CONFUSION AND XP/Time (v2)


Enantiodromos

 

Posted

CONFUSION AND XP/Time

Here I discuss in greater detail some notes I made regarding the experience yeilds one expects when using confusion powers, that were first incorporated into C-force's guide. If you're math-phobic or are able to use a herostats type program for yourself, you may want to think twice about reading this.

Also, please note: this was meant to address only the binary: confusions/no confusions, and its effect on experience. That's based on the assumptions that either you don't have to play with confusion-spamming mind controllers at all, or if you do, the mind controllers can be well-built without doing much with confusions. (Either is true.) So this attempts only to show: does it get worse if somebody's using confusions instead of holds (also, I ignore badguy buffs to your team, and the minor scatter or cluster you get confusion, depending on your skill level and luck.)

Compared with a mind controller contributing with a Brawl/Levitate/AirSuperiority chain, confusions are WAAAAY better.

HOW EXPERIENCE WORKS GENERALLY

Where:
X = Experience value of (average) spawn
D = fraction of the spawn's damage done by you, where 1 = 100%.


Under normal circumstances, everybody who damages a mob "takes" a share of that mob's experience yeild (X), proportional to the damage they do (D). That means if you and an unteamed stranger each do half the damage to a mob, and it's defeated, you each get half the experience value. Earned experience (Xp) is equal to the mobs's experience value times the fraction of damage you do to it.

Xp = DX

Ordinarily, you do all the damage to a mob, so D=1, in which case:

Xp= X (Duh!)


HOW EXPERIENCE OVER TIME WORKS GENERALLY

Where:
F = ("Fight") time each spawn takes to fully defeat
S = ("Search") time it takes after each fight to find the next fight
T = ("Time") fight plus search time = F+S


Experience per time (Xppt) is equal to experience value of the spawn (X) dividied by total time (T). Total time is a function of how long it normally takes to defeat a spawn (F) plus time between fights (S):

[color]Xppt = X/T = X/(F+S)[/color]


HOW CONFUSION EXPERIENCE WORKS

Where:
Cmd = fraction of spawn's damage done by confused mobs (where 1=100%) = 1-D


The experience value of mobs damaged by other confused mobs (Cx) is equal to the mob's base experience value times the fraction of the mob's damage once you EXCLUDE three quarters of the damage confused mobs did (Cmd) from the mob's base damage.

Cx = DX / (D + (Cmd/4)) = 4DX / (3D + 1)


HOW CONFUSION FIGHT DURATION WORKS

Fight durations depend directly on how much damage has to be dealt to finish the enemy. When they've been damaged by other, confused mobs, the amount of time fights last (Cf) is the normal fight time (F) times the fraction of the damage you yourself have to do (D).

Cf = time a spawn takes to defeat, if it's been damaged by confused mobs = FD.

***Note: F here is the duration of a NORMAL fight, that is, without confusion in play.


HOW CONFUSION EXPERIENCE OVER TIME WORKS

Total time with confusions in play (Ct) is a function of how long it takes to defeat a spawn with confusions in play (Cf), plus downtime between fights (S).

Ct = confused fight plus search time Cf+S = FD+S

Experience per time with confused mobs doing some damage (Cxppt) is equal to experience value of the spawn damaged by confused mobs (Cx) dividied by total time with confusions in play (Ct).

Just like ...Xppt = X/T... similarly:

Cxppt = Cx/Ct = 4DX / ((FD+S)*(3D + 1))



DOING AT LEAST AS WELL

What we really want to do is compare normal experience per time (Xppt) to experience per time when confusions are in play (Cxppt). We can find a "break even" point by setting the two figures equal to each other. The break even point is should be a statement about when your experience over time WITH confusions (Cxppt) is the SAME as your experience per time WITHOUT them (Xppt).

Break-even point rule: Xppt = Cxppt


<font class="small">Code:[/color]<hr /><pre>

Xppt = Cxppt
X/(F + S) = 4DX / ((DF+S)*(3D + 1)) divide by X
1/(F + S) = 4D / ((DF+S)*(3D + 1)) invert
(F + S) = ((DF+S)*(3D + 1)) / 4D multiply out
(F + S) = (3DDF+3DS+DF+S) / 4D multiply both sides by 4D
4DF+4DS = 3DDF+3DS+DF+S subtract (DF +4DS +3DDF) both sides
3DF-3DDF = S-DS divide both sides by (1-D)
3DF = S

</pre><hr />



WHAT DOES S=3DF MEAN?

So, here's our statement about when your experience over time (Cxppt) WITH confusions is the SAME as your experience per time WITHOUT them (Xppt):

S = 3DF

In english:

Time it takes after each fight to find the next fight is equal to three times the fraction of the spawn's damage done by you, times the time each spawn takes to fully defeat (normally).

If you start the next fight faster than that, your XPS is better with confuse than without. If you find them slower, experience over time gets worse.


WHAT DOES THAT PROVE?

Nothing, directly, since we have to be able to judge for ourselves whether we can meet the rule above. But we CAN safely say that if our downtime between fights is short enough, confusion is really helping with XP, to say nothing of being a hold with lots of fringe benefits.

In my experience, groups with any real interest in decent XP over time will EASILY be doing better than the S=3DF rule above. If the group can't manage S=3DF on the average, it's probably because they're doing something IN PREFERENCE to decent XP/time.

