1.3 Engineering Economic Analysis

1.3 Engineering Economic Analysis

Transportation projects involve substantial capital investments with long-term benefits and costs. Engineering Economic Analysis provides the mathematical framework to evaluate project viability, compare alternative designs, and determine the optimal allocation of public funds.

Time Value of Money (TVM)

The fundamental premise of engineering economics is that a dollar today is worth more than a dollar in the future because of its earning capacity (interest). In economic calculations, cash flows are mapped to specific points in time using the following variables:

• $P$: Present worth or present value (at time $t = 0$).
• $F$: Future worth or future value (at time $t = n$).
• $A$: Uniform end-of-period payment or receipt in a series of $n$ periods.
• $G$: Uniform arithmetic gradient increment per period.
• $i$: Interest rate per period.
• $n$: Number of interest periods.

Standard Interest Factors and Formulations

The NCEES PE Civil Reference Handbook lists standard interest factors using the functional notation $(X/Y, i, n)$, representing the factor by which you multiply $Y$ to find $X$.

Economic Analysis Metrics

Transportation alternatives are evaluated and compared using four main metrics:

1. Present Worth (PW)

Present Worth represents the net value of all cash flows discounted to the present time. The formula is: $PW = \sum_{t=0}^{n} CF_t \times (P/F, i, t)$ A positive Net Present Worth ($PW > 0$) indicates that the project is economically viable. When comparing mutually exclusive alternatives, the alternative with the highest PW (or lowest PW of cost) is selected.

2. Capitalized Cost (CC)

Capitalized Cost is the present worth of a project or asset that has an infinite life (e.g., land acquisition, tunnels, bridges, flood control structures). It represents the amount of money required today to purchase the asset and fund its maintenance and replacement in perpetuity. $CC = P + \frac{A}{i} + \frac{RC}{(1 + i)^k - 1}$ Where:

• $P$: Initial construction cost.
• $A$: Annual maintenance cost (capitalized as a perpetuity: $PW = A/i$).
• $RC$: Periodic replacement cost incurred every $k$ years.

3. Equivalent Uniform Annual Cost (EUAC)

For projects with different design lives, comparing Present Worth is invalid unless a common study period is used. Instead, engineers use EUAC to convert all capital costs, annual maintenance, salvage values, and replacement costs into an equivalent annual series over the design life ($n$): $EUAC = P \times (A/P, i, n) + A - S \times (A/F, i, n)$ Where:

• $S$: Salvage value at the end of the design life (year $n$).
• The salvage value is subtracted because it represents a cash inflow (or cost reduction) at the end of the project.

4. Benefit-Cost Ratio (B/C)

Public projects are often evaluated using the Benefit-Cost Ratio, which compares the benefits to the public against the costs incurred by the agency. The cash flows are converted to either Present Worth ($PW$) or annual worth ($AW$).

• Conventional B/C Ratio: $B/C = \frac{PV(\text{Benefits})}{PV(\text{Initial Capital}) + PV(\text{O\&M}) - PV(\text{Salvage})}$
• Modified B/C Ratio: $B/C = \frac{PV(\text{Benefits}) - PV(\text{O\&M})}{PV(\text{Initial Capital}) - PV(\text{Salvage})}$ Where $O&M$ represents Operating and Maintenance costs.
• Decision Rule: A project is economically justified if $B/C \ge 1.0$.

Nominal vs. Effective Interest Rates

When compounding occurs more frequently than once a year (e.g., monthly, semi-annually), the effective interest rate is higher than the nominal rate. Schedulers and estimators must convert the nominal annual rate ($r$) to the effective annual interest rate ($i_e$): $i_e = \left(1 + \frac{r}{m}\right)^m - 1$ Where $m$ is the number of compounding periods per year.

Benefit-Cost Ratio Example Walkthrough

Consider a proposed highway bypass project with the following estimates:

• Initial Construction Cost: $12,000,000
• Annual Operation & Maintenance: $80,000/year
• Road User Travel Time Savings (Benefits): $1,500,000/year
• Accident Reduction Benefits: $200,000/year
• Project Design Life: 20 years
• Salvage Value: $1,500,000
• Interest Rate: 5% per year

First, we calculate the interest factors:

• $(P/A, 5%, 20) = \frac{(1.05)^{20} - 1}{0.05(1.05)^{20}} \approx 12.4622$
• $(P/F, 5%, 20) = (1.05)^{-20} \approx 0.3769$

Second, we calculate the present values of benefits and costs:

• PV of Benefits = $($1,500,000 + $200,000) \times (P/A, 5%, 20) = $1,700,000 \times 12.4622 = $21,185,740$
• PV of Initial Cost = $$12,000,000$
• PV of O&M = $$80,000 \times (P/A, 5%, 20) = $80,000 \times 12.4622 = $996,976$
• PV of Salvage = $$1,500,000 \times (P/F, 5%, 20) = $1,500,000 \times 0.3769 = $565,350$

Third, we compute the Conventional B/C Ratio: $B/C_{\text{conventional}} = \frac{\$21,185,740}{\$12,000,000 + \$996,976 - \$565,350} = \frac{\$21,185,740}{\$12,431,626} \approx 1.70$

We compute the Modified B/C Ratio: $B/C_{\text{modified}} = \frac{\$21,185,740 - \$996,976}{\$12,000,000 - \$565,350} = \frac{\$20,188,764}{\$11,434,650} \approx 1.77$

Since both B/C ratios are greater than 1.0, the project is highly economically justified. The modified ratio is slightly higher because it subtracts O&M costs from public benefits in the numerator rather than adding them to capital costs in the denominator.