CELE Engineering Mathematics — Engineering EconomyDetailed Explanation
This is the "office hours" version of Engineering Economy for the CELE 2026. No shortcuts, no hand-waving — just a full unpacking of why Professional Regulation Commission (PRC) — Board of Civil Engineering cares about each concept and how the Engineering Mathematics section items tend to play out on exam day. Read this once, then hit the practice questions with real understanding.
Exam context
Professional Regulation Commission (PRC) — Board of Civil Engineering runs the Civil Engineer Licensure Examination on May and November 2026. Its Engineering Mathematics section sits under a "Core" weighting, and Engineering Economy is the 10th chapter in the 10-chapter CELE Engineering Mathematics rotation. The CELE passing mark is 70% weighted average, no sub-test below 50%, and the most recent 2026 paper drew about a meaningful share of questions from Engineering Mathematics.
Engineering Economy - Detailed Explanation
Engineering Economy is a high-yield topic in the PRC Civil Engineer Licensure Examination, consistently appearing in both the Mathematics and Engineering Sciences portions. It applies the time value of money — the principle that a peso today is worth more than a peso in the future — to compare engineering alternatives, evaluate investments, compute depreciation, and decide between design options. Mastery of this chapter means being able to set up and solve interest, annuity, and depreciation problems quickly and accurately under exam conditions. This review covers every major concept tested on the board: simple and compound interest, nominal versus effective rates, ordinary annuities, perpetuities, straight-line and accelerated depreciation, and economic comparison methods. All worked examples use Philippine Peso (₱) and SI units, and are structured in the board-style format you will encounter on exam day.
Concepts
Simple Interest
Simple interest is interest computed only on the original principal, with no compounding. The interest earned each period is constant. The governing formula is F = P(1 + in), where P is the principal (present worth), i is the interest rate per period, n is the number of periods, and F is the future worth (maturity value). The total interest earned is I = Pin. Simple interest is used in short-term loans, treasury bills, and promissory notes. In engineering economy problems, always verify whether the problem states 'simple' or 'compound' before choosing your formula — many board examinees lose points by applying the wrong model.
Examples
The interest is ₱18,000, computed only on the original ₱80,000. Note that no interest-on-interest is added because this is simple interest. The maturity value is ₱98,000.
Scenario
A civil engineer borrows ₱80,000 at 9% simple interest per year for 2.5 years to purchase surveying equipment. Find the total amount due at maturity.
Solution
Given: P = ₱80,000; i = 0.09 per year; n = 2.5 years I = Pin = 80,000 × 0.09 × 2.5 = ₱18,000 F = P + I = 80,000 + 18,000 = ₱98,000
Rearrange I = Pin to solve for n. The money grows by ₱3,000 per year under simple interest, so it takes 5 years to accumulate ₱15,000 in interest.
Scenario
How long (in years) will it take for ₱50,000 to grow to ₱65,000 at 6% simple interest per year?
Solution
Given: P = ₱50,000; F = ₱65,000; i = 0.06 I = F - P = 65,000 - 50,000 = ₱15,000 I = Pin → n = I / (Pi) = 15,000 / (50,000 × 0.06) = 15,000 / 3,000 = 5 years
Applications
- Short-term government securities (T-bills) in the Philippines use simple interest
- Promissory notes and short-term commercial loans
- Interim billing periods in construction contracts
- Quick estimation when compounding periods are very short
Misconceptions
- Confusing simple interest with compound interest — simple interest NEVER earns interest on previously earned interest
- Using annual rate with monthly periods without converting — always match units
- Thinking the interest changes each period — in simple interest it is always Pin (constant)
- Forgetting that F = P + I, not F = I alone
Related Concepts
- Compound Interest
- Present Worth
- Nominal and Effective Interest Rates
- Discount Rate
Common Exam Questions
Example
P = ₱25,000, i = 8% per year, n = 3 years → F = 25,000(1 + 0.08×3) = 25,000(1.24) = ₱31,000
Approach
Direct substitution into F = P(1 + in). Ensure i and n are in the same unit.
Question Type
Find F given P, i, n
Example
P = ₱10,000, F = ₱13,500, n = 3.5 years → i = 3,500/(10,000×3.5) = 0.10 = 10% per year
Approach
Rearrange I = Pin → i = I/(Pn) = (F-P)/(Pn)
Question Type
Find the interest rate given P, F, n
Example
P = ₱20,000, F = ₱26,000, i = 5% per year → n = 6,000/(20,000×0.05) = 6 years
Approach
Rearrange to n = (F-P)/(Pi)
Question Type
Find n given P, F, i
Key Points To Remember
- Formula: F = P(1 + in); Interest I = Pin
- Interest is earned ONLY on the original principal — it does NOT compound
- n and i must be in the same time unit (e.g., both monthly or both annual)
- Simple interest is linear: book value and interest grow at a constant rate
- Ordinary simple interest uses 360-day year; exact simple interest uses 365-day year
- For board exams, 'simple interest' always means the non-compounding formula
Compound Interest and Present Worth
In compound interest, interest earned in each period is added to the principal, so subsequent periods earn interest on the accumulated amount — interest on interest. This is the standard model for most engineering economy problems on the board exam. The future worth formula is F = P(1+i)^n, where (1+i)^n is called the single-payment compound amount factor, often written (F/P, i%, n). To find the present worth P of a future amount F, rearrange: P = F(1+i)^(-n). The factor (1+i)^(-n) is the single-payment present worth factor, written (P/F, i%, n). These factors are tabulated in engineering economy handbooks, and on the board exam you will either compute them directly or use given table values. Key strategy: identify what is known (P or F), what is unknown, and solve algebraically.
