4.2 Engineering Economics, Time Value of Money, Cash Flows, Depreciation, and Capital Costing

4.2 Engineering Economics, Time Value of Money, Cash Flows, Depreciation, and Capital Costing

Engineering economics equips Geodetic Engineers with quantitative tools to evaluate capital investments in surveying technology, bid competitively on cadastral and infrastructure mapping contracts, and manage professional firm finances. Every capital decision—whether purchasing an RTK-GNSS receiver, investing in a drone LiDAR system, or bidding on a multi-year LGU cadastral project—depends on understanding cash flow dynamics and financial equivalency.

---

1. Time Value of Money and Cash Flow Dynamics

Money possesses time value because capital can earn interest over time. A peso received today is worth more than a peso received a year from today.

Cash Flow Diagrams (CFD)

Cash flow diagrams visually summarize receipts and disbursements along a horizontal time axis:

• Vertical Upward Arrow (+): Cash Inflow (Revenue, savings, salvage value).
• Vertical Downward Arrow (-): Cash Outflow (Initial purchase cost, maintenance, operating expense).
• End-of-Period Convention: Transactions occurring during a interest period are assumed to occur at the end of that period.

Interest Formulas and Compound Interest Factors

Let $P$ = Present sum of money, $F$ = Future sum of money after $n$ periods, $A$ = End-of-period uniform annual payment, $i$ = interest rate per period, $n$ = number of compounding periods.

Nominal vs. Effective Interest Rates

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

$$\ni_{eff} = \left(1 + \frac{r}{m}\right)^m - 1$$

---

2. Depreciation Methods for Surveying Equipment

Depreciation is the non-cash allocation of an asset's historical acquisition cost over its estimated economic service life. Let $B$ = Initial cost (basis), $S$ = Estimated salvage value, $n$ = Useful life in years, $t$ = Specific year ($1 \le t \le n$), $D_t$ = Depreciation charge in year $t$, $BV_t$ = Book Value at end of year $t$.

1. Straight-Line Depreciation (SLD)

Depreciation is constant each year:

$$D_t = \frac{B - S}{n}, \quad BV_t = B - t \cdot D_t$$

2. Double Declining Balance (DDB)

Accelerated method using rate $d = \frac{2}{n}$. Salvage value $S$ is ignored in initial annual calculations, but $BV_t$ cannot drop below $S$:

$$D_t = d \cdot BV_{t-1} = \frac{2}{n} \cdot BV_{t-1}, \quad BV_t = B(1 - d)^t$$

3. Sum-of-the-Years'-Digits (SYD)

Accelerated method based on digit sum $SOYD = \frac{n(n+1)}{2}$:

$$D_t = (B - S) \left[ \frac{n - t + 1}{SOYD} \right], \quad BV_t = B - \sum_{k=1}^t D_k$$

---

3. Project Evaluation & Capital Costing Metrics

Net Present Value (NPV / PW)

Converts all cash inflows and outflows to time $t=0$ using the Minimum Attractive Rate of Return (MARR):

$$NPV = PW(Benefits) - PW(Costs) = \sum_{t=0}^n \frac{CF_t}{(1 + MARR)^t}$$

• Accept project if $NPV \ge 0$.

Benefit-Cost Ratio (B/C Ratio)

Used primarily for public infrastructure and government cadastral projects:

$$B/C = \frac{PW(\text{Benefits})}{PW(\text{Costs})} = \frac{AW(\text{Benefits})}{AW(\text{Costs})}$$

• Accept project if $B/C \ge 1.0$.

Internal Rate of Return (IRR)

The break-even interest rate $i^*$ at which $NPV = 0$ or $PW(Benefits) = PW(Costs)$:

$$\sum_{t=0}^n \frac{CF_t}{(1 + i^*)^t} = 0$$

• Accept project if $IRR \ge MARR$.

---

4. Cost Estimation & Fee Structure for Geodetic Engineering Contracts

Professional surveying fees under standard Board of Geodetic Engineering pricing frameworks compound several cost centers:

1. Direct Labor Costs ($DLC$): Salaries and per diems for Geodetic Engineer (Team Leader), Instrumentman, Rodmen, Chainmen, and GIS Encoders.
2. Direct Non-Labor Expenses ($DNE$): Boundary monuments, wooden stakes, fuel, vehicle maintenance, travel allowances, permits, and agency data fees (e.g., NAMRIA/LMB record copies).
3. Equipment Depreciation / Rental Costs ($EQ$): Hourly or daily cost allocation for Total Stations, GNSS receivers, and UAVs.
4. Overhead / Indirect Costs ($O/H$): Office rent, utilities, insurance, administrative staff, accounting (typically 30% to 50% of Direct Labor).
5. Contingencies & Professional Profit Margin ($P$): Margin applied to total base cost (typically 15% to 25%).

$$\text{Total Fee} = (DLC + DNE + EQ + O/H) \times (1 + \text{Profit \& Contingency \%})$$

---

Worked Calculation Examples

Worked Example 4.2.1: Capital Recovery Factor & EUAC of GNSS Equipment

Problem: A geodetic surveying firm buys an RTK GNSS Receiver unit for $P = \text{PHP }450,000$. It has an estimated service life $n = 6 \text{ years}$ and a residual salvage value $S = \text{PHP }50,000$. Annual calibration and maintenance expenses are $A_{maint} = \text{PHP }25,000/\text{year}$. If the firm's MARR is $i = 12\%$, calculate the Equivalent Uniform Annual Cost (EUAC).

