Engineering economics

Example: classifying a cash flow

A machine costs $10,000 to purchase, generates $3,000 per year in savings for 5 years, and has a salvage value of $1,000 at the end of year 5.

• The $10,000 purchase cost is a present value (P) - it occurs at $t=0$.
• The $3,000/year savings form an annuity (A) - a constant cash flow from $t=1$ through $t=5$.
• The $1,000 salvage value is a future value (F) - a one-time cash flow at $t=5$.

Year-end accounting convention

To keep calculations consistent, we assume cash flows occur at the end of each year. This is the year-end accounting convention.

The equations in this section rely on the time value of money: money today can grow over time if it earns interest. That growth occurs through compounding at a rate of return.

In engineering economic analysis, the rate of return is often the MARR (minimum attractive rate of return). In the formulas, MARR is represented by $i$. MARR is the lowest rate of return an investor is willing to accept, given the investment’s risk and the opportunity to earn returns elsewhere.

Two compounding models are used:

1. Discrete compounding: interest is added at regular time intervals.
2. Continuous compounding: the limit as the number of compounding periods becomes infinitely large ($n\to \infty$).

Note: For continuous compounding, gradient series formulas are not included because they are not meaningful in that case.

The equations below convert money values across time using these variables:

• $P$ - present worth (single lump sum).
• $A$ - an annuity (constant installment at the end of every period).
• $F$ - future worth (single lump sum).
• $G$ - gradient series starting at zero for the first period and increasing by constant increment G every period.
• $i$ - effective interest rate per compounding period (often the MARR). This is often stated as a percentage, but you must use a decimal value in the equations.
• $n$ - the number of compounding intervals.

Single payment compound - converts to F, given P

Symbol: $F=P(F/P,i,n)$

• Discrete compounding:

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

• Continuous compounding:

  $$F=P{e}^{in}$$

Single payment present worth - converts to P, given F

Symbol: $P=F(P/F,i,n)$

• Discrete compounding:

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

• Continuous compounding:

  $$P=F{e}^{-in}$$

Uniform series sinking fund - converts to A, given F

Symbol: $A=F(A/F,i,n)$

• Discrete compounding:

  $$A=F\cdot \frac{i}{(1+i{)}^n-1}$$

• Continuous compounding:

  $$A=F\cdot \frac{i}{e^{in}-1}$$

Capital recovery - converts to A, given P

Symbol: $A=P(A/P,i,n)$

• Discrete compounding:

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

• Continuous compounding:

  $$A=P\cdot \frac{i{e}^{in}}{e^{in}-1}$$

Uniform series compound - converts to F, given A

Symbol: $F=A(F/A,i,n)$

• Discrete compounding:

  $$F=A\cdot \frac{(1+i{)}^n-1}{i}$$

• Continuous compounding:

  $$F=A\cdot \frac{e^{in}-1}{i}$$

Uniform series present worth - converts to P, given A

Symbol: $P=A(P/A,i,n)$

• Discrete compounding:

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

• Continuous compounding:

  $$P=A\cdot \frac{e^{in}-1}{i{e}^{in}}$$

Uniform gradient present worth - converts to P, given G

Symbol: $P=G(P/G,i,n)$

• Discrete compounding:

  $$P=G\cdot \left(\frac{(1+i{)}^n-1}{i^2(1+i{)}^n}-\frac{n}{i(1+i{)}^n}\right)$$

Uniform gradient future worth - converts to F, given G

Symbol: $F=G(F/G,i,n)$

• Discrete compounding:

  $$F=G\cdot \frac{1}{i}\left(\frac{(1+i{)}^n-1}{i}-n\right)$$

Uniform gradient uniform series - converts to A, given G

Symbol: $A=G(A/G,i,n)$

• Discrete compounding:

  $$A=G\cdot \left(\frac{1}{i}-\frac{n}{(1+i{)}^n-1}\right)$$

Interest rate tables are based on these relationships. For detailed tables, refer to the FE Handbook.

Nonannual compounding

Interest can be compounded more frequently than once per year (for example, monthly or quarterly). The interest rate tables are built around a specific time period, so you need the interest rate per period to use them correctly.

• Effective annual rate of 6%, converted to a monthly rate: solve $(1+i_M{)}^{12}=1.06$ for $i_M=0.00487$ ($0.487\%$). Using this rate with $n=36$ months gives the same $(P/F)$ factor, $0.8396$, as using $6\%$ directly with $n=3$ years.

• Nominal annual rate of 6%, converted to a monthly rate: simply divide by 12, giving $i_M=0.5\%$ ($0.005$). This produces a different $(P/F,0.005,36)$ factor, $0.8356$, because a nominal rate doesn’t already account for compounding.

• Effective rate: already reflects a full year of compounding - converting it to a per-period rate preserves the same equivalent value.

• Nominal rate: does not reflect compounding - dividing it by the number of periods per year changes the equivalent value.

