Overview

Engineering Economics · FE Reference Handbook section

Engineering Economics

0 formulas

10 exam-style examples

~45 min

All Engineering Economics lectures

Learning objectives

What you must be able to do before leaving this section.

This chapter section covers Overview within Engineering Economics. Read it the way you would read a textbook chapter: the theory first so the relations mean something, then every equation with its use and its trap, then 10 fully worked examples with the arithmetic shown line by line, and finally a self-check you should be able to answer without notes.

Explain, in your own words, what overview describes physically and when it applies.
State every one of the 0 relations the handbook lists here and name each symbol with its unit.
Select the correct relation from the wording of an exam stem within 20 seconds.
Carry a complete solution from givens to a "most nearly" answer with the correct unit.
Recognise the distractors generated by the unit trap: i per period must match n periods.

Lecture

Why this section exists. Overview is the part of Engineering Economics that lets you connect an alternative being compared over a study period to a number you can defend. Before any equation is useful you must be able to picture the physical situation it describes; the schematic below is that picture.

How the theory is built. The handbook prints results, not derivations. Each relation in this section comes from one governing principle applied to the idealised system: state the principle, impose the stated assumptions, and the printed equation follows. Knowing which assumption each relation rests on is what lets you reject a wrong answer choice in seconds.

How it is examined. Items from this page are written as one cash-flow diagram converted with one factor. Roughly two thirds are direct substitution, one third require one intermediate quantity from a neighbouring relation, and a small number are conceptual — testing whether you know the assumption, not the arithmetic.

The habit that earns the points. Unit discipline. i per period must match n periods. Every relation below is dimensionally consistent only when that rule is honoured, and the distractor set is deliberately built from candidates who ignored it. Write the unit next to every number you substitute, every time.

How to study this page. Read the theory, then cover the formula cards and try to reproduce each relation from its description. Then work the examples with the solution hidden, revealing one line at a time. Finish with the self-check questions; if you cannot answer one, return to the matching formula card.

Three engineers in hard hats and safety vests reviewing drawings on a truck tailgate. — Photo 1. Where this shows up in practice: overview.
Capstone Studio instructional photograph

Engineering Economics — Overview: reference schematic for orienting the symbols used in this section.

Theory, developed

Read this before the equations — it is what makes them memorable.

The physical situation

Every item from this section describes an alternative being compared over a study period. Sketch it before you compute — a labelled sketch with the givens on it converts a wordy stem into a solvable problem and exposes the quantity the examiner left out on purpose.

The governing principle

The 0 relations on this page are consequences of one principle applied to that idealised system. Identify which quantity is conserved, balanced, or defined, and the correct equation follows without memorisation.

Assumptions and limits of validity

Each printed relation carries silent assumptions — linearity, steady state, uniformity, small deformation, or standard conditions, depending on the subject. Conceptual exam items are written by violating exactly one of these, so read the sentence above the equation as carefully as the equation itself.

Solution procedure you should automate

1) Read the last sentence of the stem to identify the requested quantity. 2) Locate the relation on this page whose left-hand side is that quantity. 3) Tabulate the givens with units and mark the missing symbol. 4) If a symbol is missing, find the one relation that produces it. 5) Rearrange symbolically, substitute once, evaluate, and round only at the end.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

Factor Name Converts Symbol Formula
Single Payment
to F given P (F/P, i%, n) (1 + i)n
Compound Amount
Single Payment
to P given F (P/F, i%, n) (1 + i) –n
Present Worth
Uniform Series i
to A given F (A/F, i%, n)
Sinking Fund (1 + i )n − 1
i (1 + i )
Capital Recovery to A given P (A/P, i%, n)
(1 + i ) − 1
Uniform Series
to F given A (F/A, i%, n)
(1 + i )n − 1
Compound Amount i
Uniform Series (1 + i )n − 1
to P given A (P/A, i%, n)
Present Worth i (1 + i )
Uniform Gradient (1 + i )n − 1 − n
to P given G (P/G, i%, n)
Present Worth i 2 (1 + i ) i (1 + i )
n n

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

This section is conceptual; there are no equations to memorise.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Capitalized cost of a perpetual public works asset — Overview

A bridge deck costs $644,000 to build and $32,000 per year to maintain forever. At 5.5% interest, determine the capitalized cost.

Given

First cost = $644,000
A = $32,000/yr
i = 5.5%

Find

Capitalized cost of the asset

Start with the thinking

Capitalized cost is the present worth of a cash flow that continues indefinitely.
For a perpetuity P = A/i, so a lower interest rate raises the capitalized cost sharply.

