United Nations Universe

OMath | Topic 2 | Saving resources for big projects

Einswerz — Sun, 27 May 2012 20:32:54 GMT

Previous topic: Mines and resources

Say you want to build something big (or anything, really, it doesn't matter much once you get to the later stages of the game). Chances are you won't have the necessary resources on hand.

A first strategy would be to simply wait until you have the resources you'll need have been accumulated by themselves (through production and/or raiding). That could be long, but hey, you need those resources, right?

Actually, there is another way to proceed: invest in mines. No, really: if you play smart, you can spend resources AND get your big project ready sooner.
It's not that surprising, when you think about it: improve your mines, and you'll gain resources faster. You do need to be careful, however, and play smart. And bring out your calculator, too.

Ok, we're going to use some fictional numbers (because 1- I'm too lazy to actually search for a real example; and 2- it'll make things easier for a simple example). Don't worry, I'll try and make them close to actual in-game values.

Situation: You have 600,000 metal, 380,000 crystal and 190,000 deuterium, and your hourly productions are 6,000, 4,000 and 2,000 (metal, crystal and deuterium, respectively). You wish, for some reason, to build a level 13 Shipyard. That costs 1,638,400 metal, 819,200 crystal and 409,600 deuterium.
If you choose to wait it out, accumulating those resources will take

(1,638,400 - 600,000) / 6,000 = 173.07 hours for the metal
(819,200 - 380,000) / 4,000 = 109.8 hours for the crystal
(409,600 - 190,00) / 2,000 = 109.8 hours for the deuterium

(I'll use these colors to refer to each kind of resources in the remainder of this article.)

Obviously, the long part is waiting for the metal. If your metal mine was a level higher, however, it would produce 7,000 per hour and gaining that million units would only take 148.34 hours (about 6 days) instead of a week. Saving a day, awesome!
It's not that easy, however. Upgrading that mine isn't free, you know. That would cost about 90,000 + 22,000, which means that saving up your resources will now take 161.2 and 115.3 hours.
And there's more. Improving your mine takes time. By the time your upgrade is finished, you'll have accumulated some resources, so the amount you wish to save will have decreased. Is that a good thing? In this particular case, not necessarily.

The impact of improving production is directly proportional to the amount of resources you wish to save. Thus, the lesser that amount, the lesser that impact.

Back to our example. Say that improving that mine will take 11.2 hours. Before that time, the old production rate is still used, which translates into a gain of 67,200. Therefore, the total time before you have the metal you want when improving your metal mine is : 11.2 + (1,638,400 + 90,000 - 600,000 - 67,200) / 7,000 = 162.8.
Ok, that's a only difference of 0.6 hours, or 36 minutes, but still. If it had taken 2 days and a half to improve the mine, it would have done much more of a difference, and it probably wouldn't have been a good idea to improve the mine. But it's still worth it, so you should improve that mine (while it's still useful; the more you wait, the less effective this technique becomes).

Now, that's a lot of computing. Isn't there an easier way to know if it's to my advantage to improve mines while saving for big projects?
The answer is, sadly, no.

If the resource whose saving time you wish to reduce is metal or crystal and you already have enough of the other resources, you can use the following criteria:

Divide cost of upgrade by production gain (not the new production, only the amount by which the old production increases). If the result is greater than the time required to save for the project, do not improve. Otherwise, improve.

It's not 100% accurate (because it does not take other resources into account), but most of the time it works just fine. However, if the mine you think about improving is a Deuterium Synthesizer, or if you want the real impact, you'll have to use the long way (steps 1 through 5 apply to each of the three resources).

Calculate the time required to gain the resources for you project without improving your mines.
Calculate the resources you want: the ones for the project plus the ones to improve the mine.
Add the resources you would gain in the time required to improve the mine to your current resources
Substract the sum in step 3 from the sum in step 2, then divide by new (post-upgrade) hourly production
Add the upgrade time to the result in step 4.
Consider only the highest time obtained in step 1 and the highest time obtained in step 5. If the one from step 5 is smaller, improve the mine. Other wise, don't.

Next topic: Going faster

OMath | Topic 1 | Mines and resources

Einswerz — Mon, 21 May 2012 15:01:16 GMT

Previous topic : Introduction

Resources are the basis on which everything in OGame is based. Simply put: whatever you want to do, you need resources to do it. It is only fitting, then, that OMath starts with the basics.

How does one acquire resources?

