CELE Engineering Mathematics — Engineering EconomyRevision Notes
Final-week revision notes for Engineering Economy. If you have already studied the full chapter, this page is your go-to refresher before sitting the CELE. Compact, high-yield, and aligned with what Professional Regulation Commission (PRC) — Board of Civil Engineering tests in the Engineering Mathematics subtest.
Exam context
The Civil Engineer Licensure Examination is conducted by Professional Regulation Commission (PRC) — Board of Civil Engineering and is scheduled for May and November 2026. The Engineering Mathematics subtest is marked as "Core" in the official pattern, and Engineering Economy appears in position 10th of 10 in the CELE Engineering Mathematics review rotation. Passing mark: 70% weighted average, no sub-test below 50%. Recent CELE 2026 papers have drawn roughly a meaningful share of questions from this subject.
Engineering Economy - Revision Notes
Engineering Economy is a high-yield topic in the PRC Civil Engineer Licensure Examination, appearing consistently in the Engineering Mathematics component. It applies the time value of money to compare engineering alternatives — a skill every practicing civil engineer needs when evaluating project costs, equipment replacement, and infrastructure financing. This revision note covers all board-exam-critical subtopics: simple and compound interest, nominal vs. effective rates, annuities and perpetuities, depreciation methods, and economic comparison techniques. Mastery of these concepts and their associated formulas — applied correctly with consistent period-interest rate matching — is essential for exam success.
Sections
Formulas
Example
₱20,000 at 9% simple interest for 3 years: F = 20,000[1 + (0.09)(3)] = 20,000(1.27) = ₱25,400.
Formula
F = P(1 + in)
Variables
F = future worth (₱); P = principal or present worth (₱); i = interest rate per period (decimal); n = number of periods
Application
Use for simple interest problems — short-term loans, promissory notes, or problems that explicitly state 'simple interest'.
Example
₱10,000 at 8% compounded annually for 5 years: F = 10,000(1.08)^5 = 10,000(1.4693) = ₱14,693.
Formula
F = P(1 + i)^n
Variables
F = future worth (₱); P = present worth (₱); i = effective interest rate per compounding period (decimal); n = total number of compounding periods
Application
The cornerstone formula for compound interest. Use whenever interest is compounded. Solve for P by rearranging: P = F(1+i)^(-n).
Example
12% compounded monthly: i_eff = (1 + 0.12/12)^12 − 1 = (1.01)^12 − 1 = 1.12683 − 1 = 12.683% per year.
Formula
i_eff = (1 + r/m)^m − 1
Variables
i_eff = effective annual interest rate (decimal); r = nominal annual interest rate (decimal); m = number of compounding periods per year
Application
Convert a nominal rate to an effective annual rate before comparing alternatives with different compounding frequencies.
Example
12% compounded continuously: i_eff = e^0.12 − 1 = 1.1275 − 1 = 12.75% per year.
Formula
i_eff = e^r − 1
Variables
i_eff = effective annual rate; r = nominal rate; e = Euler's number (2.71828); applicable when compounding is continuous.
Application
Used for continuous compounding problems. Also yields F = Pe^(rn) for continuous compounding.
Exam Tips
- Always draw a cash-flow diagram — label P at t=0, F at t=n, and each payment A at its respective period end.
- Memorize: (1.08)^5 ≈ 1.4693, (1.10)^5 ≈ 1.6105, (1.12)^10 ≈ 3.1058 as common board-exam values.
- For 'compounded monthly' problems: periodic rate = r/12, total periods = years × 12.
- If the problem gives both a nominal rate and asks for effective rate — always apply i_eff = (1+r/m)^m − 1.
- Double-check the number of periods: '5 years compounded quarterly' means n = 20 periods, not 5.
Key Points
- Money has earning power over time; a peso today is worth more than a peso in the future.
- Simple interest accrues only on the principal; compound interest accrues on principal plus accumulated interest.
- For compound interest, F = P(1+i)^n is the most fundamental formula in Engineering Economy.
- Present worth (PW) discounts a future amount back to today using P = F(1+i)^(-n).
- Interest rate i and number of periods n must always match the same compounding period — this is the single most-tested pitfall.
- When comparing investments with different compounding frequencies, always convert to effective annual rates first.
Definitions
Term
Principal (P)
Definition
The initial sum of money invested or borrowed; also called Present Worth or Present Value.
Importance
Starting point of all time-value calculations; always place P at time zero on the cash-flow diagram.
