GELE Mathematics — Engineering EconomyStudy Notes

The time value of money (TVM) is the principle that money available today is worth more than the same amount in the future because it can be invested and earn interest. Every engineering project involves cash flows occurring at different times; to compare them fairly, we must convert all amounts to a common point in time. **Why TVM Matters in Civil Engineering:** When a contractor bids on a bridge project with construction costs spread over 5 years, or when a municipality evaluates the life-cycle cost of water treatment equipment, TVM calculations determine whether investments make financial sense. **Simple Interest:** Simple interest applies interest only to the principal amount, not to accumulated interest. The formula is: $$F = P(1 + in)$$ where: - $F$ = future amount (in Philippine Pesos, ₱) - $P$ = principal (present amount, ₱) - $i$ = interest rate per period (as a decimal) - $n$ = number of periods Example: ₱50,000 borrowed at 6% simple interest for 3 years yields: $$F = 50,000(1 + 0.06 \times 3) = 50,000(1.18) = ₱59,000$$ Total interest earned: ₱9,000 (constant each year). **Compound Interest:** Compound interest applies interest to both the principal and previously accumulated interest. This is the standard in engineering economy because it reflects real investment scenarios. $$F = P(1 + i)^n$$ The inverse (present worth) discounts a future amount back to today: $$P = F(1 + i)^{-n} = \frac{F}{(1+i)^n}$$ The term $(1 + i)^n$ is the compound interest factor, often written as $(F/P, i, n)$ in engineering tables. **Board-Style Problem 1: Compound Interest Calculation** A civil engineering firm invests ₱200,000 in equipment for a construction project. If the investment grows at 9% compounded annually over 6 years, what is the future value? **Solution:** Given: $P = ₱200,000$, $i = 0.09$, $n = 6$ $$F = P(1 + i)^n = 200,000(1.09)^6$$ Calculate $(1.09)^6$: $(1.09)^2 = 1.1881$ $(1.09)^3 = 1.2950$ $(1.09)^6 = (1.09)^3 \times (1.09)^3 = 1.2950 \times 1.2950 = 1.6771$ $$F = 200,000 \times 1.6771 = ₱335,420$$ The equipment investment will be worth ₱335,420 after 6 years. **Board-Style Problem 2: Finding Present Worth** A contractor must pay ₱500,000 for construction permits and licenses in 4 years. What is the equivalent present amount if the discount rate is 8% compounded annually? **Solution:** Given: $F = ₱500,000$, $i = 0.08$, $n = 4$ $$P = F(1 + i)^{-n} = 500,000(1.08)^{-4}$$ Calculate $(1.08)^{-4} = \frac{1}{(1.08)^4}$: $(1.08)^2 = 1.1664$ $(1.08)^4 = 1.1664 \times 1.1664 = 1.3605$ $$P = \frac{500,000}{1.3605} = ₱367,556$$ The present worth of ₱500,000 due in 4 years is ₱367,556.

In practice, interest is often quoted as a nominal (stated) annual rate but compounded more frequently than annually (monthly, quarterly, or daily). We must convert to an effective annual rate to make fair comparisons between different investment options. **Formula for Effective Annual Rate:** $$i_{\text{eff}} = \left(1 + \frac{r}{m}\right)^m - 1$$ where: - $r$ = nominal annual rate (as a decimal) - $m$ = number of compounding periods per year - $i_{\text{eff}}$ = effective annual rate **Common Compounding Frequencies:** - Annually: $m = 1$ - Semi-annually: $m = 2$ - Quarterly: $m = 4$ - Monthly: $m = 12$ - Daily: $m = 365$ When dealing with monthly or quarterly interest, always convert periods to match compounding frequency before using compound interest formulas. **Board-Style Problem 3: Nominal vs. Effective Rate Comparison** A project fund offers two options: - Option A: 12% nominal compounded annually - Option B: 11.5% nominal compounded monthly Which option is better? Calculate the effective annual rates. **Solution:** Option A: $$i_{\text{eff,A}} = 1 + 0.12 = 0.12 \text{ or } 12\%$$ Option B: $$i_{\text{eff,B}} = \left(1 + \frac{0.115}{12}\right)^{12} - 1 = (1 + 0.00958)^{12} - 1 = (1.00958)^{12} - 1$$ Calculate $(1.00958)^{12}$: $(1.00958)^{12} ≈ 1.1216$ $$i_{\text{eff,B}} = 1.1216 - 1 = 0.1216 \text{ or } 12.16\%$$ **Conclusion:** Option B (11.5% monthly) yields an effective rate of 12.16%, which is better than Option A's 12%.

