Article
【不想看过程的直接在结尾看结论】
Since the useful equipment only includes the following five parameters:
○1Miss chance reduction
○2Increased critical chance
○3Increased maximum damage
○4Increased damage
○5Chance to avoid damage
We only consider the equipment of these five parameters.
Given the assumption that your limit is a fixed constant every day, as we know, your expected damage is given by the following formula:
Your expected damage = Your base damage*(1+small damage+0.6*max damage)*(1+critical probability)/(1-avoid probability)*(0.875+miss probability).
Here your base damage only depends on your rank, strength and other things such as your MU order. Anyway, your equipment has nothing to do with it so we just need to maximum the ratio of your expected damage to your base damage, that is,
F=(1+small damage+0.6*max damage)*(1+critical probability) *(0.875+miss probability)/(1-avoid probability).
To simplify the notation, we denote it as:
F=(1+SD+0.6*MD)*(1+CP)*(0.875+MP)/(1-AP) (1)
Since the value of MD is always twice as the value of SD, and absolutely 0.6*2*x>x, we have the following Theorem.
Theorem 1: In the sense of expectation, MD (Max damage) is better than SD (small damage) in any case.
Now we just need to consider the equipment including MD, CP, MP, AP. So equation (1) becomes:
F=(1+0.6*MD)*(1+CP)*(0.875+MP)/(1-AP)=0.875*[(1+0.6*MD)*(1+CP)*(1+1.143*MP)/(1-AP)]
Then as we see, taking q6 equipment as an example, here ,
every 0.0825MP is equivalent to every 0.07CP, to every 0.07AP, and to every 0.14MD
Thus the problem coverts to:
Maximum { F(MD,CP,MP,AP)=(1+0.6*MD)*(1+CP)*(1+1.143*MP)/(1-AP) },
such that MD/0.14+AP/0.07+CP/0.07+MP/0.0825=16. (3)
Here 16 is because you have 8 equipment and every equipment has two parameters contribution.
These problem seems difficult to solve, and since log(x) is monotonous to x, it is equivalent to maximum log(F), that is,
Maximum { log[F(MD,CP,MP,AP)]=log(1+0.6*MD)+log(1+CP)+log(1+1.143*MP)-log(1-AP) },
such that MD/0.14+AP/0.07+CP/0.07+MP/0.0825=16. (4)
Using the knowledge of linear optimization and Lagrange dual theory (shadow price), we have:
Theorem 2: To maximize log[F(MD,CP,MP,AP)], it must have:
∂logF/∂MD*0.14=∂logF/∂AP*0.07=∂logF/∂CP*0.07=∂logF/∂MP*0.0825.
However, ∂logF/∂MD*0.14=0.084/(1+0.6*MD) , ∂logF/∂AP*0.07=0.07/(1-AP) , ∂logF/∂CP*0.07=0.07/(1+cp), ∂logF/∂MP*0.0825=0.0943/(1+1.143*MP).
We find that 0.07/(1-AP)>0.07>0.07/(1+cp), it means AP is always better than CP.
So we assume there is only three parameters and the equation converts to 0.084/(1+0.6*MD)=0.07/(1-AP)=0.0943/(1+1.143*MP), and MD/0.14+AP/0.07+MP/0.0825=16.
Let AP=0.4, substituting it into 0.084/(1+0.6*MD)=0.07/(1-AP)=0.0943/(1+1.143*MP), we have MD<0 and MP<0. It means ∂logF/∂AP*0.07 is always larger than ∂logF/∂MD*0.14 and ∂logF/∂MP*0.0825 in this situation! So it optimal to make it the largest, 0.4!
Then we solve the equation systems, 0.084/(1+0.6*MD)=0.07/(1+CP)=0.0943/(1+1.143*MP), and MD/0.14+CP/0.07+MP/0.0825=10.29. We have MP=0.42>0.125, so MP=0.125.
Then we just need to solve: 0.084/(1+0.6*MD)=0.07/(1+CP), MD/0.14+CP/0.07=8.77.
We get CP=0.224, MD=0.779.
So we have the first conclusion:
If you have all Q6 equipment, your optimal choice is to get miss=0, avoid=0.4, max=0.779 and critical =0.224! (Without considering pain dealer)
Though we just give a particular example (all Q6), you can use the same method to compute your optimal equipment.
Now we give the simple criterion:
Compare your (0.084)/(1+0.6*Max), (0.07)/(1-Avoid), (0.07)/(1+critical), (0.0943)/(1+1.143*Miss),
If the largest is (0.07)/(1-Avoid), make it as large as you can, if not, find the largest one, then improve it! （Then it will be smaller）This is suitable for situation without pain dealer.
Compare your (0.084)/(1+0.6*Max), (0.07)/(1-Avoid), (0.14)/(1+2*critical), (0.0943)/(1+1.143*Miss), If the largest is (0.07)/(1-Avoid), make it as large as you can, if not, find the largest one, then improve it! This is suitable for situation with pain dealer.
p.s. Here the critical and avoid includes your basis critical and avoid.
不想看过程的直接在这里看结论：
（1）输出不用重拳出击：
计算下列四个值 (0.084)/(1+0.6*Max), (0.07)/(1-Avoid), (0.07)/(1+critical), (0.0943)/(1+1.143*Miss),
如果最大的是 (0.07)/(1-Avoid)， 把Avoid补到最大，如果不是它，哪项大补哪项！
（2）用重拳出击输出：
计算下列四个值 (0.084)/(1+0.6*Max), (0.07)/(1-Avoid), (0.14)/(1+2*critical), (0.0943)/(1+1.143*Miss)
如果最大的是 (0.07)/(1-Avoid)， 把Avoid补到最大，如果不是它，哪项大补哪项！
注：这里的暴击率，免体率包含了基础暴击免体
V+S+留名 每人0.2g, 新人穷，没办法，不限制人数。
【V+S for 0.2g】
更简单明了的给7个Q6＋一个Q5土豪看的结论：
用Q5把miss消掉，2.5个Q6补满免体，然后
重拳出击：最优为，2个暴击+2.5个大伤
没有重拳出击，最优为，1个暴击+3.5个大伤略优于1.5个暴击+3个大伤，相差不大
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