Term
Nominal Interest Rate (r)
Definition
The stated annual interest rate that does not account for the effect of compounding within the year. Given as '12% compounded monthly.'
Importance
Must be converted to effective rate before use in formulas when compounding is more frequent than the stated period.
Term
Effective Interest Rate (i_eff)
Definition
The actual annual rate that produces the same future value as the nominal rate when compounding is considered. Always ≥ nominal rate.
Importance
Use this rate when comparing alternatives with different compounding frequencies on a per-annum basis.
Term
Compounding Period
Definition
The interval at which interest is calculated and added to the balance. Common: annually (m=1), semi-annually (m=2), quarterly (m=4), monthly (m=12), daily (m=365).
Importance
Determines the periodic rate i = r/m to use in compound interest formulas.
Section Title
1. Time Value of Money — Simple and Compound Interest
Common Mistakes
- Using the nominal annual rate directly in F = P(1+i)^n when compounding is monthly — always use i = r/m and n = total months.
- Mixing annual interest rates with monthly periods (or vice versa) without conversion.
- Forgetting that for continuous compounding, use F = Pe^(rn), not F = P(1+i)^n.
- Using simple interest when the problem says 'compounded' — watch for keyword cues.
- Computing (1.08)^5 incorrectly by multiplying 1.08 × 5 instead of raising to a power.
Formulas
Example
₱1,000 deposited at end of each year for 5 years at 10%: F = 1,000 × [(1.10)^5 − 1]/0.10 = 1,000 × 6.105 = ₱6,105.
Formula
F = A × [(1+i)^n − 1] / i
Variables
F = future worth of the annuity series; A = uniform end-of-period payment; i = interest rate per period; n = number of periods
Application
Find the accumulated future value of a series of equal periodic deposits. The bracketed factor is called the uniform series future worth factor (F/A, i%, n).
Example
₱1,000 per year for 5 years at 10%: P = 1,000 × [(1.10)^5 − 1]/[0.10 × (1.10)^5] = 1,000 × 3.7908 = ₱3,791.
Formula
P = A × [(1+i)^n − 1] / [i(1+i)^n]
Variables
P = present worth of the annuity series; A = uniform end-of-period payment; i = interest rate per period; n = number of periods
Application
Find the present value equivalent of a series of equal future payments — used for loan analysis, equipment leasing, and present-worth comparisons.
Example
A scholarship fund pays ₱50,000 per year forever at 8% interest: P = 50,000/0.08 = ₱625,000 needed today.
Formula
P_perpetuity = A / i
Variables
P = present worth of the perpetuity; A = uniform payment per period; i = interest rate per period
Application
Use when a stream of payments continues indefinitely — endowments, scholarships, or capitalized costs with infinite life.
Example
Loan of ₱500,000 at 12% compounded monthly for 5 years: i = 1%, n = 60. A = 500,000 × [0.01(1.01)^60]/[(1.01)^60 − 1] = 500,000 × 0.02224 = ₱11,122/month.
Formula
A = P × [i(1+i)^n] / [(1+i)^n − 1]
Variables
A = uniform periodic payment; P = present worth (loan amount); i = periodic interest rate; n = number of payment periods
Application
Loan amortization — find the equal monthly or annual payment required to repay a loan. This is the capital recovery formula.
Exam Tips
- The P/A factor is also called the 'uniform series present worth factor' — know it by name as board exams use factor notation.
- Memorize: P/A at 10% for 5 years = 3.7908; F/A at 10% for 5 years = 6.1051.
- For loan problems, A is the equal payment; use the capital recovery (A/P) formula directly.
- Perpetuity shortcut: if n > 50 periods and i ≥ 8%, the annuity present worth ≈ A/i (the perpetuity formula is a close approximation).
- Always verify: P × (1+i)^n should equal F computed from the F/A formula for the same annuity — use this as a check.
Key Points
- An annuity is a series of equal payments A made at equal time intervals for n periods.
- Ordinary annuity (most common in board exams): payments occur at the END of each period.
- Annuity-due: payments occur at the BEGINNING of each period; multiply ordinary annuity factors by (1+i).
- The future worth factor (F/A) converts a series of payments to a single future lump sum.
- The present worth factor (P/A) converts a series of payments to a single present lump sum.
- A perpetuity is an annuity with an infinite number of periods; P = A/i.
- Deferred annuity: the first payment starts after a delay of k periods; discount the P/A result further by (1+i)^(-k).
Definitions
Term
Ordinary Annuity
Definition
An annuity where payments occur at the end of each compounding period. This is the default assumption unless stated otherwise.
