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3 years ago

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On a piece of paper or on a device with a touch screen, hand write the solution to the following problem. Then photograph or sav

e the file in .pdf form and submit it on this page. You would like to buy a house for $1,000,000. You put $200,000 down, and then get a mortgage for the rest at 4%, compounded monthly. What is the difference in the What is the difference in the monthly payment if you amortize the loan over 30 years vs. 15 years

1 answer:

aleksandr82 [10.1K]3 years ago

6 0

Answer:

The difference in monthly payment is:

= $2,098.18.

Explanation:

a) Data and Calculations:

Cost of the Mortgage House = $1,000,000

Down payment = $200,000 or 20%

Mortgage interest rate = 4%

Period of Mortgage amortization = 30 or 15

From an online financial calculator:

Monthly Pay: $3,819.32

House Price $1,000,000.00

Loan Amount $800,000.00

Down Payment $200,000.00

Total of 360 Mortgage Payments $1,374,956.05

Total Interest $574,956.05

Mortgage Payoff Date Apr. 2051

Monthly Pay: $5,917.50

House Price $1,000,000.00

Loan Amount $800,000.00

Down Payment $200,000.00

Total of 180 Mortgage Payments $1,065,150.61

Total Interest $265,150.61

Mortgage Payoff Date Apr. 2036

Monthly payment for 15 years = $5,917.50

Monthly payment for 30 years = 3,819.32

Difference in monthly payment = $2,098.18

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Answer:

Following are the solution to this question:

Explanation:

Assume that $r_1$ will be a 12-month for the spot rate:

$\to 1.25 \% \times \frac{100}{2} \times 0.99 + \frac{(1.25\% \times \frac{100}{2}+100)}{(1+\frac{r_1}{2})^2}=98\\\\\to \frac{1.25}{100} \times \frac{100}{2} \times 0.99 + \frac{(\frac{1.25}{100} \times \frac{100}{2}+100)}{(1+\frac{r_1}{2})^2}=98\\\\\to \frac{1.25}{2} \times 0.99 + \frac{(\frac{1.25}{2} +100)}{(1+\frac{r_1}{2})^2}=98\\\\\to 0.61875 + \frac{( 0.625 +100)}{(\frac{2+r_1}{2})^2}=98\\\\\to 0.61875 + \frac{( 100.625)}{(\frac{2+r_1}{2})^2}=98\\\\\to 0.61875 + \frac{402.5}{(2+r_1)^2}=98\\\\$

$\to 0.61875 + \frac{402.5}{(2+r_1)^2}=98\\\\\to 0.61875 -98 = \frac{402.5}{(2+r_1)^2}\\\\\to -97.38125= \frac{402.5}{(2+r_1)^2}\\\\\to (2+r_1)^2= \frac{402.5}{ -97.38125}\\\\\to (2+r_1)^2= -4.13\\\\ \to r_1=3.304\%$

Assume that $r_2$ will be a 18-month for the spot rate:

$\to 1.5\% \times \frac{100}{2} \times 0.99+1.5\% \times \frac{100}{2} \times \frac{1}{(1+ \frac{3.300\%}{2})^2}+\frac{(1.5\% \times \frac{100}{2}+100)}{(1+\frac{r_2}{2})^3}=97\\\\\to \frac{1.5}{100} \times \frac{100}{2} \times 0.99+\frac{1.5}{100} \times \frac{100}{2} \times \frac{1}{(1+ \frac{\frac{3.300}{100}}{2})^2}+\frac{(\frac{1.5}{100} \times \frac{100}{2}+100)}{(1+\frac{r_2}{2})^3}=97\\\\$

$\to \frac{1.5}{2} \times 0.99+\frac{1.5}{2}\times \frac{1}{(1+ \frac{\frac{3.300}{100}}{2})^2}+\frac{(\frac{1.5}{2} +100)}{(1+\frac{r_2}{2})^3}=97\\\\\to 0.7425+0.75 \times \frac{1}{(1+ \frac{\frac{3.300}{100}}{2})^2}+\frac{(0.75 +100)}{(1+\frac{r_2}{2})^3}=97\\\\\to 1.4925 \times \frac{1}{(1+0.0165)^2}+\frac{(100.75 )}{(1+\frac{r_2}{2})^3}=97\\\\\to 1.4925 \times \frac{1}{(1.033)}+\frac{(100.75 )}{(1+\frac{r_2}{2})^3}=97\\\\$

$\to 1.4925 \times 0.96+\frac{(100.75 )}{(1+\frac{r_2}{2})^3}=97\\\\\to 1.4328+\frac{(100.75 )}{(1+\frac{r_2}{2})^3}=97\\\\\to 1.4328-97= \frac{(100.75 )}{(1+\frac{r_2}{2})^3}\\\\\to -95.5672= \frac{(100.75 )}{(1+\frac{r_2}{2})^3}\\\\\to (1+\frac{r_2}{2})^3= -1.054\\\\\to r_2=3.577\%$

Assume that $r_3$ will be a 18-month for the spot rate:

$\to 1.25\% \times \frac{100}{2} \times 0.99+1.25\% \times \frac{100}{2} \times \frac{1}{(1+\frac{3.300\%}{2})^2}+1.25\%\times\frac{100}{2} \times \frac{1}{(1+\frac{3.577\%}{2})^3}+(1.25\% \times \frac{\frac{100}{2}+100}{(1+\frac{r_3}{2})^4})=96\\\\$

to solve this we get $r_3=3.335\%$

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