Question

 

Introduction

You are a financial planner and a new client, Kristina came to your office with the following question: How much should she save annually given her goals?

• Kristina is saving for two goals: one is to fund her child Katryna’s college education and the other is to retire.
• Katryna will begin college in exactly 18 years, and Kristina will retire on the same day.
• College costs for state institutions are currently running at $6,000 per year, and have historically increased at a rate of 5% per year.
• Kristina will pay for a 4-year undergraduate education.
• Kristina estimates she will live 30-year past her retirement, during which time she would like to pay herself a $60,000 per year salary (first payment occurs at the end of year 18, or beginning of year 19).
• On the day she retires, Kristina would like to pay cash for a BMW Z4 convertible or whatever similar model BMW produces at that time. A Z4 cost $55,000 and BMW has increased price at a rate of 7% per year.
• Kristina will save annually beginning in exactly one year—18 equal payments spread over time.
• The date of the last savings installment coincides with the first tuition payment (and the day of Kristina retirement).
• Kristina estimates an 8% expected return on all investments over the period.
• How much does she have to save each year to accomplish the above goals?

Step 7: Create a table showing all the additions and subtraction to the savings accounts and the value at the end of each year. (Hint: the value should be close to zero at the end) 25 points

Answer 1

let the retirement day be the reference point (18 years from today)

The cost of college in year 1 (18th year) = 6000*1.05^18 = 14439.72

The cost of college in year 2 (19th year) = 6000*1.05^19 =15161.70

The cost of college in year 3 (20th year)= 6000*1.05^20 =15919.79

The cost of college in year 4(21st year) = 6000*1.05^21 =16715.78

Cost of education (PV) at the time of retirement

= 6000*1.05^18 + 6000*1.05^19/1.08 + 6000*1.05^20/1.08^2+ 6000*1.05^21/1.08^3

=$55396.50

Value of Retirement income = 60000 + 60000/1.08 + ..+ 60000/1.08^29

=60000/0.08*(1-1/1.08^30) * 1.08

=$729504.36   

Cost of BMW at retirement = 55000*1.07^18 = $185896.28 (paid at end of 18th year)

So, total balance required in the savings account at the time of retirement

= 55396.50+729504.36+185896.28 = $970797.14

Amount deposited every year X is given by

X/0.08*(1.08^18-1) = 970797.14

=> 37.4502437 *X =970797.14

=> X = 25922.32

Kristina has to save $25922.32 every year for the next 18 years

The table showing all the additions and subtraction to the savings accounts and the value at the end of each year is attached as image

At the end of year 18, withdrawals = Education cost + BMW cost + salary

At the end of year 19, withdrawals = Education cost + salary

At the end of year 20, withdrawals = Education cost + salary

At the end of year 21, withdrawals = Education cost + salary

Year 22 onwards till end of year 47 (when last salary is withdrawn) , only salary amount of 60000

One can see that at the end of 47th year, balance is nearly 0 (not equal to zero because of rounding off)  

Interest is earned at 8% on beginning balance