Importance
Most board exam annuity problems assume ordinary annuity. Always verify timing before applying formulas.
Term
Annuity-Due
Definition
An annuity where payments occur at the beginning of each period. Its present worth = Ordinary P/A × (1+i); its future worth = Ordinary F/A × (1+i).
Importance
Watch for phrases like 'beginning of year,' 'first payment today,' or 'payments in advance.'
Term
Deferred Annuity
Definition
An annuity whose first payment is delayed by k periods beyond the usual start. The standard P/A factor is computed for n periods, then discounted k additional periods.
Importance
Common board exam variant: 'first payment at end of year 3' — compute P/A for the remaining n years, then multiply by (1+i)^(-2) if k=2 deferral.
Term
Gradient Series
Definition
A series where payments increase (or decrease) by a constant amount G each period. Present worth P = A(P/A,i,n) + G(P/G,i,n) where P/G = [(1+i)^n − in − 1]/[i²(1+i)^n].
Importance
Appears in board exams as maintenance costs that increase by a fixed amount each year.
Section Title
2. Annuities and Perpetuities
Common Mistakes
- Applying F/A when the question asks for P/A, and vice versa — always identify what the unknown is.
- Using annual interest rate with annual n when the problem specifies monthly deposits at a nominal annual rate — convert to monthly i and monthly n.
- Forgetting to account for deferred periods — if the annuity starts at year 3, there is a 2-year deferral to discount.
- Confusing perpetuity P = A/i with annuity formula — perpetuity has no n term.
- For annuity-due, forgetting to multiply the result by (1+i) to account for the one-period advance.
Formulas
Example
Machine: C = ₱100,000, S = ₱10,000, n = 5 yr. d_SL = (100,000 − 10,000)/5 = ₱18,000/yr. BV_3 = 100,000 − 3(18,000) = ₱46,000.
Formula
d_SL = (C − S) / n
Variables
d_SL = annual straight-line depreciation (₱/yr); C = first cost or initial cost (₱); S = salvage or scrap value (₱); n = useful life (years)
Application
Compute uniform annual depreciation. Book value at year t: BV_t = C − (d_SL × t). Equal depreciation every year.
Example
C = ₱250,000, S = ₱25,000, n = 8. SYD = 8(9)/2 = 36. Year 1: d_1 = (250,000−25,000)(8/36) = 225,000(0.2222) = ₱50,000.
Formula
d_SYD,t = (C − S) × [n − t + 1] / SYD
Variables
d_SYD,t = SYD depreciation in year t; SYD = n(n+1)/2 = sum of years digits; t = year number (1 to n); C = cost; S = salvage
Application
Accelerated method — higher depreciation in early years. Compute SYD = n(n+1)/2 first, then apply the declining year-fraction.
Example
C = ₱100,000, S = ₱10,000, n = 5. k = 2/5 = 0.40. Year 1: d_1 = 0.40 × 100,000 = ₱40,000. BV_1 = ₱60,000. Year 2: d_2 = 0.40 × 60,000 = ₱24,000. BV_2 = ₱36,000.
Formula
d_DB,t = k × BV_(t-1)
Variables
d_DB,t = declining balance depreciation in year t; k = depreciation rate (for DDB, k = 2/n); BV_(t-1) = book value at beginning of year t
Application
Applied to the current (declining) book value each year — not to (C−S). For Double Declining Balance, k = 2/n. Stop when BV reaches S.
Example
Using the DDB example above: BV_2 = 100,000(1 − 0.40)^2 = 100,000(0.36) = ₱36,000. Confirms the year-by-year result.
Formula
BV_t = C(1 − k)^t
Variables
BV_t = book value at end of year t; C = initial cost; k = fixed declining balance rate; t = year
Application
Direct formula for book value in declining balance method without computing each year individually. Useful when n is large.
Exam Tips
- For SYD in year 1: d_1 = (C−S) × n/SYD — this gives the largest single-year depreciation of all three methods.
- Total depreciation over all years must equal (C−S) for SL and SYD — use this to verify your work.
- If a board problem asks for 'the depreciation rate using declining balance given salvage and life,' use: k = 1 − (S/C)^(1/n).
- BV at end of life under SL = S exactly. Verify this: BV_n = C − n × (C−S)/n = S. Always true.
- SYD year 1 depreciation > SL annual depreciation > SYD year n depreciation — know this ordering for comparison questions.
