Engineering Economy: Interest, Annuities, and Depreciation

Engineering economy is the part of the MSTE paper where every problem is really the same problem: money at one point in time is being compared with money at another, and the interest rate is the exchange rate between the two. Learn a short list of factors, and get in the habit of drawing a cash-flow diagram before you compute, and these become some of the fastest and most reliable points on the board. This lesson carries each factor to an exact peso figure so you can check your own arithmetic against a worked number and catch the specific slips the examiners bait you with.

Cash-flow diagrams and the time value of money

A peso today is worth more than a peso next year, because today's peso can earn interest in the meantime. The cash-flow diagram is the picture that keeps this straight. Draw a horizontal line for time, with the present at t = 0 and each period marked off along it. By convention an upward arrow is a receipt (cash coming in) and a downward arrow is a disbursement (cash going out). The interest rate i per period is written beside the diagram, not drawn as an arrow. The single most common source of wrong answers is a mismatch between the period of the interest rate and the period of the cash flows, so always make n count the same units as i.

Simple and compound interest

Simple interest is charged on the original principal only. With principal P, rate per period i, and n periods, the interest is I = P i n and the total future amount is F = P(1 + i n). Compound interest is charged on the running balance, so interest itself earns interest: F = P(1 + i)^n. Over a single period the two agree; beyond that, compounding always pulls ahead, which is why lenders quote it.

Worked example (simple): Borrow PHP 40,000 at 9% simple interest for 5 years. The interest is I = 40,000(0.09)(5) = PHP 18,000, so the amount repaid is F = 40,000(1 + 0.45) = PHP 58,000. Worked example (compound): Invest PHP 50,000 at 8% compounded annually for 3 years. Then F = 50,000(1.08)^3 = 50,000(1.259712) = PHP 62,985.60, against the simple-interest figure of 50,000(1.24) = PHP 62,000. The gap is the interest earned on interest.

Nominal and effective interest rates

A nominal rate r is a yearly headline that is actually compounded m times a year, so the rate that truly acts each period is r/m. The effective annual rate is the single yearly rate that produces the same growth once the intra-year compounding is accounted for:

i_eff = (1 + r/m)^m - 1

As m grows the effective rate rises toward the continuous-compounding limit e^r - 1. The lesson for the board is simple: a nominal rate compounded more than once a year is always worth more than its face value, and the choices will offer you both.

Worked example: A loan is quoted at 12% compounded monthly. The monthly rate is r/m = 0.12/12 = 0.01, so i_eff = (1.01)^12 - 1 = 1.126825 - 1 = 0.126825, or 12.68%. The following table shows the same nominal 12% under different compounding frequencies.

Compounding	Periods per year m	Periodic rate r/m	Effective annual rate
Annually	1	12.00%	12.00%
Semiannually	2	6.00%	12.36%
Quarterly	4	3.00%	12.55%
Monthly	12	1.00%	12.68%
Daily	365	0.0329%	12.75%
Continuously	infinite	approaches 0	12.75%

The effective rate never falls below the nominal rate, and it climbs only slowly once you pass monthly compounding.

Present worth and future worth of a single payment

Two factors move a single lump sum along the time line. To carry a present amount forward, F = P(1 + i)^n; to bring a future amount back, P = F(1 + i)^(-n). The quantity (1 + i)^(-n) is the single-payment present-worth factor, and it is always less than 1, which is exactly what "discounting" means.

Worked example: How much must you deposit today at 10% compounded annually to have PHP 100,000 in 5 years? P = 100,000/(1.10)^5 = 100,000/1.61051 = PHP 62,092.13. Discounting only one year by mistake gives 100,000/1.10 = PHP 90,909.09, and compounding forward instead of back gives PHP 161,051, both of which will appear in the choices.