Strategies to obtain resources can be classified in two categories: offensive strategies (using fleets to take resources from other players) and defensive strategies (building mines to produce your own). This specific article, as its name suggest, focuses on the latter.

Say you want to produce your own resources. You'll need mines for that (for simplicity's sake, we take "mines" to include the Deuterium Synthesizer). And developing mines cost resources.

All mines follow the following pattern. If L is a mine's current level, the cost to improve it to level L+1 will be given by C = B * R^L, where B is the base cost (the cost to build level 1 of a mine), R is a certain ratio (1.5 for a Metal Mine or Deuterium Synthesizer and 1.6 for a Crystal Mine), and C is the final cost. This formula is actually applied twice: once for the metal cost, and the other for the crystal cost. Although both use the same ratio R, each have a different base cost B, which yields different final costs C.
Consider now the cost C' to upgrade a level L+1 Metal Mine. We have

C' = B * 1.5^(L+1) = B * 1.5^L * 1.5 = C * 1.5 = 1.5 * C

This simple calculation gives us an interesting result: each time a Metal Mine is upgraded, the cost for the next upgrade is increased by 50% of the old value (this also holds for a Deuterium Synthesizer, but the increase is actually 60% for a Crystal Mine). And this applies to both metal and crystal.

Let's put that aside and look at production for a moment.

The formula for a mine's production is trickier.
P = B * L * 1.1^L
Here, we see that all three types of mines use the same ratio (1.1). Again, L is the current level of the mine and B is the base amount produced (the amount produced by a level 1 mine; note that the value of B depends on a planet's average temperature for a Deuterium Synthesizer).
The key difference is that the production is multiplied by the current level of the mine.
Think about this. Mines take place: each upgrade takes up 1 field. In other words, mines use up a number of fields equal to their levels. So, what is the average production per field of a mine? Well, we already know the sum (the total production P), so we only need to divide by the number of fields (the current level L). We then find that the production-per-field Pf is
Pf = B * 1.1^L
Isn't that interesting? This means that both the upgrade cost of a mine and its production per field used create geometric sequences. This would allow us, for example, to calculate the total amount of resources required to upgrade any mine to any arbitrary level (and the increase in production that represents, but the game can already to that for us).

Trick question: should I improve a level 18 mine on my home planet or a level 15 mine on a colony?
To answer it, we will consider the upgrade cost not as a per-level cost, but as a per-additionnal-unit-of-resource-per-hour cost; put simply, we want to know which of these two upgrades gives the most production for the least resources. Of course, this will be dependent on the current level of the mine. This special cost is simply obtained by dividing the production increase by the upgrade cost. (We will replace the B in the cost equation by D in order to avoid ambiguity, and add some fancy colours that help us read it all.)

(B * (L+1) * 1.1^(L+1) - B * L * 1.1^L) / (D * R^L)
= ( (B * 1.1^L) / (D * R^L) ) * (1.1 * (L+1) - L)
= (B / D) * (1.1^L / R^L) * (1.1 * L + 1.1 - L)
= (B / D) * (1.1 / R)^L * (0.1 L + 1.1)

Okay, that looks like gibberish. We have a constant (read: independent of the current level) term, B/D, that we can safely ignore. What's left is (1.1/R)^L and (0.1 L + 1.1). We know that either R = 1.5 (Metal Mine or Deuterium Synthesizer) or R = 1.6 (Crystal Mine). This has the effect that 1.1/R is always smaller than 1. This is very important.
How? Because we then have that the limit of (1.1/R)^L is 0 (multiply a number between 0 and 1 with itself enough times and it will essentially become 0) [we use a limit here because we wish to calculate a long-term behaviour as the value of L increase]. However, the limit of (0.1 L + 1.1) is infinity (0.1 times infinity is still infinity, and adding 1.1 to that doesn't make any real difference). What happens when you have 0 * infinity?
First you scratch your head. Then you make some weird calculations that imply finding some derivatives. Finally, you find that (in this particular case), 0 * infinity = 0. This isn't always true, and depends greatly on the context. In our current situation, this means that (1.1/R)^L will decrease much faster than (0.1 L + 1.1), so that their product will decrease as L increase.

So what does this all means? Well, each time you increase the level of a mine, you spend a certain number of units of resources to increase the production of the mine by a certain number of units per hour. However, we just calculated that the production increase divided by the cost comes closer to zero as the level increase. In simpler terms, the higher a mine's level, the more expensive each unit-per-hour becomes when you improve that mine's level.