Examples
Each year the fund grows by 10% of its current value. After 7 years the amount is nearly doubled. Notice (1.10)^7 ≈ 1.9487; Rule of 72 estimates doubling at n = 72/10 = 7.2 years — consistent.
Scenario
A construction firm deposits ₱500,000 in a fund earning 10% compounded annually. Find the value of the fund after 7 years.
Solution
Given: P = ₱500,000; i = 0.10; n = 7 F = P(1+i)^n = 500,000(1.10)^7 (1.10)^7 = 1.9487 F = 500,000 × 1.9487 = ₱974,350
Discounting moves future money back in time. The present worth is only ₱113,480 — you would need to invest only this amount today at 12% to have ₱200,000 in 5 years.
Scenario
What is the present worth of ₱200,000 due 5 years from now at 12% compounded annually?
Solution
Given: F = ₱200,000; i = 0.12; n = 5 P = F(1+i)^(-n) = 200,000(1.12)^(-5) (1.12)^5 = 1.7623 → (1.12)^(-5) = 1/1.7623 = 0.5674 P = 200,000 × 0.5674 = ₱113,480
Since both obligations are expressed in future values at different times, each is discounted separately to time zero, then added. This is the principle of superposition in engineering economy.
Scenario
A contractor must pay ₱350,000 in 4 years and ₱500,000 in 7 years. What single amount today (at 8% compounded annually) is equivalent to these two obligations?
Solution
P₁ = 350,000(1.08)^(-4) = 350,000/1.3605 = ₱257,257 P₂ = 500,000(1.08)^(-7) = 500,000/1.7138 = ₱291,763 Total P = 257,257 + 291,763 = ₱549,020
Applications
- Computing loan balances and maturity values on construction equipment financing
- Evaluating the present cost of future maintenance obligations
- Life-cycle cost analysis of civil infrastructure
- Government bonds and infrastructure bonds (Republic of the Philippines RTBs)
Misconceptions
- Using annual interest rate directly when compounding is monthly or quarterly — must convert to the rate per period
- Forgetting to raise (1+i) to the power n — students sometimes just multiply P × i × n (simple interest error)
- Confusing (F/P) and (P/F) factors — F/P makes money grow, P/F shrinks it
- Not counting periods correctly — if money is invested NOW and withdrawn at end of year 5, n = 5
Related Concepts
- Simple Interest
- Nominal and Effective Interest Rates
- Annuities
- Present Worth Analysis
Common Exam Questions
Example
P = ₱10,000, i = 8%, n = 5 → F = 10,000(1.08)^5 = 10,000(1.4693) = ₱14,693
Approach
Apply F = P(1+i)^n directly. Compute (1+i)^n with a calculator or interpolate from tables.
Question Type
Find F given P, i, n (compound)
Example
F = ₱50,000, i = 12%, n = 8 → P = 50,000/(1.12)^8 = 50,000/2.4760 = ₱20,194
Approach
Apply P = F(1+i)^(-n). Divide F by (1+i)^n.
Question Type
Find P given F, i, n
Example
P = ₱30,000, F = ₱60,000, i = 6% → n = ln(2)/ln(1.06) = 0.6931/0.05827 = 11.9 ≈ 12 years
Approach
Rearrange and use logarithms: n = ln(F/P)/ln(1+i); i = (F/P)^(1/n) - 1
Question Type
Find i or n (solving exponential equation)
Key Points To Remember
- Future worth: F = P(1+i)^n
- Present worth (discounting): P = F(1+i)^(-n) = F/(1+i)^n
- i must be the interest rate per compounding period, and n the number of compounding periods
- The single-payment factors: (F/P,i,n) = (1+i)^n and (P/F,i,n) = (1+i)^(-n)
- For monthly compounding at annual rate r: i = r/12 per month, n = months
- Doubling time approximation: n ≈ 72/i% (Rule of 72) — useful for quick sanity checks
Nominal vs. Effective Interest Rates
Many real-world loans and investments compound more frequently than annually — monthly, quarterly, or even daily. The nominal rate r is the stated annual rate, and m is the number of compounding periods per year. The actual rate earned or paid per year is the effective annual interest rate, computed as: i_eff = (1 + r/m)^m - 1. This is the true cost of borrowing or the true return on investment on an annual basis. For exam problems, always convert to the effective rate (or to the rate per period i = r/m) before plugging into F = P(1+i)^n or annuity formulas. Continuous compounding is a limiting case: i_eff = e^r - 1.
Examples
Although the stated rate is 12%, monthly compounding causes the effective annual yield to be 12.683%. This difference matters when comparing loan offers from different banks.