Key Points
- Depreciation allocates the cost of an asset over its useful life; it reduces book value annually.
- Three main methods tested: Straight-Line (SL), Sum-of-Years-Digits (SYD), and Declining Balance (DB).
- Salvage value S (scrap value) is the estimated residual value at end of useful life — always subtract it for SL and SYD.
- Book value BV_t = initial cost minus total accumulated depreciation at year t.
- SYD and DB are accelerated methods — higher depreciation in early years, lower in later years.
- Double Declining Balance (DDB) uses a rate of 2/n applied to the current book value; salvage may not be deducted initially but BV must not fall below S.
- MACRS (US system) is rarely tested in Philippine boards — focus on SL, SYD, and DB.
Definitions
Term
Book Value (BV_t)
Definition
The asset's value on the accounting records at the end of year t. BV_t = C − (accumulated depreciation through year t). Always ≥ salvage value.
Importance
Board exams frequently ask for BV at a specific year — know how to compute it for all three methods.
Term
Salvage Value (S)
Definition
The estimated market value of an asset at the end of its useful life. Also called residual value or scrap value.
Importance
Critical for SL and SYD (subtract from C). For DB/DDB, salvage sets the lower bound for BV but is not subtracted initially.
Term
Sum of Years Digits (SYD)
Definition
The denominator used in SYD depreciation: SYD = n(n+1)/2. For n=5: SYD = 5(6)/2 = 15. The numerator for year t is (n−t+1).
Importance
Memorize SYD = n(n+1)/2. Quickly verify: for n=8, SYD = 8×9/2 = 36; year 1 fraction = 8/36, year 2 = 7/36, etc.
Term
Accelerated Depreciation
Definition
Methods (SYD, DB/DDB) that allocate greater depreciation in early years and less in later years — opposite of straight-line.
Importance
Accelerated methods result in lower taxable income (and lower taxes) in early years — a financial advantage for asset owners.
Section Title
3. Depreciation Methods
Common Mistakes
- Using C instead of (C−S) as the depreciable amount in the straight-line formula.
- In SYD, using the wrong numerator: year t uses digit (n−t+1), not t. Year 1 gets the LARGEST fraction.
- In DDB, allowing book value to drop below salvage value — you must stop depreciating at BV = S.
- Confusing the declining balance rate k with the straight-line rate 1/n — DDB rate is 2/n, not 1/n.
- Adding salvage value back at the end of DDB calculations — salvage is a floor for BV, not added to it.
Formulas
Example
A road project: PW of benefits = ₱50M, PW of costs = ₱38M. B/C = 50/38 = 1.32 ≥ 1.0 → project is justified.
Formula
B/C = PW of Benefits / PW of Costs
Variables
B/C = benefit-cost ratio (dimensionless); PW of Benefits = present worth of all benefits received; PW of Costs = present worth of all costs incurred including initial investment
Application
Public project justification. If B/C ≥ 1.0, the project is economically justified. For multiple alternatives, use incremental B/C analysis.
Example
Bridge: P = ₱5,000,000; annual maintenance A = ₱100,000; i = 8%. CC = 5,000,000 + 100,000/0.08 = 5,000,000 + 1,250,000 = ₱6,250,000.
Formula
CC = P + A/i
Variables
CC = capitalized cost (₱); P = initial (first) cost (₱); A = uniform annual cost or recurring cost (₱/yr); i = annual interest rate (decimal)
Application
Economic justification of long-lived infrastructure (bridges, dams). The A/i term represents the capitalized equivalent of recurring annual costs.
Example
PW = ₱50,000 over 10 years at 10%: AW = 50,000 × [0.10(1.10)^10]/[(1.10)^10 − 1] = 50,000 × 0.16275 = ₱8,137/yr.
Formula
AW = PW × [i(1+i)^n] / [(1+i)^n − 1]
Variables
AW = equivalent uniform annual worth; PW = present worth of all cash flows; i = interest rate per period; n = number of periods
Application
Convert a PW value to its equivalent annual cost or benefit. Compare alternatives on an annual cost basis — useful when study periods differ.
Example
Alt A: ₱200,000 fixed + ₱50/unit. Alt B: ₱80,000 fixed + ₱110/unit. Break-even: 200,000 + 50x = 80,000 + 110x → 120,000 = 60x → x = 2,000 units.
Formula
Break-even: Cost_A = Cost_B
Variables
Set the total cost equations of two alternatives equal and solve for the unknown variable (units, years, or quantity).