An annuity is a series of equal payments (or receipts) at regular intervals. In civil engineering, annuities model regular maintenance costs, uniform annual revenues, or regular loan payments. **Types of Annuities:** - **Ordinary Annuity (End-of-Period):** Payments occur at the end of each period; most common in practice - **Annuity Due (Beginning-of-Period):** Payments occur at the beginning of each period; less common This section focuses on ordinary annuities. **Future Worth of Ordinary Annuity:** If equal payments $A$ are made at the end of each period for $n$ periods at interest rate $i$: $$F = A \frac{(1+i)^n - 1}{i}$$ The term $\frac{(1+i)^n - 1}{i}$ is the future worth factor, often written as $(F/A, i, n)$. **Present Worth of Ordinary Annuity:** $$P = A \frac{(1+i)^n - 1}{i(1+i)^n}$$ The term $\frac{(1+i)^n - 1}{i(1+i)^n}$ is the present worth factor, written as $(P/A, i, n)$. **Key insight:** The present worth of an annuity converts a series of uniform payments into a single lump sum today. **Perpetuity (Infinite Annuity):** If an annuity continues indefinitely: $$P = \frac{A}{i}$$ Perpuities model endowment funds or permanent infrastructure maintenance costs. **Board-Style Problem 4: Future Worth of Ordinary Annuity** A construction company sets aside ₱50,000 at the end of each year for equipment replacement. If the fund earns 7% compounded annually, how much will accumulate after 10 years? **Solution:** Given: $A = ₱50,000$, $i = 0.07$, $n = 10$ $$F = A \frac{(1+i)^n - 1}{i} = 50,000 \frac{(1.07)^{10} - 1}{0.07}$$ Calculate $(1.07)^{10}$: $(1.07)^5 ≈ 1.4026$ $(1.07)^{10} ≈ 1.9672$ $$F = 50,000 \frac{1.9672 - 1}{0.07} = 50,000 \frac{0.9672}{0.07} = 50,000 \times 13.817 = ₱690,850$$ After 10 years, the equipment replacement fund will contain ₱690,850. **Board-Style Problem 5: Present Worth of Ordinary Annuity** A civil engineering consultant receives a contract promising ₱300,000 at the end of each year for 5 years. What is the present worth of this income stream if the discount rate is 10% compounded annually? **Solution:** Given: $A = ₱300,000$, $i = 0.10$, $n = 5$ $$P = A \frac{(1+i)^n - 1}{i(1+i)^n} = 300,000 \frac{(1.10)^5 - 1}{0.10(1.10)^5}$$ Calculate $(1.10)^5 ≈ 1.6105$ $$P = 300,000 \frac{1.6105 - 1}{0.10 \times 1.6105} = 300,000 \frac{0.6105}{0.16105} = 300,000 \times 3.7908 = ₱1,137,240$$ The present worth of receiving ₱300,000 annually for 5 years is ₱1,137,240. **Board-Style Problem 6: Perpetuity Calculation** A university receives a donation to endow a scholarship providing ₱100,000 annually forever. If the endowment earns 6% compounded annually, what is the required donation amount? **Solution:** Given: $A = ₱100,000$ (annual payment, perpetual), $i = 0.06$ $$P = \frac{A}{i} = \frac{100,000}{0.06} = ₱1,666,667$$ A donation of ₱1,666,667 today will support ₱100,000 scholarships forever at 6% return.

