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Calculate required savings for school fees, accounting for inflation and investment returns.
This calculator determines how much you need to save each year to cover future school fees. It accounts for fee inflation, investment returns, and ensures you'll always have sufficient funds when payments are due, even with multiple children starting at different times.
The chart below shows how the required annual contribution changes across different combinations of fee inflation and investment return rates. The x-axis varies inflation around your input value, while each coloured line represents a different return rate. This helps you understand how sensitive your savings plan is to these uncertain assumptions: if returns are lower or inflation higher than expected, you'll need to save more.
The table provides a year-by-year breakdown showing your annual contribution, fee payments, investment returns, and running account balance. The row marked with an asterisk (*) is the binding constraint year, the point where your account balance is tightest (reaches zero). This is the year that determines your minimum required contribution.
The collapsible section at the bottom provides a rigorous derivation of the calculation methodology, including the balance evolution equations, geometric series simplifications, and the liquidity constraint that ensures your account never goes negative.
| Year | Contribution | Fee Payment | Investment Return | End Balance |
|---|
Money has time value: £1 today is worth more than £1 tomorrow because you can invest it and earn a return. This means we cannot directly add or compare amounts at different points in time; we must first move them to a common reference point.
Compounding (moving forward in time): If you invest £\(P\) today at annual return \(r\), after one year you have:
After \(t\) years, the balance grows to:
Discounting (moving backward in time): If you need £\(F\) in \(t\) years, the amount you must set aside today (the present value) is:
We have \(K\) children, where child \(k\) starts school in year \(Y_k\) (measured from today). Each child attends for \(n\) years. We also allow for a delay \(D \geq 0\) before saving begins, and an initial starting budget \(S \geq 0\). The core parameters are:
Timeline transformation: When \(D > 0\), we shift the timeline so that "Year 1" is when saving begins. This requires transforming the inputs:
All subsequent formulas use these transformed values \(A'\), \(Y'_k\), and \(S'\). For simplicity, we drop the primes and write \(A\), \(Y_k\), and \(S\) with the understanding that they represent the values at the time saving begins. We then define:
School fees grow with inflation from year 1. In calendar year \(t\), the fee for one child in school is:
Child \(k\) is in school during years \(Y_k, Y_k+1, \ldots, Y_k+n-1\). We define an indicator function:
The total fees in year \(t\) sum across all children in school that year:
where \(N_t = \sum_{k=1}^{K} \mathbf{1}_k(t)\) is the number of children in school during year \(t\).
Example: Two children with \(Y_1 = 2\) and \(Y_2 = 5\), each attending for \(n = 3\) years:
| Year \(t\) | Child 1 | Child 2 | \(N_t\) | Fee \(F_t\) |
|---|---|---|---|---|
| 1 | — | — | 0 | 0 |
| 2 | Year 1 | — | 1 | \(A(1+g)^1\) |
| 3 | Year 2 | — | 1 | \(A(1+g)^2\) |
| 4 | Year 3 | — | 1 | \(A(1+g)^3\) |
| 5 | — | Year 1 | 1 | \(A(1+g)^4\) |
| 6 | — | Year 2 | 1 | \(A(1+g)^5\) |
| 7 | — | Year 3 | 1 | \(A(1+g)^6\) |
Let \(I_t \in \{0, 1\}\) indicate whether we contribute in year \(t\), and let \(C\) be the annual contribution amount (what we solve for).
Mode 1 (save only before school starts):
Mode 2 (continue contributing during school):
Let \(B_t\) denote the account balance at the end of year \(t\). The timing within each year is:
This gives the recurrence relation:
We solve the recurrence by expanding it step by step. With starting budget \(S\):
The pattern becomes clear. At year \(t\):
We define the compounded starting budget:
This is the value at year \(t\) of the initial savings, growing at the investment return rate.
The accumulated contribution factor:
This is the future value at year \(t\) of £1 contributed in each contributing year.
