The amount to save annually should be approximately $18,365.09.

[jdulude]

You have just turned 24, and you intend to start saving for your retirement. You plan to retire in 42 years when you turn 66. During your retirement you would like to have an annual income of $120,000 per year for the next 28 years (until age 94).
-Calculate how much has to be in your account before the first withdrawal at age 67.
-Calculate how much would have to save annually between now and age 67 in order to finance your retirement income and to fill that account.
Make the following assumptions:
-Assume that the relevant compounded interest rate is 4.0 percent for all 70 years
- You make the first payment today and the last payment on the day you turn 66.
-You make the first withdrawal when you turn 67 and the last withdrawal when you turn 94.

k=0.04 n=
These are the type of question that will be on the test and I need to practice
Please help me get on the right path I am sure it is an annuity formula, but I need help solving it

 

[jdulude]

I can't find a question in the book to help with this problem

the formula I am using is

A[(1-(1+k)^(-n))/k]

 

[MarkFL]

Without knowing what formula(s) you are to use, this is how I would approach the problem from a purely algebraic perspective.

I would first let $A$ be the account balance needed at the time of the first withdrawal at age 67. I would observe that there will be $94-67+1=28$ annual withdrawals of \$120,000 made from the account, and we want the account balance to be \$120,000 at the time of the last withdrawal, i.e., we want a balance of \$0 after the last withdrawal.

On your 67th birthday, at the time just before the first withdrawal, the account balance is $A$. Then you make the withdrawal $W$, and the balance is $A-W$.

Assuming interest is compounded annually at an annual rate of $r$, the account balance on your 68th birthday right before the withdrawal is:

\(\displaystyle (1+r)(A-W)\)

And after the withdrawal it is:

\(\displaystyle (1+r)(A-W)-W\)

Continuing in this manner, we will find the account balance right before the $n$th withdrawal is:

\(\displaystyle (1+r)^nA-W\sum_{k=1}^n(1+r)^{k}=(1+r)^nA-W\left(\frac{(1+r)((1+r)^n-1)}{r} \right)\)

Hence, equating this to $W$, and solving for $A$, we find:

\(\displaystyle (1+r)^nA-W\left(\frac{(1+r)((1+r)^n-1)}{r} \right)=W\)

\(\displaystyle A=\frac{W\left((1+r)^{n+1}-1 \right)}{r(1+r)^n}\)

Now plug in the given data:

\(\displaystyle W=120000,\,r=0.04,\,n=28\)

What do you find?

Can you now use a similar method to solve for $S$, the annual savings required to get this account balance?

 

[jdulude]

A=1333332.42

 

[jdulude]

I appreciate the help but I have to go to the math lab, because there is something I am missing here and I need someone to just show me where I am going wrong. thanks for the help anyways.

 

[MarkFL]

When I plug those numbers into the formula I gave, I get (to the nearest penny):

\(\displaystyle A\approx2119567.59\)

Until you show what you have tried, we can't really show you where you are going wrong.

To answer the second question, I would let $S$ be the amount annually saved and deposited into the account. You will make $66-24+1=43$ annual deposits.

On your 24th birthday, after the deposit, we have:

\(\displaystyle A=S\)

On your 25th birthday, after the deposit you have:

\(\displaystyle A=(1+r)S+S\)

Continuing, we find on the $n$th birthday after the 24th birthday after making the deposit, we have:

\(\displaystyle A=S\sum_{k=0}^n(1+r)^k=S\frac{(1+r)^{n+1}-1}{r}\)

Equating this to the value of $A$ we found earlier, and letting $n_S$ be the number deposits and $n_W$ be the number of withdrawals after retirement, we have:

\(\displaystyle \frac{W\left((1+r)^{n_W+1}-1 \right)}{r(1+r)^{n_W}}=S\frac{(1+r)^{n_S+1}-1}{r}\)

Solving for $S$, we find:

\(\displaystyle S=\frac{W\left((1+r)^{n_W+1}-1 \right)}{(1+r)^{n_W}\left((1+r)^{n_S+1}-1 \right)}\)

Plugging in the given data, I find:

\(\displaystyle S\approx18365.09\)