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Duff

BsB is king, HD oridecon/safe cert is also king

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Hey guys,

penguin here, I have a degree in math and decided to solve the question, what is the optimal way to refine?

so i wrote two programs, outlined below, to simulate 1 million refines and what the averages of 2 different methods are

 

the results are expected

 

First method, use BSB/old elu on +7 to +8, use HD oridecon otherwise (solveRates), second method refine BSB -> old ori all the way through (solveRates2)

solveRates() ;
       n counter,ln,refine,success,iteration
       
s refine=7
       
f ln=1:1:1000 q:refine=10  d
       . i refine=7 s success=$$roll(50)
       . 
i refine=8 s success=$$roll(30)
       . 
i refine=9 s success=$$roll(19)
       . 
i 'success,refine>7 s refine=refine-1
       . 
e  i success s refine=refine+1
       . 
i refine=7 s counter=counter+1
       . 
s iteration=iteration+1
       
q counter_"|"_iteration

solveRates2() ;
       n counter,ln,refine,success,iteration
       
s refine=7
       
f ln=1:1:1000 q:refine=10  d
       . i refine=7 s success=$$roll(50)
       . 
i refine=8 s success=$$roll(50)
       . 
i refine=9 s success=$$roll(30)
       . 
i refine=7 s counter=counter+1
       . 
i refine=8 s counter=counter+2
       . 
i refine=9 s counter=counter+4
       . 
i success s refine=refine+1
       . 
s iteration=iteration+1
       
q counter_"|"_iteration
       
;
roll(value) ;
       n roll
       
s roll=$r(100)
       
i roll<value q 1
       
q 0

 

i put these guys in a wrapper to iterate 1 million, these are the averages

BsB used from solveRates method: 25.593615

BsB used from solveRates2 method: 19.328085

total iterations solveRates method: 49.350849

total iterations solveRates2 method: 7.328922

 

I think its right because BSB all the way through is solvable, you have (2)(1) + 2(2) + 3.3333(4) = 19.333 which is what my averages are converging to

 

BUT WAIT, what if we used safe refines instead of BSB and old ori ALL THE WAY THROUGH???  So use HD oridecon to go higher than +7, if we drop to +6 safe refine back to +7

solveRates3() ;
    n counter,ln,refine,success,iteration
    s refine=7
    f ln=1:1:1000 q:refine=10  d
    . i refine=7 s success=$$roll(30)
    . i refine=8 s success=$$roll(30)
    . i refine=9 s success=$$roll(19)
    . i 'success,refine>6 s refine=refine-1
    . e  i success s refine=refine+1
    . i refine=6 s counter=counter+1,refine=7
    . s iteration=iteration+1
    q counter_"|"_iteration

iterating, we get 

Safe Refine +7 needed: 30.947449

HD Ori needed: 67.090516

assume

BsB = 10,000,000

HD Ori = 1,500,000

Safe Refine = 4,000,000

Old Ori = 4,000,000

Refine to +10 using BsB/old ori = 222,720,000 zanels to refine

Refine to +10 using BsB/old ori/HD ori = trash, infinity zanel don't do it

Refine to +10 using Safe Ticket/HD ori = 224,395,000 zanels to refine

 

Takeaways:

1. BsB + old ori is better than BsB at 7 and HD ori onward, even in terms of BsB used

2. Safe Refine +7 + HD ori is actually a reasonably cost effective method to refine something to +10

3. HD ori should never be used outside of refine event

4. Optimal cost efficiency during refine event is safe to 7 to preserve +7, use HD ori to get to +9, use BsB/old to get from 9 to 10

Edited by Duff
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forgot to mention the rates are calculated assuming a +10% refine event - but same logic applies (though numbers would be higher on average)

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Quote

 

Further optimizing i wanted to see what safe cert/HD ori looks like if we only take it to +9 (because +10 % sucks), results:

Safe Cert = 7.769085

HD ori = 14.460976

this seems considerably better than BsB/old ori in terms of cost:

Safe/HD ori = 52,730,000

BsB/Old Ori = 10m x 6 + 4m x 4 = 76,000,000

Edited by Duff

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