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Jim Hsu
Deep Core Mining Inc.
Posted - 2005.12.03 01:52:00 - [1]
 

Edited by: Jim Hsu on 09/12/2005 18:39:56
Edited by: Jim Hsu on 09/12/2005 18:39:27
Edited by: Jim Hsu on 07/12/2005 18:30:30
Edited by: Jim Hsu on 07/12/2005 00:11:57
Edited by: Jim Hsu on 06/12/2005 21:54:30
Edited by: Jim Hsu on 04/12/2005 00:42:26
Edited by: Jim Hsu on 03/12/2005 01:52:32

UPDATE: Mr.M has his calculator on. It's superior to mine. check www.evegeek.com.

UPDATE: Check this.

http://www.eve-files.com/media/12/stacking-real.zip

------------------------------------

I'm currently looking at the stacking equation for Sisi. I have made considerable progress, but there are errors.

I will use this guy's data, for instance. Blame him if the data's wrong.

Originally by: "Harry Voyager"

Raw data for Sisi Module stacking penalty

Fitting: Tempest with D650
Base RoF: 2.87 sec
Base Damage Mod: 2.8875x

Fitting: Gyro I's: DMx1.07, RoFx0.925

1x
RoF: 2.66
DM: 3.089265

2x
RoF: 2.48
DM: 3.27759283745

3x
RoF: 2.38
DM: 3.40850258315

4x
RoF: 2.33
DM: 3.47601431929

5x
RoF: 2.31
DM: 3.50180455707

6x
RoF: 2.30
DM: 3.50915618132



I took the damage mod into account.

Let f(x) equal a function that expresses the relative amount that performance is degraded on a single mod considering the number of modules stacked. For example, an f(2)=0.85 means that the 2nd mod has a modifier of 85% of the mod itself (If the modifier was 1.07x, it would be (1.07-1)*0.85 + 1 = 1.0595x.) f(1)=1 as we all know.

I postulated that the total modded value is equal to value*(mod*f(1))*(mod*f(2))*(mod*f(3))...(mod*f(n)). Where n is the number of modules fitted.

Doing this, and using the above data, I obtained the following values:

f(1) = 1
f(2) = 0.8708860
f(3) = 0.5705831
f(4) = 0.2829552
f(5) = 0.1059927
f(6) = 0.0299912

I obtained the values first by figuring out the damage mod for the additional mod (by dividing the total value for (N+1) mods by the total value for (N) mods [e.g. 3.408502583/3.277592837). That gives the amount that the (N+1)th mod will change it by. Then, I divided that value by the modifier (in this case, 1.07x), and got a decimal from 0 to 1, posted above.

Doing this, I saw a definite gaussian/logistic pattern to the points. I tried fitting it, and obtained:

1.08115690
--------------------------------------------- + 0.0015763657
(1 + 0.0231600863 * E^(1.207799629x)

This is what f(x) is supposed to be equal to.

Obviously, this is weird and messy. The R value is 99.96%, which indicates that this is a very close, but not exact match.

Now, questions.

1) Why is it not exact, but very, very close?
2) Are there any other values being modified? Do I have to multiply the mod by a constant, for example?
3) Does this apply for other examples?

I think I'll rest for now.

Jim Hsu
Deep Core Mining Inc.
Posted - 2005.12.03 02:21:00 - [2]
 

Edited by: Jim Hsu on 03/12/2005 02:21:26
This is what the graph of number of mods vs. f(x) looks like.

http://www.eve-files.com/media/12/test.GIF

Without getting technical, basically this is saying that you get quite severe penalties for fitting after 3 mods. Technically, the inflection point of the graph is at x=2.928, and that the derivative of this function looks like a classic bell curve.

Altrex Stoppel
Sniggerdly
Pandemic Legion
Posted - 2005.12.03 03:30:00 - [3]
 

Hmm wow that drop is quite drastic for stacking...should really change the game good job CCP

hylleX
Herrow Inc
Posted - 2005.12.03 04:08:00 - [4]
 

bleh ugh

Vishnej
Demonic Retribution
Pure.
Posted - 2005.12.03 04:29:00 - [5]
 

A different way of looking at it:

Right now, the formula is (mod1*mod2*mod3...*modn)^((1/n)^0.25)
Or, to shorten it, mod1*mod2*mod3..*modn ^ X
I'll remove the mod1*mod2 etc and just focus on X.
Currently, X = ((1/n)^0.25)
The new stacking penalty is not of the exact same form with altered constants, that much I know from playing around with them.

