14.8: Annuities and Loans

Last updated
Save as PDF

Page ID: 110128

$ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $

$ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $

$ \newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$

( \newcommand{\kernel}{\mathrm{null}\,}\) $ \newcommand{\range}{\mathrm{range}\,}$

$ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$

$ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$

$ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$

$ \newcommand{\Span}{\mathrm{span}}$

$ \newcommand{\id}{\mathrm{id}}$

$ \newcommand{\Span}{\mathrm{span}}$

$ \newcommand{\kernel}{\mathrm{null}\,}$

$ \newcommand{\range}{\mathrm{range}\,}$

$ \newcommand{\RealPart}{\mathrm{Re}}$

$ \newcommand{\ImaginaryPart}{\mathrm{Im}}$

$ \newcommand{\Argument}{\mathrm{Arg}}$

$ \newcommand{\norm}[1]{\| #1 \|}$

$ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$

$ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\AA}{\unicode[.8,0]{x212B}}$

$ \newcommand{\vectorA}[1]{\vec{#1}} % arrow$

$ \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$

$ \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $

$ \newcommand{\vectorC}[1]{\textbf{#1}} $

$ \newcommand{\vectorD}[1]{\overrightarrow{#1}} $

$ \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} $

$ \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} $

$ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $

$ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $

$\newcommand{\avec}{\mathbf a}$ $\newcommand{\bvec}{\mathbf b}$ $\newcommand{\cvec}{\mathbf c}$ $\newcommand{\dvec}{\mathbf d}$ $\newcommand{\dtil}{\widetilde{\mathbf d}}$ $\newcommand{\evec}{\mathbf e}$ $\newcommand{\fvec}{\mathbf f}$ $\newcommand{\nvec}{\mathbf n}$ $\newcommand{\pvec}{\mathbf p}$ $\newcommand{\qvec}{\mathbf q}$ $\newcommand{\svec}{\mathbf s}$ $\newcommand{\tvec}{\mathbf t}$ $\newcommand{\uvec}{\mathbf u}$ $\newcommand{\vvec}{\mathbf v}$ $\newcommand{\wvec}{\mathbf w}$ $\newcommand{\xvec}{\mathbf x}$ $\newcommand{\yvec}{\mathbf y}$ $\newcommand{\zvec}{\mathbf z}$ $\newcommand{\rvec}{\mathbf r}$ $\newcommand{\mvec}{\mathbf m}$ $\newcommand{\zerovec}{\mathbf 0}$ $\newcommand{\onevec}{\mathbf 1}$ $\newcommand{\real}{\mathbb R}$ $\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$ $\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$ $\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$ $\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$ $\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$ $\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$ $\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$ $\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$ $\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$ $\newcommand{\laspan}[1]{\text{Span}\{#1\}}$ $\newcommand{\bcal}{\cal B}$ $\newcommand{\ccal}{\cal C}$ $\newcommand{\scal}{\cal S}$ $\newcommand{\wcal}{\cal W}$ $\newcommand{\ecal}{\cal E}$ $\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$ $\newcommand{\gray}[1]{\color{gray}{#1}}$ $\newcommand{\lgray}[1]{\color{lightgray}{#1}}$ $\newcommand{\rank}{\operatorname{rank}}$ $\newcommand{\row}{\text{Row}}$ $\newcommand{\col}{\text{Col}}$ $\renewcommand{\row}{\text{Row}}$ $\newcommand{\nul}{\text{Nul}}$ $\newcommand{\var}{\text{Var}}$ $\newcommand{\corr}{\text{corr}}$ $\newcommand{\len}[1]{\left|#1\right|}$ $\newcommand{\bbar}{\overline{\bvec}}$ $\newcommand{\bhat}{\widehat{\bvec}}$ $\newcommand{\bperp}{\bvec^\perp}$ $\newcommand{\xhat}{\widehat{\xvec}}$ $\newcommand{\vhat}{\widehat{\vvec}}$ $\newcommand{\uhat}{\widehat{\uvec}}$ $\newcommand{\what}{\widehat{\wvec}}$ $\newcommand{\Sighat}{\widehat{\Sigma}}$ $\newcommand{\lt}{<}$ $\newcommand{\gt}{>}$ $\newcommand{\amp}{&}$ $\definecolor{fillinmathshade}{gray}{0.9}$

1. $A=200+.05(200)=\$ 210$

3. $\mathrm{I}=200 . \mathrm{t}=13 / 52(13 \text { weeks out of } 52 \text { in a year }) . \mathrm{P}_{0}=9800$

$200=9800(\mathrm{r})(13 / 52) \mathrm{r}=0.0816=8.16 \%$ annual rate

5. $P_{10}=300(1+.05 / 1)^{10(1)}=\$ 488.67$

$P_{20}=2000(1+.03 / 12)^{20(12)}=\$ 3641.51$ in 20 years
$3641.51-2000=\$ 1641.51$ in interest

9. $P_{8}=P_{0}(1+.06 / 12)^{8(12)}=6000 \cdot P_{0}=\$ 3717.14$ would be needed

11.

