A loan of $450,000 is amortized over 30 years with payments at
the end of each month and an interest rate of 6.3%, compounded
monthly.
Use Excel to create an amortization table showing, for each of the
360 payments, the beginning balance, the interest owed, the
principal, the payment amount, and the ending balance.
a) Find the amount of each payment. $
b) Find the total amount of interest paid during the first 15
payments. $
Suppose that payment number 6 is skipped and the interest owed for
month 6 is added to the balance. Payments then resume as usual for
the remainder of the 30 years.
c) Find the balance owing at the end of month 6.
$
d) Find the balance remaining after the 360th payment.
$
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Due to character limit I am attaching the images for amortization schedule.
PMT= | PV*i | ||||
1-(1+i)^-n | |||||
PV= | Loan amount | 450,000.00 | |||
n= | no. of payments | 360.00 | |||
i | 6.3%/12 | 0.00525 | |||
PMT= | 450000*0.00525 | ||||
1-(1+0.00525)^-360 | |||||
PMT= | 2362.5 | ||||
1-(1.00525)^-360 | |||||
PMT= | 2362.5 | ||||
1-(1.00525)^-360 | |||||
PMT= | 2362.5 | ||||
1-0.151820550212591 | |||||
PMT= | 2362.5 | ||||
0.84817945 | |||||
PMT= | 2785.377553 | ||||
PMT= | 2,785.38 | round to 2 decimal places | |||
Interest paid during first 15 months: | |||||
Sum of interest owed from the amortization table period 1 to 15:$ 35,199. | |||||
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