Step-by-step solution

Formula
$C C = F i r s t cos t + A / i$
Perpetuity term — A/i = $32,000/0.055 = $581,818
Substituting — CC = $644,000 + $581,818
Evaluate — CC = $1,225,818

Answer: Capitalized cost ≈ $1,225,818

Why the other options are there

$645,760 (multiplied instead of divided)
$581,818 (omitted the first cost)

Reference: FE Reference Handbook — Engineering Economics → Overview

Example 2

Capitalized cost of a perpetual public works asset — Overview (2)

A bridge deck costs $505,000 to build and $9,000 per year to maintain forever. At 6.0% interest, determine the capitalized cost.

Given

First cost = $505,000
A = $9,000/yr
i = 6.0%

Find

Capitalized cost of the asset

Start with the thinking

Capitalized cost is the present worth of a cash flow that continues indefinitely.
For a perpetuity P = A/i, so a lower interest rate raises the capitalized cost sharply.

Step-by-step solution

Formula
$C C = F i r s t cos t + A / i$
Perpetuity term — A/i = $9,000/0.060 = $150,000
Substituting — CC = $505,000 + $150,000
Evaluate — CC = $655,000

Answer: Capitalized cost ≈ $655,000

Why the other options are there

$505,540 (multiplied instead of divided)
$150,000 (omitted the first cost)

Reference: FE Reference Handbook — Engineering Economics → Overview

Example 3

Capitalized cost of a perpetual public works asset — Overview (3)

A bridge deck costs $624,000 to build and $33,000 per year to maintain forever. At 7.0% interest, determine the capitalized cost.

Given

First cost = $624,000
A = $33,000/yr
i = 7.0%

Find

Capitalized cost of the asset

Start with the thinking

Capitalized cost is the present worth of a cash flow that continues indefinitely.
For a perpetuity P = A/i, so a lower interest rate raises the capitalized cost sharply.

Step-by-step solution

Formula
$C C = F i r s t cos t + A / i$
Perpetuity term — A/i = $33,000/0.070 = $471,429
Substituting — CC = $624,000 + $471,429
Evaluate — CC = $1,095,429

Answer: Capitalized cost ≈ $1,095,429

Why the other options are there

$626,310 (multiplied instead of divided)
$471,429 (omitted the first cost)

Reference: FE Reference Handbook — Engineering Economics → Overview

Example 4

Capitalized cost of a perpetual public works asset — Overview (4)

A bridge deck costs $837,000 to build and $32,000 per year to maintain forever. At 7.0% interest, determine the capitalized cost.

Given

First cost = $837,000
A = $32,000/yr
i = 7.0%

Find

Capitalized cost of the asset

Start with the thinking

Capitalized cost is the present worth of a cash flow that continues indefinitely.
For a perpetuity P = A/i, so a lower interest rate raises the capitalized cost sharply.

Step-by-step solution

Formula
$C C = F i r s t cos t + A / i$
Perpetuity term — A/i = $32,000/0.070 = $457,143
Substituting — CC = $837,000 + $457,143
Evaluate — CC = $1,294,143

Answer: Capitalized cost ≈ $1,294,143

Why the other options are there

$839,240 (multiplied instead of divided)
$457,143 (omitted the first cost)

Reference: FE Reference Handbook — Engineering Economics → Overview

Example 5

Capitalized cost of a perpetual public works asset — Overview (5)

A bridge deck costs $566,000 to build and $44,000 per year to maintain forever. At 4.0% interest, determine the capitalized cost.

Given

First cost = $566,000
A = $44,000/yr
i = 4.0%

Find

Capitalized cost of the asset

Start with the thinking

Capitalized cost is the present worth of a cash flow that continues indefinitely.
For a perpetuity P = A/i, so a lower interest rate raises the capitalized cost sharply.

Step-by-step solution

Formula
$C C = F i r s t cos t + A / i$
Perpetuity term — A/i = $44,000/0.040 = $1,100,000
Substituting — CC = $566,000 + $1,100,000
Evaluate — CC = $1,666,000

Answer: Capitalized cost ≈ $1,666,000

Why the other options are there

$567,760 (multiplied instead of divided)
$1,100,000 (omitted the first cost)

Reference: FE Reference Handbook — Engineering Economics → Overview

Example 6

Capitalized cost of a perpetual public works asset — Overview (6)

A bridge deck costs $681,000 to build and $8,000 per year to maintain forever. At 5.5% interest, determine the capitalized cost.

Given

First cost = $681,000
A = $8,000/yr
i = 5.5%

Find

Capitalized cost of the asset

Start with the thinking

Capitalized cost is the present worth of a cash flow that continues indefinitely.
For a perpetuity P = A/i, so a lower interest rate raises the capitalized cost sharply.

