9.1 Engineering Economics & Depreciation

9.1 Engineering Economics & Depreciation

Engineering Economics provides quantitative mathematical tools to evaluate financial feasibility, asset replacement, and capital expenditure options. For the Registered Electrical Engineer (REE) Licensure Examination under Engineering Sciences and Allied Subjects (ESAS), candidates must master cash flow equivalence, interest formulas, asset depreciation methods, and capital project comparison techniques.

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1. Time Value of Money & Cash Flow Equivalence

The fundamental principle of engineering economics is that a unit of money available today is worth more than the same unit received in the future due to its earning capacity (interest).

Interest Definitions

• Simple Interest ($I$): Interest earned strictly on the principal amount: $I = P \cdot i \cdot n$ $F = P(1 + i \cdot n)$ where $P$ is principal, $i$ is interest rate per period, $n$ is number of periods, and $F$ is future amount.
• Compound Interest: Interest earned on both principal and accumulated past interest: $F = P(1 + i)^n$

Single Payment & Uniform Series Interest Factors

Nominal vs. Effective Interest Rates

When interest is compounded $m$ times per year at nominal annual rate $r$:

• Interest rate per compounding period ($i$): $i = \frac{r}{m}$
• Effective Annual Interest Rate ($i_e$): The actual annual rate earned under sub-annual compounding: $i_e = \left(1 + \frac{r}{m}\right)^m - 1$
• Continuous Compounding ($m \to \infty$): $i_e = e^r - 1$ $F = P e^{r t}$

Cash Flow Gradient Series

• Arithmetic Gradient ($G$): Cash flow increases by constant magnitude $G$ each period ($A_k = (k-1)G$): $P_G = \frac{G}{i} \left[ \frac{(1+i)^n - 1}{i(1+i)^n} - \frac{n}{(1+i)^n} \right]$
• Geometric Gradient ($g$): Cash flow increases by constant percentage rate $g$ each period ($A_k = A_1 (1+g)^{k-1}$): $P = \frac{A_1 [1 - (1+g)^n (1+i)^{-n}]}{i - g} \quad (\text{for } i \neq g)$ $P = \frac{n A_1}{1 + i} \quad (\text{for } i = g)$

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2. Depreciation Methods & Asset Book Value

Depreciation represents the systematic reduction in accounting value of a physical asset over its estimated useful life due to wear, tear, age, or technological obsolescence.

Common Terminology

• $C_0$ = Initial installed capital cost (first cost)
• $C_L$ = Salvage value (resale/scrap value at end of useful life)
• $n$ = Useful life in years
• $k$ = Specific year number ($1, 2, \dots, n$)
• $d_k$ = Depreciation charge during year $k$
• $D_k$ = Cumulative depreciation through year $k$
• $BV_k$ = Book value at end of year $k$ ($BV_k = C_0 - D_k$)

Summary of Standard Depreciation Methods

1. Straight-Line Method (SLM): Constant annual depreciation charge. $d = \frac{C_0 - C_L}{n}$ $D_k = k \cdot d = k \left(\frac{C_0 - C_L}{n}\right)$ $BV_k = C_0 - k \cdot d$

2. Sinking Fund Method (SFM): Assumes annual depreciation charges accumulate in an interest-bearing fund at rate $i$. $d = (C_0 - C_L) \left[ \frac{i}{(1+i)^n - 1} \right] = (C_0 - C_L) (A/F, i%, n)$ $D_k = d \left[ \frac{(1+i)^k - 1}{i} \right] = d (F/A, i%, k)$ $BV_k = C_0 - D_k$

3. Sum-of-the-Years'-Digits (SYD) Method: Accelerated depreciation allocating larger charges in early years.

   • Sum of digits: $S = \frac{n(n+1)}{2}$
   • Depreciation in year $k$: $d_k = \left( \frac{n - k + 1}{S} \right) (C_0 - C_L)$
   • Cumulative depreciation at year $k$: $D_k = \left[ \frac{k(2n - k + 1)}{2 S} \right] (C_0 - C_L)$

4. Declining Balance Method (DBM) / Matched Rate: Constant percentage rate $K$ applied to decaying book value.

   • Fixed rate $K$ derived from exact salvage value: $K = 1 - \sqrt[n]{\frac{C_L}{C_0}}$
   • Depreciation in year $k$: $d_k = K \cdot BV_{k-1} = K \cdot C_0 (1 - K)^{k-1}$
   • Book value at year $k$: $BV_k = C_0 (1 - K)^k$

5. Double Declining Balance Method (DDBM): Accelerated declining balance using twice the straight-line rate ($K = \frac{2}{n}$), ignoring initial salvage value until book value reaches $C_L$. $d_k = \left(\frac{2}{n}\right) BV_{k-1}$

6. Service Output / Unit-of-Production Method: Depreciation based on total operating hours or units produced $T$. $\text{Depreciation per unit } = \frac{C_0 - C_L}{T}$ $d_k = (\text{Units in year } k) \times \left(\frac{C_0 - C_L}{T}\right)$

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3. Capital Budgeting & Economic Comparison Criteria

When evaluating competing electrical engineering investments or equipment options, candidates must apply standardized evaluation methods.

Present Worth (PW) / Net Present Value (NPV)

Converts all cash inflows and outflows to present values at minimum attractive rate of return (MARR) $i$:

$\text{NPV} = \sum_{k=0}^n R_k (1+i)^{-k} - \sum_{k=0}^n C_k (1+i)^{-k}$

• Decision Rule: Accept project if $\text{NPV} \ge 0$.

