GELE Mathematics — Engineering EconomySummary

Money has different value at different times. A peso today is worth more than a peso in the future because it can earn interest. The fundamental equation is F = P(1 + i)^n for compound interest, where F is future worth, P is present worth, i is the periodic interest rate, and n is the number of periods. This principle underlies all engineering economy decisions.

Simple interest: F = P(1 + in) — interest is calculated only on principal each period. Compound interest: F = P(1 + i)^n — interest earns interest. For example, ₱10,000 at 8% simple interest for 5 years yields F = 10,000(1 + 0.08 × 5) = ₱14,000. With compound interest: F = 10,000(1.08)^5 = ₱14,693. The difference (₱693) is earned on previously accumulated interest.

Nominal rate r is the stated annual rate; effective rate i_eff accounts for compounding frequency. With m compoundings per year: i_eff = (1 + r/m)^m − 1. Example: 12% nominal compounded monthly gives i_eff = (1 + 0.12/12)^12 − 1 = (1.01)^12 − 1 = 0.1268 or 12.68%. This difference is critical when comparing loans or investments with different compounding frequencies.

An annuity is a series of equal payments A at regular intervals. Ordinary annuity has payments at end of period; annuity-due has payments at beginning. Future worth: F = A[(1+i)^n − 1]/i. Present worth: P = A[(1+i)^n − 1]/[i(1+i)^n]. Example: Deposits of ₱1,000 annually for 5 years at 10% give F = 1,000[(1.10)^5 − 1]/0.10 = 1,000(6.105) = ₱6,105 at the end of year 5.

A perpetuity is an annuity with infinite duration (n → ∞). The present worth simplifies to P = A/i. Example: An endowment paying ₱50,000 annually forever at 8% interest requires principal P = 50,000/0.08 = ₱625,000. This represents the sustainable yield from capital investments in public infrastructure.

Annual depreciation d = (C − S)/n, where C is original cost, S is salvage value, and n is useful life. Book value after t years: BV_t = C − dt. Example: Equipment costing ₱100,000 with ₱10,000 salvage over 5 years depreciates ₱18,000/year. After 3 years, BV = 100,000 − 3(18,000) = ₱46,000. The book value declines linearly; this method is simplest and most commonly used in Philippines per accounting standards.

Accelerated depreciation method. The denominator is the sum of useful life years: SYD = n(n+1)/2. For a 5-year asset, SYD = 15. Year 1 depreciation: d_1 = 5/15 × (C − S); Year 2: d_2 = 4/15 × (C − S), etc. Example: ₱100,000 cost, ₱10,000 salvage, 5-year life. Year 1: d = (5/15)(90,000) = ₱30,000. This front-loads depreciation for tax advantages.

Accelerated method using a fixed rate applied to remaining book value each year: BV_t = C(1 − d)^t, where d is the decline rate (often d = 2/n for double-declining-balance, DDB). Example: ₱100,000 cost, 5-year life, DDB uses d = 2/5 = 0.40. Year 1: depreciation = 0.40(100,000) = ₱40,000, BV = ₱60,000. Year 2: depreciation = 0.40(60,000) = ₱24,000, BV = ₱36,000. The depreciation decreases each year as the base shrinks.

Convert all costs and benefits to a common point in time (usually today) using the discount rate. For alternative A: PW_A = −initial cost + PV(all future benefits) − PV(all future costs). Choose the alternative with the least negative PW (or greatest positive PW). Example: Two equipment options: Option 1 costs ₱50,000 now with ₱5,000 annual savings for 10 years at 10% interest. PW = −50,000 + 5,000 × [(1.10)^10 − 1]/[0.10(1.10)^10] = −50,000 + 5,000(6.145) = −₱19,275. If Option 2 yields a less negative PW, it is economically superior.

Convert all costs and benefits to an equivalent uniform annual payment over the project life. AW = PW × [i(1+i)^n]/[(1+i)^n − 1], the inverse of the annuity present-worth factor. Example: A project with PW = −₱100,000 over 8 years at 12% has AW = −100,000 × [0.12(1.12)^8]/[(1.12)^8 − 1] = −100,000 × 0.2013 = −₱20,130 annual equivalent cost. Projects with different lives can be compared directly using AW.

Find the discount rate i* where present worth equals zero: PW = 0. This is the internal rate of return (IRR). For simple cases, solve algebraically; for complex cash flows, use iteration or financial calculator. Example: Investment of ₱100,000 returning ₱30,000/year for 5 years. Set PW = −100,000 + 30,000 × [(1+i*)^5 − 1]/[i*(1+i*)^5] = 0 and solve for i* ≈ 15.24%. If the required return exceeds 15.24%, reject the project.

Calculate the ratio of present worth of benefits to present worth of costs: B/C = PW(benefits)/PW(costs). Accept the project if B/C ≥ 1. Example: Infrastructure project with benefits ₱500,000 and costs ₱400,000 has B/C = 1.25, indicating ₱1.25 of benefit per peso of cost. Used extensively in public works evaluation per RA 544 (Professional Regulation of Engineers). For comparing alternatives, choose the one with highest B/C among those with B/C ≥ 1.

Find the point (cost, quantity, time, or interest rate) where two alternatives have equal economic performance. Example: Machine A costs ₱100,000 with ₱5,000/year maintenance; Machine B costs ₱80,000 with ₱7,000/year maintenance. Annual cost of A: AC_A = 100,000(A/P, i, n) + 5,000. Annual cost of B: AC_B = 80,000(A/P, i, n) + 7,000. Set AC_A = AC_B to find break-even. Common in manufacturing decisions (make vs. buy, lease vs. own).

Annuity formulas are derived from repeated application of the compound interest formula F = P(1+i)^n. The annuity future-worth factor is the sum of geometric series of compounded payments. Understanding this relationship helps derive formulas when memory fails during the exam.

When compounding frequency does not match payment frequency (e.g., monthly payments with quarterly compounding), convert the effective rate for the payment period before applying annuity formulas. Ignoring this causes systematic errors in multi-period problems.

Accelerated depreciation (SYD, DB) reduces taxable income more in early years, creating larger tax shields that increase project NPV. The choice of depreciation method directly affects the discount rate used in economic comparisons when tax impacts are material.

These three methods are mathematically equivalent for project selection. A project that has positive PW will have positive AW and an IRR exceeding the discount rate. The choice of method is driven by communication and problem context, not mathematical rigor. PW is most general; AW best for recurring decisions; IRR best for intuitive 'return' metrics.

B/C = PW(benefits)/PW(costs). A B/C ratio of 1.5 means benefits are 50% more than costs, equivalent to PW = 0.5 × PW(costs). Both lead to identical project ranking, but B/C is preferred in public-sector reporting per RA 544.

Break-even identifies the critical value of a parameter (cost, volume, interest rate) where decisions reverse. Sensitivity analysis examines how changes in multiple parameters affect project viability. Break-even is a special case of sensitivity analysis for a single parameter.

Perpetuities model the perpetual value of public assets (parks, monuments, permanent infrastructure). The formula P = A/i shows that sustainable annual benefit A requires capital investment of A/i. This underpins long-term infrastructure funding decisions and environmental impact valuation.