But the sum total of the X, the second part of the equation, whatever has taken its place, in this case ends up being:
1 mod: 0.998218924----0.975609756 (actual value 1, indicates rounding error somewhat)
2 mods: 0.934419309----0.938986964
3 mods: 0.812161159----0.806621411
4 mods: 0.678084198----0.677019285
5 mods: 0.561874882----0.566454012
6 mods: 0.471874706----0.483039455

These numbers are: ((sample/basestat)^(1/n)-1)/(basebonus-1)
The first figures are for Dmod, the second for RoF - the second have more rounding and are therefore less accurate.

So the values above are for X: if you want to compute what 6 of the same mods gives you, take (modifier^6)^(0.471874706)

Next step, if anyone wants to take it: Take new tests with varying quality modules (a tech 2 mod and a tech 1, for example), and figure out a way to compute whether the exponent X is the same as here.

Jim Hsu
Deep Core Mining Inc.
Posted - 2005.12.03 04:36:00 - [6]
 

Edited by: Jim Hsu on 03/12/2005 04:36:33
The following is data from shield hardening a scorp with kinetic hardeners.

Hardeners used: Tech 1 (50% active, 1% passive)
Base kin resist: 40%

http://www.eve-files.com/media/12/8hardener-sisi.zip

To sum it up, here are my results, based on number of hardeners activated:

Calculated----- (actual)------- deviation
0= 41.7036% 41.70161420% -0.00198935%
1= 70.7023% 70.55637081% -0.14597168%
2= 83.1570% 83.20542905% 0.04839982%
3= 87.9755% 87.92789721% -0.04760049%
4= 89.6213% 89.60641989% -0.01491866%
5= 90.1430% 90.14679777% 0.00377118%
6= 90.3065% 90.29164064% -0.01489737%
7= 90.3281% 90.32213645% -0.00596034%
8= 90.3496% 90.32704425% -0.02256338%

Jim Hsu
Deep Core Mining Inc.
Posted - 2005.12.03 04:42:00 - [7]
 

Originally by: Vishnej
A different way of looking at it:

Right now, the formula is (mod1*mod2*mod3...*modn)^((1/n)^0.25)
Or, to shorten it, mod1*mod2*mod3..*modn ^ X
I'll remove the mod1*mod2 etc and just focus on X.
Currently, X = ((1/n)^0.25)
The new stacking penalty is not of the exact same form with altered constants, that much I know from playing around with them.

But the sum total of the X, the second part of the equation, whatever has taken its place, in this case ends up being:
1 mod: 0.998218924----0.975609756 (actual value 1, indicates rounding error somewhat)
2 mods: 0.934419309----0.938986964
3 mods: 0.812161159----0.806621411
4 mods: 0.678084198----0.677019285
5 mods: 0.561874882----0.566454012
6 mods: 0.471874706----0.483039455

These numbers are: ((sample/basestat)^(1/n)-1)/(basebonus-1)
The first figures are for Dmod, the second for RoF - the second have more rounding and are therefore less accurate.

So the values above are for X: if you want to compute what 6 of the same mods gives you, take (modifier^6)^(0.471874706)

Next step, if anyone wants to take it: Take new tests with varying quality modules (a tech 2 mod and a tech 1, for example), and figure out a way to compute whether the exponent X is the same as here.


Interesting, that also looks strikingly similar to a logistic graph, and it also "almost fits but not quite". Perhaps there is another factor involved? Perhaps it is simply client error?

Jim Hsu
Deep Core Mining Inc.
Posted - 2005.12.03 05:01:00 - [8]
 

Potential problem spotted.

In my formula, switching the order of stacking around (with varying attribute mods) changes the resist. This should not happen.

Now what?

Jim Hsu
Deep Core Mining Inc.
Posted - 2005.12.03 05:10:00 - [9]
 

I shall now answer my own question. :)

Dmg mods stack with the largest one first (e.g. for a setup with 2x 25% invuln fields and one 50% hardener), the 50% hardener is applied first, then the invuln fields. EVEN IF the invln fields are activated first.

