Straight Line

Engineering Economics · FE Reference Handbook section

Engineering Economics

1 formulas

10 exam-style examples

~47 min

All Engineering Economics lectures

Learning objectives

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

This chapter section covers Straight Line 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 straight line describes physically and when it applies.
State every one of the 1 relation 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. Straight Line 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: straight line.
Capstone Studio instructional photograph

Engineering Economics — Straight Line: 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 1 relation 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.

Notation used in this section

Dj	Quantity produced by "Dj = n" — read its definition and unit from the handbook line directly above the equation.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

C - Sn

Core formulas for this FE topic

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

Worked exam-style examples

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

Example 1

Straight-line depreciation and book value — Straight Line

Equipment costing $329,000 has a salvage value of $14,000 after 8 years. Find the annual depreciation and the book value at the end of year 4.

Given

C = $329,000
S = $14,000
n = 8 yr
Year = 4

Find

D and BV

Start with the thinking

Straight line spreads the depreciable base evenly.
The depreciable base excludes salvage.

Step-by-step solution

Annual depreciation
$D = (C - S) / n$
Substituting — D = ($329,000 − $14,000)/8 = $39,375 per year
Book value
$B V_{j} = C - j D$
Substituting — BV = $329,000 − 4($39,375) = $171,500

Answer: D = $39,375/yr; BV(4) = $171,500

Why the other options are there

D = $41,125 (salvage not deducted)
BV = $14,000 (final-year value reported)

Reference: FE Reference Handbook — Engineering Economics → Straight Line

Example 2

Straight-line depreciation and book value — Straight Line (2)

Equipment costing $229,000 has a salvage value of $22,000 after 5 years. Find the annual depreciation and the book value at the end of year 3.

Given

C = $229,000
S = $22,000
n = 5 yr
Year = 3

Find

D and BV

Start with the thinking

Straight line spreads the depreciable base evenly.
The depreciable base excludes salvage.

Step-by-step solution

Annual depreciation
$D = (C - S) / n$
Substituting — D = ($229,000 − $22,000)/5 = $41,400 per year
Book value
$B V_{j} = C - j D$
Substituting — BV = $229,000 − 3($41,400) = $104,800

Answer: D = $41,400/yr; BV(3) = $104,800

Why the other options are there

D = $45,800 (salvage not deducted)
BV = $22,000 (final-year value reported)

Reference: FE Reference Handbook — Engineering Economics → Straight Line

Example 3

Straight-line depreciation and book value — Straight Line (3)

Equipment costing $274,000 has a salvage value of $10,000 after 5 years. Find the annual depreciation and the book value at the end of year 4.

Given

C = $274,000
S = $10,000
n = 5 yr
Year = 4

Find

D and BV

Start with the thinking

Straight line spreads the depreciable base evenly.
The depreciable base excludes salvage.

Step-by-step solution

Annual depreciation
$D = (C - S) / n$
Substituting — D = ($274,000 − $10,000)/5 = $52,800 per year
Book value
$B V_{j} = C - j D$
Substituting — BV = $274,000 − 4($52,800) = $62,800

Answer: D = $52,800/yr; BV(4) = $62,800

Why the other options are there

D = $54,800 (salvage not deducted)
BV = $10,000 (final-year value reported)

Reference: FE Reference Handbook — Engineering Economics → Straight Line

Example 4

Straight-line depreciation and book value — Straight Line (4)

Equipment costing $181,000 has a salvage value of $10,000 after 10 years. Find the annual depreciation and the book value at the end of year 3.

Given

C = $181,000
S = $10,000
n = 10 yr
Year = 3

Find

D and BV

Start with the thinking

Straight line spreads the depreciable base evenly.
The depreciable base excludes salvage.

Step-by-step solution

Annual depreciation
$D = (C - S) / n$
Substituting — D = ($181,000 − $10,000)/10 = $17,100 per year
Book value
$B V_{j} = C - j D$
Substituting — BV = $181,000 − 3($17,100) = $129,700

Answer: D = $17,100/yr; BV(3) = $129,700

Why the other options are there

D = $18,100 (salvage not deducted)
BV = $10,000 (final-year value reported)

Reference: FE Reference Handbook — Engineering Economics → Straight Line

Example 5

Straight-line depreciation and book value — Straight Line (5)

Equipment costing $85,000 has a salvage value of $8,000 after 14 years. Find the annual depreciation and the book value at the end of year 2.

Given

C = $85,000
S = $8,000
n = 14 yr
Year = 2

Find

D and BV

Start with the thinking

Straight line spreads the depreciable base evenly.
The depreciable base excludes salvage.

