An investor deposits 50,000 into a new savings account, which earns an annual effective discount rate of 3.2%. The investor then withdraws an amount X at the end of every two-year period. The balance at the end of 12 years, just after the withdrawal, is 45,000.

Determine which of the following is an equation of value that can be used to solve for X.

• [[math]]\frac{45,000}{(1.032)^{12}}+X \sum_{k=1}^7 \frac{1}{(1.032)^{2(k-1)}}=50,000[[/math]]
• [[math]]\frac{45,000}{(1.032)^{12}}+X \sum_{k=1}^6 \frac{1}{(1.032)^{2 k}}=50,000[[/math]]
• [[math]]\quad 45,000(0.968)^{12}+X \sum_{k=1}^6(0.968)^{2 k}=50,000[[/math]]
• [[math]]\frac{45,000}{(1.032)^{12}}+X \sum_{k=1}^6(0.968)^{2 k}=50,000[[/math]]
• [[math]]\quad 45,000(0.968)^{12}+X \sum_{k=1}^6 \frac{1}{(1.032)^{2 k}}=50,000[[/math]]

Copyright 2023 . The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Solution: C

Since the annual effective discount rate is [math]3.2 \%[/math], the present value of an amount is calculated by multiplying it by a discounting factor of [math](1-0.032)^t=(0.968)^t[/math], where [math]t[/math] is the number of years since the deposit.

At time [math]t=0[/math], an initial deposit of 50,000 is made just after the balance is 0 (the account is new just before the deposit). The withdrawals are then [math]X[/math] at each of times [math]t=2,4,6,8,10,12[/math] or equivalently at time [math]t=2 k[/math] for each whole number [math]k[/math] from 1 to 6 inclusive.

Then to make the final balance 0 , an additional withdrawal of 45,000 at time [math]t=12[/math] would be needed.

Since the net present value of the cash flows (withdrawals minus deposits) must be zero, in a time period from a zero balance to another zero balance, we have

[[math]] \begin{aligned} & 45,000(0.968)^{12}+X \sum_{k=1}^6(0.968)^{2 k}-50,000=0 \\ & 45,000(0.968)^{12}+X \sum_{k=1}^6(0.968)^{2 k}=50,000 \end{aligned} [[/math]]

Copyright 2023 . The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.