From: flebber Date: 2010-11-07T13:20:25+09:00 Subject: Re: Create a class - ideas 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"