Scenario
A bank offers 12% nominal interest compounded monthly. What is the effective annual interest rate?
Solution
Given: r = 12% = 0.12; m = 12 (monthly) i_eff = (1 + r/m)^m - 1 = (1 + 0.12/12)^12 - 1 = (1 + 0.01)^12 - 1 = (1.01)^12 - 1 = 1.12683 - 1 = 0.12683 = 12.683%
The quarterly compounding inflates the effective rate from 18% to 19.25%. On the board exam, this type of conversion is a prerequisite for any subsequent annuity or present-worth calculation.
Scenario
What is the effective annual rate for 18% nominal compounded quarterly?
Solution
r = 0.18; m = 4 i_eff = (1 + 0.18/4)^4 - 1 = (1.045)^4 - 1 (1.045)^4 = 1.1925 i_eff = 0.1925 = 19.25%
Continuous compounding gives the maximum possible effective rate for a given nominal rate. Use e^r - 1 when the problem states 'compounded continuously.'
Scenario
A contractor takes a loan at 9% compounded continuously. Find the effective annual rate.
Solution
r = 0.09 i_eff = e^r - 1 = e^0.09 - 1 = 1.09417 - 1 = 0.09417 = 9.417%
Applications
- Comparing mortgage rates from different Philippine banks (BDO, BPI, Metrobank) with different compounding conventions
- Credit card rates quoted as monthly rates (convert to effective annual for true cost)
- Construction equipment lease vs buy analysis when compounding periods differ
- Pag-IBIG and SSS loan rate comparisons
Misconceptions
- Using nominal rate directly in F = P(1+i)^n when compounding is not annual
- Thinking nominal and effective rates are the same — they are only equal when m = 1
- Forgetting to subtract 1 in i_eff = (1 + r/m)^m - 1
- Confusing continuous compounding formula e^r with (1 + r/m)^m for finite m
Related Concepts
- Compound Interest
- Annuities
- Loan Amortization
- Present Worth Analysis
Common Exam Questions
Example
r = 10%, quarterly → i_eff = (1 + 0.10/4)^4 - 1 = (1.025)^4 - 1 = 10.38%
Approach
Apply i_eff = (1 + r/m)^m - 1 directly
Question Type
Find effective annual rate from nominal rate and compounding frequency
Example
i_eff = 12%, monthly → r = 12[(1.12)^(1/12) - 1] = 12[1.00949 - 1] = 12(0.00949) = 11.39%
Approach
Rearrange: r = m[(1 + i_eff)^(1/m) - 1]
Question Type
Find nominal rate given effective rate
Example
P = ₱20,000, r = 8% compounded quarterly, n = 3 years → i = 8%/4 = 2%, n = 12 quarters → F = 20,000(1.02)^12 = ₱25,364
Approach
Compute i_eff, then use as i in F = P(1+i)^n with n in years
Question Type
Use effective rate in a compound interest problem
Key Points To Remember
- Nominal rate r = stated annual rate; m = compounding frequency per year
- Rate per period: i = r/m (this is what goes into F = P(1+i)^n)
- Effective annual rate: i_eff = (1 + r/m)^m - 1
- Continuous compounding: i_eff = e^r - 1 ≈ r + r²/2 + ...
- The effective rate is ALWAYS ≥ the nominal rate (equality only when m = 1)
- To compare loans with different compounding periods, ALWAYS convert to effective annual rates
Annuities
An annuity is a series of equal (uniform) payments A made at regular intervals for n periods. The most common type on the board exam is the ordinary annuity (also called annuity-immediate), where payments occur at the END of each period. The two key formulas are: (1) Future Worth F = A × [(1+i)^n - 1]/i, where the bracketed term is the uniform-series compound amount factor (F/A, i%, n); and (2) Present Worth P = A × [(1+i)^n - 1]/[i(1+i)^n], where the bracketed term is the uniform-series present worth factor (P/A, i%, n). A perpetuity is an infinite annuity (n → ∞), giving P = A/i. The annuity-due has payments at the beginning of each period; multiply ordinary annuity P or F by (1+i) to convert. Board exam problems frequently require you to identify whether a problem is an annuity or single-payment scenario, then select the correct factor.
Examples
The future worth accumulates all deposits plus compound interest to the time of the last deposit. The present worth discounts all deposits back to time zero — if you invested ₱3,791 today at 10%, you could make five annual withdrawals of ₱1,000.
Scenario
A structural engineer deposits ₱1,000 at the end of each year for 5 years at 10% compounded annually. Find the future and present worth of the annuity.
Solution
Given: A = ₱1,000; i = 0.10; n = 5 Future Worth: F = A[(1+i)^n - 1]/i = 1,000 × [(1.10)^5 - 1]/0.10 (1.10)^5 = 1.6105 F = 1,000 × (1.6105 - 1)/0.10 = 1,000 × 0.6105/0.10 = 1,000 × 6.105 = ₱6,105 Present Worth: P = A[(1+i)^n - 1]/[i(1+i)^n] = 1,000 × (0.6105)/[0.10 × 1.6105] = 1,000 × 0.6105/0.16105 = 1,000 × 3.7908 = ₱3,791
The perpetuity formula is the limiting case of the annuity formula as n → ∞. Depositing ₱625,000 at 8% generates exactly ₱50,000 per year in perpetual interest without touching the principal — a common application in public infrastructure endowment funds.