Annuities: ordinary, due, deferred, and perpetual

An annuity is a series of equal payments A spread over equal periods. The ordinary annuity pays at the end of each period, and its two factors are

P = A[1 - (1 + i)^(-n)] / i (present worth) F = A[(1 + i)^n - 1] / i (future worth)

Worked example (ordinary): Year-end payments of PHP 10,000 for 6 years at 8% have present worth P = 10,000[1 - (1.08)^(-6)]/0.08 = 10,000(4.622873) = PHP 46,228.73 and future worth F = 10,000[(1.08)^6 - 1]/0.08 = 10,000(7.335930) = PHP 73,359.30.

An annuity due pays at the beginning of each period, so every payment sits one period earlier and earns one extra period of interest. You get its worth by multiplying the ordinary-annuity result by (1 + i). Worked example (due): the same payments paid at the start of each year have present worth 46,228.73(1.08) = PHP 49,927.03.

A deferred annuity does not begin until after a waiting period. Value the series with the ordinary-annuity formula as of one period before its first payment, then discount that single equivalent amount back to time zero. Worked example (deferred): five year-end payments of PHP 10,000 at 10% whose first payment falls at the end of year 4 (a three-year deferment). At the end of year 3 the series is worth 10,000[1 - (1.10)^(-5)]/0.10 = 10,000(3.790787) = PHP 37,907.87; discounting three years, P = 37,907.87/(1.10)^3 = 37,907.87/1.331 = PHP 28,480.75.

A perpetuity is an annuity that never ends, and its present worth collapses to a single clean ratio, P = A/i. Worked example (perpetuity): a fund that must pay PHP 60,000 every year forever at 6% requires P = 60,000/0.06 = PHP 1,000,000 today. Multiplying by i instead of dividing, which gives PHP 3,600, is the trap.

Gradients: arithmetic and geometric

Real cash flows often grow rather than staying level. An arithmetic gradient grows by a constant peso amount G each period (for example 2,000, then 2,500, then 3,000). Split such a series into a uniform base A plus the gradient part, and add their present worths. The gradient present-worth factor is

(P/G, i, n) = (1/i) [ ((1 + i)^n - 1) / (i(1 + i)^n) - n/(1 + i)^n ].

Worked example (arithmetic): costs of PHP 2,000, 2,500, 3,000, and 3,500 over 4 years at 10% have base A = 2,000 and G = 500. The uniform part is 2,000(P/A, 10%, 4) = 2,000(3.169865) = PHP 6,339.73, and the gradient part is 500(P/G, 10%, 4) = 500(4.378107) = PHP 2,189.05, for a total present worth of PHP 8,528.78.

A geometric gradient grows by a constant percentage g each period. Starting from a first payment A1 at the end of period 1, and provided g is not equal to i,

P = A1 [1 - ((1 + g)/(1 + i))^n] / (i - g).

(When g equals i the formula collapses to P = A1 n / (1 + i).) Worked example (geometric): first-year revenue of PHP 100,000 growing 5% per year for 4 years, discounted at 10%, has present worth P = 100,000[1 - (1.05/1.10)^4]/(0.10 - 0.05) = 100,000[1 - 0.830207]/0.05 = 100,000(3.395860) = PHP 339,586. Treating the revenue as a level annuity of 100,000, which ignores the growth, gives only PHP 316,987.

Depreciation: straight line, SYD, and declining balance

Depreciation spreads the loss in value of an asset, from its first cost FC down toward its salvage value SV over a useful life of n years, across the accounting periods. Three methods appear on the board.

Method	Annual charge	Book value after k years	Pattern
Straight line (SL)	(FC - SV)/n	FC - k(FC - SV)/n	level every year
Sum of years digits (SYD)	[(n - k + 1)/SYD](FC - SV), SYD = n(n + 1)/2	FC minus accumulated charges	front-loaded
Declining balance (DB)	rate x book value at start of year	FC(1 - rate)^k	front-loaded, ignores salvage in the rate

Worked example (SL): a machine costs PHP 800,000 with salvage PHP 80,000 and an 8-year life. The depreciable base is 720,000, so the level charge is 720,000/8 = PHP 90,000 per year, and the book value after 3 years is 800,000 - 3(90,000) = PHP 530,000. Forgetting to subtract salvage and writing 800,000/8 = PHP 100,000 is the standard error.