And as anyone can see, such a group has no business complaining: "Confusion is hurting our XP!" Because of course, if the group wasn't doing other things that hurt XP/time more, the confusion itself would be *helping* XP/time. If you're not trying to get decent XP, you ought to have no complaints about not getting the best possible XP.

HOW MUCH BETTER?

Remember, none of this speaks to *how much* better the XP would be if you were getting in more than your 2.7%.

To generalize, Cxppt/Xppt -1 is the % bonus to XP you get (Cb).

Cb= Cxppt/CXppt = (4D(F+S) / ((FD+S)*(3D + 1)))-1

Let's take a couple cases.

Ocb (optimum Cb) = highest value for Cb obtainable with a D between 0 and 1.

Od (optimum D) = D between 0 and 1 that obtains highest value for Cb.


Where
F=120
S=10

Cb ~= (26D / (18DD +7.5D+1)) -1
Ocb= +62.5%
Od = .25

Where
F=30
S=30

Cb= 8D/(3DD+4D+1) -1
Ocb ~= +7.2%
Od ~=.58



THE HOLY GRAIL: FINDING Od

For a general rule re: (Od) based on F and S, what you do is find the derivative of (Cxppt/Xppt -1), then set it equal to zero.

(derivative with respect to D)(Cxxpt/Xppt -1) =

(4(F+S)(3FDD + 3SD - 5FD -F -3S -1))/((FFDD+SS+FSD)(9DD+6D+1) + FSD(9DD+6D))


At the present time, the author can find no way to simplify this rule when set to 0.

Note: When I use the notation, for example, DD, I mean D*D, or D^2. DDD then is D^3, etc. I tried to make each distinct variable start with a capital letter. e.g., Cxppt is not C*x*p*p*t. It's a single variable.


Choosing a Controller V2 | Splattrollers | Plant/Rad | Fire/Storm | Mind/Emp & Mind/Rad
Weird Controller Powers | Conf & XP/Time | Controller Damage
Being a Healer | The word Necessary | Natural Concept Characters

 

Posted

Fantastic guide, while I am a 'math-phobic' I still excel at math when I want to, your explanations were well written. 5 Stars.


 

Posted

Interesting analysis. I followed the construction and your reasoning, but honestly I am not 100% sure how to interpret or apply the model (sometimes I overlook things).

The search for the Holy Grail is always an elusive one. Looking at the expression that you are trying to evaluate at the end, I think all that you need to do is break it apart, and you can evaluate it. I am not sure if the result is meaningful or not, but...

4(F+S)(3FDD + 3SD - 5FD -F -3S-1)
--------------------------------------------------- = 0
(FFDD+SS+FSD)(9DD+6D+1) + FSD(9DD+6D)


This is only going to be true if:

(1) (F+S)=0

and/or

(2) (3FDD + 3SD - 5FD -F -3S-1)=0

and/or

(3) (FFDD+SS+FSD)(9DD+6D+1) + FSD(9DD+6D) --&gt; Infinity


Clearly (1) makes no sense, as it happens when F=-S, so that will never happen unless you become really, really fast or time starts running backward during the mission.

(3) will never approach Infinity unless you are staying in your mission maps for too long, so that leaves us with only one option.


(2) (3FDD + 3SD - 5FD -F -3S-1)=0

If you sort out the terms on the order of D and factor:

3FDD + 3SD -5FD -F -3S -1 = 0

3FDD + D(3S - 5F) -F -3S -1 = 0

This is a quadratic in D (aD^2 + bD +c)--grouping visually:

(3F)DD + (3S - 5F) - (F + 3S + 1)


Grinding through the quadratic formula, you find that

-(3S - 5F) +/- SQRT((3S - 5F)^2 + 12F(F + 3S + 1))
-------------------------------------------------- = D
6F


Now take that and see if it jives with your expectations and relate it back to the rest of your work.


 

Posted

Is is bad that I completely comprehend the mathematical backing but can only vaguely grasp the exact way that this corresponds to in regards to experience?

I always hated word problems. Curse you well Calculus...curse you!

Good guide, by the way, and thank you for pointing out that, excluding external factors, that XP gain v. XP loss when using confusion is minimal.


 

Posted

Wait wait wait...you mean to tell me that I've just encountered a real world use for the Quadratic Forumla I endlessly slaved over 25 years ago on the complete assurance from my math teacher that Yes! indeed I would find it useful in my day-to-day life!??!

But dang if I can't figure out just what the heck is going on in the above posts where I saw the Q-word...

I guess it will just suffice for me and my group that "the purply-head-dotted guys aren't hitting me" is a good thing, like it's always been!

PS: Ya'll should try 3-slotting for recharge...Sandy can machinegun Deceives on whole groups, to the cheers of her friends as they wade in and decimate spawns with little to no reprisals!

-Sandolphan


"When heroes fail, the Angels will save you."

MASTERMIND NUMERIC KEYPAD PET CONTROLS
HAMIDON NUKE RAID GUIDE

 

Posted

You know what I find funny? Considering the trivial case of mobs that are too low for you to even get experience.

In that case, the value of X becomes 0. Then if you apply that to the break-even point, you find that S = 3DF becomes undefined; which I suppose is another way of saying that there is no break-even point under those conditions.

And that agrees with my original assertion in the trivial case!