Mt217 Unit 4 Assignment

In: Business and Management

Submitted By salvimom
Words 405
Pages 2
John buys a house and pays it back in 5 years. The house is worth $150,000. The current rate is 6% and he expects rates to go up 1% every year. What does his amortization table look like?
Year 1 payment Year 2 payment Year 3 payment Year 4 payment Year 5 payment
N 5 N 4 N 3 N 2 N
I 6.0% I 7.0% I 8.0% I 9.0% I
PV -$150,000 PV -$123,391 PV -$95,600 PV -$66,152 PV
FV $0 FV $0 FV $0 FV $0 FV PMT $35,609.46 PMT $36,428.36 PMT $37,095.82 PMT $37,605.16 PMT Loan Amortization Schedule, $100,000 with variable rates
Amount borrowed: $150,000 The calculations for payment were done through excel calculations Beginning Amount (1) Payment (2) Interest (3) Repayment of Principal (4) Ending Balance (5) 150000 * 6% = 9000 interest on year 1 payment
Year 35609.46 - 9000 = 26609.46 Repayment of principal for year 1
0 $150,000.00 150000 - 26609.46 = 123390.54 Ending balance for year 1
1 $150,000.00 $35,609.46 $9,000.00 $26,609.46 $123,390.54 123390.54 * 7% = 8637.34 interest on year 2 payment
2 $123,390.54 $36,428.36 $8,637.34 $27,791.02 $95,599.52 36428.36 - 8637.34 = 27791.02 Repayment of principal for year 2
3 $95,599.52 $37,095.82 $7,647.96 $29,447.86 $66,151.66 123390.54 - 27791.02 = 95599.52 Ending balance for year 2
4 $66,151.66 $37,605.16 $5,953.65 $31,651.51 $34,500.15 95599.52 * 8% = 7647.96 interest on year 3 payment
5 $34,500.15 $37,950.17 $3,450.02 $34,500.15 $0.00 37095.82 - 7647.96 = 29447.86 Repayment of principal for year 3 95599.52 - 29447.86 = 66151.66 Ending balance for year 3 66151.66 * 9% = 5953.65 interest on year 4 payment Total payments: $184,688.96 37605.16 - 5953.65 = 31651.51 Repayment of principal for year 4 Total…...

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