So, no there is no "activation order".

Vishnej
Demonic Retribution
Pure.
Posted - 2005.12.03 05:37:00 - [10]
 

Edited by: Vishnej on 03/12/2005 05:41:36
Edited by: Vishnej on 03/12/2005 05:38:10
False.

There is no order like it highest% to lowest or first put on the ship.

Here's the stacking equation right now:

TotalModifier = (Modifier1*modifier2*modifier3...*modifierN) ^ ((1/n)^0.25)

You'll note there's no need to put them in order.

I think it's likely that the first part of the equation (mod1*mod2...) holds true, and also likely that the second part of the equation is a power of that. I've just noted that whatever is used to create that exponent, X, is purely a function of # of mods, rather than being based on modifier - the Dmod and RoF figures match up closely enough to explain the difference with rounding errors.

Most likely, find me that function X(n) to a reasonable degree of accuracy, and we'll have the stacking equation - all that will be left will be matching it up to something made with simple constants.


Vishnej
Demonic Retribution
Pure.
Posted - 2005.12.03 05:43:00 - [11]
 

And yes, there is "client error" involved - the client rounds the numbers, or simply cuts off the rest of the digits.

ELECTR0FREAK
Eye of God
Posted - 2005.12.03 05:51:00 - [12]
 

Gah, I went through all this with missiles... good luck folks.

xOm3gAx
Caldari
Stain of Mind
Posted - 2005.12.03 06:12:00 - [13]
 

Originally by: ELECTR0FREAK
Gah, I went through all this with missiles... good luck folks.


i remember =( was truly horrible (great job on your part though)

Jim Hsu
Deep Core Mining Inc.
Posted - 2005.12.03 06:17:00 - [14]
 

Edited by: Jim Hsu on 03/12/2005 06:18:50
The following equation for X(n) has an R^2 value of 0.9996. About the best I can do.

X(n) = 0.695384707 / (1 + E^(-2.89517392+0.795065316*n) ) + 0.381401531

Technical data:

Originally by: NLREG

Test
Number of observations = 6
Maximum allowed number of iterations = 500
Convergence tolerance factor = 1.000000E-010
Stopped due to: Relative function convergence.
Number of iterations performed = 21
Final sum of squared deviations = 8.1591434E-005
Final sum of deviations = -2.9228286E-012
Standard error of estimate = 0.00638715
Average deviation = 0.00339333
Maximum deviation for any observation = 0.00597077
Proportion of variance explained (R^2) = 0.9996 (99.96%)
Adjusted coefficient of multiple determination (Ra^2) = 0.9991 (99.91%)
Durbin-Watson test for autocorrelation = 3.028
Analysis completed 3-Dec-2005 00:12. Runtime = 0.01 seconds.


---- Descriptive Statistics for Variables ----

Variable Minimum value Maximum value Mean value Standard dev.
------------------ -------------- -------------- -------------- --------------
Mods 1 6 3.5 1.870829
X 0.4718747 0.9982189 0.7427722 0.20837


---- Calculated Parameter Values ----

Parameter Initial guess Final estimate Standard error t Prob(t)
------------------ ------------- ---------------- -------------- --------- -------
K 0.7 0.695384707 0.04850193 14.34 0.00483
a 1 -2.89517392 0.2982305 -9.71 0.01045
b 0.8 0.795065316 0.08336475 9.54 0.01082
c 0.1 0.381401531 0.02777393 13.73 0.00526


---- Analysis of Variance ----

Source DF Sum of Squares Mean Square F value Prob(F)
---------- ---- -------------- -------------- --------- -------
Regression 3 0.2170087 0.07233624 1773.13 0.00056
Error 2 8.159143E-005 4.079572E-005
Total 5 0.2170903



Could you clean this up? :)

PS Thanks a lot for this alternate view. It helps to simplify the excel programming sufficiently. Now ... how do hardeners work...

Jim Hsu
Deep Core Mining Inc.
Posted - 2005.12.03 06:38:00 - [15]
 

Originally by: Vishnej
A different way of looking at it:

Right now, the formula is (mod1*mod2*mod3...*modn)^((1/n)^0.25)
Or, to shorten it, mod1*mod2*mod3..*modn ^ X
I'll remove the mod1*mod2 etc and just focus on X.
Currently, X = ((1/n)^0.25)
The new stacking penalty is not of the exact same form with altered constants, that much I know from playing around with them.

