Engineering Economics: Time Value of Money

A rupee today is worth more than a rupee next year, because today's rupee can be invested to earn interest. This simple fact, the time value of money, sits at the heart of every engineering investment decision, from choosing between two machines to deciding whether a solar plant pays for itself. In this article we derive the core formulas and apply them to realistic problems, with every step shown.

Why Engineers Need Economics

Engineers rarely choose designs on technical merit alone. A pump with higher efficiency may cost more up front but save electricity for twenty years. To compare such options fairly, all cash flows must be moved to the same point in time. The tools below let you do exactly that.

Key notation

• \( P \): present worth (value now)
• \( F \): future worth (value at the end of \( n \) periods)
• \( A \): uniform amount per period (an annuity)
• \( i \): interest rate per period
• \( n \): number of periods

The Core Formulas

Single payment compound amount

Money invested at compound interest grows by a factor \( (1 + i) \) each period:

$$ F = P(1 + i)^n $$

The reverse, discounting a future sum to today, is

$$ P = \frac{F}{(1 + i)^n} $$

Uniform series (annuities)

For equal payments \( A \) at the end of each of \( n \) periods, the present worth is

$$ P = A\,\frac{(1 + i)^n - 1}{i(1 + i)^n} $$

and turning it around gives the capital recovery formula, used for loan instalments:

$$ A = P\,\frac{i(1 + i)^n}{(1 + i)^n - 1} $$

Nominal and effective rates

Banks quote nominal annual rates \( r \) compounded \( m \) times a year. The effective annual rate is

$$ i_{\text{eff}} = \left(1 + \frac{r}{m}\right)^m - 1 $$

Solved Problem 1: Growing an Investment

An engineer invests Rs 500,000 in a certificate paying 12% per year, compounded annually. What will it be worth after 5 years?

Step 1: Identify values. \( P = 500\,000 \), \( i = 0.12 \), \( n = 5 \).

Step 2: Compound factor.

$$ (1.12)^5 = 1.12 \times 1.12 \times 1.12 \times 1.12 \times 1.12 = 1.7623 $$

Step 3: Future worth.

$$ F = 500\,000 \times 1.7623 = \text{Rs } 881{,}171 $$

Sanity check. Simple interest would give \( 500\,000 \times (1 + 5 \times 0.12) = \text{Rs } 800{,}000 \). Compound interest must exceed this because interest earns interest, and it does by about Rs 81,000. The "rule of 72" says money doubles in roughly \( 72/12 = 6 \) years, so after 5 years we should be somewhat below double, consistent with 1.76 times.

Solved Problem 2: Monthly Instalment on Equipment Finance

A workshop finances a CNC machine with a loan of Rs 2,000,000 at 14% per year, compounded monthly, repaid over 5 years in equal monthly instalments. Find the instalment and the total interest paid.

Step 1: Per-period rate and number of periods.

$$ i = \frac{0.14}{12} = 0.011667\ \text{per month}, \qquad n = 5 \times 12 = 60 $$

Step 2: Growth factor.

$$ (1 + i)^{60} = (1.011667)^{60} = 2.0056 $$

Step 3: Capital recovery formula.

$$ A = 2\,000\,000 \times \frac{0.011667 \times 2.0056}{2.0056 - 1} = 2\,000\,000 \times \frac{0.023399}{1.0056} = \text{Rs } 46{,}537 $$

(More precise calculation gives Rs 46,536.50 per month.)

Step 4: Total interest.

$$ \text{Total repaid} = 60 \times 46{,}536.50 = \text{Rs } 2{,}792{,}190 $$

$$ \text{Interest} = 2{,}792{,}190 - 2{,}000{,}000 = \text{Rs } 792{,}190 $$

Sanity check. Paying the principal alone would cost \( 2\,000\,000/60 = 33{,}333 \) per month; the extra Rs 13,200 covers interest, which is reasonable since the average balance outstanding is roughly half the loan. The effective annual rate is \( (1.011667)^{12} - 1 = 14.93\% \), slightly above the quoted 14%, as monthly compounding always produces.

Solved Problem 3: Is an Energy-Saving Project Worthwhile?

A factory can install a heat recovery system for Rs 1,500,000. It saves Rs 400,000 per year in fuel for 6 years and has a salvage value of Rs 200,000 at the end. The company's minimum attractive rate of return (MARR) is 10%. Should it invest?

Step 1: Present worth of the savings.

$$ \frac{(1.1)^6 - 1}{0.1(1.1)^6} = \frac{1.771561 - 1}{0.1771561} = 4.3553 $$

$$ P_{\text{savings}} = 400\,000 \times 4.3553 = \text{Rs } 1{,}742{,}104 $$

Step 2: Present worth of the salvage.

$$ P_{\text{salvage}} = \frac{200\,000}{(1.1)^6} = \frac{200\,000}{1.771561} = \text{Rs } 112{,}895 $$

Step 3: Net present worth.

$$ NPW = -1{,}500{,}000 + 1{,}742{,}104 + 112{,}895 = \text{Rs } 354{,}999 $$

Decision. NPW is positive, so the project earns more than 10% and should be accepted.

Sanity check. The undiscounted total of savings plus salvage is Rs 2,600,000, far above the cost, but discounting rightly shrinks it. Solving for the rate that makes NPW zero gives an internal rate of return of about 17.3%, comfortably above the 10% MARR, which agrees with the positive NPW.

Comparing Alternatives

When choosing between options, engineers use one of these equivalent methods:

1. Present worth analysis: convert all cash flows to today; choose the highest NPW (or lowest cost).
2. Annual worth analysis: convert everything into an equivalent uniform yearly amount; ideal when lives differ.
3. Rate of return analysis: compare the internal rate of return with the MARR.
4. Payback period: quick but ignores the time value of money and cash flows after payback, so use it only as a screening tool.

Did You Know?

• The "rule of 72" was described by the Italian mathematician Luca Pacioli in 1494.
• At 12% interest, Rs 1 left untouched for 40 years grows to more than Rs 93.
• Many engineering licensing exams worldwide, including the US Fundamentals of Engineering exam, include a dedicated engineering economics section.

Key Takeaways

• Compound growth follows \( F = P(1 + i)^n \); discounting reverses it.
• Use the capital recovery formula for loan instalments and the annuity factor for uniform savings.
• Always match the interest rate to the compounding period.
• A positive net present worth at the MARR means an investment is worthwhile.

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