Equivalent Uniform Annual Cost (EUAC) / Annual Worth (AW)

Converts all capital expenditures, operating costs, and salvage values into an equivalent uniform annual amount over useful life $n$:

$\text{EUAC} = C_0 (A/P, i%, n) + \text{Annual O&M} - C_L (A/F, i%, n)$

• Decision Rule: Select option with the lowest EUAC.

Capitalized Cost ($CC$)

The present worth of an asset intended to provide continuous service perpetually ($n \to \infty$).

• Non-replaceable asset / perpetual stream: $CC = C_0 + \frac{A}{i}$
• Asset replaced every $k$ years perpetually with salvage value $C_L$: $CC = C_0 + \frac{A}{i} + \frac{C_0 - C_L}{(1+i)^k - 1}$

Internal Rate of Return (IRR)

The discount rate $i^*$ at which Net Present Value equals zero:

$\text{NPV}(i^*) = 0$

• Decision Rule: Accept project if $\text{IRR} \ge \text{MARR}$.

Benefit-Cost Ratio ($B/C$)

Used primarily in public works projects:

$B/C = \frac{\text{PW of Benefits}}{\text{PW of Costs}} = \frac{\text{Equivalent Annual Benefits}}{\text{Equivalent Annual Costs}}$

• Decision Rule: Accept project if $B/C \ge 1.0$.

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4. Break-Even Analysis & Payback Period

Break-Even Volume Analysis

Determines operational level where total revenues equal total production costs.

• Total Cost: $TC = FC + VC \cdot Q$
• Total Revenue: $TR = P \cdot Q$
• Break-Even Quantity ($Q_{\text{BE}}$): $Q_{\text{BE}} = \frac{FC}{P - VC}$ where $FC$ is fixed costs, $VC$ is variable cost per unit, and $P$ is selling price per unit.

Payback Period

• Simple Payback Period ($n_p$): Time required for cumulative cash inflows to equal initial investment (ignoring time value of money): $n_p = \frac{C_0}{\text{Annual Net Cash Flow}}$
• Discounted Payback Period: Time required for cumulative discounted cash flows to equal initial investment.

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Solved Board Exam Examples

Example 1: EUAC Comparison of Motor Drive Systems

Problem: An industrial plant requires an electrical motor drive. Two candidate motors are being evaluated at an interest rate of $i = 10%$ per annum over a 6-year period:

• Motor A: Initial cost $C_0 = \text{PHP }150,000$, annual maintenance $A = \text{PHP }20,000$, salvage value $C_L = \text{PHP }30,000$.
• Motor B: Initial cost $C_0 = \text{PHP }220,000$, annual maintenance $A = \text{PHP }10,000$, salvage value $C_L = \text{PHP }50,000$. Which motor should be selected based on Equivalent Uniform Annual Cost (EUAC)?

Solution:

1. Calculate Capital Recovery Factor $(A/P, 10%, 6)$: $(A/P, 10\%, 6) = \frac{0.10(1.10)^6}{(1.10)^6 - 1} = \frac{0.10(1.771561)}{0.771561} = 0.229607$
2. Calculate Sinking Fund Factor $(A/F, 10%, 6)$: $(A/F, 10\%, 6) = \frac{0.10}{(1.10)^6 - 1} = \frac{0.10}{0.771561} = 0.129607$
3. Compute EUAC for Motor A: $\text{EUAC}_A = 150,000(0.229607) + 20,000 - 30,000(0.129607)$ $\text{EUAC}_A = 34,441.05 + 20,000 - 3,888.21 = \text{PHP }50,552.84$
4. Compute EUAC for Motor B: $\text{EUAC}_B = 220,000(0.229607) + 10,000 - 50,000(0.129607)$ $\text{EUAC}_B = 50,513.54 + 10,000 - 6,480.35 = \text{PHP }54,033.19$
5. Conclusion: Select Motor A because it yields a lower EUAC by $\text{PHP }3,480.35$ per year.

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Example 2: SYD vs. DDB Depreciation of a Transformer Asset

Problem: A $500\ \text{kVA}$ distribution transformer has an installed initial cost $C_0 = \text{PHP }400,000$ and an estimated salvage value $C_L = \text{PHP }40,000$ after a useful life of $n = 5$ years. Calculate:

1. The depreciation charge for Year 3 using Sum-of-the-Years'-Digits (SYD).
2. The Book Value at the end of Year 2 using Double Declining Balance (DDB).

Solution:

1. SYD Method (Year 3):
   • Depreciable amount: $C_0 - C_L = 400,000 - 40,000 = \text{PHP }360,000$.
   • Sum of years' digits: $S = \frac{5(5+1)}{2} = 15$.
   • For Year 3 ($k = 3$), remaining life digit is $5 - 3 + 1 = 3$.
   • Depreciation charge $d_3$: $d_3 = \left(\frac{3}{15}\right) \times 360,000 = 0.20 \times 360,000 = \text{PHP }72,000$
2. DDB Method (Book Value end of Year 2):
   • DDB rate: $K = \frac{2}{n} = \frac{2}{5} = 0.40$ (or $40%$).
   • Year 1 depreciation: $d_1 = 0.40 \times 400,000 = \text{PHP }160,000$.
   • Book value end of Year 1: $BV_1 = 400,000 - 160,000 = \text{PHP }240,000$.
   • Year 2 depreciation: $d_2 = 0.40 \times 240,000 = \text{PHP }96,000$.
   • Book value end of Year 2 ($BV_2$): $BV_2 = 240,000 - 96,000 = \text{PHP }144,000$ (Alternatively: $BV_2 = C_0 (1 - K)^2 = 400,000(1 - 0.40)^2 = 400,000(0.36) = \text{PHP }144,000$.)