Valuing Irregular Savings and Withdrawal Schedules

Why Irregular Savings Plans Need a Common Date

A savings plan may combine an opening balance, deposits on selected dates, and withdrawals whenever funds are needed. Adding those amounts directly does not tell us the balance at a particular time: a deposit made earlier has earned interest for longer, while a withdrawal removes funds that otherwise would have earned interest. The consistent method is to move every transaction to one date and then combine the values there.

The method applies to personal savings, reserve funds, and other hypothetical accumulation problems. It is aligned with the SOA VEE Accounting and Finance syllabus. The focus here is valuing an uneven schedule and solving for a required transaction, rather than valuing a regular annuity.

Definition: A dated cash flow $C_j$ is the amount transacted at time $t_j$, measured in years from a stated time origin. For an account-holder's savings plan, a deposit is positive and a withdrawal is negative. The value of the schedule at comparison time $T$ is the sum of each signed cash flow accumulated or discounted to $T$.

Unless stated otherwise, assume a constant effective annual interest rate $i$, with interest compounded according to the effective annual rate. The account has no fees, taxes, or other transactions beyond those listed. Times are measured in years; the examples use whole-year dates. The account is assumed to permit the transactions shown and not to impose a separate minimum-balance restriction. If rates vary by period, or spot rates are specified, each cash flow must instead use the applicable date-specific accumulation or discount factor.

The Common-Date Valuation Rule

A dollar invested at time $t_j$ accumulates for $T-t_j$ years to $1(1+i)^{T-t_j}$ dollars at time $T$. The exponent is positive when the transaction occurs before $T$, zero when it occurs at $T$, and negative when it occurs after $T$. Thus, the value of a schedule at $T$ is:

V_T=\sum_{j=1}^{n} C_j(1+i)^{T-t_j}.

If $t_j<T$, the term accumulates the cash flow forward. If $t_j>T$, the negative exponent discounts it back to $T$. This formulation works with signed amounts: a withdrawal contributes a negative term, whether it is accumulated or discounted.

Why the Formula Holds

The formula follows from applying compound interest to one transaction at a time and then using the account's linearity with respect to its balance. A cash flow $C_j$ at time $t_j$ grows to $C_j(1+i)^{T-t_j}$ at $T$. Since balances add, the total balance is the sum of these individual contributions. A withdrawal is simply a negative contribution.

For example, if a deposit is made at time 0 and a withdrawal at time 2, the balance at time 3 can be found by accumulating both signed amounts to time 3. The result is the same as first calculating the balance just after time 2 and then accumulating that balance one more year. This consistency is why transactions can be moved to a common date without changing the account value.

Choose the comparison date.
Use the date at which the requested balance or target is stated.

Record dated amounts with signs.
Enter deposits as positive and withdrawals as negative from the account's perspective.

Move every amount to the chosen date.
Multiply by $(1+i)^{T-t_j}$, using a negative exponent for a later transaction.

Add and interpret.
Sum the terms, then check whether the result is consistent with the deposits, withdrawals, and target.

Worked Savings Schedules

Worked Example: Finding the Opening Deposit for a Target Balance

A plan must have $5,000 at time 5. It will receive a $1,200 deposit at time 2 and make a $700 withdrawal at time 4. What opening deposit $P$ is required at time 0? Assume a constant effective annual interest rate of 4%, with no fees or other transactions.

Method. Accumulate every transaction to time 5, using a positive sign for deposits and a negative sign for the withdrawal. The opening deposit accumulates for five years; the other transactions accumulate for three and one years, respectively.

$$ P(1.04)^5+1200(1.04)^3-700(1.04)=5000. $$

Calculate each known amount at time 5:

(1.04)^3=1.124864,\qquad 1200(1.124864)=1349.8368,

700(1.04)=728,\qquad (1.04)^5=1.2166529024.

Rearrange the target equation. The withdrawal reduces the account balance, so its accumulated amount is added back when isolating the opening deposit:

$$ P(1.2166529024)=5000-1349.8368+728=4378.1632. $$

Therefore,

P=\frac{4378.1632}{1.2166529024}=3598.531\ldots

Answer: The required opening deposit is $3,598.53, rounded to the nearest cent. As a check, its accumulated value is approximately $4,378.16; adding the $1,349.84 deposit and subtracting the $728.00 withdrawal gives $5,000.00 after rounding.

Common mistake avoided: Subtracting the withdrawal again when solving for $P$ would treat it as an additional deposit in the target equation. The signed equation at time 5 is $P(1.04)^5+1349.8368-728=5000$, not $P(1.04)^5+1349.8368+728=5000$.

Worked Example: Calculating the Balance After Several Transactions

A savings account receives $1,000 at time 0 and $500 at time 2. The account-holder withdraws $200 at time 3 and another $300 at time 4. Find the balance at time 5, assuming a constant effective annual interest rate of 10%, no fees, and no other transactions.

