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On April 26 2010 21:52 shoop wrote:Hm. I have some objections. Show nested quote +On April 26 2010 17:27 Arrian wrote:
Now, a very compelling pattern emerges here, one that looks like an exponential function.
Sorry, but this is nonsense. If the miners would not hold each other up, then the mean time between gas returns would obviously be inversely proportional to the number of miners; to be precise
Yes, and that is what the OP 2nd's graph clearly shows. I think you may be confused. e^x is given by:
As you can see, the exponential function's value is proportional to its previous values. Considering e^-x it is inversely proportional to its previous values. It is this property of prior dependence that makes it particularly good for modeling growth and decay.
Also see this to convince yourself.
#gas = r * #miners * #time, where the mining rate r = 0.75 gas/second
To calculate the time for a single return, substitute #gas = 4 to obtain #time = 4 / (r * #miners) = 5.33... / #miners. This imperfect model already fits the numbers quite well:
#miners | 5.33... / #miners
1 | 5.33
2 | 2.67
3 | 1.78
4 | 1.33
5 | 1.07
6 | 0.89
This is a linearization using the number of workers and the average mining rate. Of course it fits the data! Because the number of miners is always an integer you aren't interested in the data between points so there is a strong argument for a simpler model. For something more complex ( mineral collection) a more sophisticated model would be desirable. Regardless, one model doesn't invalidate the other, they should both generally confirm the underlying truth of the system.
Now obviously the miners do hold each other up, an effect that presumably gets worse when you increase the number of miners. Thus, in reality the mean time between gas returns will be larger than the amount of time predicted by the inverse proportional model. (This is exactly what happens: the predicted numbers are smaller than the measurements, especially for #miners equal to 5 or 6.)
The miners to not "hold each other up." What you are describing is a constraint on the number of workers able to collect gas at a given time. If only one worker can collect gas, then the other workers cannot collect gas so they must wait for their turn. When this is happening the gas you mine per second is increasing albeit at a decreasing rate until you reach the equilibrium point.
In contrast, in the proposed exponential model the mean time between returns drops ridiculously quickly as a function of the number of miners. For example, according to the exponential model the mean time between returns for 25 workers would be 0.00013; in other words you would collect 4/0.00013 = 31121 gas per second, while according to the inverse proportional model you would collect 25*r=18.75 gas per second, a much more reasonable figure.
The exponential model is an inversely proportional model. The end behavior of 1/x and e^-x is the same as x->oo. Besides, the modeling function is a solution over a specific domain, who cares what it does at infinity?
While an exponential model may give a reasonable fit if you just don't look at the graph beyond 6 miners, it is clearly a completely inappropriate model in this case, so you're just as well off just drawing any reasonable looking line through the data points.
I don't follow your logic.
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On April 27 2010 16:24 space_yes wrote:Show nested quote +On April 26 2010 21:52 shoop wrote:Hm. I have some objections. On April 26 2010 17:27 Arrian wrote:
Now, a very compelling pattern emerges here, one that looks like an exponential function.
Sorry, but this is nonsense. If the miners would not hold each other up, then the mean time between gas returns would obviously be inversely proportional to the number of miners; to be precise Yes, and that is what the OP 2nd's graph clearly shows. Also see this to convince yourself. I think you may be confused. e^x is given by: As you can see, the exponential function's value is proportional to its previous values. Considering e^-x it is inversely proportional to its previous values. It is this property of prior dependence that makes it particularly good for modeling growth and decay.
lol
What you've written here is nonsensical. There's no "previous values" to a real function, and the only function that's inversely proportional to it's "previous values" is 1 (given a reasonable definition of what this even means, e.g. f(x) = a/f(x-c), c>0, for all x).
If a miner does a trip in 5 seconds, then two miners do two trips in 5 seconds, and k miners do k trips in 5 seconds, so miners do k/5 trips a second. Sticking with k miners, Let's call this rate R. If k miners do R trips a second, then it takes 1/R seconds for a trip to be done. Notice how we took the inverse? This shows that they are inversely proportional. (We've just discovered the obvious concept that period is the inverse of frequency.) The relevant function here was 1/x, not e^-x.
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4(2) interests me greatly.
