In: Finance
1. You are buying a new home with a purchase price of $189,500. You have a cash down payment of $18,950 and are financing the remaining amount at an interest rate of 4.5% for 30 years. Provide the following
2. Now suppose that you have financed the new house in problem #1 for 30 years. After 10 years into the Mortgage (April 2030), you’ve decided to make extra payments to pay the house off in 10 years (for a total of 20 years)
1
a
Principal = 189,500 - 18,950 = 170,550
b
Monthly payment | = | [P × R × (1+R)^N ] / [(1+R)^N -1] | |
Using the formula: | |||
Loan amount | P | $ 170,550 | |
Rate of interest per period: | |||
Annual rate of interest | 4.500% | ||
Frequency of payment | = | Once in 1 month period | |
Numer of payments in a year | = | 12/1 = | 12 |
Rate of interest per period | R | 0.045 /12 = | 0.3750% |
Total number of payments: | |||
Frequency of payment | = | Once in 1 month period | |
Number of years of loan repayment | = | 30 | |
Total number of payments | N | 30 × 12 = | 360 |
Period payment using the formula | = | [ 170550 × 0.00375 × (1+0.00375)^360] / [(1+0.00375 ^360 -1] | |
Monthly payment | = | $ 864.15 |
Monthly payment is $864.15
c
Total interest pay: | ||
Total payments | = | 864.15 × 360 |
$ 311,094.00 | ||
Less principle amount | $ 170,550.00 | |
Interest payment- Finance charge | $ 140,544.00 |
Total interest is $140,544
2
a
Loan balance | = | PV * (1+r)^n - P[(1+r)^n-1]/r |
Loan amount | PV = | 170,550.00 |
Rate of interest | r= | 0.3750% |
nth payment | n= | 120 |
Payment | P= | 864.15 |
Loan balance | = | 170550*(1+0.00375)^120 - 864.15*[(1+0.00375)^120-1]/0.00375 |
Loan balance | = | 136,592.80 |
Current balance is $136,592.80
b
Principal is $136,592.80
c
Monthly payment | = | [P × R × (1+R)^N ] / [(1+R)^N -1] | |
Using the formula: | |||
Loan amount | P | $ 136,593 | |
Rate of interest per period: | |||
Annual rate of interest | 4.500% | ||
Frequency of payment | = | Once in 1 month period | |
Numer of payments in a year | = | 12/1 = | 12 |
Rate of interest per period | R | 0.045 /12 = | 0.3750% |
Total number of payments: | |||
Frequency of payment | = | Once in 1 month period | |
Number of years of loan repayment | = | 10 | |
Total number of payments | N | 10 × 12 = | 120 |
Period payment using the formula | = | [ 136592.8 × 0.00375 × (1+0.00375)^120] / [(1+0.00375 ^120 -1] | |
Monthly payment | = | $ 1,415.63 |
New payment is $1,415.63
d
Interest for last ten years:
Total interest pay: | ||
Total payments | = | 1415.63 × 120 |
$ 169,875.60 | ||
Less principle amount | $ 136,592.80 | |
Interest payment- Finance charge | $ 33,282.80 |
Interest for first ten years =864.15*120-(170550-136592.8) |
$ 69,740.80 | |
Total interest | $ 103,023.60 |
Total interest paid is 103,023.60
please rate.