Appreciation And Depreciation
Assets are items of value that a person or company owns. Examples include: houses, cars, land, jewellery, shares in a company, currency, and machinery.
Some assets increase in value over time. This is called appreciation. Other assets decrease in value over time. This is called depreciation (or decay).
Before You Begin: Make sure you have reviewed our Simple and Compound Interest page and our Rates of Change page. Appreciation and depreciation use the same formulas as compound interest, the only difference is that appreciation adds value (growth), while depreciation subtracts value (decay).
Why understanding rates is important for appreciation and depreciation: The rate (often written as r in formulas) tells us how fast an asset grows or shrinks in value. For example, if a house appreciates at 8% per year, the rate is 8% (0.08). If a car depreciates at 15% per year, the rate is 15% (0.15). Without the rate, you cannot predict how the value will change over time. A higher rate means faster growth (appreciation) or faster loss (depreciation). Understanding rates helps you compare different investments (e.g., which house grows faster?) or decide when to sell an asset before it loses too much value.
Appreciation (Increase in Value)
Appreciation occurs when an asset increases in value over time. This happens with assets like houses, land, gold, and sometimes currency.
The Appreciation Formula
A = P (1 + r)t
Where:
- A = Final value of the asset after appreciation
- P = Initial value of the asset (principal)
- r = Rate of appreciation per year (written as a decimal)
- t = Time (in years)
Note: This is the compound interest formula because appreciation happens on the growing value each year.
Example 1: House Appreciation
A house is bought for M500,000. Its value appreciates at 8% per year. What will it be worth after 3 years?
Step 1: Identify the values.
P = M500,000
r = 8% = 0.08
t = 3 years
Step 2: Apply the appreciation formula.
A = P (1 + r)t
A = 500,000 × (1 + 0.08)3
Step 3: Calculate the bracket first.
1 + 0.08 = 1.08
Step 4: Raise to the power of 3.
1.083 = 1.08 × 1.08 × 1.08
1.08 × 1.08 = 1.1664
1.1664 × 1.08 = 1.259712
Step 5: Multiply by the initial value.
A = 500,000 × 1.259712 = M629,856
Answer: The house will be worth M629,856 after 3 years.
Example 2: Currency Appreciation
An investor bought currency worth M10,000. The currency appreciated at 5% per year. Find its value after 4 years.
Step 1: Identify the values.
P = M10,000
r = 5% = 0.05
t = 4 years
Step 2: Apply the appreciation formula.
A = 10,000 × (1 + 0.05)4
A = 10,000 × (1.05)4
Step 3: Calculate 1.054.
1.052 = 1.1025
1.054 = 1.1025 × 1.1025 = 1.21550625
Step 4: Multiply by the initial value.
A = 10,000 × 1.21550625 = M12,155.06
Answer: The currency will be worth M12,155.06 after 4 years.
Types of Appreciation
There are different types of appreciation depending on what is increasing in value:
- Capital Appreciation: When assets like houses, land, or shares increase in value.
- Currency Appreciation: When the value of a country's currency increases compared to another currency.
Note: Both types use the same formula A = P (1 + r)t.
Depreciation (Decrease in Value)
Depreciation occurs when an asset decreases in value over time. This happens with assets like cars, machinery, electronics, and office equipment.
The Depreciation Formula
A = P (1 - r)t
Where:
- A = Final value of the asset after depreciation
- P = Initial value of the asset (principal)
- r = Rate of depreciation per year (written as a decimal)
- t = Time (in years)
Note: We use (1 - r) because the value is decreasing by r each year.
Simple Depreciation (Straight-Line Method)
Simple depreciation reduces the value by the same amount each year. This is similar to simple interest, but decreasing instead of increasing.
Simple Depreciation Formula
A = P - (P × r × t)
Or: A = P (1 - r × t)
Example 3: Simple Depreciation of a Car
A car is bought for M150,000. It depreciates at 10% per year using the simple depreciation method. What is its value after 4 years?
Step 1: Identify the values.
P = M150,000
r = 10% = 0.10
t = 4 years
Step 2: Apply the simple depreciation formula.
A = P (1 - r × t)
A = 150,000 × (1 - 0.10 × 4)
A = 150,000 × (1 - 0.4)
A = 150,000 × 0.6
Step 3: Calculate the final value.
A = M90,000
Answer: The car will be worth M90,000 after 4 years.
Check: Each year, depreciation = 150,000 × 0.10 = M15,000. Over 4 years = M60,000 lost. 150,000 - 60,000 = M90,000.
Compound Depreciation (Reducing Balance Method)
Compound depreciation reduces the value by a percentage of the current value each year. This is similar to compound interest, but decreasing instead of increasing.
Compound Depreciation Formula
A = P (1 - r)t
Example 4: Compound Depreciation of a Car
A car is bought for M150,000. It depreciates at 10% per year using the compound depreciation method. What is its value after 4 years?
Step 1: Identify the values.
P = M150,000
r = 10% = 0.10
t = 4 years
Step 2: Apply the compound depreciation formula.
A = P (1 - r)t
A = 150,000 × (1 - 0.10)4
A = 150,000 × (0.9)4
Step 3: Calculate 0.94.
0.92 = 0.81
0.94 = 0.81 × 0.81 = 0.6561
Step 4: Multiply by the initial value.
A = 150,000 × 0.6561 = M98,415
Answer: The car will be worth M98,415 after 4 years using compound depreciation.
Comparing Simple and Compound Depreciation
The table below compares the value of a M150,000 car depreciating at 10% per year using both methods:
| Year | Simple Depreciation (10%) | Compound Depreciation (10%) |
|---|---|---|
| 0 | M150,000 | M150,000 |
| 1 | M135,000 | M135,000 |
| 2 | M120,000 | M121,500 |
| 3 | M105,000 | M109,350 |
| 4 | M90,000 | M98,415 |
Observation: Compound depreciation loses value more slowly in the early years because the depreciation amount decreases as the value decreases. Simple depreciation loses the same amount each year.
Summary Table: Appreciation vs Depreciation
| Concept | Formula | Example |
|---|---|---|
| Appreciation (Compound) | A = P (1 + r)t | House: M500,000 → M629,856 at 8% over 3 years |
| Simple Depreciation | A = P (1 - r × t) | Car: M150,000 → M90,000 at 10% over 4 years |
| Compound Depreciation | A = P (1 - r)t | Car: M150,000 → M98,415 at 10% over 4 years |
Key Takeaways:
• Appreciation uses (1 + r) — value increases over time.
• Depreciation uses (1 - r) — value decreases over time.
• Simple depreciation loses the same amount each year.
• Compound depreciation loses a percentage of the current value each year.
• Always convert the percentage rate to a decimal (divide by 100).
NB: Appreciation and depreciation are real-life applications of compound interest. When buying assets like cars or houses, it is important to understand how their value changes over time. Simple depreciation is often used for accounting purposes, while compound depreciation better reflects how assets like cars actually lose value.