Calculating annuity payment

Question

I have a time payment question I'm having trouble with:

"Bill would like to save 60000 dollars for a deposit on his first home. He decides to invest his net monthly salary of 3000 dollars in a bank account that pays interest at a rate of 6% per annum compounded monthly. Bill intends to withdraw $E$ dollars at the end of each month from this account for living expenses, immediately after the interest has been paid.

Calculate the value of E if Bill is to reach his goal after 4 years."

Well, this is the equation I have applied: $60000 = 3000\cdot1.005^{48}-E(1+1.005+1.005^2+\cdots+1.005^{47}$, then applying the geometric sum formula. However, I get the question wrong. The answer is $\$1905.898$. What am I doing wrong?

Thanks for the help

Answer 1

You're trying to solve this equation:

$60000 = (3000(1.005) - E)(1 + 1.005 + ... + 1.005^{47})$.

Note that solving the problem using this method yields the correct answer of $E = 1905.898$.

We are assuming the paycheck deposit happens at the beginning of each month, so by the time he withdraws $E$, the $3000$ has grown in interest. Your formula is only taking into account one paycheck when there should be $48$ monthly paychecks, just like there are $48$ monthly withdrawals.

Answer 2

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ $\ds{r = 6/1200 =0.005\,,\quad \delta = c_{0} - E\,,\quad n = 4 \times 12 = 48\,,\quad c_{n} = c_{48} = 60000}$.

It's assumed there isn't any withdraw when the client makes the first deposit. It means that $$ {\large c_{0} = 3000} $$

\begin{align} c_{1} &= c_{0}\pars{1 + r} - E + c_{0} \\ c_{2} &= c_{1}\pars{1 + r} - E + c_{0}= c_{0}\pars{1 + r}^{2} + \delta\pars{1 + r} + \delta \\ c_{3} &= c_{2}\pars{1 + r} - E + c_{0}= c_{0}\pars{1 + r}^{3} + \delta\pars{1 + r}^{2} + \delta\pars{1 + r} + \delta \\ \vdots & \vdots\phantom{AAAAAAAAAAAAAAAAAAAAA}\vdots \\ c_{n - 1} &= c_{0}\pars{1 + r}^{n - 1} + \delta\bracks{% \pars{1 + r}^{n - 2} + \pars{1 + r}^{n - 2} + \cdots + \pars{1 + r} + 1} \\[3mm]&= c_{0}\pars{1 + r}^{n - 1} + \delta\,{\pars{1 + r}^{n - 1} - 1 \over \pars{1 + r} - 1} = c_{0}\pars{1 + r}^{n - 1} + \delta\,{\pars{1 + r}^{n - 1} - 1 \over r} \\[3mm] c_{n}&=c_{n - 1}\pars{1 + r} = c_{0}\pars{1 + r}^{n} + \delta\,{\pars{1 + r}^{n } - 1 - r\over r} \end{align}

$$ \delta = {c_{n} - c_{0}\pars{1 + r}^{n} \over \pars{1 + r}^{n} - 1 - r}\,r \quad\imp\quad E = c_{0} - \delta = c_{0} - {c_{n} - c_{0}\pars{1 + r}^{n} \over \pars{1 + r}^{n} - 1 - r}\,r $$ $$\color{#ff0000}{\large% E = \bracks{1 - {c_{n}/c_{0} - \pars{1 + r}^{n} \over \pars{1 + r}^{n} - 1 - r}\,r\,} c_{0}} $$

$$ E =\bracks{% 1 - {20 - 1.005^{48} \over 1.005^{48} - 1.005}\,0.005}\times 3000 $$

$$ \color{#0000ff}{\large E = 1941.79} $$

I checked it numerically:

/* saving_0.cc 16-nov-2013
http://math.stackexchange.com/users/85343/felix-marin

http://math.stackexchange.com/questions/569883/calculating-annuity-payment/569893#569893
*/
#include <iostream>
#include <iomanip>
using namespace std;
typedef long double ldouble;
const ldouble C0=3000,E=1941.79L,ANUAL_R_PER_CENT=6.0L;
const ldouble DELTA=C0 - E;
const ldouble MONTHLY_R_PER_ONE=ANUAL_R_PER_CENT/(100.0L*12.0L);
const ldouble R_PLUS_ONE=MONTHLY_R_PER_ONE + 1.0L;
const size_t N = 48U;

int main()
{
 ldouble actualC=C0;

 cout<<setprecision(7)<<'c'<<0<<" ="<<actualC<<endl;
 for ( size_t n = 1U ; n<N ; ++n ) {
     actualC=actualC*R_PLUS_ONE + DELTA;
     cout<<'c'<<n<<" = "<<actualC<<endl;
 }
 actualC=actualC*R_PLUS_ONE;
 cout<<'c'<<N<<" = "<<actualC<<endl;

 return 0;
}

It yields:

c0 = 3000
c1 = 4073.21
c2 = 5151.786
c3 = 6235.755
c4 = 7325.144
c5 = 8419.979
c6 = 9520.289
c7 = 10626.1
c8 = 11737.44
c9 = 12854.34
c10 = 13976.82
c11 = 15104.91
c12 = 16238.65
c13 = 17378.05
c14 = 18523.15
c15 = 19673.98
c16 = 20830.56
c17 = 21992.92
c18 = 23161.1
c19 = 24335.11
c20 = 25515
c21 = 26700.78
c22 = 27892.5
c23 = 29090.17
c24 = 30293.83
c25 = 31503.51
c26 = 32719.24
c27 = 33941.04
c28 = 35168.96
c29 = 36403.01
c30 = 37643.24
c31 = 38889.66
c32 = 40142.32
c33 = 41401.24
c34 = 42666.46
c35 = 43938
c36 = 45215.9
c37 = 46500.19
c38 = 47790.9
c39 = 49088.07
c40 = 50391.72
c41 = 51701.89
c42 = 53018.6
c43 = 54341.91
c44 = 55671.83
c45 = 57008.4
c46 = 58351.65
c47 = 59701.62
c48 = 60000.12