Solution:

1. Capital Recovery Factor $(A/P, 12\%, 6)$:

   $$(A/P, 12\%, 6) = \frac{0.12(1 + 0.12)^6}{(1 + 0.12)^6 - 1} = \frac{0.12(1.97382)}{0.97382} = \frac{0.23686}{0.97382} = 0.24323$$

2. Sinking Fund Factor $(A/F, 12\%, 6)$:

   $$(A/F, 12\%, 6) = \frac{0.12}{(1.12)^6 - 1} = \frac{0.12}{0.97382} = 0.12323$$

3. EUAC Calculation:

   $$EUAC = P(A/P, 12\%, 6) - S(A/F, 12\%, 6) + A_{maint}$$ $$EUAC = 450,000(0.24323) - 50,000(0.12323) + 25,000$$ $$EUAC = 109,453.50 - 6,161.50 + 25,000 = \text{PHP }128,292.00 / \text{year}$$

---

Worked Example 4.2.2: Depreciation Method Comparison

Problem: An aerial photogrammetry drone is purchased for $B = \text{PHP }800,000$ with $S = \text{PHP }80,000$ after $n = 5 \text{ years}$. Compute the Year 2 depreciation $D_2$ and Book Value $BV_2$ using (a) Straight-Line (SLD), (b) Double Declining Balance (DDB), and (c) Sum-of-the-Years'-Digits (SYD).

Solution:

• (a) Straight-Line Depreciation (SLD):

  $$D = \frac{800,000 - 80,000}{5} = \text{PHP }144,000 / \text{year}$$ $$D_2 = \text{PHP }144,000$$ $$BV_2 = 800,000 - (2 \times 144,000) = \text{PHP }512,000$$

• (b) Double Declining Balance (DDB): Rate $d = \frac{2}{5} = 0.40$ (40%).

  $$Year 1: \quad D_1 = 0.40 \times 800,000 = \text{PHP }320,000, \quad BV_1 = 800,000 - 320,000 = \text{PHP }480,000$$ $$Year 2: \quad D_2 = 0.40 \times 480,000 = \text{PHP }192,000, \quad BV_2 = 480,000 - 192,000 = \text{PHP }288,000$$

• (c) Sum-of-the-Years'-Digits (SYD):

  $$SOYD = \frac{5(6)}{2} = 15, \quad \text{Depreciable Basis} = 800,000 - 80,000 = \text{PHP }720,000$$ $$Year 1: \quad D_1 = 720,000 \times \frac{5}{15} = \text{PHP }240,000$$ $$Year 2: \quad D_2 = 720,000 \times \frac{4}{15} = \text{PHP }192,000$$ $$BV_2 = 800,000 - (240,000 + 192,000) = \text{PHP }368,000$$

---

Worked Example 4.2.3: Benefit-Cost Ratio for Municipal CORS Station

Problem: A local government unit (LGU) evaluates installing an automated Continuously Operating Reference Station (CORS) for municipal land titling. Initial capital expenditure is $C_0 = \text{PHP }2,500,000$. Annual operational cost is $\text{PHP }150,000/\text{year}$ for 10 years. Annual quantifiable public benefits (reduced survey dispute litigation and faster property titling turnarounds) are estimated at $\text{PHP }650,000/\text{year}$. Using $i = 10\%$ discount rate and $n = 10 \text{ years}$, calculate the Benefit-Cost ($B/C$) Ratio and verify project feasibility.

Solution:

1. Uniform Series Present Worth Factor $(P/A, 10\%, 10)$:

   $$(P/A, 10\%, 10) = \frac{(1.10)^{10} - 1}{0.10(1.10)^{10}} = \frac{2.59374 - 1}{0.259374} = 6.14457$$

2. Present Worth of Costs ($PW_C$):

   $$PW_C = 2,500,000 + 150,000(6.14457) = 2,500,000 + 921,685.50 = \text{PHP }3,421,685.50$$

3. Present Worth of Benefits ($PW_B$):

   $$PW_B = 650,000(6.14457) = \text{PHP }3,993,970.50$$

4. Benefit-Cost Ratio ($B/C$):

   $$B/C = \frac{PW_B}{PW_C} = \frac{3,993,970.50}{3,421,685.50} = 1.167$$

   Since $B/C = 1.17 > 1.0$, the project is economically justified.