But the sum total of the X, the second part of the equation, whatever has taken its place, in this case ends up being:
1 mod: 0.998218924----0.975609756 (actual value 1, indicates rounding error somewhat)
2 mods: 0.934419309----0.938986964
3 mods: 0.812161159----0.806621411
4 mods: 0.678084198----0.677019285
5 mods: 0.561874882----0.566454012
6 mods: 0.471874706----0.483039455

These numbers are: ((sample/basestat)^(1/n)-1)/(basebonus-1)
The first figures are for Dmod, the second for RoF - the second have more rounding and are therefore less accurate.

So the values above are for X: if you want to compute what 6 of the same mods gives you, take (modifier^6)^(0.471874706)

Next step, if anyone wants to take it: Take new tests with varying quality modules (a tech 2 mod and a tech 1, for example), and figure out a way to compute whether the exponent X is the same as here.


Um.. are you sure that is correct. Look at this:

Assume base: 100, modifier = 1.07x

New Value Change
Fitting 1 mod 107.006728
Fitting 2 mods 113.3868397 5.96%
Fitting 3 mods 118.0105419 4.08%
Fitting 4 mods 120.2009964 1.86%
Fitting 5 mods 120.7623671 0.47%
Fitting 6 mods 121.210488 0.37%
Fitting 7 mods 122.3799773 0.96%
Fitting 8 mods 124.3401723 1.60%

Why does it start increasing FASTER when we go down to 7 or 8? I think there is something wrong.

Vishnej
Demonic Retribution
Pure.
Posted - 2005.12.03 07:18:00 - [16]
 

It should end in a horizontal asymptote, at a certain % benefit... IE, on the 10th mod, you have to use an absurd number of decimal places to show its benefit over having 9 mods on.


Vishnej
Demonic Retribution
Pure.
Posted - 2005.12.03 07:20:00 - [17]
 

Edited by: Vishnej on 03/12/2005 07:21:26
Hardeners work with the same formula, however, you have to think a bit differently. Think in terms of a "vulnerability" stat rather than a "resistance" stat. A 20% resistance is an 80% vulnerability.

So right now, three 20% hardeners =
(0.8*0.8*0.8) ^ ((1/3)^0.25)

BTW, when I asked you to find the formula, I was implying that I didn't know it it was a logarithm, hyperbola, exponent, what.
Tweaking either constant that's there now (1 or 0.25) does not produce the right shaped result, so I presume it's in a different form.

Jim Hsu
Deep Core Mining Inc.
Posted - 2005.12.03 18:45:00 - [18]
 

Edited by: Jim Hsu on 03/12/2005 18:46:54
OMG! R = 1.0000!

The formula is ABSURDLY complicated though. Behold:

Definitions:

Let parameters p1-p5 be the following:

P1 = asymptotic minimum
P2 = asymptotic maximum
P3 = slope parameter
P4 = value at inflection point
p5 = slope parameter

And let the following hold:

c = 2*p3*p5/abs(p3+p5);
f = 1/(1+exp(-c*(p4-n)));
g = exp(p3*(p4-n));
h = exp(p5*(p4-n));
Function modvalue = p1+(p2/(1+f*g+(1-f)*h));

Where p1-p5 are the above, n is the number of mods used, and modvalue is the relative performance increase (from the base value).

I let p5 be a constant (1.07). And used the data:

0,1
1,1.069875325
2,1.135097087
3,1.180433795
4,1.203814483
5,1.212746167
6,1.215292184

(caculated by dividing the modded value by the base value *2.8875))

And obtained the following:

Originally by: nlreg

---- Final Results ----

NLREG version 6.3
Copyright (c) 1992-2005 Phillip H. Sherrod.

Mod value vs. number
Number of observations = 7
Maximum allowed number of iterations = 500
Convergence tolerance factor = 1.000000E-010
Stopped due to: Both parameter and relative function convergence.
Number of iterations performed = 6
Final sum of squared deviations = 1.7340339E-006
Final sum of deviations = -6.6613381E-016
Standard error of estimate = 0.000760271
Average deviation = 0.000440172
Maximum deviation for any observation = 0.000729685
Proportion of variance explained (R^2) = 1.0000 (100.00%)
Adjusted coefficient of multiple determination (Ra^2) = 0.9999 (99.99%)
Durbin-Watson test for autocorrelation = 2.169
Analysis completed 3-Dec-2005 12:42. Runtime = 0.01 seconds.


