Answer to Question #120609 in Financial Math for Ishal

Question

A family buys a house worth $326,000. They pay $75,000 deposit and take a mortgage for the balance at J12=9% p.a. to be amortized over 30 years with monthly payments.

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

A. Convert the interest rate J2=9% to J12 equivalent
B. Assuming that the family seeks to accept the change in interest rates, what would be their new payment based on the new interest rate?
C. 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

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

P-payment

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

$i=0.0884$

i=0.0884=8.84%

The interest rate J12 which equivalent J2 equals 8.84%

B) $0=((S50*q_{51}-P)*q_{52}-P)....*q_{360}-P$

$0=S50*q^{310}-P*q^{309}-....P*q-P$

$0=S50*q^{310}-P*(q^{309}+q^{308}*...q+1)$

$S50*q^{310}=P*\frac{1-q^{310}}{1-q}$

$S50=P*\frac{1-q^{310}}{1-q}/q^{310}$

$q_n$- certain year rate

$S50=P*\frac{1-q^n}{1-q}/q^n$

S50-sum of loan after 50 payments

P-payment

q=1.0075 (multiplier = 1+i; i=9%/12)

n- number of payments that are not made (310)

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

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

P'-value of new payments

C) Payments are made every month throughout the 30 years.

$n=30*12=360$

n-number full payments

$0=((S50*q_{359}-P)*q_{360}-P$

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

$S358=3994.21$

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)