Question

In: Finance

Scott and Linda have been saving to pay for their daughter Casie's college education. Casie just...

Scott and Linda have been saving to pay for their daughter Casie's college education. Casie just turned 10 at (t = 0), and she will be entering college 8 years from now (at t = 8). College tuition and expenses at State U. are currently $14,500 a year, but they are expected to increase at a rate of 3.5% a year. Ellen should graduate in 4 years¾if she takes longer or wants to go to graduate school, she will be on her own. Tuition and other costs will be due at the beginning of each school year (at t = 8, 9, 10, and 11). So far, Scott and Linda have accumulated $15,000 in their college savings account (at t = 0). Their long-run financial plan is to add an additional $5,000 in each of the next 4 years (at t = 1, 2, 3, and 4). Then they plan to make 3 equal annual contributions in each of the following years, t = 5, 6, and 7. They expect their investment account to earn 9%. How large must the annual payments at t = 5, 6, and 7 be to cover Casie's anticipated college costs?

a. $2,412.76

b. $2,177.51

c. $1,965.21

d. $2,292.12

e. $2,068.64

Solutions

Expert Solution

College Tuition Fees would be due at the end of Year 8, Year 9, Year 10 and Year 11. The amount due today would be $ 14500 and is expected to grow at 3.5 % per annum.

Therefore, Expense at end of Year 8 = 14500 x (1.035)^(8) = $ 19093.73

Expense at end of Year 9 = 14500 x (1.035)^(9) = $ 19762.01

Expense at end of Year 10 = 14500 x (1.035)^(10) = $ 20453.68

Expense at end of Year 11 = 14500 x (1.035)^(11) = $ 21169.56

Discounted Value of College Expenses at the end of Year 8 = 19093.73 + 19762.01 / (1.09) + 20453.68 / (1.09)^(2) + 21169.56 / (1.09)^(3) = $ 70786.25

Combined Future Value of Initial Accumulation and annual deposits upto Year 4 = 15000 x (1.09)^(8) + 5000 x (1.09)^(7) + 5000 x (1.09)^(6) + 5000 x (1.09)^(5) + 5000 x (1.09)^(4) = $ 62165.16

Let the equal annual contributions be $ K

Therefore, K x (1.09)^(3) + K x (1.09)^(2) + K x (1.09) = 70786.25 - 62165.16

K x 3.573129 = $ 8621.09

K = 8621.09 / 3.573129 = $ 2412.756 or $ 2412.76 approximately.

Hence, the correct option is (a).


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