Hi,
I am trying to calculate a monthly payment for a loan in Ruby. Does
anyone know what the equivalent for math.pow in Java is for Ruby?
@monthlypayment = @principal*@rate/(1-Math.pow(1/(1+@rate),@payments))
Ive searched everywhere and cant seem to find anything.
thanks
Johnny B
On Fri, Mar 30, 2007 at 05:43:38AM +0900, John B. wrote:
I am trying to calculate a monthly payment for a loan in Ruby. Does
anyone know what the equivalent for math.pow in Java is for Ruby?
@monthlypayment = @principal*@rate/(1-Math.pow(1/(1+@rate),@payments))
Ive searched everywhere and cant seem to find anything.
http://www.rubycentral.com/book/ref_c_float.html#Float.Arithmeticoperations
Look for ‘exponentiation’
On 3/30/07, John B. [email protected] wrote:
Does
anyone know what the equivalent for math.pow in Java is for Ruby?
@monthlypayment = @principal*@rate/(1-Math.pow(1/(1+@rate),@payments))
Is this what you are looking for?
p 4 ** 3
Harry
–
http://www.kakueki.com/ruby/list.html
Japanese Ruby List Subjects in English
John B. wrote:
Hi,
I am trying to calculate a monthly payment for a loan in Ruby. Does
anyone know what the equivalent for math.pow in Java is for Ruby?
@monthlypayment = @principal*@rate/(1-Math.pow(1/(1+@rate),@payments))
Ive searched everywhere and cant seem to find anything.
thanks
Johnny B
I think this is what you’re looking for:
irb(main):001:0> 2 ** 1
=> 2
irb(main):002:0> 2 ** 2
=> 4
irb(main):003:0> 2 ** 3
=> 8
irb(main):004:0> 2 ** 4
=> 16
Drew O. wrote:
John B. wrote:
Hi,
I am trying to calculate a monthly payment for a loan in Ruby. Does
anyone know what the equivalent for math.pow in Java is for Ruby?
@monthlypayment = @principal*@rate/(1-Math.pow(1/(1+@rate),@payments))
Ive searched everywhere and cant seem to find anything.
thanks
Johnny B
I think this is what you’re looking for:
irb(main):001:0> 2 ** 1
=> 2
irb(main):002:0> 2 ** 2
=> 4
irb(main):003:0> 2 ** 3
=> 8
irb(main):004:0> 2 ** 4
=> 16
Yes this is what i am looking for. How do i then use it to calculate
amonthly payment, i have tried quite a few things but cant get the
syntax right: @monthlypayment =
@principal*@rate/(1-**(1/(1+@rate),@payments))
So for the example below:
Suppose you finance your car with a loan of $12000 at a yearly interest
rate of 11% for four years, and make equal payments monthly. How much
will your payments have to be? Here the parameters are principal P =
$12000, interest rate i = 0.11, number of years n = 4, and number of
periods per year q = 12. Then the monthly car payment M is given by
M = Pi/[q(1-[1+(i/q)]-nq)],
= ($12000)(0.11)/[(12)(1-[1+(0.11/12)]-(4)(12))],
= $110/(1-1.009166666...-48),
= $310.15.
Anyone done this before?
Alle venerdì 30 marzo 2007, John B. ha scritto:
irb(main):003:0> 2 ** 3
So for the example below:
= $310.15.
Anyone done this before?
** is an operator, just like + or ; it’s used this way: base**exponent.
In
your case, you should write:
@principal@rate/(1-1/(1+@rate)**@payments)
I hope this helps
Stefano