Depreciation is the systematic reduction in the book value of an asset over its useful life. For tax purposes and financial reporting under Philippine accounting standards, engineers must calculate depreciation accurately. Different methods are appropriate for different assets. **Why Depreciation Matters:** - Reduces taxable income (tax benefit) - Allocates asset cost over its useful life - Estimates residual value after service life **Key Terms:** - $C$ = original cost (purchase price, ₱) - $S$ = salvage (scrap) value at end of life - $n$ = useful life (years) - $BV_t$ = book value at end of year $t$ - $d_t$ = depreciation in year $t$ **Method 1: Straight-Line Depreciation (SL)** The most common method; depreciation is constant each year: $$d = \frac{C - S}{n}$$ Book value after $t$ years: $$BV_t = C - t \cdot d = C - \frac{t(C-S)}{n}$$ **Advantages:** Simple, predictable; used widely in the Philippines **Disadvantages:** Does not reflect accelerated wear in early years **Board-Style Problem 7: Straight-Line Depreciation** A concrete pump for a construction site costs ₱800,000. Its salvage value after 8 years is ₱80,000. Calculate annual depreciation and the book value after 5 years using straight-line method. **Solution:** Given: $C = ₱800,000$, $S = ₱80,000$, $n = 8$ years Annual depreciation: $$d = \frac{C - S}{n} = \frac{800,000 - 80,000}{8} = \frac{720,000}{8} = ₱90,000$$ Book value after 5 years: $$BV_5 = C - 5 \cdot d = 800,000 - 5(90,000) = 800,000 - 450,000 = ₱350,000$$ Alternatively: $$BV_5 = 800,000 - \frac{5(800,000 - 80,000)}{8} = 800,000 - \frac{5 \times 720,000}{8} = ₱350,000$$ **Method 2: Sum-of-Years-Digits (SYD)** This method accelerates depreciation; larger charges in early years, smaller in later years. Depreciation in year $t$: $$d_t = \frac{n + 1 - t}{\text{Sum of years}} \times (C - S)$$ where Sum of years = $1 + 2 + 3 + ... + n = \frac{n(n+1)}{2}$ **Board-Style Problem 8: Sum-of-Years-Digits Depreciation** A steel bridge inspection vehicle costs ₱500,000 with a 5-year life and ₱50,000 salvage value. Calculate depreciation for each year using SYD method. **Solution:** Given: $C = ₱500,000$, $S = ₱50,000$, $n = 5$ Sum of years = $\frac{5(6)}{2} = 15$ Depreciable base = $C - S = 500,000 - 50,000 = ₱450,000$ Year 1: $d_1 = \frac{5}{15} \times 450,000 = ₱150,000$ Year 2: $d_2 = \frac{4}{15} \times 450,000 = ₱120,000$ Year 3: $d_3 = \frac{3}{15} \times 450,000 = ₱90,000$ Year 4: $d_4 = \frac{2}{15} \times 450,000 = ₱60,000$ Year 5: $d_5 = \frac{1}{15} \times 450,000 = ₱30,000$ Total: ₱450,000 (checks out) Book values: - End Year 1: $BV_1 = 500,000 - 150,000 = ₱350,000$ - End Year 2: $BV_2 = 350,000 - 120,000 = ₱230,000$ - End Year 3: $BV_3 = 230,000 - 90,000 = ₱140,000$ **Method 3: Declining Balance (DB)** Depreciation rate is applied to the remaining book value each year: $$BV_t = C(1 - d)^t$$ where $d$ is the depreciation rate (often $d = 2/n$ for double-declining balance). Depreciation in year $t$: $$\text{Depreciation}_t = BV_{t-1} \times d$$ **Board-Style Problem 9: Double-Declining Balance Depreciation** A surveying instrument costs ₱200,000 with a 4-year life and negligible salvage. Using double-declining balance method (d = 2/n), calculate depreciation for years 1 and 2, and book values. **Solution:** Given: $C = ₱200,000$, $n = 4$ Depreciation rate: $d = \frac{2}{4} = 0.50$ (50% per year) Year 1 depreciation: $$\text{Dep}_1 = 200,000 \times 0.50 = ₱100,000$$ $$BV_1 = 200,000 - 100,000 = ₱100,000$$ Year 2 depreciation: $$\text{Dep}_2 = 100,000 \times 0.50 = ₱50,000$$ $$BV_2 = 100,000 - 50,000 = ₱50,000$$ Year 3 depreciation: $$\text{Dep}_3 = 50,000 \times 0.50 = ₱25,000$$ $$BV_3 = 50,000 - 25,000 = ₱25,000$$ Year 4 depreciation: $$\text{Dep}_4 = 25,000 \times 0.50 = ₱12,500$$ $$BV_4 = 25,000 - 12,500 = ₱12,500$$ Note: Declining balance does not reach zero salvage value; a switch to straight-line may be made in later years.