Application
Find the production level or service life at which two alternatives are equally economical. Below break-even, choose lower fixed-cost alternative; above, choose lower variable-cost alternative.
Exam Tips
- For PW comparison with unequal lives: LCM of 5 and 8 = 40 years — very long! AW method avoids this: compare each alternative's AW over its own life.
- Capitalized cost problems often appear as: 'a bridge costs ₱X with ₱Y annual maintenance at i%' — apply CC = P + A/i directly.
- B/C ≥ 1.0 is the threshold for public projects; B/C < 1.0 means reject. Do not round B/C — 0.98 still means reject.
- For ROR problems, if the problem gives two i-values that bracket the unknown, use linear interpolation: i* = i1 + (PW1/(PW1−PW2)) × (i2−i1).
- In break-even problems, always define variables clearly: let x = unknown quantity, write cost equations for each alternative, set equal, and solve.
Key Points
- All alternatives must be compared over the same study period using a common economic basis.
- Three main comparison methods: Present Worth (PW), Annual Worth (AW), and Rate of Return (ROR).
- For mutually exclusive alternatives, select the one with the highest PW (or AW) if benefits are equal.
- Benefit-Cost Ratio (B/C): select the alternative if B/C ≥ 1.0; used mainly for public-sector projects.
- Break-even analysis finds the quantity or time where two alternatives have equal costs.
- Least Common Multiple (LCM) of lives is used when comparing alternatives with unequal service lives using PW.
- Capitalized cost = P + A/i for an asset with infinite (or very long) life — used for dams, bridges, roads.
- Incremental ROR analysis: compare mutually exclusive alternatives by analyzing the increment of investment.
Definitions
Term
Present Worth (PW) Method
Definition
Converts all cash flows of an alternative to their equivalent value at time zero using the appropriate discount rate. The alternative with the highest PW (least negative for cost-only) is selected.
Importance
Most versatile comparison method; foundation for all other methods. If lives differ, use LCM of lives as the study period.
Term
Annual Worth (AW) Method
Definition
Converts all cash flows to a uniform equivalent annual amount over the study period. Advantage: does not require LCM of lives — each alternative is analyzed over its own life.
Importance
Preferred method when service lives differ — eliminates the need for LCM calculation.
Term
Rate of Return (ROR)
Definition
The interest rate i* that makes the PW of all cash flows equal to zero (or PW of benefits = PW of costs). Also called Internal Rate of Return (IRR).
Importance
If ROR > MARR (Minimum Attractive Rate of Return), the investment is acceptable. Board exams test ROR by trial-and-error or interpolation.
Term
Capitalized Cost
Definition
The present worth of an alternative that is assumed to last forever (or a very long time). CC = P + A/i. Includes initial cost plus capitalized recurring costs.
Importance
Used for public infrastructure comparison. Lower CC = economically preferred alternative.
Term
MARR (Minimum Attractive Rate of Return)
Definition
The minimum ROR that management considers acceptable for an investment. Also called the hurdle rate or discount rate. Investments with ROR > MARR are pursued.
Importance
MARR is the benchmark for ROR analysis. It reflects the opportunity cost of capital for the organization.
Section Title
4. Economic Comparison of Alternatives
Common Mistakes
- Comparing alternatives with different service lives using PW without applying the LCM — always extend to the LCM of lives.
- Confusing PW (maximize) with cost analysis (minimize) — when alternatives have identical benefits, select the one with the LEAST negative PW (lowest cost).
- Using B/C > 1 for private-sector projects instead of public-sector ones — private projects typically use ROR > MARR.
- In break-even, solving for x but not checking which alternative is cheaper ABOVE and BELOW the break-even point.
- Forgetting to include salvage value as a positive cash flow at the end of the asset's life in PW calculations.
Connections
- Engineering Economy connects directly to Fluid Mechanics and Hydraulics in water infrastructure project evaluation — dam and pipeline alternatives are compared using capitalized cost and B/C ratio.
- Structural Engineering links to Engineering Economy through equipment and material cost optimization — selecting between steel and RC alternatives involves present-worth or annual-worth comparison over the structure's service life.
- The compound interest formula F = P(1+i)^n has the same mathematical structure as exponential growth in population studies (Demography) and radioactive decay in Physics — the exponent controls the rate of change.
- Annuity formulas underpin loan amortization in Construction Project Management — every construction loan repayment schedule is an application of the capital recovery (A/P) formula.
- Depreciation in Engineering Economy directly interfaces with Tax Law (NIRC of the Philippines) — the method chosen for tax reporting affects annual income tax liability of engineering firms.