Scenario
A water treatment plant requires maintenance costing ₱50,000 per year indefinitely. At 8% interest, what lump sum today should be set aside?
Solution
This is a perpetuity: P = A/i = 50,000/0.08 = ₱625,000
This is the capital recovery formula, which is the inverse of the present worth annuity formula. It finds the uniform payment A that repays a present amount P over n periods at rate i. This is exactly how Pag-IBIG and bank mortgage amortizations are computed.
Scenario
A civil engineer takes a ₱1,500,000 home loan at 9% compounded monthly for 20 years. Find the monthly amortization payment.
Solution
Given: P = ₱1,500,000; r = 9% nominal compounded monthly i = 9%/12 = 0.75% = 0.0075 per month; n = 20 × 12 = 240 months A = P × i(1+i)^n / [(1+i)^n - 1] (1.0075)^240: Use ln method: 240 × ln(1.0075) = 240 × 0.007472 = 1.7933 → e^1.7933 = 6.0076 A = 1,500,000 × 0.0075 × 6.0076 / (6.0076 - 1) = 1,500,000 × 0.045057 / 5.0076 = 1,500,000 × 0.008997 = ₱13,496 per month
Applications
- Monthly amortization of Pag-IBIG and bank housing loans
- Annual sinking fund contributions for infrastructure replacement
- Equipment lease-or-buy decisions based on present worth of lease payments
- Retirement fund accumulation for engineers (annual deposits)
Misconceptions
- Treating an annuity-due (beginning of period) as an ordinary annuity — always check timing of first payment
- Using annual i when payments are monthly without converting to monthly rate
- Forgetting that n in annuity formulas must match the payment period (not years if monthly)
- Confusing F/A (future worth factor) and P/A (present worth factor) — F > P always
Related Concepts
- Compound Interest
- Present Worth Analysis
- Nominal and Effective Rates
- Depreciation
- Economic Comparison
Common Exam Questions
Example
A = ₱2,000/month, i = 0.5%/month, n = 60 months → F = 2,000[(1.005)^60 - 1]/0.005 = 2,000(69.77) = ₱139,540
Approach
Apply F = A[(1+i)^n - 1]/i. Compute (1+i)^n first.
Question Type
Find F given A, i, n (future worth of annuity)
Example
A = ₱5,000/year, i = 10%, n = 10 → P = 5,000 × (6.1446 - 1)/(0.10 × 6.1446) = ₱30,723
Approach
Apply P = A[(1+i)^n - 1]/[i(1+i)^n]
Question Type
Find P given A, i, n (present worth of annuity)
Example
Loan P = ₱500,000, i = 1%/month, n = 36 months → A = 500,000 × 0.01(1.01)^36/[(1.01)^36 - 1]
Approach
Apply A = P × i(1+i)^n/[(1+i)^n - 1]
Question Type
Find A given P (capital recovery / amortization)
Example
Annual cost A = ₱100,000, i = 5% → P = 100,000/0.05 = ₱2,000,000
Approach
Apply P = A/i directly
Question Type
Perpetuity: find P given A and i
Key Points To Remember
- Ordinary annuity: payments at END of each period (most common on board exams)
- Future worth: F = A[(1+i)^n - 1]/i
- Present worth: P = A[(1+i)^n - 1]/[i(1+i)^n]
- Perpetuity: P = A/i (when n approaches infinity)
- Annuity-due (beginning of period): multiply P or F of ordinary annuity by (1+i)
- To find A given P: A = P × i(1+i)^n/[(1+i)^n - 1] — this is the capital recovery formula
- To find A given F: A = F × i/[(1+i)^n - 1] — this is the sinking fund formula
Depreciation Methods
Depreciation is the systematic allocation of an asset's cost over its useful life. In engineering economy, depreciation affects the book value of an asset and is critical for tax and cost analysis. The three methods tested on the PRC board exam are: (1) Straight-Line (SL): equal annual charges; (2) Sum-of-Years-Digits (SYD): accelerated, front-loaded charges; (3) Declining Balance (DB): a fixed percentage of current book value each year. Key variables: C = first cost (initial cost), S = salvage value at end of life, n = useful life in years, BV_t = book value at end of year t. The choice of method affects tax savings but NOT the total depreciation (C - S is always fully depreciated regardless of method, except DB which may not reach salvage).
Examples
Straight-line gives equal annual charges of ₱18,000. After 3 years, accumulated depreciation is ₱54,000, leaving a book value of ₱46,000. This is the simplest and most commonly tested method.
Scenario
A concrete batching plant costs ₱100,000 with ₱10,000 salvage value and a 5-year useful life. Compute the annual depreciation and book value at end of year 3 using the straight-line method.