Step-by-step solution

Formula
$C C = F i r s t cos t + A / i$
Perpetuity term — A/i = $8,000/0.055 = $145,455
Substituting — CC = $681,000 + $145,455
Evaluate — CC = $826,455

Answer: Capitalized cost ≈ $826,455

Why the other options are there

$681,440 (multiplied instead of divided)
$145,455 (omitted the first cost)

Reference: FE Reference Handbook — Engineering Economics → Overview

Example 7

Capitalized cost of a perpetual public works asset — Overview (7)

A bridge deck costs $477,000 to build and $20,000 per year to maintain forever. At 5.5% interest, determine the capitalized cost.

Given

First cost = $477,000
A = $20,000/yr
i = 5.5%

Find

Capitalized cost of the asset

Start with the thinking

Capitalized cost is the present worth of a cash flow that continues indefinitely.
For a perpetuity P = A/i, so a lower interest rate raises the capitalized cost sharply.

Step-by-step solution

Formula
$C C = F i r s t cos t + A / i$
Perpetuity term — A/i = $20,000/0.055 = $363,636
Substituting — CC = $477,000 + $363,636
Evaluate — CC = $840,636

Answer: Capitalized cost ≈ $840,636

Why the other options are there

$478,100 (multiplied instead of divided)
$363,636 (omitted the first cost)

Reference: FE Reference Handbook — Engineering Economics → Overview

Example 8

Capitalized cost of a perpetual public works asset — Overview (8)

A bridge deck costs $610,000 to build and $25,000 per year to maintain forever. At 6.0% interest, determine the capitalized cost.

Given

First cost = $610,000
A = $25,000/yr
i = 6.0%

Find

Capitalized cost of the asset

Start with the thinking

Capitalized cost is the present worth of a cash flow that continues indefinitely.
For a perpetuity P = A/i, so a lower interest rate raises the capitalized cost sharply.

Step-by-step solution

Formula
$C C = F i r s t cos t + A / i$
Perpetuity term — A/i = $25,000/0.060 = $416,667
Substituting — CC = $610,000 + $416,667
Evaluate — CC = $1,026,667

Answer: Capitalized cost ≈ $1,026,667

Why the other options are there

$611,500 (multiplied instead of divided)
$416,667 (omitted the first cost)

Reference: FE Reference Handbook — Engineering Economics → Overview

Example 9

Capitalized cost of a perpetual public works asset — Overview (9)

A bridge deck costs $357,000 to build and $39,000 per year to maintain forever. At 9.0% interest, determine the capitalized cost.

Given

First cost = $357,000
A = $39,000/yr
i = 9.0%

Find

Capitalized cost of the asset

Start with the thinking

Capitalized cost is the present worth of a cash flow that continues indefinitely.
For a perpetuity P = A/i, so a lower interest rate raises the capitalized cost sharply.

Step-by-step solution

Formula
$C C = F i r s t cos t + A / i$
Perpetuity term — A/i = $39,000/0.090 = $433,333
Substituting — CC = $357,000 + $433,333
Evaluate — CC = $790,333

Answer: Capitalized cost ≈ $790,333

Why the other options are there

$360,510 (multiplied instead of divided)
$433,333 (omitted the first cost)

Reference: FE Reference Handbook — Engineering Economics → Overview

Example 10

Capitalized cost of a perpetual public works asset — Overview (10)

A bridge deck costs $731,000 to build and $24,000 per year to maintain forever. At 9.5% interest, determine the capitalized cost.

Given

First cost = $731,000
A = $24,000/yr
i = 9.5%

Find

Capitalized cost of the asset

Start with the thinking

Capitalized cost is the present worth of a cash flow that continues indefinitely.
For a perpetuity P = A/i, so a lower interest rate raises the capitalized cost sharply.

Step-by-step solution

Formula
$C C = F i r s t cos t + A / i$
Perpetuity term — A/i = $24,000/0.095 = $252,632
Substituting — CC = $731,000 + $252,632
Evaluate — CC = $983,632

Answer: Capitalized cost ≈ $983,632

Why the other options are there

$733,280 (multiplied instead of divided)
$252,632 (omitted the first cost)

Reference: FE Reference Handbook — Engineering Economics → Overview

Self-check

Answer these without notes before moving on.

Without looking, state the relation on this page whose left-hand side is the quantity most often requested, and name every symbol in it.
Which assumption, if violated, makes the main relation of this section invalid?
Given an alternative being compared over a study period, what is the first quantity you would compute, and why that one first?
Which unit conversion in this subject most often produces a wrong answer choice, and what is its numerical factor?
Rework Example 1 above from the givens alone, without reading the solution lines.

Chapter summary

Overview contains 0 relations; you must be able to find this page in under 15 seconds.
Exam style: one cash-flow diagram converted with one factor.
Unit rule: i per period must match n periods.
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

i per period must match n periods
Answering the intermediate quantity instead of the quantity requested.
Rounding intermediate values before the final step.
Using a relation from an adjacent handbook section that shares a symbol.
Skipping the sketch — most lost points on this page start with a misread geometry.