Comparing Linear and Percentage Depreciation
A bakery buys a delivery van for $48,000. The owner is comparing two depreciation models for tax planning. Model L assumes the van loses $6,000 of value each year. Model E assumes it loses 15% of its current value each year. Both models start at $48,000 in year 0. The owner plans to replace the van when its value falls below $25,000, and wants to know how the two models compare in the first several years, when they cross, and which model gives a lower value at the replacement point. The accountant warns that 'the percentage model always loses value faster,' and the owner wants that claim checked with numbers rather than accepted. The van is used about 18,000 miles per year, so resale value matters to the trade-in price.
1. Write both depreciation functions and compute values at years 1, 3, and 4. Identify the interval in which the two models cross and explain what the crossing means in context.
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Linear: L(t) = 48000 - 6000t. Exponential: E(t) = 48000(0.85)^t.
Year 0: L = 48,000; E = 48,000. Year 1: L = 42,000; E = 48,000(0.85) = 40,800. Linear is higher by 1,200, meaning the linear model removed more value in year 1. Year 3: L = 48,000 - 18,000 = 30,000; E = 48,000(0.85)^3 = 48,000(0.614125) = 29,478. Linear is higher by 522. Year 4: L = 24,000; E = 48,000(0.85)^4 = 48,000(0.52200625) = 25,056.30. Now exponential is higher by 1,056.30.
Because the values switch order between year 3 and year 4, the models cross between t = 3 and t = 4. In context, the linear model predicts a lower value (faster dollar loss) early on, but after roughly 3.4 to 3.5 years the percentage model predicts the higher value because a 15% loss shrinks as the van becomes less valuable. The crossing is not a failure of either model; it reflects different assumptions about how depreciation works.
2. Determine which model first puts the van below $25,000 and in which year that happens. Support your answer with calculations.
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Linear: solve 48000 - 6000t < 25000. Then 23000 < 6000t, so t > 23/6, about 3.83. At year 4, L(4) = 24,000, which is already below 25,000. At year 3 it is 30,000, still above.
Exponential: solve 48000(0.85)^t < 25000, or (0.85)^t < 25000/48000 = 0.520833... At year 4, E(4) = 25,056.30, slightly above 25,000. At year 5, E(5) = 48000(0.85)^5 = 48000(0.4437053125) = 21,297.86, which is below.
So the linear model reaches the replacement threshold first, during year 4, while the exponential model does so during year 5. At year 4 the models differ by 25,056.30 - 24,000 = $1,056.30, which is a meaningful difference in trade-in value. The linear model triggers replacement about a year earlier.
3. Evaluate the accountant's warning that the percentage model always loses value faster. Use the values or yearly changes to support your judgment.
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The warning is not accurate as stated. In the first years, the percentage model actually loses more dollar value, not less: from year 0 to year 1, the linear model loses $6,000, while the exponential model loses 48,000 - 40,800 = $7,200. By year 3, exponential value is 29,478 versus 30,000, so the percentage model is still lower.
After the crossing between years 3 and 4, the pattern reverses. From year 4 to year 5, the linear model loses another $6,000, but the exponential model loses only 25,056.30 - 21,297.86 = $3,758.44. The reason is that 15% of a smaller value is a smaller dollar amount, while the linear model keeps subtracting the same $6,000.
So the percentage model initially loses value faster in dollar terms, then loses value more slowly, even though its percentage rate stays at 15%. The claim would only be true if it meant the percentage rate, not the dollar amount; as a statement about which model produces a lower value, it is false after about 3.4 years.