---- Descriptive Statistics for Variables ----

Variable Minimum value Maximum value Mean value Standard dev.
------------------ -------------- -------------- -------------- --------------
n 0 6 3 2.160247
modvalue 1 1.215292 1.145323 0.08267536


---- Calculated Parameter Values ----

Parameter Initial guess Final estimate Standard error t Prob(t)
------------------ ------------- ---------------- -------------- --------- -------
p1 0 0.848843439 0.007575417 112.05 0.00001
p2 1 0.369053789 0.008004306 46.11 0.00002
p3 0.5 0.447261067 0.02644867 16.91 0.00045
p4 0.5 0.495751353 0.05486836 9.04 0.00286


---- Analysis of Variance ----

Source DF Sum of Squares Mean Square F value Prob(F)
---------- ---- -------------- -------------- --------- -------
Regression 3 0.04100955 0.01366985 23649.80 0.00001
Error 3 1.734034E-006 5.780113E-007
Total 6 0.04101129



Observe that p5 is not in there (I made it a constant).

Basically, the function looks like a modified logistic / asymtopic. Observe that there are none of the weird problems with 7-8 mods increasing anymore

Updates coming...

Jim Hsu
Deep Core Mining Inc.
Posted - 2005.12.03 18:49:00 - [19]
 

For reference, if anyone wishes to verify the results, download NLREG at www.nlreg.com and use the following script:

Quote:

Title "Mod value vs. number";
Variable n,modvalue;
//Parameter K = 1;
//Parameter a = -2;
//Parameter b = 1;
//Parameter c = 0.8;

//Function modvalue = ( a + b * exp(-c*n+K) );

Parameters p1=0, p2=1, p3=0.5, p4=0.5, p5=0.3;
Double c, f, g, h;
c = 2*p3*1.07/abs(p3+1.07);
f = 1/(1+exp(-c*(p4-n)));
g = exp(p3*(p4-n));
h = exp(1.07*(p4-n));
Function modvalue = p1+(p2/(1+f*g+(1-f)*h));

Plot Domain = 0,10, ylabel = "Efficiency";
Data;
0,1
1,1.069875325
2,1.135097087
3,1.180433795
4,1.203814483
5,1.212746167
6,1.215292184



My impression of the program is that it's pretty powerful, but the user interface is somewhat lacking.

Grimpak
Gallente
Midnight Elites
Echelon Rising
Posted - 2005.12.03 19:23:00 - [20]
 

Originally by: Jim Hsu
Edited by: Jim Hsu on 03/12/2005 18:46:54
OMG! R = 1.0000!

The formula is ABSURDLY complicated though. Behold:

Definitions:

Let parameters p1-p5 be the following:

P1 = asymptotic minimum
P2 = asymptotic maximum
P3 = slope parameter
P4 = value at inflection point
p5 = slope parameter

And let the following hold:

c = 2*p3*p5/abs(p3+p5);
f = 1/(1+exp(-c*(p4-n)));
g = exp(p3*(p4-n));
h = exp(p5*(p4-n));
Function modvalue = p1+(p2/(1+f*g+(1-f)*h));

Where p1-p5 are the above, n is the number of mods used, and modvalue is the relative performance increase (from the base value).

I let p5 be a constant (1.07). And used the data:

0,1
1,1.069875325
2,1.135097087
3,1.180433795
4,1.203814483
5,1.212746167
6,1.215292184

(caculated by dividing the modded value by the base value *2.8875))

And obtained the following:

Originally by: nlreg

---- Final Results ----

NLREG version 6.3
Copyright (c) 1992-2005 Phillip H. Sherrod.

Mod value vs. number
Number of observations = 7
Maximum allowed number of iterations = 500
Convergence tolerance factor = 1.000000E-010
Stopped due to: Both parameter and relative function convergence.
Number of iterations performed = 6
Final sum of squared deviations = 1.7340339E-006
Final sum of deviations = -6.6613381E-016
Standard error of estimate = 0.000760271
Average deviation = 0.000440172
Maximum deviation for any observation = 0.000729685
Proportion of variance explained (R^2) = 1.0000 (100.00%)
Adjusted coefficient of multiple determination (Ra^2) = 0.9999 (99.99%)
Durbin-Watson test for autocorrelation = 2.169
Analysis completed 3-Dec-2005 12:42. Runtime = 0.01 seconds.


