Actuarial Mathematics: Annuity and Loan Practice Problems

Actuarial Science Practice Problems

Problem 11: Investment Option Comparison

11. $50,000 is invested in two options:

• Option A: Invested into a fund with an annual effective rate i.
• Option B: Purchase of an annuity-immediate with 24 level annual payments at a rate of 10%.

Option B is equivalent to Option A with an interest rate of 5%. The accumulated value for Option A is 50,000(1+i)²⁴. For Option B: * = Option A. Find i.

Problem 14: Monthly Deposits and Future Value

14. Smith deposits $1,000 at the end of each month with a nominal rate i⁽¹²⁾ = 0.12. The first deposit is on 1/31/2010, and the last is on 12/31/2034. From 2035 to 2059, the accumulated value Y is:

Problem 15: Interest Rate Changes and Balances

15. Since 6/30/2014, a deposit of $100 is made at the end of each month. During 2014-2015, the account earns a compounded monthly annual rate of 9% (0.0075 per month). For the first 9 months of 2016, i⁽¹²⁾ = 0.105, and after that, i⁽¹²⁾ = 0.12. Find the balance at 1/1/2015: 100s_7|0.0075. Find the balance for February 2017:

Problem 16: Accumulation with Variable Deposits

16. To accumulate $8,000 at the end of 3n years, deposits of $98 are made at the end of each year for n years, and $196 for the following 2n years. The effective annual rate is i. Given (1+i)ⁿ = 2, find i.

Problem 113: Savings for Future Purchases

113. Chunk plans to purchase an item in 10 years. The cost is $200 today, increasing by 4% per year. He deposits $20 at the beginning of 6 years. He also makes extra deposits of X at years 4, 5, and 6 to meet the goal. The annual rate is 10%.

Problem 120: Loan Repayment with Step-Up Payments

120. A loan of $10,000 is being repaid with 10 semi-annual payments, with the first one occurring 1.5 years after the loan is issued. The first 5 payments are K each, and the final 5 are K + 200. What is K if i⁽²⁾ = 0.06 (0.06/2 = 0.03)?

Problem 21: Monthly Loan Amortization

21. A $50,000 loan was made on Jan 1, 2010, to be repaid over 25 years at the end of each month beginning Jan 1, 2010.

• a) If i⁽²⁾ = 0.10, find the monthly payment X.
• b) Find n = 168 using X + 100 = 547. (where v is vⁿ).

Problem 24: Quarterly Deposits and Annuity Depletion

24. Jerry deposits $450 at the end of each quarterly period for 10 years. At the end of 15 years, Jerry uses the fund to make payments of Y at the beginning of each year for 4 years, after which the fund is exhausted. The effective annual rate is 7%. Find Y. (where v is at 7%).

Problem 26: Perpetuity Present Value Comparison

26. A perpetuity paying 1 at the beginning of each 6-month period has a Present Value (PV) of 20. A second perpetuity pays X every 2 years. If the rates are the same and the two PVs are equal, find X. Get j = 1/19.

Problem 231: Retirement Income with Annual Increases

231. Stan receives retirement income over 20 years at a rate of $2,000 per month, beginning every month from now. The payment increases by 5% every year. The nominal rate is 6% monthly. Calculate the Present Value. 6%/12 = 0.005; i = (1.005)¹² - 1. = 24,671 (i - r).

Problem 2312: Increasing Monthly Annuity Calculation

2312. Olga buys a 5-year increasing annuity for X. She receives $2 at the end of the first month and $4 at the end of the second month. The payments increase by $2 each month thereafter. The rate is 9% quarterly. Find X.

Problem 2313: Reinvestment of Interest Income

2313. $1,000 is deposited into account X, earning an annual rate of 6%. At the end of each year, the interest plus $100 of the principal is withdrawn. This continues for ten years. these withdrawals are deposited into account Y, which earns 9%. Find the accumulated value of account Y after 10 years.

Problem 311: Loan Amortization Schedule Analysis

311. A loan has 10 annual payments at the end of each year. The first 5 payments are $1,000 each, and the last 5 are $500 each. The annual rate is 10%.

• a) Initial amount: 1000a_5|0.1 + 500v⁵a_5|0.1 = 4,967
• b) Outstanding balance after 3rd payment: OB3 = 4,987 - 1,000s_3|0.1 = 3,302
• c) Interest and principal in 4th payment: I4 = 3,302 * 0.1 = 330; Principal = 1,000 - 330
• d) OB8: 500a_2|0.1 = 867

Problem 312: Geometrically Decreasing Loan Payments

312. A five-year loan has monthly payments at a rate of 9%. The first payment is $1,000. Each subsequent month, the payment is 2% lower than the month before. Find the balance after the 40th payment.

Problem 324: Loan Balance Threshold Analysis

324. A 5-year loan was made on July 1, 2014, with 60 level monthly payments starting August 1, 2014. The rate is i⁽¹²⁾ = 0.12. Find when the balance falls below one-half of the original amount.