Step-by-step solution

Annual depreciation
$D = (C - S) / n$
Substituting — D = ($85,000 − $8,000)/14 = $5,500 per year
Book value
$B V_{j} = C - j D$
Substituting — BV = $85,000 − 2($5,500) = $74,000

Answer: D = $5,500/yr; BV(2) = $74,000

Why the other options are there

D = $6,071 (salvage not deducted)
BV = $8,000 (final-year value reported)

Reference: FE Reference Handbook — Engineering Economics → Straight Line

Example 6

Straight-line depreciation and book value — Straight Line (6)

Equipment costing $363,000 has a salvage value of $19,000 after 8 years. Find the annual depreciation and the book value at the end of year 4.

Given

C = $363,000
S = $19,000
n = 8 yr
Year = 4

Find

D and BV

Start with the thinking

Straight line spreads the depreciable base evenly.
The depreciable base excludes salvage.

Step-by-step solution

Annual depreciation
$D = (C - S) / n$
Substituting — D = ($363,000 − $19,000)/8 = $43,000 per year
Book value
$B V_{j} = C - j D$
Substituting — BV = $363,000 − 4($43,000) = $191,000

Answer: D = $43,000/yr; BV(4) = $191,000

Why the other options are there

D = $45,375 (salvage not deducted)
BV = $19,000 (final-year value reported)

Reference: FE Reference Handbook — Engineering Economics → Straight Line

Example 7

Straight-line depreciation and book value — Straight Line (7)

Equipment costing $240,000 has a salvage value of $34,000 after 10 years. Find the annual depreciation and the book value at the end of year 4.

Given

C = $240,000
S = $34,000
n = 10 yr
Year = 4

Find

D and BV

Start with the thinking

Straight line spreads the depreciable base evenly.
The depreciable base excludes salvage.

Step-by-step solution

Annual depreciation
$D = (C - S) / n$
Substituting — D = ($240,000 − $34,000)/10 = $20,600 per year
Book value
$B V_{j} = C - j D$
Substituting — BV = $240,000 − 4($20,600) = $157,600

Answer: D = $20,600/yr; BV(4) = $157,600

Why the other options are there

D = $24,000 (salvage not deducted)
BV = $34,000 (final-year value reported)

Reference: FE Reference Handbook — Engineering Economics → Straight Line

Example 8

Straight-line depreciation and book value — Straight Line (8)

Equipment costing $285,000 has a salvage value of $40,000 after 14 years. Find the annual depreciation and the book value at the end of year 3.

Given

C = $285,000
S = $40,000
n = 14 yr
Year = 3

Find

D and BV

Start with the thinking

Straight line spreads the depreciable base evenly.
The depreciable base excludes salvage.

Step-by-step solution

Annual depreciation
$D = (C - S) / n$
Substituting — D = ($285,000 − $40,000)/14 = $17,500 per year
Book value
$B V_{j} = C - j D$
Substituting — BV = $285,000 − 3($17,500) = $232,500

Answer: D = $17,500/yr; BV(3) = $232,500

Why the other options are there

D = $20,357 (salvage not deducted)
BV = $40,000 (final-year value reported)

Reference: FE Reference Handbook — Engineering Economics → Straight Line

Example 9

Straight-line depreciation and book value — Straight Line (9)

Equipment costing $319,000 has a salvage value of $19,000 after 5 years. Find the annual depreciation and the book value at the end of year 4.

Given

C = $319,000
S = $19,000
n = 5 yr
Year = 4

Find

D and BV

Start with the thinking

Straight line spreads the depreciable base evenly.
The depreciable base excludes salvage.

Step-by-step solution

Annual depreciation
$D = (C - S) / n$
Substituting — D = ($319,000 − $19,000)/5 = $60,000 per year
Book value
$B V_{j} = C - j D$
Substituting — BV = $319,000 − 4($60,000) = $79,000

Answer: D = $60,000/yr; BV(4) = $79,000

Why the other options are there

D = $63,800 (salvage not deducted)
BV = $19,000 (final-year value reported)

Reference: FE Reference Handbook — Engineering Economics → Straight Line

Example 10

Straight-line depreciation and book value — Straight Line (10)

Equipment costing $397,000 has a salvage value of $5,000 after 12 years. Find the annual depreciation and the book value at the end of year 2.

Given

C = $397,000
S = $5,000
n = 12 yr
Year = 2

Find

D and BV

Start with the thinking

Straight line spreads the depreciable base evenly.
The depreciable base excludes salvage.

Step-by-step solution

Annual depreciation
$D = (C - S) / n$
Substituting — D = ($397,000 − $5,000)/12 = $32,667 per year
Book value
$B V_{j} = C - j D$
Substituting — BV = $397,000 − 2($32,667) = $331,667

Answer: D = $32,667/yr; BV(2) = $331,667

Why the other options are there

D = $33,083 (salvage not deducted)
BV = $5,000 (final-year value reported)

Reference: FE Reference Handbook — Engineering Economics → Straight Line

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

Straight Line contains 1 relation; 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.

Straight Line

Learning objectives

Lecture

Theory, developed

Notation used in this section

Handbook notes for this section

Core formulas for this FE topic

Dj=n

Worked exam-style examples

Self-check

Chapter summary

Common traps in this section