Solution
C = ₱100,000; S = ₱10,000; n = 5 d = (C - S)/n = (100,000 - 10,000)/5 = 90,000/5 = ₱18,000/year BV₃ = C - 3d = 100,000 - 3(18,000) = 100,000 - 54,000 = ₱46,000
SYD assigns larger depreciation in earlier years (₱30,000 in year 1 vs ₱18,000 in year 3). This front-loading reflects the greater loss of economic value early in an asset's life. The digits used in year t are (n - t + 1) out of SYD.
Scenario
Using the same machine (C = ₱100,000, S = ₱10,000, n = 5), compute the depreciation in years 1 and 3 using the Sum-of-Years-Digits method.
Solution
SYD = n(n+1)/2 = 5(6)/2 = 15 Depreciable amount = C - S = 90,000 Year 1: fraction = (5 - 1 + 1)/15 = 5/15 = 1/3 d₁ = (5/15)(90,000) = ₱30,000 Year 3: fraction = (5 - 3 + 1)/15 = 3/15 = 1/5 d₃ = (3/15)(90,000) = ₱18,000
In DDB, the rate 2/n is applied to the current book value, NOT to (C - S). This creates a geometric decline. Notice DDB does not directly incorporate the salvage value — you must stop depreciation if BV would fall below S.
Scenario
Using the Double Declining Balance method (C = ₱100,000, S = ₱10,000, n = 5), find the depreciation in years 1, 2, and 3.
Solution
k = 2/n = 2/5 = 0.40 = 40% Year 1: BV₀ = 100,000; d₁ = 0.40 × 100,000 = ₱40,000; BV₁ = 60,000 Year 2: d₂ = 0.40 × 60,000 = ₱24,000; BV₂ = 36,000 Year 3: d₃ = 0.40 × 36,000 = ₱14,400; BV₃ = 21,600 Note: BV₃ = ₱21,600 > S = ₱10,000, so depreciation continues.
Applications
- Computing book value of construction equipment for financial statements
- Tax depreciation scheduling for engineering firms (NIRC of the Philippines allows SL and DB)
- Life-cycle cost analysis including residual asset value
- Equipment replacement studies (economic life determination)
Misconceptions
- Using salvage value in DDB calculation — DDB applies the rate to book value, NOT (C-S)
- In SYD, applying digits in the wrong order — year 1 gets the HIGHEST digit
- Forgetting that BV can never go below S in any depreciation method
- Confusing book value with market value — depreciation tracks accounting value only
Related Concepts
- Economic Life of Equipment
- Annual Cost Analysis
- Tax and After-Tax Analysis
- Equipment Replacement
Common Exam Questions
Example
C = ₱250,000, S = ₱25,000, n = 8 → d = (250,000 - 25,000)/8 = ₱28,125/year
Approach
d = (C - S)/n — straightforward substitution
Question Type
Find annual depreciation using SL method
Example
C = ₱250,000, S = ₱25,000, n = 8, year 1 → SYD = 36; d₁ = (8/36)(225,000) = ₱50,000
Approach
Compute SYD = n(n+1)/2; fraction for year t = (n-t+1)/SYD; multiply by (C-S)
Question Type
Find depreciation in a specific year using SYD
Example
SL: BV₃ = 250,000 - 3(28,125) = ₱165,625
Approach
SL: BV_t = C - t×d. SYD: sum depreciation for years 1 to t and subtract from C.
Question Type
Find book value at end of year t
Key Points To Remember
- Total depreciation over life = C - S (always, for SL and SYD)
- SL: d = (C - S)/n per year; BV_t = C - t×d
- SYD: SYD = n(n+1)/2; d_t = (n - t + 1)/SYD × (C - S)
- DB: d_t = k × BV_(t-1); k is the fixed depreciation rate (k = 1/n for SL rate; k = 2/n for double declining balance)
- Double Declining Balance (DDB): k = 2/n; do NOT subtract salvage before applying the rate
- In SYD, the digits are applied in REVERSE — highest in year 1, lowest in year n
- Book value from DB may never go below salvage value
Economic Comparison of Alternatives
Engineering decisions almost always involve choosing among alternatives that differ in first cost, operating cost, life, and salvage value. The standard methods for comparison are: (1) Present Worth (PW) Analysis — convert all cash flows to time zero; choose the alternative with the highest PW (or least negative PW for cost-only problems); (2) Annual Worth (AW) Analysis — convert all costs to a uniform annual amount; choose lowest AW of costs; (3) Rate of Return (ROR) / Internal Rate of Return (IRR) — the interest rate that makes PW of benefits equal to PW of costs (PW = 0); accept if ROR ≥ MARR (Minimum Attractive Rate of Return); (4) Benefit-Cost Ratio (B/C) — B/C = PW of benefits / PW of costs; accept if B/C ≥ 1.0; (5) Break-Even Analysis — find the output or service level where two alternatives have equal cost. For alternatives with different lives, use the Least Common Multiple (LCM) of lives for PW comparison, or use Annual Worth which is life-independent.
Examples
Even though Pump B has lower annual maintenance, its much higher first cost and only modest salvage value make it more expensive in present worth terms at 10% interest. Always bring salvage values back to PW by multiplying by (P/F, i, n).
