Chapter 14, Problem 84P

To determine

The alternative that should be selected by the couple.

Answer to Problem 84P

Buying the house is an effective alternative.

Explanation of Solution

Calculation:

Alternative 1: Continue to rent the duplex home.

Determine the payment for year $1$.

$P_1=\left(\text{House}\text{rent}+\text{Basic}\text{utilities}\text{cost}\right)\times N$ ...... (I)

Here, N is the number of months per year, $P_1$ is the payment.

Substitute $\$450$ for house rent, $\$139$ for basic utilities cost and $12$ months for N.

$\begin{array}{c}{P}_1=\left(\$450+\text{\$139}\right)\times 12\\ =589\times 12\\ =\$7,068\end{array}$

Calculate the equivalent year $0$ payment in year $0$ dollars.

$P_0=P_1{\left(1+f\right)}^{-n}$ ...... (II)

Here, f is the inflation rate.

Substitute $\$7,068$ for $P_1$, $5\%$ for f and $1$ year for n in Equation (II).

$\begin{array}{c}{P}_0=\$7,068{\left(1+0.05\right)}^{-1}\\ =\$7,068{\left(1.05\right)}^{-1}\\ =\$7,068\left(0.9523\right)\\ =\$6,731.42\end{array}$

Calculate the equivalent interest rate.

$i=\left(i\text{'}+f+i\text{'}\times f\right)$ ...... (III)

Here, $i\text{'}$ is the nominal interest rate and i is the equivalent interest rate.

Substitute $15.5\%$ for $i\text{'}$ and $5\%$ for f.

$\begin{array}{c}i=\left(0.155+0.05+0.155\times 0.07\right)\\ =\left(0.205+0.01085\right)\\ =\left(0.21585\right)\times 100\\ =21.58\%\end{array}$

Calculate the present worth of house rent after $10$ years.

$P{W}_{10}=A\left(\frac{P}{A},i,n\right)$ ...... (IV)

Substitute $\$6,731.42$ for A, $21.58\%$ for i and $10$ years for n.

$\begin{array}{c}P{W}_{10}=\$6,731.42\left(\frac{P}{A},21.58\%,10\right)\\ =\$6,731.42\left(\frac{{\left(1+0.2158\right)}^{10}-1}{0.2158{\left(1+0.2158\right)}^{10}}\right)\\ =\$6,731.42\left(3.9768\right)\\ =\$26,769.74\end{array}$

Thus, the present worth of the house rent after $10$ years is $\$26,769.74$.

Alternative 1: Buying a house.

Calculate the mortgage interest rate per month.

$i_m=\frac{\text{Mortgage}\text{interest}\text{rate}}{12}$

Substitute $0.08$ for $\text{Mortgage}\text{interest}\text{rate}$.

$\begin{array}{c}{i}_m=\frac{\text{0}\text{.08}}{12}\\ =0.66\%\end{array}$

Calculate the number of payments.

$n=N\times 12$ ...... (V)

Substitute $30$ years for N in Equation (V).

$\begin{array}{c}n=30\times 12\\ =360\text{payments}\end{array}$

Calculate the down payment.

$\text{Down}\text{payment}=\text{House}\text{cost}\times \text{inflation}\text{rate}$ ...... (VI)

Substitute $\$75,000$ for house cost and $5\%$ for inflation rate.

$\begin{array}{c}\text{Down}\text{payment}=\$75,000\times \text{0}\text{.05}\\ \text{=}\$3,750\end{array}$

Calculate the closing cost in constant dollars.

$\text{Closing}\text{cost}\text{in}\text{constant}\text{dollars}=\left[\begin{array}{l}\text{House}\text{cost}\times \\ \text{interest}\text{rate}\text{of}\text{closing}\text{cost}\end{array}\right]$ ...... (VII)

Substitute $\$75,000$ for house cost and $1\%$ for interest rate at closing cost in Equation (VII).