- Break-even analysis in Engineering Economy parallels the concept in business management and connects to cost accounting, where fixed and variable costs determine minimum production for profitability.
- The effective interest rate concept connects to Finance and Banking — Philippine banks quote nominal rates (BSP-regulated), and borrowers must compute effective rates to compare actual borrowing costs across institutions.
- Rate of Return analysis connects to project management and feasibility studies required under RA 9184 (Government Procurement Reform Act) for government infrastructure projects in the Philippines.
Exam Strategy
Engineering Economy typically contributes 8–12 questions to the PRC Civil Engineer board exam Mathematics component. Prioritize compound interest (F = P(1+i)^n), annuity PW and FW factors, and straight-line depreciation — these three areas account for roughly 60% of Economy questions. Always begin by identifying: (1) what is given vs. what is asked, (2) the compounding period and matching it to the interest rate, and (3) whether payments are ordinary or annuity-due. Draw a cash-flow diagram for every problem, even in the exam room — it takes 20 seconds and prevents period-mismatch errors. For depreciation problems, write down d_SL first as a benchmark, then apply SYD or DDB if specified. In economic comparison questions, if lives are unequal, default to the Annual Worth method unless told otherwise. For multiple-choice problems, use the answer choices to back-check your calculation — if your answer is not among the options, check your period-rate consistency first (the most common source of error). Time-box Economy problems at 90 seconds each; if a problem requires long trial-and-error ROR interpolation, flag it and return after completing faster questions.
Quick Review Questions
₱50,000 will be needed 8 years from now. Find its present worth at 12% compounded annually.
Apply P = F(1+i)^(−n) with F = 50,000, i = 0.12, n = 8. Compute (1.12)^8 = 2.4760, then divide F by this factor. This is a straightforward discounting problem — the present worth is always less than the future amount.
What is the effective annual interest rate corresponding to 12% nominal compounded monthly?
With m = 12 compoundings per year: periodic rate = 12%/12 = 1% per month. Apply i_eff = (1 + r/m)^m − 1. Note that the effective rate (12.683%) always exceeds the nominal rate (12%) when m > 1.
Monthly deposits of ₱2,000 are made for 3 years at 6% compounded monthly. Find the future worth.
Convert annual rate to monthly: i = 0.5%/month, n = 36 months. Apply the F/A formula with these monthly values. Never use the annual rate of 6% with monthly periods — that is the most common error in this type of problem.
A machine costs ₱250,000 with a salvage value of ₱25,000 and a useful life of 8 years. Find the SYD depreciation in year 1.
Step 1: SYD = n(n+1)/2 = 36. Step 2: Year 1 uses the largest digit = n = 8 (numerator). Step 3: Depreciable amount = C − S = 225,000. Step 4: d_1 = 225,000 × (8/36) = ₱50,000. Compare with SL: d_SL = 225,000/8 = ₱28,125/yr — SYD year 1 is much higher, confirming it is an accelerated method.
A scholarship fund pays ₱80,000 per year forever. If i = 10%, how much must be invested today?
This is a perpetuity (infinite annuity). Apply P = A/i directly. No n is needed since n → ∞. The ₱800,000 invested at 10% generates ₱80,000 interest per year indefinitely, with the principal remaining intact.
Two alternatives: Alt A has PW = −₱180,000 and Alt B has PW = −₱220,000 over the same study period. Which is preferred?
For cost-only alternatives (no revenue difference), select the alternative with the LEAST negative PW — it represents the lower total cost in present-value terms. If benefits were included and both were positive, you would choose the higher PW.
A bridge costs ₱8,000,000 with annual maintenance of ₱150,000. At i = 8%, find the capitalized cost.
Capitalized cost assumes infinite life — typical for major infrastructure. The A/i term (₱1,875,000) is the lump sum needed today to fund all future maintenance costs in perpetuity at 8% interest. This is the present worth of all costs over infinite time.
A machine costing ₱500,000 has a DDB depreciation rate. Useful life n = 5 years. Find the book value at the end of year 2 (assume zero salvage for DDB computation).
DDB rate k = 2/n = 2/5 = 40%. Use BV_t = C(1−k)^t directly: BV_2 = 500,000 × (0.60)^2 = ₱180,000. Year-by-year verification: d_1 = 0.40(500,000) = ₱200,000 → BV_1 = ₱300,000; d_2 = 0.40(300,000) = ₱120,000 → BV_2 = ₱180,000. Both methods agree.
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