4.3 Engineering Economics & Financial Analysis
Key Takeaways
- Time value of money converts single future sums F = P(1+i)^n using the single payment compound amount factor (F/P, i, n).
- Uniform series annual cash flows relate to present worth via P = A \left[ \frac{(1+i)^n - 1}{i(1+i)^n} \right] using (P/A, i, n).
- Depreciation reduces asset book value over recovery life: Straight-Line gives constant d = \frac{C - S}{n}, whereas SYD accelerates early tax deductions.
- Net Present Value (NPV) evaluation criteria accepts projects when NPV \ge 0 at the Minimum Attractive Rate of Return (MARR).
- Capitalized cost CC = C_0 + \frac{A}{i} + \frac{RC}{(1+i)^k - 1} represents the present worth of an asset maintained indefinitely through periodic replacement.
4.3 Engineering Economics & Financial Analysis
1. Principles of Engineering Economics & Interest Formulas
Engineering Economics combines engineering decision-making with financial evaluation to select the most cost-effective alternatives over a project's lifecycle.
Simple vs. Compound Interest
- Simple Interest: Interest is charged only on the principal amount $P$:
- Compound Interest: Interest earned in each period is added to the principal to earn interest in subsequent periods: where $P$ is Present Worth, $F$ is Future Worth, $i$ is interest rate per period, and $n$ is number of compounding periods.
Nominal vs. Effective Interest Rates
When an annual nominal interest rate $r$ is compounded $m$ times per year:
- Period interest rate: $i = \frac{r}{m}$
- Total periods in $Y$ years: $n = m \cdot Y$
- Effective Annual Rate ($i_{\text{eff}}$):
- Continuous Compounding: As $m \to \infty$, the future worth becomes $F = P e^{r n}$, and $i_{\text{eff}} = e^r - 1$.
2. Discrete Cash Flow Equivalence Factors
| Equivalence Factor Name | Standard Notation | Formula |
|---|---|---|
| Single Payment Compound Amount | $(F/P, i, n)$ | $(1 + i)^n$ |
| Single Payment Present Worth | $(P/F, i, n)$ | $(1 + i)^{-n}$ |
| Uniform Series Sinking Fund | $(A/F, i, n)$ | $\frac{i}{(1 + i)^n - 1}$ |
| Uniform Series Compound Amount | $(F/A, i, n)$ | $\frac{(1 + i)^n - 1}{i}$ |
| Capital Recovery Factor | $(A/P, i, n)$ | $\frac{i(1 + i)^n}{(1 + i)^n - 1}$ |
| Uniform Series Present Worth | $(P/A, i, n)$ | $\frac{(1 + i)^n - 1}{i(1 + i)^n}$ |
| Arithmetic Gradient Present Worth | $(P/G, i, n)$ | $\frac{(1+i)^n - i n - 1}{i^2 (1+i)^n}$ |
3. Project Evaluation Methods
1. Present Worth (PW) Method
All cash inflows ($+R_t$) and outflows ($-C_t$) are discounted to $t=0$ using the Minimum Attractive Rate of Return (MARR):
Decision Criterion: Accept project if $\text{PW} \ge 0$.
2. Annual Worth (AW) Method
Converts all project cash flows into an equivalent uniform annual cash flow over the service life:
Decision Criterion: Accept project if $\text{AW} \ge 0$. Useful for comparing mutually exclusive projects with unequal service lives.
3. Internal Rate of Return (IRR / ROR)
The discount rate $i^$ at which the Present Worth of cash inflows equals the Present Worth of cash outflows (i.e., $\text{PW}(i^) = 0$):
Decision Criterion: Accept project if $\text{IRR} \ge \text{MARR}$.
4. Capitalized Cost ($CC$)
Capitalized cost is the present worth of an asset that is assumed to provide service perpetually (infinite life $n = \infty$):
where $C_0$ is initial cost, $A$ is annual operating cost, $RC$ is replacement cost every $k$ years, and $i$ is interest rate.
4. Asset Depreciation & Tax Accounting
Depreciation is the systematic allocation of an asset's initial cost $C$ over its useful recovery life $n$, leaving a salvage value $S$.
Straight-Line Method (SL)
Constant annual depreciation charge $d_t$:
Book value at end of year $t$: $BV_t = C - t \cdot d$.
Sum-of-the-Years-Digits Method (SYD)
Accelerated depreciation method where annual charge decreases linearly:
Declining Balance Method (DB / DDB)
Fixed percentage rate $k$ applied to the beginning book value $BV_{t-1}$:
- $k = \frac{1.5}{n}$ (150% Declining Balance) or $k = \frac{2.0}{n}$ (200% Double Declining Balance)
- $d_t = k \cdot BV_{t-1}$
- $BV_t = C(1 - k)^t$
5. Break-Even Analysis & Inflation
Break-Even Analysis
Determines the production volume $Q$ where Total Revenue $TR$ equals Total Cost $TC$:
where $FC$ is fixed costs, $VC$ is variable cost per unit, and $P$ is selling price per unit.
Inflation Adjustment
If market interest rate is $i$ and annual inflation rate is $f$, the real (inflation-free) interest rate $i'$ is:
An electronics firm borrows $500,000 at a nominal annual interest rate of 12% compounded monthly. What is the effective annual interest rate (EAR)?
A telecommunications company plans to install a fiber network costing $1,200,000 that generates annual net savings of $250,000 for 8 years. If the MARR is 10%, what is the Present Worth (PW) of the project? (Given: (P/A, 10%, 8) = 5.3349)
An industrial test bench costs $150,000 with an estimated salvage value of $30,000 after 5 years. Using the Sum-of-the-Years-Digits (SYD) method, what is the depreciation expense in Year 2?