Thus, the increase in production obtained from upgrading a higher-level mine is less than the increase in production obtained from spending the same amount of resources to upgrade lower-level mines. In other words, investing in a lot of small mines gives you more resources than have a few big ones.

Next topic: Saving resources for big projects

OMath | Topic 0 | Introduction

Einswerz — Wed, 16 May 2012 01:01:47 GMT

Hi! I'm Einswerz, and welcome to OMath!

Now you're probably wondering what that is.
Basically, OMath aims to be a series of articles about different aspects of OGame being cut down and analyzed from a mathematical point of view. Why? Because it grants greater insight into the workings of the game, which in turn brings additional benefits when it comes down to strategies and skill.
I'm doing this because I believe it can help members of the alliance become better players, and because I like maths (unlike 99,9% of the population). Hopefully, it'll make other people take interest in maths too, but that's not my goal.

Table of content
Topic 0 - Introduction: Basic definitions and jargon used throughout the articles
Topic 1 - Mines and resources: A first look at resource production and mine upgrades
Topic 2 - Saving resources for big projects: How improving mines can make it possible to reach resources objectives faster.

Today's Topic : Introduction
Although I plan to write OMath articles in layman's terms, my aim is also to present my reasoning and the steps I use to get my results. As such, the first topic will be dedicated to presenting a brief (and down-to-earth/not-as-abstract-as-the-rest; I hope potential math gurus will pardon my gross simplifications) overview of mathematical notions I'll use in the other articles. (Note : all of these presume a basic understanding of arithmetic and elementary algebra.)

Functions: Functions are fundamental and analysis, the mathematical discipline I'll use the most in upcoming OMath articles. Basically, if f(x) is an equation involving x, then we call f a function. We then say that f associates to each x a value f(x) that is obtained by substituting x for its value in the equation. Okay, that was abstract, so here's an example :

f(x) = 3x + 4

That's defining a function. Easy as pie. We can then evaluate (that means choose a value for x) the function f :

f(0) = 3 * 0 + 4 = 0 + 4 = 4
f(10) = 3 * 10 + 4 = 30 + 4 = 34
f(-4/3) = 3 * (-4/3) + 4 = -4 + 4 = 0

Ok, we could continue like that, but there's an awful lot of real numbers so I'll assume you understood the basics. Of course, we could define more complex functions, such as g(x) = x^2 + 4x - 2 or h(x) = 3^x. (If you're wondering, the '^' character is a stand-in for exponentiation, so 2^4 = 2 * 2 * 2 * 2 = 16. Think of it as an arrow pointing upward, because you would write the second number like so : x² = x^2.)

Sequences: That one is easy. A sequence is basically a list of numbers taken in a particular order. Usually, each number in the sequence is given by a certain function f(x) by having x take the values 0, 1, 2, 3, etc. For instance, the sequence (1, 1/2, 1/3, 1/4, ...) is the sequences of 1/x. When discussing sequences, it is customary to write a(n) instead of f(x) to represent the function generating the sequence.
To make things even more easier, the only sequences used by OGame are geometric sequences, or sequences that looks like a(n) = q*(r^n) for a certain q and r.

Limit: Although the notion of limit is a very interesting one, we will only need a very particular case of its applications. Basically, whenever I mention the limit of a function f(x) or of a sequence a(n), I refer to the behaviour of the value of f(x) or of a(n) when x or n takes very large values (as in 1,000,000+). A classic example is the limit of the sequence a(n) = 1/n, which is equal to 0. This means that, as n takes larger and larger values, 1/n will take smaller and smaller values until it "reaches" 0 when n "reaches" infinity. (Yes, you calculate the limit by evaluating the function or the sequence at infinity even though infinity is not a number.)

Series: Series are almost as easy as sequences. A series is, simply put, the sum of some terms taken from a sequence. Again, the context of OGame makes things easier for us : the only series that will have to consider will be sums of consecutive terms in a geometric sequence (a(n) = qr^n).
Now, I'll omit the explanation of the following result (because the proof is actually tedious and boring), but the series obtained by taking the s first terms in the sequence qr^n has a sum of q * (1 - r^s) / (1 - r).
Thus, we can calculate the following example. Let's consider the sequence a(n) = 500*2^n = (500; 1,000; 2,000; 4,000; ...). Say we want the sum of terms ranging from n=3 to n=7. I'll write this kind of stuff like this :

sum (500*2^n) from 3 to 7 = [sum (500*2^n) from 0 to 7] - [sum (500*2^n) from 0 to 2] = 500 * (1 - 2^8) / (1 - 2) - 500 * (1 - 2^3) / (1 - 2) = 500 * (1 - 256) / (-1) - 500 * (1 - 8) / (-1) = 500 * 255 - 500 * 7 = 127,500 - 3,500 = 124,000

Since we can only calculate the sum when starting from 0 (which gives s+1 terms if the last is n=s), we find the sum of any arbitrary range by adding the missing terms and then subtracting them.