In civil engineering practice, you often compare competing projects or designs. Economic comparison methods convert all costs and benefits to a common basis so the best alternative emerges. The PRC exam tests your ability to apply these techniques. **Three Primary Comparison Methods:** **Method 1: Present Worth (PW) Analysis** Convert all costs and benefits to present-day value. Choose the alternative with the highest present worth (lowest cost in a cost-only scenario). $$PW = \sum_{t=0}^{n} \frac{\text{Cash Flow}_t}{(1+i)^t}$$ For two alternatives with equal lives: - If revenues/benefits considered: choose highest PW - If only costs: choose lowest cost (most negative or least positive) **Method 2: Annual Worth (AW) Analysis** Convert all costs and benefits to equivalent uniform annual amounts. This is useful when alternatives have different lifespans. $$AW = PW \times \frac{i(1+i)^n}{(1+i)^n - 1}$$ The factor $\frac{i(1+i)^n}{(1+i)^n - 1}$ converts present worth to annual payments. **Method 3: Rate of Return (ROR) Analysis** Find the interest rate where costs equal benefits (or present worth equals zero). Compare the ROR to the minimum acceptable rate of return (MARR). Choose the alternative with ROR ≥ MARR; if comparing alternatives, choose the one with highest ROR (subject to investment rules). **Method 4: Benefit-Cost Ratio (B/C)** For public projects (infrastructure, utilities): $$\text{B/C} = \frac{\text{Present Worth of Benefits}}{\text{Present Worth of Costs}}$$ A project is justified if B/C ≥ 1.0. When comparing mutually exclusive alternatives, choose the one with highest B/C (or conduct incremental analysis). **Break-Even Analysis** Find the value of a variable (cost, revenue, rate) at which two alternatives are equally attractive. This identifies the threshold for decision-making. **Board-Style Problem 10: Present Worth Comparison** A municipality must choose between two street lighting systems: **System A (LED):** - Initial cost: ₱2,000,000 - Annual operating cost: ₱150,000 - Life: 10 years - Salvage value: ₱200,000 **System B (Halogen):** - Initial cost: ₱1,200,000 - Annual operating cost: ₱300,000 - Life: 5 years (then must be replaced) - Salvage value: ₱100,000 The discount rate is 8%. Which system has the lower present cost? (Assume System B is replaced with identical system for years 6–10.) **Solution:** **System A (single cycle, 10 years):** $$PW_A = -2,000,000 - 150,000 \left[ \frac{(1.08)^{10} - 1}{0.08(1.08)^{10}} \right] + 200,000(1.08)^{-10}$$ Calculate annuity factor: $(1.08)^{10} ≈ 2.1589$ $$\text{Annuity factor} = \frac{1.1589}{0.08 × 2.1589} = \frac{1.1589}{0.1727} ≈ 6.710$$ $$PW_A = -2,000,000 - 150,000(6.710) + 200,000(2.1589)^{-1}$$ $$PW_A = -2,000,000 - 1,006,500 + 200,000(0.4632)$$ $$PW_A = -2,000,000 - 1,006,500 + 92,640$$ $$PW_A ≈ -₱2,913,860$$ **System B (5-year cycle, repeated once for years 6–10):** For the first cycle (years 