Method. Accumulate each dated transaction separately to time 5. The respective accumulation periods are five, three, two, and one year.

$$ V_5=1000(1.10)^5+500(1.10)^3-200(1.10)^2-300(1.10). $$

Evaluate the powers and transaction values:

(1.10)^5=1.61051,\quad 1000(1.61051)=1610.51,

(1.10)^3=1.331,\quad 500(1.331)=665.50,

(1.10)^2=1.21,\quad 200(1.21)=242.00,\quad 300(1.10)=330.00.

Adding the signed amounts gives:

$$ V_5=1610.51+665.50-242.00-330.00=1704.01. $$

Answer: The balance at time 5 is $1,704.01. As a check, the accumulated deposits total $2,276.01 at time 5 and the accumulated withdrawals total $572.00, leaving $1,704.01.

Common mistake avoided: Adding the original deposits and subtracting the original withdrawals without interest would give $1,000, which is not the balance at time 5. Each amount must first be moved to the requested date.

Worked Example: Solving for a Required Later Deposit

A savings plan has a $2,000 deposit at time 0, a $700 deposit at time 1, and a $500 withdrawal at time 2. What additional deposit at time 4 is needed for the account to reach $4,000 at time 5? Assume a constant effective annual interest rate of 5%, with no fees or other transactions.

Method. Accumulate the known transactions to time 5, then solve for the time-4 deposit. Let $D$ be that deposit. It earns interest for one year.

$$ 2000(1.05)^5+700(1.05)^4-500(1.05)^3+D(1.05)=4000. $$

The known terms are:

$$ 2000(1.05)^5=2000(1.2762815625)=2552.563125, $$

$$ 700(1.05)^4=700(1.21550625)=850.854375, $$

$$ 500(1.05)^3=500(1.157625)=578.8125. $$

The known transactions therefore contribute:

$$ 2552.563125+850.854375-578.8125=2824.605. $$

The deposit must supply the remaining value at time 5, so:

D(1.05)=4000-2824.605=1175.395, \qquad D=\frac{1175.395}{1.05}=1119.423809\ldots

Answer: The required deposit at time 4 is $1,119.42, rounded to the nearest cent. Checking at time 5, the known transactions provide $2,824.605 and the deposit accumulates to $1,175.395, for a total of $4,000.000.

Common mistake avoided: The $1,175.395 is the amount needed at time 5, not the deposit amount at time 4. Dividing by $1.05$ discounts that shortfall one year to the date the deposit is made.

Sign and Timing Checks

The most reliable check is to write the cash-flow equation at the requested date before rearranging it. This makes the role of each transaction visible and reduces the chance of reversing a sign while solving for an unknown. A deposit at time $t$ contributes positively; a withdrawal at time $t$ contributes negatively. If the unknown is a deposit, a positive solution is consistent with needing additional funds. A negative solution means the target would instead require a withdrawal, or that the stated transaction has been described with the wrong sign.

A second useful check is the time interval in each exponent. A transaction at time 2 moved to time 5 earns interest for $5-2=3$ years, not five years. A transaction at time 5 has exponent zero when the comparison date is time 5, so its value is unchanged. If the comparison date is earlier than a transaction, the exponent is negative because the later amount must be discounted back.

The common-date method assumes that the specified interest rate applies over each period being traversed. If a problem gives different rates for different years, do not use one power of a single rate for the entire interval: use the product of the applicable period-by-period accumulation factors. If the plan includes fees or taxes, represent them as additional dated cash flows or incorporate them only through an explicitly specified net rate. A calculation that omits a stated charge does not value the actual plan.

Key takeaway: To value an irregular savings or withdrawal schedule, assign each transaction a sign, move it to the date of interest using the correct time interval, and add the resulting amounts. When a target is given, place the target and all accumulated transactions in one equation before solving for the unknown.

Check Your Understanding

For each problem, assume no fees, taxes, or unlisted transactions. Round money to the nearest cent.

An account receives $500 at time 0 and $300 at time 2, then pays a $100 withdrawal at time 3. Find its balance at time 4 if the effective annual interest rate is 6%. (Answer: $862.32)
A plan must hold $2,000 at time 3. It will receive $400 at time 1 and pay a $250 withdrawal at time 2. Find the required opening deposit at time 0 if the effective annual interest rate is 5%. (Answer: $1,573.48)
An account receives $1,000 at time 0 and pays a $200 withdrawal at time 1. What deposit at time 2 is required to produce a balance of $2,000 at time 4, at a 10% effective annual interest rate? (Answer: $663.72)
Find the value at time 2 of a $1,000 deposit at time 0 and a $300 withdrawal at time 3, using a 5% effective annual interest rate. Show why the withdrawal is discounted to time 2 rather than accumulated. (Answer: $816.79)

What You'll Learn