Scenario
Two pumps are considered for a water system. Pump A: first cost ₱80,000, annual maintenance ₱8,000, life 10 years, no salvage. Pump B: first cost ₱120,000, annual maintenance ₱5,000, life 10 years, salvage ₱20,000. At i = 10%, which is preferred using present worth?
Solution
PW of costs (lower is better): Pump A: PW_A = 80,000 + 8,000(P/A, 10%, 10) (P/A, 10%, 10) = [(1.10)^10 - 1]/[0.10(1.10)^10] = [1.5937]/[0.2594] = 6.1446 PW_A = 80,000 + 8,000(6.1446) = 80,000 + 49,157 = ₱129,157 Pump B: PW_B = 120,000 + 5,000(P/A, 10%, 10) - 20,000(P/F, 10%, 10) (P/F, 10%, 10) = 1/(1.10)^10 = 1/2.5937 = 0.3855 PW_B = 120,000 + 5,000(6.1446) - 20,000(0.3855) = 120,000 + 30,723 - 7,710 = ₱143,013 Conclusion: Pump A has lower present worth of costs (₱129,157 < ₱143,013). Select Pump A.
The B/C ratio is the primary economic justification tool for Philippine government infrastructure projects. A ratio of 1.364 means every peso spent returns ₱1.364 in benefits — a positive net contribution to society.
Scenario
A highway project has PW of benefits = ₱15,000,000 and PW of costs = ₱11,000,000. Should the project be pursued?
Solution
B/C ratio = PW of Benefits / PW of Costs = 15,000,000 / 11,000,000 = 1.364 Since B/C = 1.364 ≥ 1.0, the project is economically justified.
Capitalized cost converts all infinite future expenditures into a single present lump sum. It is used for permanent public works (bridges, dams, roads) where the service is intended to continue indefinitely.
Scenario
Capitalized cost: A bridge is to last forever. Construction cost = ₱50,000,000; annual maintenance = ₱500,000; major repairs every 10 years costing ₱5,000,000. At i = 8%, find the capitalized cost.
Solution
Capitalized cost = First cost + PW of annual costs (perpetuity) + PW of recurring costs (repeated) Component 1 (first cost): ₱50,000,000 Component 2 (annual maintenance perpetuity): PW = A/i = 500,000/0.08 = ₱6,250,000 Component 3 (₱5,000,000 every 10 years): Convert to equivalent annual cost A_repair: A_repair = F × i/[(1+i)^n - 1] (sinking fund) = 5,000,000 × 0.08/[(1.08)^10 - 1] = 5,000,000 × 0.08/1.1589 = 5,000,000 × 0.06903 = ₱345,150/year PW_repair = A_repair/i = 345,150/0.08 = ₱4,314,375 Total Capitalized Cost = 50,000,000 + 6,250,000 + 4,314,375 = ₱60,564,375
Applications
- Selecting between construction methods (e.g., cast-in-place vs precast) based on life-cycle cost
- Government infrastructure project evaluation (DPWH uses B/C analysis per RA 7160 and NEDA guidelines)
- Equipment replacement timing decisions
- Bridge vs culvert alternatives for drainage design
- Lease vs buy decisions for heavy equipment
Misconceptions
- Comparing alternatives with different lives using PW without using LCM — AW analysis avoids this problem
- Forgetting to include salvage value as a positive cash flow in PW analysis
- Using B/C < 1 to reject and not considering that benefits must always be defined consistently
- In break-even: setting up the wrong cost equations (fixed vs variable cost confusion)
Related Concepts
- Present Worth
- Annuities
- Depreciation
- Rate of Return
- MARR
Common Exam Questions
Example
Alt A: P=₱100k, A=₱10k/yr, S=0, n=5, i=10% → PW_A = 100k + 10k(P/A,10%,5) = 100k + 37,908 = ₱137,908
Approach
Compute PW of all costs (and benefits if applicable) for each alternative; choose lowest PW of costs or highest PW of net benefits
Question Type
Present worth comparison of two alternatives (equal lives)
Example
Machine A: ₱50,000 + ₱200x; Machine B: ₱30,000 + ₱300x. Break-even: 50,000 + 200x = 30,000 + 300x → x = 200 units
Approach
Set Total Cost A = Total Cost B; solve for the unknown (units, years, rate)
Question Type
Break-even analysis
Example
CC = 500,000 + 20,000/0.06 = 500,000 + 333,333 = ₱833,333
Approach
CC = First Cost + A/i (for annual recurring costs); for periodic costs, convert to annual using sinking fund, then divide by i
Question Type
Capitalized cost
Key Points To Remember
- PW analysis: choose highest PW (or least cost); use LCM of lives if alternatives have unequal lives
- AW analysis: does not require LCM — can compare alternatives with different lives directly
- ROR/IRR: accept project if IRR ≥ MARR; for mutually exclusive alternatives, use incremental ROR
- B/C ratio ≥ 1.0 justifies a project; used mainly for public projects in the Philippines
- Break-even: equate total cost equations of two alternatives and solve for the variable (units, years)
- MARR = Minimum Attractive Rate of Return — the minimum acceptable return for investment
- Capitalized cost = first cost + PW of perpetual annual costs = C + A/i (important for public infrastructure)
Practice Problems
Exact simple interest counts actual calendar days and uses a 365-day year (360 for ordinary simple interest). For exam problems, if 'exact' is stated, use 365. The answer is ₱125,993.
