Question

On 1st January 2020, a loan was issued and it was amortized over 20 years with monthly payments at a nominal interest rate of 8.4% compounded monthly. The first payment was paid on 1st February 2020. The loan is structured in such a way that the first 120 installments will be paid in equal installments of N$6000 and the next 120 installments will be paid in such a way that each succeeding monthly payment will be 3% lower than the previous month. Calculate the outstanding loan balance immediately after the 160th payment is made. What is the value of the loan today?

Answer 1

The loan has total 20*12 = 240 payments and first 120 payments are of N$6000 each and then in each of the months from month 121 to 240 , payments are 3% lesser than the previous month i.e. each payment will be 0.97 times the previous payment

Monthly interest rate = 8.4%/12 = 0.007

Present value of the loan = 6000/0.007*(1-1/1.007^120)+ 6000*0.97/1.007^121+6000*0.97^2/1.007^122+.....+6000*0.97^120/1.007^240

= 486020.82 + 6000*0.97/1.007^121*(1-(0.97/1.007)^120)/(1-0.97/1.007)

=486020.82+67343.38

=N$ 553364.20 which is the value of the loan today

Loan balance immediately after 160th payment

= value of loan payments from month 161-240 at the time of month 160

= 6000*0.97^41/1.007 +6000*0.97^42/1.007^2 +6000*0.97^43/1.007^3 +...+6000*0.97^120/1.007^80

= 6000*0.97^41/1.007*(1-(0.97/1.007)^80)/(1-0.97/1.007)

=N$ 44186.80