Worked example (SYD): for the same machine, SYD = 8(9)/2 = 36. The first-year charge is (8/36)(720,000) = PHP 160,000 and the second year is (7/36)(720,000) = PHP 140,000, so the book value after 2 years is 800,000 - 160,000 - 140,000 = PHP 500,000. SYD front-loads: its first-year charge dwarfs the straight-line PHP 90,000.

Worked example (declining balance): equipment costing PHP 500,000 over a 5-year life under double declining balance uses rate = 2/n = 0.40. The book value after 2 years is FC(1 - rate)^2 = 500,000(0.60)^2 = 500,000(0.36) = PHP 180,000. Note the first-year charge alone is 500,000(0.40) = PHP 200,000, so book value after year 1 is PHP 300,000.

Rate of return and benefit-cost ratio

The rate of return (ROR) measures the yearly earning power of invested capital. The quick form is ROR = annual net profit / capital invested; the exact form is the internal rate of return, the interest rate at which the present worth of all benefits equals the present worth of all costs (net present worth = 0). Worked example: an investment of PHP 500,000 that yields an annual net profit of PHP 90,000 has ROR = 90,000/500,000 = 0.18, or 18%.

For public projects the test is the benefit-cost ratio, B/C = present worth of benefits / present worth of costs, and a project is economically justified when B/C is at least 1. Worked example: a project with PHP 12,000,000 of benefits and PHP 10,000,000 of costs, both on a present-worth basis, has B/C = 12/10 = 1.20, which is above 1, so the project is justified.

Capitalized cost

Capitalized cost is the present sum that would fund an asset and its upkeep forever. For a first cost FC with annual maintenance M in perpetuity, it is the first cost plus the present worth of the maintenance perpetuity:

CC = FC + M/i (add RC/[(1 + i)^k - 1] if the asset is also replaced every k years at cost RC).

Worked example: a structure that costs PHP 5,000,000 to build and needs PHP 100,000 per year of maintenance forever, evaluated at 8%, has CC = 5,000,000 + 100,000/0.08 = 5,000,000 + 1,250,000 = PHP 6,250,000. Using 10% by mistake would drop the maintenance term to 100,000/0.10 = 1,000,000 and understate the answer by PHP 250,000.

Break-even analysis

The break-even point is the output at which total revenue exactly equals total cost, so profit is zero. With a fixed cost FC, a selling price p per unit, and a variable cost v per unit, the break-even quantity is

units = FC / (p - v).

Below it the operation loses money, above it the operation profits. Worked example: a plant with fixed cost PHP 300,000 per month, selling price PHP 250 per unit, and variable cost PHP 100 per unit breaks even at 300,000/(250 - 100) = 300,000/150 = 2,000 units per month.

Exam-day strategy

• Draw the cash-flow diagram first and make the interest period match the payment period; if a rate is nominal and compounding is not annual, convert to the effective rate before you touch an annuity factor.
• Read whether interest is "simple" or "compounded": both answers are placed in the choices, and past a single period the compound figure is always the larger.
• Match the annuity to its timing. End-of-period is ordinary; beginning-of-period is an annuity due, which is the ordinary result times (1 + i); a waiting period means a deferred annuity, valued one period before the first payment and then discounted; and forever means a perpetuity, P = A/i.
• On depreciation, straight line is level, while SYD and declining balance both front-load. If an item asks for the first-year charge and the choices differ, the front-loaded methods give the larger number, and remember that declining balance ignores salvage inside its rate.
• For public-project items compute B/C and accept when it is at least 1; for a capitalized cost, add the first cost to the maintenance perpetuity M/i, and never forget the first cost itself.
• For break-even, divide the fixed cost by the contribution margin p - v, not by the price alone; dividing by the price is the most common miss.