Problem
PROBLEM 1 (Simple Interest): A promissory note for ₱120,000 at 7.5% simple interest is made on January 1 and is due September 1 of the same year (exact simple interest, non-leap year). Find the maturity value.
Solution
Number of days from Jan 1 to Sep 1 = 31(Jan) + 28(Feb) + 31(Mar) + 30(Apr) + 31(May) + 30(Jun) + 31(Jul) + 31(Aug) = 243 days n = 243/365 years I = Pin = 120,000 × 0.075 × (243/365) = 120,000 × 0.075 × 0.6658 = ₱5,993 F = P + I = 120,000 + 5,993 = ₱125,993
Always convert to the rate per compounding period (3%/quarter) and the number of periods (16 quarters). The future worth is ₱80,052 — significantly higher than if compounded annually [(1.12)^4 = 1.5735, giving only ₱78,675].
Problem
PROBLEM 2 (Compound Interest): ₱50,000 is invested at 12% nominal interest compounded quarterly for 4 years. Find the future worth.
Solution
r = 12% nominal, m = 4 (quarterly) i = r/m = 12%/4 = 3% per quarter n = 4 years × 4 quarters/year = 16 quarters F = P(1+i)^n = 50,000(1.03)^16 (1.03)^16: (1.03)^4 = 1.12551; (1.12551)^4 = 1.60103 F = 50,000 × 1.60103 = ₱80,052
The key is to equate the effective annual rates. A rate of 12% semi-annual ≈ 11.71% monthly nominal — these two rates are economically equivalent (same effective annual rate of 12.36%). This type of equivalence problem is common on the board.
Problem
PROBLEM 3 (Effective Rate): Find the nominal interest rate compounded monthly that is equivalent to 12% compounded semi-annually.
Solution
First find i_eff for 12% compounded semi-annually: i_eff = (1 + 0.12/2)^2 - 1 = (1.06)^2 - 1 = 1.1236 - 1 = 0.1236 = 12.36% Now find the nominal rate r compounded monthly with the same effective rate: (1 + r/12)^12 - 1 = 0.1236 (1 + r/12)^12 = 1.1236 1 + r/12 = (1.1236)^(1/12) = 1.1236^0.08333 ln(1.1236) = 0.11653 → 0.11653/12 = 0.009711 → e^0.009711 = 1.009758 r/12 = 0.009758 r = 0.11710 = 11.71% nominal compounded monthly
Monthly deposits require monthly rate and monthly periods. The future worth of ₱78,672 compares to total deposits of 2,000 × 36 = ₱72,000 — the difference of ₱6,672 is the interest earned. This is the basis for SSS and Pag-IBIG monthly contribution projections.
Problem
PROBLEM 4 (Annuity — Monthly Deposits): Monthly deposits of ₱2,000 are made for 3 years at 6% nominal interest compounded monthly. Find the future worth.
Solution
A = ₱2,000/month; r = 6%, m = 12 → i = 0.5%/month = 0.005 n = 3 × 12 = 36 months F = A[(1+i)^n - 1]/i = 2,000 × [(1.005)^36 - 1]/0.005 (1.005)^36: ln(1.005) = 0.004988; 36 × 0.004988 = 0.17957; e^0.17957 = 1.19668 F = 2,000 × (1.19668 - 1)/0.005 = 2,000 × 0.19668/0.005 = 2,000 × 39.336 = ₱78,672
SYD front-loads depreciation. By end of year 4 (halfway through the 8-year life), the asset has depreciated by ₱162,500 out of ₱225,000 total — about 72% of total depreciation in the first 50% of its life. For SL, 4 years would give only 50% of total depreciation.
Problem
PROBLEM 5 (SYD Depreciation): A surveying instrument costs ₱250,000 with a salvage value of ₱25,000 and a useful life of 8 years. Using SYD, find the depreciation in year 1 and the book value at end of year 4.
Solution
C = ₱250,000; S = ₱25,000; n = 8 Depreciable amount = C - S = ₱225,000 SYD = n(n+1)/2 = 8(9)/2 = 36 Year 1: d₁ = (8/36)(225,000) = ₱50,000 Year 2: d₂ = (7/36)(225,000) = ₱43,750 Year 3: d₃ = (6/36)(225,000) = ₱37,500 Year 4: d₄ = (5/36)(225,000) = ₱31,250 Total depreciation, years 1–4 = 50,000 + 43,750 + 37,500 + 31,250 = ₱162,500 BV₄ = C - accumulated depreciation = 250,000 - 162,500 = ₱87,500
Capitalized cost analysis for perpetual structures first converts all periodic costs to an equivalent annual amount, then capitalizes that annual amount using P = A/i. This method is standard for DPWH and NEDA infrastructure project evaluations in the Philippines.
