Finding your annuity, Finding your reverse annuity

[yumito]

~TLDR at the bottom~

Homework Statement

The Big question:

Suppose you can put $250/month into a retirement account that pays a 6% (annualized) return on investment.
When you reach 65 years of age, how much will you have in your retirement account? [“Annualized” refers to “on a yearly basis”
—this is why you see things like “i/12” in formulas (or “i/n” in general). 6% annually would correspond to 6%/12 = 0.5% per month.]

At this point when you retire, you move your money into a low-risk account that pays 3% (annualized) return on investment.
If you plan to live for 15 more years, how much can you draw out each month to live on?

Variables Account 1:

P=250
R=.005
N=12
T=46

Variables Account 2:

A[SUB]1[/SUB]=155123.83
R=.0025
N=12
T=15

Homework Equations

Annuity Formula(Given):

A=P[(1+(r/n))[SUP]nt[/SUP])-1]
     ______________
           (r/n)

The Attempt at a Solution

For the savings account it was easy.
My Attempt:

A=250((1+(.005/12))^((12)(46))-1)/(.005/12)
A=$155123.83

(This answer was received using my age of 19)

After this point it shows my attempts at solving the second part of the question.
If you know how to solve it skip this next very long and very tedious section.
If you are trying to solve it and am not sure how,
the next part may inspire you or just confuse you.
__________________________________________________________________________
The second part is tricky. I must find the compound interest while removing money.
My First Attempt:
Solve for P using annuity
Variables:

A=155123.83
R=.0025
T=15
N=12

Formula:

155123.83=P[(1+(.0025/12))^((12)(15))-1]
     ______________
           (.0025/12)

P=$12912.18

This cannot be true if A is the annuity than P would be what you add per month to get A.
____________________________________________________________________________
My Seccond Attempt:
Sum of a Geometric Sequence
Variables:

A[SUB]1[/SUB]=155123.83
R=1.0025
N=180

Formula:

((155123.83)(1-1.0025^180))
____________________
(1-1.0025)

I felt this would give me the end amount but without the money removed each month which is unhelpful :(.
_________________________________________________________________________
My Third Attempt:
Indicated Sum
Variables:

X=180
I=155123.83
F=(i-x)(1-1.0025^180))/(1-1.0025)

Assuming:
X
$\Sigma$ F
I
Formula:

180
Ʃ (i-x)(1.0025)
155123.83

Which i see no way i would be abel to solve for x this was also a dead end :(
__________________________________________________________________
My Final Attempt:
Recursion
Variables:

A[SUB]1[/SUB]=155123.83
A[SUB]N[/SUB]=A[SUB]n-1[/SUB]-x)(1.0025)
N>=180

Formula:

A[SUB]N[/SUB]=A[SUB]n-1[/SUB]-x)(1.0025)
A[SUB]1[/SUB]=155123.83
A[SUB]2[/SUB]=(155123.83-x)(*1.0025)

But as you see i cannot solve for x without knowing what A₁₈₀ equals. and i cannot find what A₁₈₀
equals without X.
___________________________________________________________________________

I'm sure there is a way to do it that i am just missing. As you can see i did try...
I do not necessarily need the answer just a push in the right direction.
And if you get the answer show your work so i do not lose my mind wondering.
___________________________________________________________________________
TLDR: Answer the problem and show work ;)
TLDR EXTENDED:

$250/month into a retirement account that pays a
6% (annualized) return on investment.
When I am 65 the amount in the account will be A.
A=$155123.83

you put the money from the first account into another account. this account pays a
3% (annualized) return on investment. At age 65 If you plan to live for 15 more years,
how much can you draw out each month to live on?

 

[yumito]

I figured the way to find it would be

A=P[(1+(r/n)nt)-1]
______________
(r/n)
x=Amount in the account
P=Amount you remove each month.
R=3% (annualized)
n=12

but P would initially have to be the amount deposited. than the amount in the account subtracted from the amount needed to live on.

My brain hurts

 

[alan2]

A(180) =0. Think of what you are doing. You have a sum of money deposited on your 65th birthday. Each month you earn interest, then withdraw some fixed amount, call it m. At the end of 180 months you make the last withdrawal of m, leaving 0 in the account. Write out the account balance for the first couple of months and you will see that what you have is the original principal growing minus a monthly annuity. Decide whether you are withdrawing at the beginning or end of the month. Set this quantity equal to 0 and solve for the monthly withdrawal.

 

[yumito]

like this?

A[SUB]N[/SUB]=An-1-m)(1.0025)
A[SUB]1[/SUB]=155123.83
A[SUB]2[/SUB]=(155123.83-m)(*1.0025)
A[SUB]180[/SUB]=0

i just don't know how to solve for m in a situation like this.

