Therefore, the size of the payments is $2562.81.
b.
Total payment of the loan = 2562.81 × 4 = $10,251.24
Therefore, the total interest paid in the loan period = 10,251.24 – 10,000 = $251.24
c.
Contracting an amortization schedule:
Payment
Number
Amount Paid
Interest portion
Principal
Portion
Principal
Balance
0



$10,000.00
1
$2562.81
$100.00
$2462.81
7527.19
2
2562.81
75.37
2487.44
5049.75
3
2562.81
50.50
2512.31
2537.44
4
2562.81
25.37
2537.44
0
Exercise 12.1, Solution 5:
a.
This is an ordinary simple annuity because:
•
Payments are made at the end of each payment period (annually)
•
Compounding period (annually) = payment period (annually)
n
= 1 payment/year × 5 years = 5 annual payments.
i = j
= 7.54% = 0.0754
The amount of the loan = 750,400(10.4) = $450.240.00
n
1 (1+ i)
i
PV = PMT
450,240 = PMT
Solving for
PMT
, we get
PMT
= $111,401.7157…
N
I/Y
P/Y
C/Y
PV
PMT
FV
5
7.54
1
1
450,240
?
0
From the calculator computations shown, we get the
PMT
= 111,401.7157
Therefore, the size of the payments is $111,401.72.
b.
Total payment of the loan = 111,401.72 × 5 = $557,008.60
c.
Contracting an amortization schedule:
Payment
Number
Amount Paid
Interest portion
Principal
Portion
Principal
Balance
0



$450,240.00
1
$111,401.72
$33,948.10
$77,453.62
372,786.38
2
111,401.72
28,108.09
83,293.63
289,492.75
3
111,401.72
21,827.75
89,573.97
199,918.78
4
111,401.72
15,073.88
96,327.84
103,590.94
5
111,401.70
7810.76
103,590.94
0.00