I believe it could be worked as a middle gear for gas collection. All you'd have to do is set 3 workers to one gas until your 2nd geyser is set up and then take one off that gas. Then just put him on the nearest mineral patch after he's returned his gas and take another worker off the other far edge mineral patch after it delivers its crystal and drop him in that gas to increase gas collection rate while not hurting your resource collection rate.
Then you can fill your gas out to 3 workers each as you see fit.
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On April 27 2010 17:40 DarkChrono wrote:Show nested quote +On April 27 2010 16:24 space_yes wrote:On April 26 2010 21:52 shoop wrote:Hm. I have some objections. On April 26 2010 17:27 Arrian wrote:
Now, a very compelling pattern emerges here, one that looks like an exponential function.
Sorry, but this is nonsense. If the miners would not hold each other up, then the mean time between gas returns would obviously be inversely proportional to the number of miners; to be precise Yes, and that is what the OP 2nd's graph clearly shows. Also see this to convince yourself. I think you may be confused. e^x is given by: As you can see, the exponential function's value is proportional to its previous values. Considering e^-x it is inversely proportional to its previous values. It is this property of prior dependence that makes it particularly good for modeling growth and decay. lol What you've written here is nonsensical. There's no "previous values" to a real function, and the only function that's inversely proportional to it's "previous values" is 1 (given a reasonable definition of what this even means, e.g. f(x) = a/f(x-c), c>0, for all x). If a miner does a trip in 5 seconds, then two miners do two trips in 5 seconds, and k miners do k trips in 5 seconds, so miners do k/5 trips a second. Sticking with k miners, Let's call this rate R. If k miners do R trips a second, then it takes 1/R seconds for a trip to be done. Notice how we took the inverse? This shows that they are inversely proportional. (We've just discovered the obvious concept that period is the inverse of frequency.) The relevant function here was 1/x, not e^-x.
I am trying to use easy to understand terms. Something inversely proportional is given by 1/x.The OP function is of the form be^(ax) + c = e^-x = 1/e^x. Look at the definition of the exponential function. You must not understand something. With respect to "previous values" I'm referring to the last value for x i.e. a Maclaurin series polynomial of degree 5 (or whatever we need for that interval of workers and accuracy). The exponential function is used to model over a specific domain so how is that not a real valued function?
Here is the Taylor series expansion for e^x with a = 0:
Look very hard at that series before you post claiming nonsense and wikipedia exponential function until you understand.
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On April 27 2010 17:40 DarkChrono wrote:
The relevant function here was 1/x, not e^-x.
obviously you've not taken calculus 
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I think he has the mean time confused with the rate of change of the mean time.
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On April 27 2010 18:14 hacpee wrote:
I think he has the mean time confused with the rate of change of the mean time.
lol
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On April 27 2010 18:17 space_yes wrote:Show nested quote +On April 27 2010 18:14 hacpee wrote:
I think he has the mean time confused with the rate of change of the mean time. lol
Just try to graph it yourself. You see that it is in fact 1/x. Do a thought experiment.
Lets say the interval to return gas with 1 worker is 1. The interval to return gas with two workers is 1/2. The interval to return gas with 3 workers is 1/3. The interval to return gas with 4 workers is 1/4. You see the pattern? 1/x is f(x) and 1,2,3,4,5,6,7,8 is x. The numbers match.
Usually, you get an exponential or inverse exponential when dy/dt=-y. I'm just not seeing which variable, as you increase the rate of change of it, will decrease the variable. Then again, its been 3 months since I did modeling with differential equations so I might not be the best person to see the relationship.
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All this math hurts my head. But I would like to object to the blub thing in that this well written article containing good information does not actually talk about gas USAGE at all.
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On April 27 2010 18:24 hacpee wrote:Show nested quote +On April 27 2010 18:17 space_yes wrote:On April 27 2010 18:14 hacpee wrote:
I think he has the mean time confused with the rate of change of the mean time. lol Just try to graph it yourself. You see that it is in fact 1/x. Do a thought experiment. Lets say the interval to return gas with 1 worker is 1. The interval to return gas with two workers is 1/2. The interval to return gas with 3 workers is 1/3. The interval to return gas with 4 workers is 1/4. You see the pattern? 1/x is f(x) and 1,2,3,4,5,6,7,8 is x. The numbers match.
Consider the Taylor series where a = 0 of degree 1.
If you do not know what this is do not respond to my post.
