Another algebra equation needed

  • Thread starter Thread starter Karl Thompson
  • Start date Start date
K

Karl Thompson

Once again, I need a general equation that will solve for "A". Can someone
tell me what that would be please?


K = -1000
R = 0.0083333
N=5

A = (K*(1+R)^-1) + (K*(1+R)^-2) + (-A*(1+R)^-3) + ... + (A*(1+R)^-N)

FYI: In this case I know A equals -40,554.920

Please note (don't know if it's important) that "A" can be negated one or
more times..

TIA
 
(-A*(1+R)^-3) + ... + (A*(1+R)^-N)

at what point in the above sequence does -A become +A
 
Oh, sorry.

There is only one occurance of -A.

My assumption is though, that once I see how the equation is rewritten, that
I'll be able to figure out how make the necessary modifications myself for
1) different values of N and 2) occassions when there may be more than one
occurance of -A.
 
In the second half of the equation, is it A or K that is
being multiplied by (1+R) ^ -N?

I can help if it is K....
 
Thank you. It is "A".

However, maybe this example is a little better in that it's simplier(?):

K = 1000
R = 0.0083333

A = (K*(1+R)^-1) + (-A*(1+R)^-2)

in this case A = 499.98
 
A = (K*(1+R)^-1) + (K*(1+R)^-2) + (-A*(1+R)^-3) + ... + (-A*(1+R)^-N)


A = (K*(1+R)^-1) + (K*(1+R)^-2) + A*((-1*(1+R)^-3) + ... + (-1*(1+R)^-N))

A - A*((-1*(1+R)^-3) + ... + (-1*(1+R)^-N)) = (K*(1+R)^-1) + (K*(1+R)^-2)

A * ( 1 - ((-1*(1+R)^-3) + ... + (-1*(1+R)^-N))) = (K*(1+R)^-1) +
(K*(1+R)^-2)

A = ((K*(1+R)^-1) + (K*(1+R)^-2)) / ( 1 - ((-1*(1+R)^-3) + ... +
(-1*(1+R)^-N)))

The answer for A with your numbers is -506.206107674368, checked by
substituting back in the original. So if you know the answer is something
else, then the formula must not be given correctly.
 
A = (K*(1+R)^-1) + (-A*(1+R)^-2)
A - (-A*(1+R)^-2) = (K*(1+R)^-1)
A * ( 1 + (R+1)^-2) = (K*(1+R)^-1)

A = (K*(1+R)^-1) / ( 1 + (R+1)^-2)


That does solve to 499.982783099181 with your numbers.
 
Thank you.

I agree, except for one thing. Sometimes A is negated, and sometimes it
isn't.

I see if I modify your divisor to this (2 neg signs removed) then the result
is the expected
-40554.92:

1 - ( (-1*(1+R)^-3) + (1*(1+R)^-4) + (1*(1+R)^-5) )


This is great. Thanks.

Now to write the VBA code to do this....
 
Here's my 2 cents. If the first two power terms use k, the 3rd power term
uses -A, and the 4th, 5th, and above use +A, then it appears to me that the
equation may be something like this...

A = (k*r*(1 + r)^(1 + n)*(2 + r))/((1 + r)^3 + (1 + r)^n*(-1 + r*(2 + r)*(1
+ r + r^2)))

With k = -1000, r=.0083333, and n=5, I get
-40555.08210206875
If your only approximate number 'r' of .0083333 is really .008333333333333,
(or more exactly 1/120), then I get your solution of -40554.92209880997

HTH
 
Thanks Tom, Dana and Keyt for your help.

There turns out to be one more variation to this that I'm hoping one of you
can help me with.

Given this:

K = -1000
K2 = 500.0011
R = 1/120 = 0.0083333333..
N=5

K2 = (K*(1+R)^-1) + (K*(1+R)^-2) + (-A*(1+R)^-3) +
(+A*(1+R)^-4) + (+A*(1+R)^-N)

Again, how do I solve for A?

In this case A equals 2,602.01

TIA,
Karl
 
K2 = (K*(1+R)^-1) + (K*(1+R)^-2) + (-A*(1+R)^-3) + (+A*(1+R)^-4) +
(+A*(1+R)^-5)

K2 = (K*(1+R)^-1) + (K*(1+R)^-2) + ((-A*(1+R)^-3) + (+A*(1+R)^-4) +
(+A*(1+R)^-5))

K2 - ((-A*(1+R)^-3) + (+A*(1+R)^-4) + (+A*(1+R)^-N)) = (K*(1+R)^-1) +
(K*(1+R)^-2)

K2 - A * ((-(1+R)^-3) + ((1+R)^-4) + ((1+R)^-5)) = (K*(1+R)^-1) +
(K*(1+R)^-2)


A * ((+(1+R)^-3) - ((1+R)^-4) - ((1+R)^-5)) = -(K*(1+R)^-1) - (K*(1+R)^-2)
+ K2

A = ( -(K*(1+R)^-1) - (K*(1+R)^-2) + K2)/((+(1+R)^-3) - ((1+R)^-4) -
((1+R)^-5))
 
If n could be 5,10, 20, etc...(with +a being used for the 4th power term and
above) then here's another idea...

Sub Demo()
Dim k, k2, z, n, answer

k = -1000
k2 = 500.0011
'z=1+r
z = 1 + (1 / 120)
n = 5

answer = ((z - 1) * z ^ (n + 1) * (k2 * z ^ 2 - k * (z + 1))) / ((z - 2)
* z ^ n + z ^ 3)
Debug.Print answer
End Sub

returns:
-2602.00995082074

HTH.
 
Back
Top