---- Descriptive Statistics for Variables ----

Variable Minimum value Maximum value Mean value Standard dev.
------------------ -------------- -------------- -------------- --------------
n 0 6 3 2.160247
modvalue 1 1.215292 1.145323 0.08267536


---- Calculated Parameter Values ----

Parameter Initial guess Final estimate Standard error t Prob(t)
------------------ ------------- ---------------- -------------- --------- -------
p1 0 0.848843439 0.007575417 112.05 0.00001
p2 1 0.369053789 0.008004306 46.11 0.00002
p3 0.5 0.447261067 0.02644867 16.91 0.00045
p4 0.5 0.495751353 0.05486836 9.04 0.00286


---- Analysis of Variance ----

Source DF Sum of Squares Mean Square F value Prob(F)
---------- ---- -------------- -------------- --------- -------
Regression 3 0.04100955 0.01366985 23649.80 0.00001
Error 3 1.734034E-006 5.780113E-007
Total 6 0.04101129



Observe that p5 is not in there (I made it a constant).

Basically, the function looks like a modified logistic / asymtopic. Observe that there are none of the weird problems with 7-8 mods increasing anymore

Updates coming...



I just had a brain fart by looking at the first 5 lines.


most defenitely, numbers are not my thingEmbarassed

Jim Hsu
Deep Core Mining Inc.
Posted - 2005.12.03 20:44:00 - [21]
 

Edited by: Jim Hsu on 03/12/2005 20:47:06
The formula works perfectly when all the mods are the same, but what if they are different?

Um.. let multipler = (modvalue1*modvalue2 ... modvaluen) ^ (1/n). Will that work? In effect, if the modvalues are the same, this will be (modvalue^n)^(1/n), which is what we wanted. If they are different, it will take an "average" of the values.

I will try to verify this in game, if it works or not.

Jim Hsu
Deep Core Mining Inc.
Posted - 2005.12.03 20:55:00 - [22]
 

Alternatively, I can try to do some multiple regression (ick) with the data if it really doesn't work.

E.g.

0,1.07,1
1,1.07,1.069875325
2,1.07,1.135097087
3,1.07,1.180433795
4,1.07,1.203814483
5,1.07,1.212746167
6,1.07,1.215292184
0,1.085,1
1,1.085,?
2,1.085,?
3,1.085,?
4,1.085,?
5,1.085,?
6,1.085,?
0,1.035,1
1,1.035,?
2,1.035,?
3,1.035,?
4,1.035,?
5,1.035,?
6,1.035,?


Jim Hsu
Deep Core Mining Inc.
Posted - 2005.12.03 22:57:00 - [23]
 

Edited by: Jim Hsu on 03/12/2005 23:25:59
Edited by: Jim Hsu on 03/12/2005 23:17:52
Upon further consideration, the previus formula was WAY too complicated. The following is a logistic function of four parameters:

Originally by: nlreg

NLREG version 6.3
Copyright (c) 1992-2005 Phillip H. Sherrod.

Mod value vs. number
Number of observations = 7
Maximum allowed number of iterations = 500
Convergence tolerance factor = 1.000000E-010
Stopped due to: Relative function convergence.
Number of iterations performed = 7
Final sum of squared deviations = 6.1563129E-006
Final sum of deviations = -2.9433123E-011
Standard error of estimate = 0.00143252
Average deviation = 0.000798255
Maximum deviation for any observation = 0.00134958
Proportion of variance explained (R^2) = 0.9998 (99.98%)
Adjusted coefficient of multiple determination (Ra^2) = 0.9997 (99.97%)
Durbin-Watson test for autocorrelation = 2.202
Analysis completed 3-Dec-2005 16:52. Runtime = 0.00 seconds.


