Create a class - ideas

[flebber]

Evening

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

What I want to test is the value of X needed to equal Per from X's
maximum allocation down.

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?

 

[Robert Klemme]

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

 

[flebber]

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.


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?













Sorry you lost me somewhere along the path.  Also it's tea time right now...

Cheers

        robert

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.

 

[flebber]

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

 

[Mike Cargal]

=20
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.
=20
Main =3D> Per - where main is the lowest value it can be greater than
per.
=20
Cheers
=20
Sayth
=20

The wording is pretty vague at several points, but I'm trying to work =
through it to make sense of it...

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?


Mike Cargal

(e-mail address removed)
http://blog.mikecargal.com

 

[Mike Cargal]

=20
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.
=20
Main =3D> Per - where main is the lowest value it can be greater than
per.
=20
Cheers
=20
Sayth
=20

Something's not right with your formulas (I suspect)

and X =3D X*P

so... main =3D =20
R*(X*P)
----------- * 100
P =20

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...
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
=3D=3D=3D=3D=3D=3D=3D
r =3D 5.0
p =3D 10.0
per =3D 190.0
x =3D 0.8

begin
main =3D ((r*(x*p))/p)*100 =20
lastSuccess =3D x if main >=3D per=20
x -=3D 0.05
end while main >=3D per
puts "#{lastSuccess*100.0}%"=20
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
=3D=3D=3D=3D=3D=3D
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...
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
=3D=3D=3D=3D=3D=3D
r =3D 5.0
p =3D 10.0
per =3D 190.0
max_x =3D 0.8

x =3D 0.05
const =3D r*100 # simplified without X
x +=3D 0.05 while (x*const < per) && (x <=3D max_x)
puts x <=3D max_x ? "#{lastSuccess*100.0}%" : "no answer"
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
=3D=3D=3D=3D=3D=3D


Mike Cargal

(e-mail address removed)
http://blog.mikecargal.com

 

[flebber]

Something's not right with your formulas (I suspect)


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

(e-mail address removed)://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 isa 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 at http://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

 

[flebber]

Thanks for looking at this for me.


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"

 

[John Mair]

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 }

This WHILE loop will not work because the condition is only evaluated
once. If you want it to work you need to turn the condition into a
lambda, like so:

def WHILE(cond)
loop do
return if !cond.()
yield
end
end

and use like this:

i = 0
WHILE(->{i < 3}) { puts i; i+=1 }

 

[Robert Klemme]

This WHILE loop will not work because the condition is only evaluated
once. If you want it to work you need to turn the condition into a
lambda, like so:

def WHILE(cond)
loop do
return if !cond.()
yield
end
end

and use like this:

i = 0
WHILE(->{i< 3}) { puts i; i+=1 }

Where's the point in doing this when there are "while" loops already
built into the language?

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.

Cheers

robert

 

[John Mair]

Where's the point in doing this when there are "while" loops already

built into the language?
robert

fun

 

[Mike Cargal]

=20
=20
=20
=20
=20

On Nov 6, 2010, at 7:00 PM, flebber wrote: =20
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...

=20
Actually after re-reading your post I get it. All but one small bit.
=20
In your formula how does the ? "#{lastSuccess*100.0}%" bit work? what
data does lastSuccess pull.
=20
x =3D 0.05
const =3D r*100 # simplified without X
x +=3D 0.05 while (x*const < per) && (x <=3D max_x)
puts x <=3D max_x ? "#{lastSuccess*100.0}%" : "no answer"
=20

my bad...

r =3D 5.0
p =3D 10.0
per =3D 190.0
max_x =3D 0.8

x =3D 0.05
const =3D r*100 # simplified without X
x +=3D 0.05 while (x*const < per) && (x <=3D max_x)
puts x <=3D max_x ? "#{x*100.0}%" : "no answer"

Mike Cargal

(e-mail address removed)
http://blog.mikecargal.com

 

[flebber]

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

(e-mail address removed)://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...

 

[flebber]

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.


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.


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.

 

[flebber]

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