From: flebber <flebber.crue@...>
Date: 2010-11-10T10:55:14+09:00
Subject: Re: Create a class - ideas

On Nov 10, 12:44 pm, flebber <flebber.c...@gmail.com> wrote:
> On Nov 8, 9:40 pm, flebber <flebber.c...@gmail.com> wrote:
>
> > On Nov 8, 3:33 am, Mike Cargal <m...@cargal.net> wrote:
>
> > > On Nov 7, 2010, at 12:20 AM, flebber wrote:
>
> > > > On Nov 7, 12:13 pm, flebber <flebber.c...@gmail.com> wrote:
> > > >> On Nov 7, 11:06 am, Mike Cargal <m...@cargal.net> wrote:
>
> > > >>> On Nov 6, 2010, at 7:00 PM, flebber wrote:
>
> > > >>>> On Nov 7, 9:45 am, flebber <flebber.c...@gmail.com> wrote:
> > > >>>>> On Nov 7, 2:24 am, Robert Klemme <shortcut...@googlemail.com> 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...
>
> This is where I was headed with it....any suggestions apprecaited.
>
> Bank = $500.00 # later to be a running total calculated
>
> def ROIcalc
> 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"
> end
>
> def pool
> if Bank > 200 then Bank * 0.05
> Else 10.00
> end
>
> # Scenario 1
> a = ROIcalc(per = 190, max_x = 0.8, puts "What ratio of return do you
> expect?" r = gets.chomp, pool)
> b = ROIcalc(per = 200, max_x = 0.4, r = 12.00, pool)
> # output result if valid
> # or return no valid result choose another scenario or end.
>
> This is where I was headed.
>
> Bank = $500.00 # later to be a running total calculated
>
> def ROIcalc
> 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"
> end
>
> def pool
> if Bank > 200 then Bank * 0.05
> Else $10.00
> end
>
> # Scenario 1
> a = ROIcalc(per = 190, max_x = 0.8, puts "What ratio of return do you
> expect?" r = gets.chomp, pool)
> b = ROIcalc(per = 200, max_x = 0.4, r = 12.00, pool)
> # output result if valid
> # or return no valid result choose another scenario or end.

upated.

Bank = $500.00 # later to be a running total calculated

def ROIcalc
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"
end

def pool
if Bank > 200 then Bank * 0.05
Else $10.00
end

# Scenario 1
a = ROIcalc(per = 190, max_x = 0.8, puts "What ratio of return do you
expect?" r = gets.chomp, pool)
b = ROIcalc(per = 200, max_x = 0.4, r = 12.00, pool)

if a + b > 10 then err = 10.00 - ( a + b) puts "Your exceeding your
pool by" + err.to_s
else a + b < 10 then readd = 10.00 - ( a + b)
		# then use readd to reassign extra units in ratio 75/25 to a and b
end