0–5): $$PW_{B,1} = -1,200,000 - 300,000(3.993) + 100,000(0.6806)$$ Annuity factor for 5 years at 8%: $$(1.08)^5 ≈ 1.4693$$ $$\text{Annuity factor} = \frac{1 - (1.08)^{-5}}{0.08} = \frac{1 - 0.6806}{0.08} ≈ 3.993$$ $$PW_{B,1} = -1,200,000 - 300,000(3.993) + 100,000(0.6806)$$ $$PW_{B,1} = -1,200,000 - 1,197,900 + 68,060 = -₱2,329,840$$ For the replacement in years 5–10 (discount back 5 years): $$PW_{B,2} = \frac{-1,200,000 - 300,000(3.993) + 100,000(0.6806)}{(1.08)^5}$$ $$PW_{B,2} = \frac{-2,329,840}{1.4693} ≈ -₱1,585,530$$ Total for System B: $$PW_B = PW_{B,1} + PW_{B,2} = -2,329,840 - 1,585,530 = -₱3,915,370$$ **Conclusion:** System A has PW = –₱2,913,860, while System B has PW = –₱3,915,370. System A is better because it costs less in present-worth terms (smaller magnitude of negative value). **Board-Style Problem 11: Annual Worth Comparison (Different Life Cycles)** Instead of comparing PW over a 10-year study period, convert each system to annual equivalent cost (AEC). **Solution:** **System A (10-year life):** $$AW_A = PW_A × \left[ \frac{i(1+i)^n}{(1+i)^n - 1} \right] = -2,913,860 × \left[ \frac{0.08(1.08)^{10}}{(1.08)^{10} - 1} \right]$$ Capital recovery factor: $$CRF = \frac{0.08(2.1589)}{2.1589 - 1} = \frac{0.1727}{1.1589} ≈ 0.1490$$ $$AW_A = -2,913,860 × 0.1490 ≈ -₱434,160/\text{year}$$ **System B (5-year life, repeated):** For the first 5-year cycle: $$AW_B = -2,329,840 × \left[ \frac{0.08(1.08)^5}{(1.08)^5 - 1} \right]$$ CRF for 5 years: $$CRF = \frac{0.08(1.4693)}{1.4693 - 1} = \frac{0.1175}{0.4693} ≈ 0.2505$$ $$AW_B = -2,329,840 × 0.2505 ≈ -₱583,330/\text{year}$$ **Conclusion:** System A has a lower annual equivalent cost (–₱434,160/yr vs. –₱583,330/yr), so System A is again preferred. **Board-Style Problem 12: Benefit-Cost Ratio for Public Infrastructure** A water treatment plant expansion project has: - Initial cost: ₱50,000,000 - Annual operation and maintenance: ₱2,000,000 - Useful life: 20 years - Salvage value: ₱5,000,000 - Annual benefits (water supply reliability): ₱8,000,000 - Discount rate (MARR): 6% Calculate the benefit-cost ratio. Is the project justified? **Solution:** **Present Worth of Benefits:** $$PW_B = \sum_{t=1}^{20} \frac{8,000,000}{(1.06)^t}$$ Using annuity factor for 20 years at 6%: $$(1.06)^{20} ≈ 3.2071$$ Annuity factor = $\frac{(1.06)^{20} - 1}{0.06(1.06)^{20}} = \frac{3.2071 - 1}{0.06 × 3.2071} ≈ 11.470$ $$PW_B = 8,000,000 × 11.470 = ₱91,760,000$$ **Present Worth of Costs:** $$PW_C = 50,000,000 + 2,000,000 × 11.470 - 5,000,000(1.06)^{-20}$$ Discount salvage: $(1.06)^{-20} = 1/3.2071 ≈ 0.3118$ $$PW_C = 50,000,000 + 22,940,000 - 5,000,000(0.3118)$$ $$PW_C = 50,000,000 + 22,940,000 - 1,559,000$$ $$PW_C = ₱71,381,000$$ **Benefit-Cost Ratio:** $$\text{B/C} = \frac{91,760,000}{71,381,000} ≈ 1.286$$ **Conclusion:** B/C = 1.286 > 1.0, so the project is justified. For every peso of cost, the project generates ₱1.286 in benefits.