---- Descriptive Statistics for Variables ----

Variable Minimum value Maximum value Mean value Standard dev.
------------------ -------------- -------------- -------------- --------------
n 0 6 3 2.160247
modvalue 1 1.215292 1.145323 0.08267536


---- Calculated Parameter Values ----

Parameter Initial guess Final estimate Standard error t Prob(t)
------------------ ------------- ---------------- -------------- --------- -------
K 1 1.2194265 0.001462092 834.03 0.00001
a -2 -0.334306464 0.01655995 -20.19 0.00027
b 1 0.964036977 0.03971412 24.27 0.00015
c 0.8 0.667902481 0.1019214 6.55 0.00723


---- Analysis of Variance ----

Source DF Sum of Squares Mean Square F value Prob(F)
---------- ---- -------------- -------------- --------- -------
Regression 3 0.04100513 0.01366838 6660.66 0.00001
Error 3 6.156313E-006 2.052104E-006
Total 6 0.04101129




Current working formula:

modvalue = 1.2194265 + (-0.334306464 * (mod1*mod2*mod3...modn)/(1+exp(0.964036977*(n-0.667902481))));

PS This is also broken. A big problem I'm having is that the models all fit for mod=1.07, but nothing else. I have to reconsider..

keepiru
Omega Fleet Enterprises
Executive Outcomes
Posted - 2005.12.03 22:58:00 - [24]
 

IF that's the actual formula forum-fu maths on dmg mods is gonna be pretty tough, heh.

Jim Hsu
Deep Core Mining Inc.
Posted - 2005.12.03 23:57:00 - [25]
 

Edited by: Jim Hsu on 04/12/2005 00:08:15
Fit this:

0,1.07,1
1,1.07,1.07
2,1.07,1.135097087
3,1.07,1.180433795
4,1.07,1.203814483
5,1.07,1.212746167
6,1.07,1.215292184
0,1.077,1
1,1.077,1.077
2,1.077,1.149075251
3,1.077,1.199559759
4,1.077,1.225695224
0,1.098,1
1,1.098,1.098
2,1.098,1.191520786

Um.. yea. A 3d logistic decay thing in both directions. (huh??!!)

If this does not fit (I'm not very confident...), then I'll have to go back to my original idea...

Jim Hsu
Deep Core Mining Inc.
Posted - 2005.12.04 00:42:00 - [26]
 

Working but unelegant version of stacking:

http://www.eve-files.com/media/12/stackwork-beta1.zip

Things to note:

1. Edit values in bordered areas, don't edit any other values.
2. Put in the mods in descending order (highest dmg mod in 1st)
3. The values are inexact, but very close.

Jim Hsu
Deep Core Mining Inc.
Posted - 2005.12.04 03:58:00 - [27]
 

Generally:

3 mods is the "sweet spot".

Example statistics for a Gyro II:

ROF mod

Base: 100

Fitting 1 mod 89.5 -10.50%
Fitting 2 mods 81.41714826 -18.58%
Fitting 3 mods 76.45498308 -23.55%
Fitting 4 mods 74.21879518 -25.78%
Fitting 5 mods 73.42019139 -26.58%
Fitting 6 mods 73.15929612 -26.84%
Fitting 7 mods 73.0715301 -26.93%
Fitting 8 mods 73.03670702 -26.96%



Damage mod

Base: 100

Fitting 1 mod 110 10.00%
Fitting 2 mods 119.4611726 19.46%
Fitting 3 mods 126.3953157 26.40%
Fitting 4 mods 129.9161386 29.92%
Fitting 5 mods 131.2474856 31.25%
Fitting 6 mods 131.6916589 31.69%
Fitting 7 mods 131.8421205 31.84%
Fitting 8 mods 131.9019595 31.90%



Jim Hsu
Deep Core Mining Inc.
Posted - 2005.12.04 20:06:00 - [28]
 

Edited by: Jim Hsu on 04/12/2005 20:09:33
Edited by: Jim Hsu on 04/12/2005 20:06:09
Vishnej:

After thinking some more, I doubt you can take a simple function and multiply and/or exponentiate the (mod1*mod2*...modn). It would have to be automatically sorted, and likewise applied to each mod individually. (As per my first example).

Consider the following:

I have 8 50% hardeners in a scorp. Hardeners have an active resist of 50% and a passive resist of 1%.

Let's say I activate one hardener.