And the accumulated fee value:
This is the future value at year \(t\) of all fees paid through year \(t\).
Substituting into our expanded recurrence gives the closed-form balance equation:
In practice, \(S_t\), \(A_t\), and \(V_t\) are computed iteratively using recurrence relations:
The fundamental requirement is that the account balance must never go negative:
Substituting the closed-form balance (6) into constraint (10):
The starting budget \(S_t\) effectively reduces the fee burden. Constraint (11) must hold for every year \(t\) where \(A_t > 0\). Therefore, the minimum contribution is:
The outer max ensures \(C_{\min} \geq 0\) (if the starting budget covers all fees, no contributions are needed). The year \(t^*\) where the inner maximum is achieved is the binding constraint year.
A naive present-value approach sets contributions equal to fees in present value terms, which is equivalent to enforcing only the terminal constraint \(B_T = 0\) from (10):
Comparing (13) with our solution (12), the naive approach fails when an intermediate year \(t < T\) has:
In this case, using \(C_{\text{naive}}\) in equation (6) gives \(B_t = S_t + C_{\text{naive}} \cdot A_t - V_t < 0\), so the account goes negative, requiring borrowing.
When does this happen? From equations (8) and (9), the ratio \((V_t - S_t) / A_t\) increases when fees are paid (adding to \(V_t\)) and decreases when contributions are made without corresponding fees (adding to \(A_t\)). An intermediate year becomes binding when:
Consider:
Fee schedule:
| Year | Children in school | Fee \(F_t\) |
|---|---|---|
| 1 | 0 | £0 |
| 2 | 3 (children 1,2,3) | £30,000 |
| 3 | 3 | £30,000 |
| 4 | 3 | £30,000 |
| 5 | 0 | £0 |
| 6 | 1 (child 4) | £10,000 |
| 7 | 1 | £10,000 |
| 8 | 1 | £10,000 |
Computing \(A_t\) and \(V_t\) using recurrences (8) and (9) with \(r = 0.05\) and \(S = 0\):
| Year \(t\) | \(A_t\) | \(V_t\) | \(V_t / A_t\) |
|---|---|---|---|
| 1 | 1.000 | 0 | 0 |
| 2 | 2.050 | 30,000 | 14,634 |
| 3 | 3.153 | 61,500 | 19,513 |
| 4 | 4.310 | 94,575 | 21,943 |
| 5 | 5.526 | 99,304 | 17,974 |
| 6 | 6.802 | 114,269 | 16,799 |
| 7 | 8.142 | 129,982 | 15,965 |
| 8 | 9.549 | 146,481 | 15,340 |
Applying equation (12) with \(S = 0\), the maximum ratio occurs at year 4 (value £21,943), not year 8. This is the binding constraint year \(t^* = 4\). Using the naive approach (13) would give \(C_{\text{naive}} = 15{,}340\), but substituting into (6) gives \(B_4 = 15{,}340 \times 4.310 - 94{,}575 = -28{,}470 < 0\), so the account goes negative.
The correct minimum contribution from (12) is \(C_{\min} = £21{,}943\) per year, ensuring \(B_t \geq 0\) for all \(t\) as required by (10).
The ratio \(\rho = \frac{1+g}{1+r}\) (appearing in the fee calculations) determines the relationship between fee growth and investment returns:
In Mode 1, once contributions stop (\(I_t = 0\)), examining equations (8) and (9) shows that \(A_t\) only grows by compounding while \(V_t\) grows by compounding plus new fees. Thus \((V_t - S_t) / A_t\) can only increase (when fees are paid) or stay constant (during gaps), so the binding constraint in (12) is always the terminal year. The naive approach (13) suffices.
In Mode 2, ongoing contributions (\(I_t = 1\)) add to \(A_t\) each year, which can cause the ratio \((V_t - S_t) / A_t\) to decrease during gap years or when fees drop. This is precisely when an intermediate year can become the binding constraint in (12), and the liquidity-aware calculation becomes essential.