$P_{30}=\frac{200\left((1+0.03 / 12)^{30(12)}-1\right)}{0.03 / 12}=\$ 116,547.38$
$200(12)(30)=\$ 72,000$
$\$ 116,547.40-\$ 72,000=\$ 44,547.38$ of interest

13.

$P_{30}=800,000=\frac{d\left((1+0.06 / 12)^{30(12)}-1\right)}{0.06 / 12} \mathrm{d}=\$ 796.40$ each month
$\$ 796.40(12)(30)=\$ 286,704$
$\$ 800,000-\$ 286,704=\$ 513,296$ in interest

15.

$P_{0}=\frac{30000\left(1-(1+0.08 / 1)^{-25(1)}\right)}{0.08 / 1}=\$ 320,243.29$
$30000(25)=\$ 750,000$
$\$ 750,000-\$ 320,243.29=\$ 429,756.71$

17. $P_{0}=500,000=\frac{d\left(1-(1+0.06 / 12)^{-20(12)}\right)}{0.06 / 12} \mathrm{d}=\$ 3582.16$ each month

19.

$P_{0}=\frac{700\left(1-(1+0.05 / 12)^{-30(12)}\right)}{0.05 / 12}=$ a $\$ 130,397.13$ loan

$700(12)(30)=\$ 252,000$

$\$ 252,200-\$ 130,397.13=\$ 121,602.87$ in interest

21. $P_{0}=25,000=\frac{d\left(1-(1+0.02 / 12)^{-48}\right)}{0.02 / 12}=\$ 542.38$ a month

23.

Down payment of $10 \%$ is $\$ 20,000$, leaving $\$ 180,000$ as the loan amount

$P_{0}=180,000=\frac{d\left(1-(1+0.05 / 12)^{-30(12)}\right)}{0.05 / 12} \mathrm{d}=\$ 966.28 \mathrm{amonth}$

$P_{0}=180,000=\frac{d\left(1-(1+0.06 / 12)^{-30(12)}\right)}{0.06 / 12} \mathrm{d}=\$ 1079.19$ a month

25. First we find the monthly payments:

$P_{0}=24,000=\frac{d\left(1-(1+0.03 / 12)^{-5(12)}\right)}{0.03 / 12} \cdot d=\$ 431.25$

Remaining balance: $P_{0}=\frac{431.25\left(1-(1+0.03 / 12)^{-2(12)}\right)}{0.03 / 12}=\$ 10,033.45$

27. $6000(1+0.04 / 12)^{12 N}=10000$

$(1.00333)^{12 N}=1.667$

$\log \left((1.00333)^{12 N}\right)=\log (1.667)$

$12 N \log (1.00333)=\log (1.667)$

$N=\frac{\log (1.667)}{12 \log (1.00333)}=$ about 12.8 years

29. $3000=\frac{60\left(1-(1+0.14 / 12)^{-12 N}\right)}{0.14 / 12}$

$3000(0.14 / 12)=60\left(1-(1.0117)^{-12 N}\right)$

$\frac{3000(0.14 / 12)}{60}=0.5833=1-(1.0117)^{-12 N}$

$0.5833-1=-(1.0117)^{-12 N}$

$-(0.5833-1)=(1.0117)^{-12 N}$

$\log (0.4167)=\log \left((1.0117)^{-12 N}\right)$

$\log (0.4167)=-12 N \log (1.0117)$

$N=\frac{\log (0.4167)}{-12 \log (1.0117)}=$ about 6.3 years

31. First 5 years: $P_{5}=\frac{50\left((1+0.08 / 12)^{5(12)}-1\right)}{0.08 / 12}=\$ 3673.84$

Next 25 years: $3673.84(1+.08 / 12)^{25(12)}=\$ 26,966.65$

33. Working backwards, $P_{0}=\frac{10000\left(1-(1+0.08 / 4)^{-10(4)}\right)}{0.08 / 4}=\$ 273,554.79$ needed at retirement. To end up with that amount of money, $273,554.70=\frac{d\left((1+0.08 / 4)^{15(4)}-1\right)}{0.08 / 4}$. He’ll need to contribute $d=\$ 2398.52$ a quarter.

Search

Text Color

Text Size

Margin Size

Font Type