Question

A family buys a house worth $326,000. They pay $75,000 deposit and take a mortgage for the Find the loan outstanding after making 20 payments?
Find the principal repaid in the 21st payment?

Fill out the loan amortization schedule for the first 5 loan payments. What do you notice about the composition of the payment amount?
balance at J12=9% p.a. to be amortized over 30 years with monthly payments.

Find the value of the mortgage on their house?
Find the value of the monthly payment?

Suppose that after making 50 payments, the interest rate changes to J2=9% p.a.

Convert the interest rate J2=9% to J12 equivalent
Assuming that the family seeks to accept the change in interest rates, what would be their new payment based on the new interest rate?
Assuming that the family seeks to continue their initial monthly payment, calculated in part I, how many full payments would be required to pay off the loan and what would be the final concluding smaller payment one period later?

Answer 1

$P=S*\frac{i*(1+i)^n}{(1+i)^n-1}$

P-payments

S-sum of mortgage

$P=(326,000-75,000)*\frac{0.09/12*(1+0.09/12)^{360}}{(1+0.09/12)^{360}-1}=2019.6$

the loan outstanding after making 20 payments is:

$S20= 2019.6*\frac{1-1.0075^{340}}{1-1.0075}/1.0075^{340}$

$S20=248,053.15$

S20-loan outstanding after 20 payments

S21-sum of principal repaid in the 21st payment

$S21=248,053.15*1.0075=249,913.55$

The loan amortization schedule for the first 5 loan payments.

payment balance

1) 2019.6 250,862.9

2) 2019.6 250,724.77

3) 2019.6 250,585.61

4) 2019.6 250,445.4

5) 2019.6 250,304.14

I noticed the share of interest payments becomes less.

The value of the mortgage on their house =251,000

The value of the monthly payment =2019.6

$(1+0.09/2)^2=(1+i/12)^{12}$

$i=0.0884=8.84$

The interest rate J12 which equivalent J2 equals 8.84%

$S50=2019.6*\frac{1-1.0075^{310}}{1-1.0075}/1.0075^{310}=242,719.45$

S50-sum of loan after 50 payments

$P'=242,719.45*\frac{0.09/2*(1+0.09/2)^{310/6}}{(1+0.09/2)^{310/6}-1}=5,684.01$

P'-value of new payments

$n=30*12=360$

n-number full payments

$(S358*1.0075-2019.6)*1.0075-2019.6=0$

$S358=3994.21$

S358-sum remaining after 358 payments

$Pl=3994.21*1.0075^2=4054.35$

PL-last payment (the final concluding smaller payment one period later)