Derivatives: Calculus. Big stuff. I plan on touching a bit of calculus (to get derivatives of functions, no integrals or other scary stuff), but that's going into advanced math stuff that most people won't follow. To keep it simple, the idea of a derivative is to measure how "fast" a function grows around a certain x. For instance, the function f(x) = 2x always increases at the same rate, but the function g(x) = x^2 increases faster when x increases. An intuitive way of thinking it is to ask yourself the question : "If I'm at currently at x (so my function has the value f(x)), how far away (or how much do I need to add) is f(x+1)?". Keeping the g(x) = x^2 example, we see that going from x=1 to x-2 increases the value of g(x) from 1 to 4, or by +3; however, going from x=11 to x=12 increases the value of g(x) from 121 to 144, or by +23.

Alright, that should be the end of my introduction topic. I wish to apologize to those who think it to be too much maths; I promise the next one will be better (or not as bad, at least). I'm aware that some of the stuff might seem too abstract for some, but hopefully the next topic will be more interesting (having a real concrete approach should help in that regard). I'll probably post it soon not to keep you all waiting too much.

Next topic: Mines and resources.

Pandemonium Games

Sergen3 — Mon, 30 Apr 2012 21:59:33 GMT

My name is Adam T Self CEO of Pandemonium Games i am looking for testers for my proto type add me on facebook to test. The game is going to be a mix between magic the gathering, risk, chess, and skyrim please help me test my game thank you all. :)

Getting to know Ogame and TUN

Moeskitoe — Fri, 27 Apr 2012 02:17:49 GMT

Hi Gang,

I think it was about two months ago that I started ogame and joined TUN as soon as the tutorial told me I should pick an alliance. What a lucky pick! I like to ask a lot of questions and so far no has seemed to mind. I closed my account on hydra because I started getting farmed my second day on. I can't imagine I was worth farming at that stage. Anyhow, I've never been attacked in any of the universes where I'm a TUN member and I'm sure the TUN tag by my planets has a lot to do with that.

I'd like to propose that a primer or tip sheet be created for TUN members to help with some of the frequently asked questions. I know there are lots of resources out there for answering questions but the ones I have found have only been kinda useful. I play ogame because the concept is cool and you guys are cool. The game layout and mechanics and documentation are not very good I don't think and the advertising surrounding the resource material is really annoying. To me advertising in a library is just rude. Advertising in game is one thing but posing ads as if they were pure information is a bit of a sacrilege. I'll write the doc if you guys will help me with the facts, assuming others think it a good idea. Combat and tactics are just way over my head.

Regards,

Moe

Ogame

Aleslosh — Wed, 29 Feb 2012 17:32:04 GMT

It's hard to believe that I have been involved with this game for about a year now... Where has the time gone? What started with a friend saying "Hey, check out Ogame" has ended up being a hobby that consumes about 5 to 6 hours every week! I never thought of myself as a gamer, but I do enjoy building things. This game has given me the chance to do just that - and in a much bigger fashion than I ever imagined!

I now have the United Nations in 5 Universes (Barym, Draco, Fornax, Gemini and Hydra) with a sixth (Andromeda) UN soon to be added. This makes for approximately 100 persons under the UN banner. Of course, I can't manage all this myself - I am merely good at providing the tools to do so. Every Universe has a core group of players with a very high participation in the operation of the alliance. My thanks to each and every one of them!

What are my goals for this Multi-Universe Alliance? I would like to see us in the Top 10 in every category in every Universe. A little much? maybe... maybe not. We are already top ten in 2 universes - Gemini and Hydra. I'd also like to see a total of 500 persons be a part of the UN by the end of 2012. I think this is very doable!

With this in mind, I will probably be opening a UN in Cappella, Electra and Universe 7 (all US servers) by the end of June, 2012 unless something happens earlier to force my hand quicker. I am currently playing in these universes but am part of other alliances. I am in all Universes under the name of Aleslosh. Look me up!