the second it wants me to take the antiderivative in the expliantion
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2023.12.05 21:30 BDady Is there a nice way on conceptually understanding the FTOC part 2?
2023.11.23 16:34 hpxvzhjfgb Things taught in high school math classes that are false or incompatible with real math
2023.11.15 02:33 andrewl_ What is the name for this simple and stateful approach to integration? Why isn't it taught?
a = {1, 1, 1, 1, 1, 1, 1, 1, 1, 1}And velocity would just accumulate acceleration, creating:
v = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}And position, starting at 0, would just accumulate velocity, creating:
p = {1, 3, 6, 10, 15, 21, 28, 36, 45, 55}I think this simple act of accumulating list values is integration, and it's happening without complicated symbols, just addition.
a = {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1} v = {0.5, 1.0, 1.5, 2.0, 2.5, 3.0, 3.5, 4.0, 4.5, 5.0, 5.5, 6.0, 6.5, 7.0, 7.5, 8.0, 8.5, 9.0, 9.5, 10.0} p = {0.25, 0.75, 1.5, 2.5, 3.75, 5.25, 7.0, 9.0, 11.25, 13.75, 16.5, 19.5, 22.75, 26.25, 30.0, 34.0, 38.25, 42.75, 47.5, 52.5}And the resulting graph is even closer to the true integral! Increasing to 100 samples and it's nearly dead on.
f = [6]*10 f = accumulate(f, add) f = accumulate(f, add) f = accumulate(f, add)Which looks even simpler in this form:
c, d, e, f = 6, 0, 0, 0 for _ in range(10): d += c e += d f += eThat this works was evidently obvious to everyone except me. I think I would have benefited had my teacher started with "Let me show you just how simple integration can be." and after the demonstration continued "Now let's formalize it a bit with this technique called RS."
2023.11.09 18:10 felixnice2 PDE problem
2023.11.09 18:10 felixnice2 PDE problem