By your formula, I would get (1.5*1.01^7)=1.6082. Taking that to be "approximately" equal to a multiplier of 1.5, I would get (1.5*1.01^7)^x(8) = 1.5, so x(8) has to be 0.853403 (by using logs). But then if I activate all of my hardeners, that would be (1.5^8)^0.853403 = 15.929960. (n=8 for both examples of course.) There is NO WAY that that can ever be correct. Multiplying by (K) or (K^n) or (K^(n+c)) or (K^(n*c)) before the mod stacking doesn't help, I've verified. The reason is that n does not change, so you're just multiplying by a constant in all cases.

Conversely, If we let the (1.5^8)^x = (whatever it is), we would get an absurdly small value for x(8) when we apply it to (1.5*1.07^7).

Recall:

Originally by: blog

Also the dreaded problem of stacking super and lamer modifiers giving zero and negative values has been fixed (like the crazy tracking computer modified by cormack himself and a basic tracking enhancer could do if you got unlucky when dragging the modules to your ship).





Denrace
Amarr
PURE Legion
Pure.
Posted - 2005.12.04 20:13:00 - [29]
 

This game was much better back before people knew math.

Oh wait, haven't I heard that line before? Laughing


Anywho, good job on the equation chaps!

Vishnej
Demonic Retribution
Pure.
Posted - 2005.12.05 01:47:00 - [30]
 

Edited by: Vishnej on 05/12/2005 01:50:50
Originally by: Jim Hsu
Edited by: Jim Hsu on 04/12/2005 20:09:33
Edited by: Jim Hsu on 04/12/2005 20:06:09
Vishnej:

After thinking some more, I doubt you can take a simple function and multiply and/or exponentiate the (mod1*mod2*...modn). It would have to be automatically sorted, and likewise applied to each mod individually. (As per my first example).

Consider the following:

I have 8 50% hardeners in a scorp. Hardeners have an active resist of 50% and a passive resist of 1%.

Let's say I activate one hardener.

By your formula, I would get (1.5*1.01^7)=1.6082. Taking that to be "approximately" equal to a multiplier of 1.5, I would get (1.5*1.01^7)^x(8) = 1.5, so x(8) has to be 0.853403 (by using logs). But then if I activate all of my hardeners, that would be (1.5^8)^0.853403 = 15.929960. (n=8 for both examples of course.) There is NO WAY that that can ever be correct. Multiplying by (K) or (K^n) or (K^(n+c)) or (K^(n*c)) before the mod stacking doesn't help, I've verified. The reason is that n does not change, so you're just multiplying by a constant in all cases.

Conversely, If we let the (1.5^8)^x = (whatever it is), we would get an absurdly small value for x(8) when we apply it to (1.5*1.07^7).

Recall:

Originally by: blog

Also the dreaded problem of stacking super and lamer modifiers giving zero and negative values has been fixed (like the crazy tracking computer modified by cormack himself and a basic tracking enhancer could do if you got unlucky when dragging the modules to your ship).








You need to get down basic, pre-stacking penalty resistance stacking before you even try something like this with a fancy math program.

50% hardener + 50% hardener = 75% hardener [1-((1-0.5)*(1-0.5)=0.25) = 0.75]
NOT 1.5*1.5 = 225%

It's best to think of it as tolerance rather than a resistance. A 20% hardener tolerates 80% of that damagetype to go through. So for two hardeners, 0.8*0.8 = 0.64 tolerence for that damagetype.

It fits into the same stacking penalty as everything else, and it's multiplicative, meaning there is NO ORDER TO IT WHATSOEVER.
Take a 55%, a 20%, and a 10% hardener. 0.45*0.8*0.9 = 0.324 = 0.8*0.45*0.9

It's called a commutative property, and it's one thing that WILL be part of whatever stacking penalty is tossed on.

Here's how the stacking penalty CURRENTLY works with resistances:
Take the same 55%, 20%, and 10% resistance hardeners. They equal 0.45, 0.8, and 0.9 "tolerence" hardeners. So (0.45 * 0.8 * 0.9) ^ ((1/3)^0.25) = 0.288 tolerence = 71.2% resistance. This is how the game works right now, verified by all manner of testing.

The stacking penalty is the same equation for almost all manner of positive and negative mods - it's proven itself very flexible.

There IS an F(n), whether it's the type of equation your program is searching for or not, which will describe the new stacking penalty thusly:

(mod1*mod2*mod3...*modn) ^ F(n)


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