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On April 27 2010 18:36 space_yes wrote:Show nested quote +On April 27 2010 18:24 hacpee wrote:On April 27 2010 18:17 space_yes wrote:On April 27 2010 18:14 hacpee wrote:
I think he has the mean time confused with the rate of change of the mean time. lol Just try to graph it yourself. You see that it is in fact 1/x. Do a thought experiment. Lets say the interval to return gas with 1 worker is 1. The interval to return gas with two workers is 1/2. The interval to return gas with 3 workers is 1/3. The interval to return gas with 4 workers is 1/4. You see the pattern? 1/x is f(x) and 1,2,3,4,5,6,7,8 is x. The numbers match. Consider the Taylor series where a = 0 of degree 1. If you do not know what this is do not respond to my post.
a=0 of degree 1? What notation are you using?
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On April 27 2010 18:40 hacpee wrote:Show nested quote +On April 27 2010 18:36 space_yes wrote:On April 27 2010 18:24 hacpee wrote:On April 27 2010 18:17 space_yes wrote:On April 27 2010 18:14 hacpee wrote:
I think he has the mean time confused with the rate of change of the mean time. lol Just try to graph it yourself. You see that it is in fact 1/x. Do a thought experiment. Lets say the interval to return gas with 1 worker is 1. The interval to return gas with two workers is 1/2. The interval to return gas with 3 workers is 1/3. The interval to return gas with 4 workers is 1/4. You see the pattern? 1/x is f(x) and 1,2,3,4,5,6,7,8 is x. The numbers match. Consider the Taylor series where a = 0 of degree 1. If you do not know what this is do not respond to my post. a=0 of degree 1? What notation are you using?
You have not had calculus. There is only one parameter for a Taylor series and the degree is the number of times you take a linear approximation. Open a calculus textbook or use wikipedia.
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Look, to be clear, I'm not trying to talk over anyone here I'm merely responding to someone claiming I was spouting nonsense and unfortunately its hard to argue with someone who doesn't know what you're talking about. I hope you can empathize with that.
I tried to make my original post as clear to the non-math person as possible. It is not my intent to come across as a math genius here but I want to be clear that both 1/x and 1/e^x = e^-x (over a specific domain) are inversely proportional.
FYI: The Taylor series for e^-x where a = 0 is 1 + 1/x . There's a 1 but hey, we always fix the constant
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On April 27 2010 18:52 space_yes wrote:
Look, to be clear, I'm not trying to talk over anyone here I'm merely responding to someone claiming I was spouting nonsense and unfortunately its hard to argue with someone who doesn't know what you're talking about. I hope you can empathize with that.
I tried to make my original post as clear to the non-math person as possible. It is not my intent to come across as a math genius here but I want to be clear that both 1/x and 1/e^x = e^-x (over a specific domain) are inversely proportional.
Ok imagine this. If you have one worker, you return gas at one second. If you have two workers, you return gas in .5 seconds. If you have 3 workers, you return gas in .333 seconds. Does that make sense? Now plot y=1/x in matlab and plot the points (1,1), (2,.5), (3,.3333), (4,.25), etc.
Remember what we're plotting. We're plotting time between return of gas vs number of workers.
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On April 27 2010 18:59 hacpee wrote:Show nested quote +On April 27 2010 18:52 space_yes wrote:
Look, to be clear, I'm not trying to talk over anyone here I'm merely responding to someone claiming I was spouting nonsense and unfortunately its hard to argue with someone who doesn't know what you're talking about. I hope you can empathize with that.
I tried to make my original post as clear to the non-math person as possible. It is not my intent to come across as a math genius here but I want to be clear that both 1/x and 1/e^x = e^-x (over a specific domain) are inversely proportional. Ok imagine this. If you have one worker, you return gas at one second. If you have two workers, you return gas in .5 seconds. If you have 3 workers, you return gas in .333 seconds. Does that make sense? Now plot y=1/x in matlab and plot the points (1,1), (2,.5), (3,.3333), (4,.25), etc. Remember what we're plotting. We're plotting time between return of gas vs number of workers.
See my edit.
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The OP's 2nd graph was time to collect x number workers. You can use an exponential fit for that. I'm not saying you should, but I am saying you can.
Regardless, as I stated previously, you're only dealing with integer values for workers so you don't need the time for gas collection between worker values i.e. 1.5 workers so using an exponential model on such a small domain isn't necessary which others have already point out in their own words.
