From: flebber Date: 2010-11-08T19:45:20+09:00 Subject: Re: Create a class - ideas On Nov 8, 3:33 am, Mike Cargal wrote: > On Nov 7, 2010, at 12:20 AM, flebber wrote: > > > > > > > > > > > On Nov 7, 12:13 pm, flebber wrote: > >> On Nov 7, 11:06 am, Mike Cargal wrote: > > >>> On Nov 6, 2010, at 7:00 PM, flebber wrote: > > >>>> On Nov 7, 9:45 am, flebber wrote: > >>>>> On Nov 7, 2:24 am, Robert Klemme wrote: > > >>>>>> On 06.11.2010 10:19, flebber wrote: > > >>>>>>> I am trying to create a class. I am struggling to figure the best flow > >>>>>>> to get the maths side to work. > > >>>>>>> So say that > > >>>>>>> R is a float given by user input > >>>>>>> P is a Total amount(Pool) > >>>>>>> Per is a variable % > >>>>>>> X is a variable that is a percentage of P defined by a maximum > >>>>>>> allocation. > > >>>>>>> So main = (( R * X)/P)*100 > > >>>>>> First of all you should get your variables right.  Variable "Per" does > >>>>>> not show up in the formula and "main" is not mentioned in the list. > > >>>>>> The meaning of the formula is totally unclear to me.  From what you gave > >>>>>> you are calculating the fraction (R/P) multiply it with 100 (so you > >>>>>> actually get (R/P) percent and now you multiply with another percentage > >>>>>> (X).  So you have a percentage of a percentage. > > >>>>>>> What I want to test is the value of X needed to equal Per from X's > >>>>>>> maximum allocation down. > > >>>>>> Can you write down a formula that contains all variables in your list > >>>>>> and point at the fixed ones (constants), user inputs and variables you > >>>>>> want to resolve? > > >>>>>>> What i am thinking but cant get right > > >>>>>>> Say > > >>>>>>> R = 5 > >>>>>>> P = 10 > >>>>>>> Per = 190 > >>>>>>> X = max 80% of P > > >>>>>>> For X in main = Per ( Closest whole number or half that equals closest > >>>>>>> to but greater than Per) > >>>>>>> main = (( 5 * 8)/10)*100 > > >>>>>>> So in example intially main equalled 400%. And answer I would want to > >>>>>>> resolve it to is X = 4 which is 200% as 3.50 equals 180%. > > >>>>>>> Any ideas? > > >>>>>> Sorry you lost me somewhere along the path.  Also it's tea time right now... > > >>>>>> Cheers > > >>>>>>         robert > > >>>>>> -- > >>>>>> remember.guy do |as, often| as.you_can - without endhttp://blog.rubybestpractices.com/ > > >>>>> So I want to check by changing X when in forumla "main" that it is => > >>>>> than "per" > > >>>>> In simple terms I want to calculate units needed to reach a rate of > >>>>> return, X represents the variable units and per is the ROI(return of > >>>>> investment rate I would deaire to acheive), R is the ratio of return > >>>>> and P is a pool or base amount, I am using base 10 to start off with. > > >>>>> I am trying to test X for a value, the only constraint on X is that it > >>>>> cannot exceed 80% of the P or Pool amount. > >>>>> So if I set per = 190% > > >>>>> R = 5 > >>>>> P = 10 > >>>>> Per = 190 > >>>>> X = max 80% of P > > >>>>> For X in main >= per > > >>>>> main = (( 5 * X)/10)*100 >= 190% > > >>>>> so for X = 80% of P or 8 base units > > >>>>> main = (( 5 * 8)/10)*100 > > >>>>> which would test out as > > >>>>> main >= per > > >>>>> 400 => 190 - > > >>>>> So when X is 8 units the main section is greater than per but its not > >>>>> the closest whole unit to per. > > >>>>> So when X = 40% or 4 base units > > >>>>> main = (( 5 * 4)/10)*100 > >>>>> main >= per > >>>>> 200 >= 190 > > >>>>> this is the largest unit in 0.5 increments that remains greater than > >>>>> Per of 190 so I would want X once tested to resolve to this. > > >>>>> I hope that made sense. > > >>>> So how do I best get X to run a loop in 0.5 increments until it > >>>> reaches the closest value that makes the left side of an equation > >>>> greater or equal to the right. But where it is the lowest value that > >>>> is greater than or equal to the the right. > > >>>> Main => Per - where main is the lowest value it can be greater than > >>>> per. > > >>>> Cheers > > >>>> Sayth > > >>> Something's not right with your formulas (I suspect) > > >>>>>>> main = (( R * X)/P)*100 > > >>> and X = X*P > > >>> so... main =   > >>> R*(X*P) > >>> ----------- * 100 > >>>       P         > > >>> the P's cancel out... > >>> R*X*100 > > >>> varying P will not change the results of your calculations... > > >>> here's what I believe that you're asking for... > >>> ================================ > >>> r = 5.0 > >>> p = 10.0 > >>> per = 190.0 > >>> x = 0.8 > > >>> begin > >>>         main = ((r*(x*p))/p)*100   > >>>         lastSuccess = x if main >= per > >>>         x -= 0.05 > >>> end while main >= per > >>> puts "#{lastSuccess*100.0}%" > >>> =============================== > >>> note: you can change p al day long and always get the same answer > > >>> there are probably more "rubified" ways to express this. > >>> I've tried to maintain the approach you've stated.  However, I would simplify the equation first, and since the last X that succeeds as you decrement is the same thing as the first that succeeds as you're going up, I'd probably turn t into something like... > >>> =============================== > >>> r = 5.0 > >>> p = 10.0 > >>> per = 190.0 > >>> max_x = 0.8 > > >>> x = 0.05 > >>> const = r*100 # simplified without X > >>> x += 0.05 while (x*const < per) && (x <= max_x) > >>> puts x <= max_x ? "#{lastSuccess*100.0}%" : "no answer" > >>> =============================== > > >>> Mike Cargal > > >>> m...@cargal.nethttp://blog.mikecargal.com > > >> Thanks for looking at this for me. > > >>> You've really lost me when 80% turns into 8 (or 8 base units).  What is a base unit? And why would it be equal to 10%? > >>> Are you looking to vary X from 80% down by 5% increments? > > >> This test is the first of 4 I plan to make into one program. For each > >> their while be a maximum allocation so in this case 80% so may be 40% > >> etc. The pool in future will vary but I am using base 10 while I write > >> it(hoping it would be clearer for another person reading it). So that > >> means that X has a max allocation of units if 40% was the maximum > >> allocation and base 10 then X would be 4 units and I would want my > >> loop to test X from 0 to 4 in 0.5 increments. > > >> I am going to need more time to read your solution as I haven't got it > >> first read. I had started to look athttp://www.rubyist.net/~slagell/ruby/iterators.html > > >> and a solution flow similar to > > >> ruby> def WHILE(cond) > >>     |   return if not cond > >>     |   yield > >>     |   retry > >>     | end > >>    nil > >> ruby> i=0; WHILE(i<3) { print i; i+=1 } > >> 012   nil- Hide quoted text - > > >> - Show quoted text - > > > Actually after re-reading your post I get it. All but one small bit. > > > In your formula how does the ? "#{lastSuccess*100.0}%" bit work? what > > data does lastSuccess pull. > > > x = 0.05 > > const = r*100 # simplified without X > > x += 0.05 while (x*const < per) && (x <= max_x) > > puts x <= max_x ? "#{lastSuccess*100.0}%" : "no answer" > > my bad... > > r = 5.0 > p = 10.0 > per = 190.0 > max_x = 0.8 > > x = 0.05 > const = r*100 # simplified without X > x += 0.05 while (x*const < per) && (x <= max_x) > puts x <= max_x ? "#{x*100.0}%" : "no answer" > > Mike Cargal > > m...@cargal.nethttp://blog.mikecargal.com > I still think we first need to get the math correct before we can come > up with solutions.  Formulas I have seen in this thread look like they > could be solved with some simple transformations and do not need any > nested intervals or similar approximation algorithms.  So far I find the > problem description quite confusing. Essentially without formulas all I am calculating is how many units at a specified rate of return it would take to reach a percentage of rate of return. Only additionally I have specified a maximum unit allocation. There would ultimately be several options with different return rates and max allocations from each pool. A pool is a percentage subset of a bank. > x = 0.05 > const = r*100 # simplified without X > x += 0.05 while (x*const < per) && (x <= max_x) > puts x <= max_x ? "#{x*100.0}%" : "no answer" Mikes solution definitely works for working units(X)..thank you. I need to set constants and obtain user inputs for different scenarios and call it to the function. > could be solved with some simple transformations and do not need any I need to lookup what a transformation is...