13.2 Depreciation and the Time Value of Money
Key Takeaways
- Depreciation is a non-cash allocation of a past capital spend; it changes cash flow only through its effect on tax, and book value is not market value
- Straight line charges (P − S)/n each year, while double declining balance applies a rate of 2/n to book value and ignores salvage until book value reaches it
- Sum-of-years-digits weights the depreciable amount by remaining life using a digit sum of n(n+1)/2
- Present worth of a uniform series uses P = A[(1+i)ⁿ − 1]/[i(1+i)ⁿ]; adding undiscounted future cash flows is the classic error
- Land and working capital are never depreciated; working capital is invested at start-up and recovered at the end of the project
What depreciation is, and what it is not
Depreciation is the systematic allocation of the cost of a capital asset over its useful life. Three properties decide most exam questions:
- It is a non-cash charge. The money left when the compressor was bought; depreciation only spreads that historical spend across the years the asset earns.
- It reduces taxable income, which is why the method chosen changes cash flow through tax even though depreciation itself moves no cash.
- Book value is not market value. Book value is simply original cost minus accumulated depreciation; the resale price of a used exchanger is set by the market, not by the ledger.
Land is not depreciated. Working capital is not depreciated—it is recovered at the end of the project.
| Symbol | Meaning |
|---|---|
| P | Initial installed cost (first cost) |
| S | Salvage value at end of life |
| n | Useful life in years |
| D_k | Depreciation charge in year k |
| BV_k | Book value at end of year k |
The four methods you should be able to apply
Straight line
D = (P − S) / n — the same charge every year, and book value falls along a straight line to exactly S.
Declining balance and double declining balance
A fixed rate is applied to the current book value, so the charge shrinks each year:
D_k = R × BV_(k−1), with R = 2/n for double declining balance
Two rules that stems love to test: the salvage value does not enter the rate calculation, but depreciation stops once book value reaches salvage—you never depreciate below S.
Sum-of-years-digits
Weight the depreciable amount by remaining life. For life n, the digit sum is n(n+1)/2, and:
D_k = [(n − k + 1) / (n(n+1)/2)] × (P − S)
Units of production
D per unit = (P − S) / total expected units, then multiply by the units actually produced. This suits a campaign plant whose life is set by throughput rather than by calendar.
Worked comparison
A compressor costs P = QAR 12,000,000, has salvage S = QAR 2,000,000, and a life of n = 10 years.
Straight line: D = (12,000,000 − 2,000,000) / 10 = QAR 1,000,000 per year. Book value after four years = 12,000,000 − 4,000,000 = QAR 8,000,000.
Double declining balance at R = 2/10 = 0.20:
| Year | Opening BV | D_k = 0.20 × BV | Closing BV |
|---|---|---|---|
| 1 | 12,000,000 | 2,400,000 | 9,600,000 |
| 2 | 9,600,000 | 1,920,000 | 7,680,000 |
| 3 | 7,680,000 | 1,536,000 | 6,144,000 |
| 4 | 6,144,000 | 1,228,800 | 4,915,200 |
Sum-of-years-digits, digit sum = 10 × 11 / 2 = 55, so year 1 = (10/55) × 10,000,000 = QAR 1,818,182.
Year-one comparison: straight line 1,000,000; sum-of-years-digits 1,818,182; double declining balance 2,400,000. All three are accelerated relative to straight line except straight line itself, and straight line and sum-of-years-digits both total exactly (P − S) over the full life. The choice changes the timing of the tax shield, not the total amount depreciated.
Time value of money
Money available now is worth more than the same amount later, because it can earn a return in the meantime. Four relationships cover almost everything this exam asks.
| Relationship | Formula | Use it when |
|---|---|---|
| Future worth of a present sum | F = P(1 + i)ⁿ | Growing one lump sum forward |
| Present worth of a future sum | P = F / (1 + i)ⁿ | Discounting one future amount back |
| Present worth of a uniform series | P = A · [(1 + i)ⁿ − 1] / [i(1 + i)ⁿ] | Valuing equal annual savings or payments |
| Capital recovery (annualised cost) | A = P · i(1 + i)ⁿ / [(1 + i)ⁿ − 1] | Turning a capital sum into an equivalent annual charge |
Simple interest (F = P(1 + i·n)) appears only in elementary stems; unless a question says "simple," assume compound.
Worked examples
Compounding forward. QAR 1,000,000 invested at 8% for 5 years: F = 1,000,000 × 1.08⁵ = 1,000,000 × 1.4693 = QAR 1,469,328
Discounting back. A QAR 5,000,000 payment due in 6 years at 10%: P = 5,000,000 / 1.1⁶ = 5,000,000 / 1.7716 = QAR 2,822,371
Note how brutal discounting is: a payment six years out is worth barely more than half its face value at a 10% rate. This is why long-dated benefits rarely rescue a weak project.
Valuing an annual saving. An energy-integration project saves QAR 800,000 per year for 10 years; the hurdle rate is 10%: Series present-worth factor = (1.1¹⁰ − 1) / [0.1 × 1.1¹⁰] = (2.5937 − 1) / 0.25937 = 6.1446 P = 800,000 × 6.1446 = QAR 4,915,656
So the savings stream justifies spending up to about QAR 4.9 million today—not the QAR 8 million you get by adding ten years of savings undiscounted. Confusing the sum of future cash flows with their present worth is the classic error.
Annualising a capital cost. The same factor run backwards turns a QAR 4,915,656 investment into an equivalent annual charge of 4,915,656 / 6.1446 = QAR 800,000 per year, which is exactly how a capital-recovery charge is compared against an annual operating saving.
Quick sanity checks
- Raising the interest rate lowers every present worth and raises every future worth.
- Doubling time is roughly 72 / i(%): at 8%, money doubles in about nine years.
- Nominal versus effective: if a rate is quoted per annum but compounded monthly, the effective annual rate is (1 + i/12)¹² − 1, always slightly higher than the nominal figure.
Traps to drill
| Trap | Correction |
|---|---|
| Subtracting salvage before applying the declining-balance rate | Declining balance applies the rate to book value; salvage only sets the floor |
| Depreciating below salvage value | Stop once BV reaches S |
| Treating book value as resale value | Book value is an accounting figure, not a market price |
| Adding undiscounted future cash flows | Discount every flow to a common point before comparing |
| Using the number of payments as the exponent when compounding is not annual | Match the period of i and n: monthly rate with months, annual with years |
| Calling depreciation a cash outflow in a cash-flow table | It is non-cash; it affects cash only through tax |
A compressor costs QAR 12,000,000, has a salvage value of QAR 2,000,000, and a ten-year life. What is the year-one depreciation under straight line and under double declining balance?
An energy-integration project will save QAR 800,000 per year for ten years. At a hurdle rate of 10 percent, what is the maximum justified investment today?
Which statement about depreciation is correct for engineering-economics questions?