$\begin{array}{c}\text{Closing}\text{cost}\text{in}\text{constant}\text{dollars}=\$75,000\times \text{0}\text{.01}\\ \text{=}\$750\end{array}$

Write the formula to Calculate the monthly payment.

$A=\left(\text{House}\text{cost}-\text{Down}\text{payment}\right)\left(\frac{A}{P},i_m,n\right)$ ...... (VIII)

Here, A is the monthly payment.

Substitute $\$75,000$ for house cost, $\$3750$ for down payment, $0.66\%$ for $i_m$ and $360$ for n in Equation (VIII).

$\begin{array}{c}A=\left(\$75,000-\$3750\right)\left(\frac{A}{P},0.66\%,360\right)\\ =\left(\$75,000-\$3750\right)\left(\frac{0.00667{\left(1+0.00667\right)}^{360}}{{\left(1+0.00667\right)}^{360}-1}\right)\\ =\$71,250\left(0.00734\right)\\ =\$523\end{array}$

Calculate the mortgage balance after $10$ year comparison period.

$A\text{'}=A\left(\frac{P}{A},i_m,n\right)$ ...... (IX)

Substitute $\$523$ for A, $0.66\%$ for $i_m$ and $240$ for n in Equation (IX).

$\begin{array}{c}A\text{'}=\$523\left(\frac{P}{A},0.66\%,240\right)\\ =\$523\left(\frac{{\left(1+0.0067\right)}^{240}-1}{0.0067{\left(1+0.0067\right)}^{240}}\right)\\ =\$523\left(119.51\right)\\ =\$62,504\end{array}$

Calculate the total amount paid in payments.

$n=A\times N\times 12$ ...... (X)

Here, N is number of years.

Substitute $\$523$ for A and $10$ years for N in Equation (X).

$\begin{array}{l}n=\$523\times 10\times 12\\ =\$62,760\end{array}$

Calculate the principal repayments.

$\text{Principal}\text{repayments}=\left(\text{Home}\text{cost}-\text{Down}\text{payment}\right)-A\text{'}$ ...... (XI)

Substitute $\$75,000$ for home cost, $\$3750$ for down payment and $\$62,504$ for $A\text{'}$ in Equation (XI).

$\begin{array}{c}\text{Principal}\text{repayments}=\left(\$75,000-\$3750\right)-\$62,504\\ =\$8,746\end{array}$

Calculate the interest payments.

$\text{Interest}\text{payments}=n-\text{Principal}\text{repayments}$ ...... (XII)

Substitute $\$62,760$ for n and $\$8,746$ for $\text{Principal}\text{repayments}$ in Equation (XII).

$\begin{array}{c}\text{Interest}\text{payments}=\$62,760-\$8,746\\ =\$54,014\end{array}$

Calculate the monthly tax saving.

$\text{Monthly}\text{tax}\text{saving}=\left(\begin{array}{l}\text{A}\times \text{Average}\text{interest}\text{rate}\text{on}\text{loan}\text{payment}\\ \times \text{Marginal}\text{income}\text{tax}\end{array}\right)$ ...... (XIII)

Substitute $\$523$ for A, $0.8772$ for $\text{Average}\text{interest}\text{rate}\text{on}\text{loan}\text{payment}$, $0.3$ for Marginal tax rate in Equation (XIII).

$\begin{array}{c}\text{Monthly}\text{tax}\text{saving}=\left(523\times \text{0}\text{.877}\times \text{0}\text{.3}\right)\\ =\$138\end{array}$

Calculate the cost of mortgage after tax.

$\text{Cost}\text{of}\text{mortgage}\text{after}\text{the}\text{tax}=A-\text{Monthly}\text{tax}\text{saving}$ ...... (XIV)

Substitute $\$523$ for A and $\$138$ for monthly tax saving in equation (XIV).