The PRC Civil Engineer Licensure Examination frequently tests engineering economy concepts in scenarios resembling real-world project decisions. This section highlights typical exam question patterns and mistakes to avoid. **Typical Exam Question Formats:** 1. **Multi-step Compound Interest Problem** - Initial investment → compound at varying rates → future worth or present worth - Requires careful period counting and rate conversion 2. **Loan/Mortgage Calculation** - Given: loan amount, rate, term - Find: monthly/annual payment, total interest paid - Common: matching period to compounding frequency 3. **Equipment Replacement Analysis** - Compare two equipment options over different lifespans - Annual worth method often simplest - Requires careful handling of salvage values and repeated cycles 4. **Project Economic Justification** - Present worth, benefit-cost ratio, or rate of return - Typical for infrastructure projects - Pay attention to when benefits/costs occur 5. **Depreciation and Tax Impact** - Calculate book value using SL, SYD, or DB methods - Link to taxable income and after-tax analysis (if tested) **Common Exam Mistakes to Avoid:** **Mistake 1: Confusing Nominal and Effective Rates** - **Error:** Using nominal 12% compounded monthly as if it were 12% annually - **Impact:** Significantly underestimates future value or overestimates present worth - **Fix:** Always convert to effective annual rate or use monthly rate with monthly periods Example: ₱100,000 at 12% nominal compounded monthly - WRONG: F = 100,000(1.12)^5 = ₱176,234 (assumes annual compounding) - RIGHT: F = 100,000(1.01)^60 = ₱181,940 (monthly rate, 60 months) **Mistake 2: Mismatching Periods** - **Error:** Using annual rate with quarterly periods, or vice versa - **Impact:** All subsequent calculations are wrong - **Fix:** Verify that interest rate period matches payment period **Mistake 3: Sign Convention in Cash Flow** - **Error:** Not consistent with outflows (negative) vs. inflows (positive) - **Impact:** Present worth calculation gives wrong sign or magnitude - **Fix:** Define outflows as negative, inflows as positive; verify total is reasonable **Mistake 4: Forgetting Salvage Value in Depreciation** - **Error:** SL formula d = C/n instead of d = (C – S)/n - **Impact:** Over-depreciation; book value goes negative - **Fix:** Always subtract salvage value before dividing by life **Mistake 5: Annuity Timing Confusion** - **Error:** Treating annuity due as ordinary annuity (or vice versa) - **Impact:** Present/future worth off by one period - **Fix:** Verify whether payments occur at beginning (annuity due) or end (ordinary) of period **Mistake 6: Not Considering Repeated Cycles** - **Error:** Comparing 5-year and 10-year alternatives without aligning study periods - **Impact:** Unfair comparison; shorter-life alternative appears better - **Fix:** Use annual worth to handle different-life comparisons; or repeat short-life alternative **Practical Advice for Exam Day:** 1. **Read carefully:** Identify whether problem asks for F, P, A, or comparison result 2. **Define periods:** Explicitly state n (number of periods) and verify it matches compounding frequency 3. **Convert rates:** If compounding frequency ≠ payment frequency, convert to match 4. **Set up formula:** Write the formula with all known values substituted before calculating 5. **Verify reasonableness:** Does the answer make intuitive sense? (FV should exceed PV, lower discount rate → higher PW, etc.) 6. **Show work:** In board exam format, show all steps; partial credit available **Real-World Engineering Context:** In professional practice (as required by RA 544 for Civil Engineers), you apply these concepts to: - **Infrastructure projects:** Comparing life-cycle costs of bridge designs, pavement types - **Building systems:** HVAC, electrical, structural options over 20–50 year lifespans - **Water/Wastewater:** Treatment plant expansions, pipeline replacements - **Transportation:** Road tolling, light rail feasibility, parking garage projects - **Energy:** Solar panels, LED lighting, energy efficiency upgrades