Formula

$$F=P{\left(1+\frac{i}{m}\right)}^{n\cdot m}$$

Where:

• $F$ = Future value
• $P$ = Present value
• $i$ = Annual interest rate
• $m$ = Compounding periods per year
• $n$ = Number of years

Example: nonannual compounding

You invest $1,000 at 8% compounded quarterly for 3 years. What is the future value?

$$F=1000{\left(1+\frac{0.08}{4}\right)}^{3\cdot 4}=1000(1.02{)}^{12}=1268.24$$

Answer: $1,268.24

Capitalized cost

Capitalized cost is used when a project is assumed to continue indefinitely (an infinite life). It expresses the project’s cost as a single present-worth amount.

Formula

$$CC=P+\frac{A}{i}$$

Where:

• $P$ = Initial cost
• $A$ = Annual cost
• $i$ = Interest rate

The FE Reference Handbook writes only the perpetuity term, as capitalized costs $=P=A/i$. There $P$ is the present worth of the perpetual annual series, not the first cost, so a project’s capitalized cost adds its first cost to that amount.

Example: capitalized cost

A project costs $50,000 initially and $2,000 per year to maintain, at an interest rate of 5%. What is the capitalized cost?

$$CC=50000+\frac{2000}{0.05}=50000+40000=90000$$

Answer: $90,000

Equivalent uniform annual cost (EUAC)

Converts all costs into an equivalent annual amount, including the effect of a salvage value recovered at the end of the asset’s life.

Formula

$$EUAC=P(A/P,i,n)+A-S(A/F,i,n)$$

Where:

• $P$ = Initial cost
• $A$ = Annual operating cost
• $S$ = Salvage value
• $(A/P,i,n)$ = Capital recovery factor
• $(A/F,i,n)$ = Sinking fund factor

Example: equivalent uniform annual cost

A machine costs $20,000, has annual operating costs of $2,000, and a salvage value of $4,000 after 5 years. At $i=8\%$, the capital recovery factor $(A/P,8\%,5)=0.2505$ and the sinking fund factor $(A/F,8\%,5)=0.1705$.

$$EUAC=20000(0.2505)+2000-4000(0.1705)=5010+2000-682=6328$$

Answer: $6,328/year

Depreciation

Depreciation models how an asset’s value is allocated (or reduced) over its useful life.

General formula (straight-line)

$$D=\frac{P-S}{n}$$

Where:

• $D$ = Annual depreciation
• $P$ = Initial cost
• $S$ = Salvage value
• $n$ = Useful life

Modified accelerated cost recovery system (MACRS)

Used in U.S. tax code for accelerated depreciation. It assumes a half-year convention: for example, an asset classified as 5-year property is depreciated at 20% in year 1. Because the first and last years each count as half a year, 5-year property is recovered over six tax years. The FE Reference Handbook prints the full MACRS recovery-rate table, so the rates are looked up rather than memorized.

Tax issues

Key concepts

• Depreciation is tax-deductible.

• Taxable income:

  $$TI=Revenue-Expenses-Depreciation$$

• Taxes:

  $$\text{Tax}=TI\times \text{Tax rate}$$

Bonds

A bond’s value today is the present worth of its coupon payments plus the present worth of its face value paid at maturity.

Bond valuation

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

Where:

• $P$ = Bond price
• $C$ = Coupon payment
• $F$ = Face value
• $i$ = Market interest rate
• $n$ = Maturity

Example: bond valuation

A bond has a $1,000 face value, pays a $50 annual coupon, matures in 2 years, and the market interest rate is 6%. What is the bond’s price today?

$$P=\frac{50}{1.06}+\frac{50}{1.06^2}+\frac{1000}{1.06^2}=47.17+44.50+890.00=981.67$$

Answer: $981.67

Carry at least four decimal places through each intermediate term and round only the final answer - rounding the coupon or face-value terms too early can shift the total enough to match a wrong answer choice.

Break-even analysis

Break-even analysis finds the output level where total revenue equals total cost.

Formula

$$Q=\frac{F}{P-V}$$

Where:

• $Q$ = Break-even quantity
• $F$ = Fixed cost
• $P$ = Price per unit
• $V$ = Variable cost per unit

Example: break-even quantity

A product sells for $25 per unit and costs $15 per unit to produce, with $8,000 in fixed costs. What is the break-even quantity?

$$Q=\frac{8000}{25-15}=800$$

Answer: 800 units

Benefit-cost (B/C) analysis

Benefit-cost analysis compares the present worth of benefits to the present worth of costs.

Formula

$$B/C=\frac{\text{Present worth of benefits}}{\text{Present worth of costs}}$$

• If $B/C\ge 1$ (equivalently, $B-C\ge 0$), the project is acceptable.

Example: benefit-cost ratio

A public project has a present worth of benefits of $450,000 and a present worth of costs of $300,000. What is the B/C ratio?

$$B/C=\frac{450000}{300000}=1.5$$

Answer: $B/C=1.5$, which is greater than 1, so the project is acceptable.