$\begin{array}{c}\text{Cost}\text{of}\text{mortgage}\text{after}\text{the}\text{tax}=523-\text{138}\\ \text{=\$385}\end{array}$

Calculate the sale amount of the property after $10$ years.

$\text{Sale}\text{amount}\text{of}\text{property}=\text{Home}\text{cost}\left(\frac{F}{P},i,n\right)$ ...... (XV)

Substitute $\$75,000$ for home cost and $6\%$ for i in Equation (XV).

$\begin{array}{c}\text{Sale}\text{amount}\text{of}\text{property}=\$75,000\left(\frac{F}{P},6\%,10\right)\\ =\$75,000{\left(1+0.06\right)}^{10}\\ =\$134,314\end{array}$

Calculate the net income from the sale.

$\text{Net}\text{income}\text{from}\text{the}\text{sale}=\left(\begin{array}{l}\text{Sale}\text{amount}\text{of}\text{the}\text{property}\\ -\left(\begin{array}{l}\text{Commission×}\\ \text{sale}\text{amount}\text{of}\text{the}\text{property}\end{array}\right)-A\text{'}\end{array}\right)$ ...... (XVI)

Substitute $\$134,314$ for sale amount and $5\%$ for Commission, $\$62,504$ for $A\text{'}$ in Equation (XVI).

$\begin{array}{c}\text{Net}\text{income}\text{from}\text{the}\text{sale}=\left(\$134,314-\left(\text{0}\text{.05×}\$134,314\right)-\$62,504\right)\\ =\$134,314-6716-62504\\ =\$65,094\end{array}$

Calculate the present worth of home cost of owning house for year $1$.

$P{W}_1=\left[\begin{array}{l}\left(\text{Down}\text{payment}+\text{Closing}\text{cost}\right)+\\ \left(\text{Cost}\text{of}\text{mortgage}\times 12\right)\left(\frac{P}{A},i\text{'},n\right)\\ +\left(\text{Constant}\text{dollar}\text{utilities}\times \text{12}\right)\left(\frac{P}{A},i,n\right)\\ +\left(\text{Constant}\text{dollar}\text{insurance}\text{and}\text{maintenance}\times 12\right)\left(\frac{P}{A},i,n\right)\\ -\text{Net}\text{income}\text{from}\text{the}\text{sale}\left(\frac{P}{F},i\text{'},n\right)\end{array}\right]$ ...... (XVII)

Substitute $\$3,750$ for down payment, $\$750$ for closing costs, $\$385$ for $\text{Cost}\text{of}\text{mortgage}$, $0.155$ for $i\text{'}$, $10$ years for n, $0.10$ for i and $\$65,094$ for $\text{Net}\text{income}\text{from}\text{the}\text{sale}$ in Equation (XVII).

$\begin{array}{c}P{W}_1=\left[\begin{array}{l}\left(\text{3750}+\text{750}\right)+\left(\text{385}\times 12\right)\left(\frac{{\left(1+0.155\right)}^{10}-1}{0.155{\left(1+0.155\right)}^{10}}\right)\\ +\left(\text{160}\times \text{12}\right)\left(\frac{{\left(1+0.10\right)}^{10}-1}{0.10{\left(1+0.10\right)}^{10}}\right)\\ +\left(\text{50}\times 12\right)\left(\frac{{\left(1+0.10\right)}^{10}-1}{0.10{\left(1+0.10\right)}^{10}}\right)\\ -\text{65,094}\left(\frac{1}{{\left(1+0.155\right)}^{10}}\right)\end{array}\right]\\ =\left[\begin{array}{l}4500+4620\left(4.925\right)+\left(1920\right)\left(6.144\right)+\left(600\right)\left(6.144\right)\\ -65,094\left(0.2366\right)\end{array}\right]\\ =\$27,335.14\end{array}$

Thus, the present worth of cost of owning the house is $\$27,335.14$.

Conclusion:

Thus, buying the house is an affective alternative.