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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[(1+i)n−1i(1+i)n]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=C−Snd = \frac{C - S}{n}, whereas SYD accelerates early tax deductions.

  • Net Present Value (NPV) evaluation criteria accepts projects when NPV ≥0\ge 0 at the Minimum Attractive Rate of Return (MARR).

  • Capitalized cost CC=C0+Ai+RC(1+i)k−1CC = 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 PP: I=P⋅i⋅n,F=P(1+i⋅n)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 PP is Present Worth, FF is Future Worth, ii is interest rate per period, and nn is number of compounding periods.

Nominal vs. Effective Interest Rates

When an annual nominal interest rate rr is compounded mm times per year:

  • Period interest rate: i=rmi = \frac{r}{m}
  • Total periods in YY years: n=m⋅Yn = m \cdot Y
  • Effective Annual Rate (ieffi_{\text{eff}}): ieff=(1+rm)m−1i_{\text{eff}} = \left(1 + \frac{r}{m}\right)^m - 1
  • Continuous Compounding: As m→∞m \to \infty, the future worth becomes F=PernF = P e^{r n}, and ieff=er−1i_{\text{eff}} = e^r - 1.

2. Discrete Cash Flow Equivalence Factors

Equivalence Factor NameStandard NotationFormula
Single Payment Compound Amount(F/P,i,n)(F/P, i, n)(1+i)n(1 + i)^n
Single Payment Present Worth(P/F,i,n)(P/F, i, n)(1+i)−n(1 + i)^{-n}
Uniform Series Sinking Fund(A/F,i,n)(A/F, i, n)i(1+i)n−1\frac{i}{(1 + i)^n - 1}
Uniform Series Compound Amount(F/A,i,n)(F/A, i, n)(1+i)n−1i\frac{(1 + i)^n - 1}{i}
Capital Recovery Factor(A/P,i,n)(A/P, i, n)i(1+i)n(1+i)n−1\frac{i(1 + i)^n}{(1 + i)^n - 1}
Uniform Series Present Worth(P/A,i,n)(P/A, i, n)(1+i)n−1i(1+i)n\frac{(1 + i)^n - 1}{i(1 + i)^n}
Arithmetic Gradient Present Worth(P/G,i,n)(P/G, i, n)(1+i)n−in−1i2(1+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 (+Rt+R_t) and outflows (−Ct-C_t) are discounted to t=0t=0 using the Minimum Attractive Rate of Return (MARR):

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

Decision Criterion: Accept project if PW≥0\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 AW≥0\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∗i^* at which the Present Worth of cash inflows equals the Present Worth of cash outflows (i.e., PW(i∗)=0\text{PW}(i^*) = 0):

∑t=0nRt−Ct(1+i∗)t=0\sum_{t=0}^n \frac{R_t - C_t}{(1 + i^*)^t} = 0

Decision Criterion: Accept project if IRR≥MARR\text{IRR} \ge \text{MARR}.

4. Capitalized Cost (CCCC)

Capitalized cost is the present worth of an asset that is assumed to provide service perpetually (infinite life n=∞n = \infty):

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

where C0C_0 is initial cost, AA is annual operating cost, RCRC is replacement cost every kk years, and ii is interest rate.


4. Asset Depreciation & Tax Accounting

Depreciation is the systematic allocation of an asset's initial cost CC over its useful recovery life nn, leaving a salvage value SS.

Straight-Line Method (SL)

Constant annual depreciation charge dtd_t:

dt=d=C−Snd_t = d = \frac{C - S}{n}

Book value at end of year tt: BVt=C−t⋅dBV_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=(C−S)×n−t+1SOYDd_t = (C - S) \times \frac{n - t + 1}{\text{SOYD}}

Declining Balance Method (DB / DDB)

Fixed percentage rate kk applied to the beginning book value BVt−1BV_{t-1}:

  • k=1.5nk = \frac{1.5}{n} (150% Declining Balance) or k=2.0nk = \frac{2.0}{n} (200% Double Declining Balance)
  • dt=k⋅BVt−1d_t = k \cdot BV_{t-1}
  • BVt=C(1−k)tBV_t = C(1 - k)^t

5. Break-Even Analysis & Inflation

Break-Even Analysis

Determines the production volume QQ where Total Revenue TRTR equals Total Cost TCTC:

TR=P⋅Q,TC=FC+VC⋅QTR = P \cdot Q, \quad TC = FC + VC \cdot Q QBE=FCP−VCQ_{\text{BE}} = \frac{FC}{P - VC}

where FCFC is fixed costs, VCVC is variable cost per unit, and PP is selling price per unit.

Inflation Adjustment

If market interest rate is ii and annual inflation rate is ff, the real (inflation-free) interest rate i′i' is:

(1+i)=(1+i′)(1+f)  ⟹  i=i′+f+i′f(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)?

A

12.00%

B

12.36%

C

12.68%

D

13.04%

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)

A

-$133,725

B

$800,000

C

$333,725

D

$133,725

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?

A

$24,000

B

$40,000

C

$32,000

D

$48,000

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