Depreciation (AI HL)
Depreciation models an asset - a car, a machine, a laptop - losing a fixed percentage of its value every year. It's the mirror image of compound interest: instead of a balance growing, a value shrinks, but the underlying exponential model is identical. This page is a narrow slice of the wider Financial Maths topic, focused just on the depreciation formula and the exam patterns built around it.
11 questions on this sub-topic.
The formula you need
Covered under IB syllabus reference SL1.4: financial applications of geometric sequences and series, including compound interest and annual depreciation. Deriving the formula is never examined - only applying it.
Depreciation
\[FV = PV\left(1-\tfrac{r}{100}\right)^n\]
Same structure as compound interest, but the rate subtracts from 1 instead of adding to it. \(PV\) is the original value, \(r\) the annual rate of depreciation, \(n\) the number of years.
Value lost
\[\text{Value lost} = PV - FV\]
If a question asks how much value an asset has lost rather than what it's now worth, find \(FV\) first, then subtract it from \(PV\).
The depreciation formula isn't listed separately in the formula booklet - it's the compound interest formula with a negative rate. See Financial Maths for GDC finance-solver setup.
Worked examples
A car bought for $28,000 depreciates at 18% per year.
(a) Find its value after 3 years.
(b) Find the first year its value falls below $10,000.
Worked solution
(a) \(V = 28000(0.82)^3\) M1
\(\approx $15\,438.78.\) A1
(b) \((0.82)^n < 0.3571 \Rightarrow n > \dfrac{\ln 0.3571}{\ln 0.82}\) M1
\(\approx 5.19.\) A1
Year \(6.\) A1
A machine costing $20,000 can be depreciated using either:
Method A (straight-line): the value decreases by a fixed $2500 per year.
Method B (reducing balance): the value decreases by 15% of its current value each year.
(a)(i) Find the value under Method A after 4 years.
(a)(ii) Find the value under Method B after 4 years.
(a)(iii) State which method gives the higher value after 4 years.
Worked solution
(a)(i) \(V_A = 20000 - 2500(4)\) M1
\(V_A = $10\,000.\) A1
(a)(ii) \(V_B = 20000(0.85)^4\) M1
\(V_B \approx $10\,440.13.\) A1
(a)(iii) Since \($10\,440.13 > $10\,000,\) R1
Method B (reducing balance) gives the higher value after 4 years. A1
Common mistakes
- Adding instead of subtracting the rate. Depreciation uses \(\left(1-\tfrac{r}{100}\right)\), not \(\left(1+\tfrac{r}{100}\right)\) - copying the compound interest formula without flipping the sign gives a value that grows instead of shrinks.
- Answering "value lost" with the new value instead of the difference. If the question asks how much the asset has lost, subtract \(FV\) from \(PV\) - don't just state \(FV\) on its own.
- Rounding too early when solving for \(n\). Questions like "find the first year the value falls below..." need an inequality solved with logs, then rounded up to the next whole year - rounding the intermediate ratio first can nudge the answer into the wrong year.
Ready to practise properly?
11 depreciation questions, marked instantly like the real exam.
Quick answers
What is the depreciation formula in IB AI HL?
\(FV = PV\left(1-\tfrac{r}{100}\right)^n\), where \(PV\) is the original value, \(r\) is the annual rate of depreciation as a percentage, and \(n\) is the number of years.
Is depreciation given as its own formula in the formula booklet?
No - it isn't listed separately. It's the same compound interest formula, \(FV = PV\left(1+\tfrac{r}{100}\right)^n\), with a negative rate substituted in.