Or is it more like

0=m((1+(.0025/12))^((12)(15))-1)/(.0025/12)

 

[alan2]

I guess what I'm saying is that once you understand the general process and how to sum a geometric series, then you can do any variation of an annuity. I think that you're trying to shove numbers through a formula until it works and your code is confusing me.

let A(n)=amount in account at time n, 0<n<180
i=monthly rate =0.0025
m=monthly withdrawal

So let's assume this is an ordinary annuity with withdrawals at the end of the month. If you want withdrawals at the beginning of each month then you can adjust the process.

A(0)=155,123.83 in your case but we can start with any A(0).

Now think of the process. At the end of the first month you have earned interest in the amount of iA(0) and then withdrawn m. Your balance at the end of the first month is

A(1)=(1+i)A(0)-m

Now repeat for the second month

A(2)=(1+i)A(1)-m={(1+1)^2}A(0)-(1+i)m-m

Continue this process and sum the resulting geometric series. You find

A(180)=0={(1+i)^180}A(0)-(m/i){(1+i)^180-1}

I set A(180)=0 because I wanted my money to last for exactly 15 years. Now I can solve for the monthly payment m. I can solve any problem involving any stream of payments and withdrawals by writing it out to see what happens. If I try to solve a problem by forcing it to fit some formula then I really don't know what I'm doing and I might get it wrong. Change the payments to the beginning of the month and one formula doesn't work (although I could shift it). By the way, I find m just over $1,071.

I do have one question. Are you in an interest theory class or are you in an algebra class where they just give you formulas and you don't know how to sum a geometric series?

 

[yumito]

(Yes its a precalc class)
i understand geometric series in terms of a formula, that's the way i was taught it.
i was taught geometric series is:

a[SUB]n[/SUB]=a[SUB]1[/SUB](r)[SUP]n-1[/SUP]

A_n being the result of a part of a series.
A₁ being the first step in a series.
R being the common ratio.

but from what your saying the equation to find m would be.

0=((1+.0025)[SUP]180[/SUP])155123.83-(m/.0025)((1+.0025)[SUP]180[/SUP]-1)

solve for m?
attempt to get m by itself:
(m/.0025)=((1.0025)^180)155123.83-((1.0025)^179)
(m/.0025)=243144.44887108957942
m=607.86112217772394855

this not being the sane answer you received what did i do wrong?

another attempt at what you said.
0=(1.5674)155123.83-(m/.0025)(1.5635)
0=243141.091142-(m/.0025)(1.5635)
-243141.091142=(-m/-.0025)(1.5635)
-155510.7714=(-m/-.0025)
388.7769=-m
m=-388.7769

:/

 

[yumito]

i just am not sure i understand lol

 

[Ray Vickson]

i just am not sure i understand lol

I get something very different from you. For the amount in the account at age 65 (starting from age 19) I get A = 738,250.83, not your value A1 = 155,123.83. I used a monthly growth factor r = 1.005 for a total of 46*12 = 552 months, giving $A = 250 \times \sum_{n=1}^{552} r^n = 738250.83 .$

RGV

 

[alan2]

Thanks Ray. I never even checked his principal at age 65, I just went from his amount. My trusty financial calculator tells me $734,577.94 at age 65.

But you have to check your algebra Yumito, your solution for m is all wrong. My answer above is correct if I assumed that your principal at age 65 was correct. After you make the correction that Ray pointed out you should find m=$5,073.

 

[Ray Vickson]

Thanks Ray. I never even checked his principal at age 65, I just went from his amount. My trusty financial calculator tells me $734,577.94 at age 65.

But you have to check your algebra Yumito, your solution for m is all wrong. My answer above is correct if I assumed that your principal at age 65 was correct. After you make the correction that Ray pointed out you should find m=$5,073.

Since your trusty financial calculator and my computational package (Maple) give different answers, we must be computing different things or using different data, or something. Could you please say _explicitly_ what formula you are calculating, and what data you use?

RGV

 

[yumito]

you guys are going about it wrong. this is not actual financial information. The formula i set as (given) is the formula i am suppose to use...

 

[Ray Vickson]

you guys are going about it wrong. this is not actual financial information. The formula i set as (given) is the formula i am suppose to use...

Sorry, but that is nonsense. If the data you gave is accurate, and if your description of the problem is correct, then your numerical results are wrong. There must be something you are not telling us, or else you are using the formulas incorrectly.

Personally, I do not like the use of "canned" formulas; I much prefer to work the problem from first principles, stating exactly what calculations are involved. I generally do not trust "financial" packages, etc., without thoroughly testing them first before use, or reading reviews and tests of them.

RGV

 

[alan2]

I'm sure you're correct Ray, I didn't check. You used beginning of period. I used end so he didn't have a gap in his stream. I'm embarrassed that I took his earlier work for granted, I should have noticed that it made no sense. I think what's occurring is that he's given some formulas without explanation. It makes me crazy that this is how math is taught.