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.
Last updated: July 2026

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 $P$: I=Pin,F=P(1+in)I = P \cdot i \cdot n, \quad F = P(1 + i \cdot n)
  • Compound Interest: Interest earned in each period is added to the principal to earn interest in subsequent periods: F=P(1+i)nF = P(1 + i)^n 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 ($i_{\text{eff}}$): ieff=(1+rm)m1i_{\text{eff}} = \left(1 + \frac{r}{m}\right)^m - 1
  • 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 NameStandard NotationFormula
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):

PW=t=0nRtCt(1+MARR)t\text{PW} = \sum_{t=0}^n \frac{R_t - C_t}{(1 + \text{MARR})^t}

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:

AW=PW×(A/P,i,n)\text{AW} = \text{PW} \times (A/P, i, n)

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., $\text{PW}(i^) = 0$):

t=0nRtCt(1+i)t=0\sum_{t=0}^n \frac{R_t - C_t}{(1 + i^*)^t} = 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 $n = \infty$):

CC=C0+Ai+RC(1+i)k1CC = C_0 + \frac{A}{i} + \frac{RC}{(1 + i)^k - 1}

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 $d_t$:

dt=d=CSnd_t = d = \frac{C - S}{n}

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:

SOYD=n(n+1)2\text{SOYD} = \frac{n(n + 1)}{2}

dt=(CS)×nt+1SOYDd_t = (C - S) \times \frac{n - t + 1}{\text{SOYD}}

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 $TC$:

TR=PQ,TC=FC+VCQTR = P \cdot Q, \quad TC = FC + VC \cdot Q

QBE=FCPVCQ_{\text{BE}} = \frac{FC}{P - VC}

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:

(1+i)=(1+i)(1+f)    i=i+f+if(1 + i) = (1 + i')(1 + f) \implies i = i' + f + i' f

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Engineering Project Cash Flow Diagram and Evaluation Workflow
Book Value Trajectory ($100k Initial Cost, $10k Salvage, 5-Yr Life)
Test Your Knowledge

An electronics firm borrows $500,000 at a nominal annual interest rate of 12% compounded monthly. What is the effective annual interest rate (EAR)?

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Test Your Knowledge

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)

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Test Your Knowledge

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?

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