Problem
PROBLEM 6 (Capitalized Cost): A road is to be maintained permanently. Resurfacing costs ₱5,000,000 every 5 years. Annual routine maintenance is ₱200,000. Initial construction = ₱30,000,000. At i = 8%, compute the capitalized cost.
Solution
Component 1: Initial construction = ₱30,000,000 Component 2: Annual maintenance (perpetuity) PW₂ = A/i = 200,000/0.08 = ₱2,500,000 Component 3: Resurfacing ₱5,000,000 every 5 years Convert to equivalent annual cost (sinking fund to accumulate ₱5,000,000 in 5 years): A_resurface = F × i/[(1+i)^n - 1] = 5,000,000 × 0.08/[(1.08)^5 - 1] (1.08)^5 = 1.46933; [(1.08)^5 - 1] = 0.46933 A_resurface = 5,000,000 × 0.08/0.46933 = 400,000/0.46933 = ₱852,228/year PW₃ = A_resurface/i = 852,228/0.08 = ₱10,652,850 Capitalized Cost = 30,000,000 + 2,500,000 + 10,652,850 = ₱43,152,850
Exam Preparation Tips
- KNOW YOUR FACTORS: Memorize the six standard engineering economy factors — (F/P), (P/F), (F/A), (P/A), (A/F), (A/P) — and their formulas. Board exams frequently ask you to compute these without tables.
- UNIT CONSISTENCY: Before solving any problem, confirm that i and n are in the SAME time unit. Monthly payments require monthly i and n in months. This is the #1 source of errors.
- NOMINAL VS EFFECTIVE: Whenever the problem mentions 'compounded monthly/quarterly/semi-annually,' immediately compute i = r/m and n = (years × m). Do this as Step 1 before any other calculation.
- IDENTIFY THE PROBLEM TYPE FIRST: Read the problem and classify it: single payment (F=P(1+i)^n), uniform series (annuity), gradient series, or perpetuity. Wrong formula selection wastes time and marks.
- SALVAGE VALUE SIGN CONVENTION: In present worth analysis, salvage is a POSITIVE (benefit) cash flow. Subtract PW of salvage from PW of costs. Students frequently forget to include salvage or assign it the wrong sign.
- ANNUITY TIMING CHECK: If the first payment is at time 0 (not time 1), it is an annuity-due. Multiply the ordinary annuity PW or FW by (1+i) to account for the one-period shift.
- SYD DIGIT ORDER: The highest digit is ALWAYS assigned to Year 1. For n=5, digits assigned are: Yr1=5, Yr2=4, Yr3=3, Yr4=2, Yr5=1 (each divided by SYD=15).
- USE LOGARITHMS FOR UNKNOWN EXPONENTS: To find n when F and P are known: n = ln(F/P)/ln(1+i). To find i: i = (F/P)^(1/n) - 1. Practice these on your calculator.
- B/C FOR PUBLIC PROJECTS: In Philippine engineering practice, B/C ≥ 1.0 justifies government infrastructure. B = present worth of all societal benefits; C = present worth of all costs. Do not subtract costs from the numerator.
- BREAK-EVEN SETUP: Identify which variable is being solved for (units, years, or rate). Write separate total cost equations for each alternative, set them equal, and solve algebraically.
- CAPITALIZED COST = FIRST COST + A/i: For any perpetual structure, annual recurring costs are capitalized as A/i. For periodic costs, first convert to annual equivalent using (A/F) factor, then divide by i.
- DOUBLE-CHECK WITH ESTIMATES: After solving, check with a rough estimate. Rule of 72: money doubles in approximately 72/i% years. Use this to verify compound interest answers are in the right ballpark.
- MARR IS THE THRESHOLD: If ROR > MARR → accept. If ROR < MARR → reject. For mutually exclusive alternatives, the highest ROR does not necessarily mean the best choice — use incremental analysis.
- PRACTICE WITH PRC PAST BOARD PROBLEMS: The PRC Engineering Mathematics board exam consistently tests: (a) compound interest with non-annual compounding, (b) ordinary annuity future/present worth, (c) straight-line and SYD depreciation, and (d) present worth comparison. Focus your review time on these four areas.
In summary
Engineering Economy is one of the most formula-rich yet conceptually straightforward topics on the PRC Civil Engineer Licensure Examination. Success comes from three habits: (1) correctly identifying the problem type — single payment, annuity, depreciation, or comparison — before touching your calculator; (2) rigorous unit consistency — matching the interest rate period to the payment period every single time; and (3) mastering the five or six core formulas so thoroughly that you can apply them in under two minutes per problem under exam pressure. Review the six engineering economy factors [(F/P), (P/F), (F/A), (P/A), (A/F), (A/P)], practice the depreciation digit sequences for SYD, and always draw a cash flow timeline for comparison problems. Filipino engineers who practice ten to fifteen board-style problems per topic before the exam consistently report that Engineering Economy becomes one of their strongest scoring areas. Use the worked examples and practice problems in this chapter as your template — replicate the solution format on exam day, and you will not only get the right answer but demonstrate clear, organized engineering reasoning that earns full credit.
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