FIN 534
Chapter 4 Homework
Loramie Johnson
January 26, 2013

1. (b) If the loan were amortized over 10 years rather than 7 years, and if the interest rate were the same in either case, the first payment would include more dollars of interest under the 7-year amortization plan. This statement is true. The first payment would income more dollars in interest since interest in each period is calculated by multiplying the loan balance at the beginning of the year by the interest rate. At the end of the payment schedule, the principal is less therefore the interest paid is less than the first payment.

2. (b) If you have a series of cash flows, and CF0 is negative but each of the following CFs is positive, you can solve for I, but only if the sum of the undiscounted cash flows exceeds cost. This statement is correct. Equation 4-10 of irregular cash flow can illustrate mathematically. Even if the CF0 is negative but the succeeding cash flows are positive and with value greater than the cost or present value, the value if I can be calculated.

3. (d) 0.30%. Midwest Bank will be charging 0.30% lower on effective annual rate.
EAR = [ 1 + (0.065/12)]12 – 1.0 = 0.37%
EAR = [ 1 + (0.07/ 1)]1 – 1.0 = 0.07%

4. (a) $3,726.00.
FVA46 = 2500 {[(1 + 0.08)46/0.08] – [1/0.08]} = 1,046,075

1046075 = PMT {[(1 + 0.08)41/0.08] – [1/0.08]} = 3725.60 = 3726.00

5. (c) $2,292.12
Total money needed for 4 year tuition:
14500 * {(1.035)^8} + 14500 * {(1.035)^9} + 14500 {(1.035)^10} + 14500 {(1.035)^11} = $80479.00

Tuition at t=8:
14500 * {(1.035)^7} = 18448.05

Tuition at t=9:
14500 * {(1.035)^8} = 19093.73

Tuition at t=10:
14500 * {(1.035)^9} = 19762.01

Tuition at t=11:
14500 * {(1.035)^10} = 20453.68

Accumulations on Savings A. Start-up money:
15000 * (1.09^7) = 27420.59 B. From $5000 contributions from t=1 to t=4
(5000*1.09^7) + (5000*1.09^6) + (5000*1.09^5) + (5000*1.09^4) = 32276.70 C. Total = 59697.30

Cash Flows while in college t | Fund on hand | Tuition | Remaining funds after tuition is paid for the year | Result {Year’s end * 1.09} | 8 | 59697.31 | 18448.05 | 41249.26 | 44961.6934 | 9 | 44961.69 | 19093.73 | 25867.963 | 28196.07967 | 10 | 27196.08 | 19762.01 | 7434.07 | 8103.1363 | 11 | 8103.1363 | 20453.68 | -12350.5437 | |

Total amount that needs to saved from t=5 to t=7:
12350.5437 / (1.09^3) = 9536.8858

3 Equal Contributions:
X*(1.09^3) + X*(1.09^2) + X*(1.09^1) = 9536.8858
X = 2292.12