Question

The XYZ Company has borrowed $100,000. Payments will be made over a four-year period (first payment at the end of the first year). The bank charges interest of 0.20 per year.

a. The annual payment will be ________.

b. The debt amortization schedule is Amount owed (beginning of period) Interest Principal 1 $100,000 2 3 4

c. If there are five payments with the first payment made at the moment of borrowing, the annual payment will be ________.

Answer 1

a)

EMI = P*i*(1+i)^n/[{(1+i)^n}-1]

Where,

P = Principal = 100000

i= Interest Rate = 0.2

n= Number of periods = 4

Therefore, EMI = 100000*0.2*(1+0.2)^4/[{(1+0.2)^4}-1]

= 20000*(2.0736)/[1.0736] = $38628.91

b)

Amortization Schedule:

Period	Opening Principal (previous closing)	Interest (opening*0.2)	Installment	Principal Repayment (installment-interest)	Closing Principal (opening-principal repayment)
1	100000	20000	38628.91	18628.91	81371.09
2	81371.09	16274.218	38628.91	22354.692	59016.398
3	59016.398	11803.2796	38628.91	26825.6304	32190.7676
4	32190.7676	6438.15352	38628.9211	32190.7676	0

c)

Let, EMI be x. Therefore, Principal will be 100000-x

Therefore, x = (100000-x)*0.2*(1+0.2)^4/[{(1+0.2)^4}-1]

x = (20000-0.2x)*(2.0736)/[1.0736]

x = [20000-0.2x]*1.9314456

x = 38628.91 - 0.386289x

1.386289x = 38628.91

Therefore, New EMI = x = 38628.91/1.386289 = $27864.98