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On April 27 2010 19:00 space_yes wrote:Show nested quote +On April 27 2010 18:59 hacpee wrote:On April 27 2010 18:52 space_yes wrote:
Look, to be clear, I'm not trying to talk over anyone here I'm merely responding to someone claiming I was spouting nonsense and unfortunately its hard to argue with someone who doesn't know what you're talking about. I hope you can empathize with that.
I tried to make my original post as clear to the non-math person as possible. It is not my intent to come across as a math genius here but I want to be clear that both 1/x and 1/e^x = e^-x (over a specific domain) are inversely proportional. Ok imagine this. If you have one worker, you return gas at one second. If you have two workers, you return gas in .5 seconds. If you have 3 workers, you return gas in .333 seconds. Does that make sense? Now plot y=1/x in matlab and plot the points (1,1), (2,.5), (3,.3333), (4,.25), etc. Remember what we're plotting. We're plotting time between return of gas vs number of workers. See my edit.
1/x! from 2-infinity is 1/2, 1/6,1/12, etc. Nothing to do with 1/2, 1/3,1/4,1/5 etc.
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On April 27 2010 19:03 space_yes wrote:
The OP's 2nd graph was time to collect x number workers. You can use an exponential fit for that. I'm not saying you should, but I am saying you can.
Regardless, as I stated previously, you're only dealing with integer values for workers so you don't need the time for gas collection between worker values i.e. 1.5 workers so using an exponential model on such a small domain isn't necessary which others have already point out in their own words.
Try to model it. Thats just my advice.
y=e^-x is what you're saying the function is. What is y? What is x? How does that relate to dy/dx=-y(because that is the differential equation you need to solve for y=e^-x). Does it make sense?
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On April 27 2010 19:05 hacpee wrote:Show nested quote +On April 27 2010 19:00 space_yes wrote:On April 27 2010 18:59 hacpee wrote:On April 27 2010 18:52 space_yes wrote:
Look, to be clear, I'm not trying to talk over anyone here I'm merely responding to someone claiming I was spouting nonsense and unfortunately its hard to argue with someone who doesn't know what you're talking about. I hope you can empathize with that.
I tried to make my original post as clear to the non-math person as possible. It is not my intent to come across as a math genius here but I want to be clear that both 1/x and 1/e^x = e^-x (over a specific domain) are inversely proportional. Ok imagine this. If you have one worker, you return gas at one second. If you have two workers, you return gas in .5 seconds. If you have 3 workers, you return gas in .333 seconds. Does that make sense? Now plot y=1/x in matlab and plot the points (1,1), (2,.5), (3,.3333), (4,.25), etc. Remember what we're plotting. We're plotting time between return of gas vs number of workers. See my edit. 1/x! from 2-infinity is 1/2, 1/6,1/12, etc. Nothing to do with 1/2, 1/3,1/4,1/5 etc.
The series expansion is not a factorial. That was my exclamation point. I fixed the comment to make this more obvious.
For clarity the series is:
1 + 1/x
Go to the wikipedia page on the exponential functions and read it. Then Google how to use an exponential function to model growth or decay. If you haven't had calculus so you don't understand what a Taylor series then I can't help you 
My question to you: is 1/x^2 inversely proportional?
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On April 27 2010 19:08 hacpee wrote:Show nested quote +On April 27 2010 19:03 space_yes wrote:
The OP's 2nd graph was time to collect x number workers. You can use an exponential fit for that. I'm not saying you should, but I am saying you can.
Regardless, as I stated previously, you're only dealing with integer values for workers so you don't need the time for gas collection between worker values i.e. 1.5 workers so using an exponential model on such a small domain isn't necessary which others have already point out in their own words. Try to model it. Thats just my advice. y=e^-x is what you're saying the function is. What is y? What is x? How does that relate to dy/dx=-y(because that is the differential equation you need to solve for y=e^-x). Does it make sense?
![[image loading]](http://www.teamliquid.net/staff/Arrian/gasexpfunction.png)
y = interval between returns
x = number of workers
Look at the fitting function at the bottom of the graph it is of the form y = be^(ax). I'm not going to try and model it. Someone already did
Have you convinced yourself that y = f(x) = 1/x^2 is inversely proportional yet? Consider:
u = x^2 -> y = f(u) = 1/u
Now consider 1/e^x over a specific domain...
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