Repairing prove. [Heavy load picture]

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Repairing prove. [Heavy load picture]

Postby stsky0001 on Sat Mar 08, 2008 7:39 pm

Hello everyone.........

Many ppl argue about "which is better between half dur repair or 1dur repair". Today I'll prove it!!!!

First... as U know the equation of repairing is

Image

FL Alway CUT all the less than 1....

This graph will show u. The area of the colored bar is the dur of item U use it.(1bar = using per 1time repair)

Example. DL sword dur 2000/2000.

This graph show using till dur 1/xxxx then repair it.

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This graph show using till half of max dur then repair it.(white bar is max dur at each time using)

Image

...............

by the way. we can draw few equation to explain the dur u use each time..
(Y=dur u can use each time n=number of time u repair it Y0=Start dur)

1dur repairing..
Image

half dur repairing..
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draw the eqution as a graph line....

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and
Image


Is it looklike we can integrate the equation to find the area under the line?
yes we can... but if we do that way. many error will occur. (see the pic below)

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and
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positive error (white) not equal to negative error(black)... then if we find integral of the equation our result will missmatch from the fact.

(And I not sure if I use integral ... all of ready will understand it)

hmmm wht should we do to find the sum of dur we can use?
If we not use calculus to find it..... I'll use sequence and series to find the result!!!!

(Continue next reply)
Last edited by stsky0001 on Sat Mar 08, 2008 9:09 pm, edited 10 times in total.
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Re: Half dur repair VS. 1dur repair

Postby stsky0001 on Sat Mar 08, 2008 7:46 pm

As we know..

common sequence equation.
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and common series equation.
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Leave it here for now.... let see below

summation of dur we can use is. (1dur repair)
Image
then
Image
we use series equation to tranform something.....
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and finally we got
Image

and

summation of dur we can use is. (half use repair)
Image
then
Image
we use series equation to tranform something.....
Image
and finally we got
Image
Last edited by stsky0001 on Sun Mar 09, 2008 12:15 am, edited 7 times in total.
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Re: Half dur repair VS. 1dur repair

Postby stsky0001 on Sat Mar 08, 2008 7:47 pm

Assume as U want to stop repair at below 100 afer last repair thwn u use it till it broke and start dur is 2000/2000.

Calculate by equation I give u above in rep0.

if u work in 1durleft repair
U can use 6times by repair it 5 times.

if u work in halfdurleft repair
U can use 12times by repair it 11 times.

take 6 and 12 into the summation finding equation.

total dur u can use by 1durleft reparing is 3932.

total dur u can use by half durleft reparing is 3913.
(use half dur 11 time repair 11time then use till broke (full dur) 1 time)

hmm...

U see there's not too much difference.

anyway FL alway cut all less then 1.

that mean... many time u repair many dur u lost by cutting.

....................................

ok now..... I prove it alrdy. U think wht's better? :)

p.s. If I miss something or do something wrong. I'll apologize for it all.

thx to read my messy topic.
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Re: Repairing prove. [Heavy load picture]

Postby Multivira on Sun Mar 09, 2008 1:48 am

in conclusion:

as long as you repair at even numbers, its all good ^^
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Re: Repairing prove. [Heavy load picture]

Postby Shukster on Sun Mar 09, 2008 6:25 am

For math haters:

After generating a program that does this, I have come up with the following results:
Code: Select all
Repairing every 1/10 of the duration down:
Duration left: 9
Total duration utilized: 3932

Repairing every 1/9 of the duration down:
Duration left: 8
Total duration utilized: 3938
.
.
.
Repairing every 1/4 of the duration down:
Duration left: 3
Total duration utilized: 3970

Repairing every 1/3 of the duration down:
Duration : 2
Total duration utilized: 3975

Repairing every 1/2 of the duration down:
Duration left: 1
Total duration utilized: 3985

Repairing when 1 duration left:
Duration left: 1
Total duration utilized: 3989


This does not consider repairing only at even values, which is probably why anything other than 1/2 is relatively less efficient. Repairing every 2 duration-used will yield the best efficiency, but isn't very practical. In conclusion, as long as you repair at even numbers, it should be relatively close. I don't think it is very accurate, but it tells us that the less you repair the better, but not by much.
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