Modified Accelerated Cost Recovery System (MACRS)

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 Modified Accelerated Cost Recovery System (MACRS) 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 modified accelerated cost recovery system (macrs) 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. Modified Accelerated Cost Recovery System (MACRS) 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: modified accelerated cost recovery system (macrs).
Capstone Studio instructional photograph

Engineering Economics — Modified Accelerated Cost Recovery System (MACRS): 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 = (factor) C" — 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.

A table of MACRS factors is provided below.

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 — Modified Accelerated Cost Recovery System (MACRS)

Equipment costing $344,000 has a salvage value of $20,000 after 5 years. Find the annual depreciation and the book value at the end of year 2.

Given

C = $344,000
S = $20,000
n = 5 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 = ($344,000 − $20,000)/5 = $64,800 per year
Book value
$B V_{j} = C - j D$
Substituting — BV = $344,000 − 2($64,800) = $214,400

Answer: D = $64,800/yr; BV(2) = $214,400

Why the other options are there

D = $68,800 (salvage not deducted)
BV = $20,000 (final-year value reported)

Reference: FE Reference Handbook — Engineering Economics → Modified Accelerated Cost Recovery System (MACRS)

Example 2

Straight-line depreciation and book value — Modified Accelerated Cost Recovery System (MACRS) (2)

Equipment costing $116,000 has a salvage value of $33,000 after 14 years. Find the annual depreciation and the book value at the end of year 3.

Given

C = $116,000
S = $33,000
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 = ($116,000 − $33,000)/14 = $5,929 per year
Book value
$B V_{j} = C - j D$
Substituting — BV = $116,000 − 3($5,929) = $98,214

Answer: D = $5,929/yr; BV(3) = $98,214

Why the other options are there

D = $8,286 (salvage not deducted)
BV = $33,000 (final-year value reported)

Reference: FE Reference Handbook — Engineering Economics → Modified Accelerated Cost Recovery System (MACRS)

Example 3

Straight-line depreciation and book value — Modified Accelerated Cost Recovery System (MACRS) (3)

Equipment costing $198,000 has a salvage value of $25,000 after 14 years. Find the annual depreciation and the book value at the end of year 3.

Given

C = $198,000
S = $25,000
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 = ($198,000 − $25,000)/14 = $12,357 per year
Book value
$B V_{j} = C - j D$
Substituting — BV = $198,000 − 3($12,357) = $160,929

Answer: D = $12,357/yr; BV(3) = $160,929

Why the other options are there

D = $14,143 (salvage not deducted)
BV = $25,000 (final-year value reported)

Reference: FE Reference Handbook — Engineering Economics → Modified Accelerated Cost Recovery System (MACRS)

Example 4

Straight-line depreciation and book value — Modified Accelerated Cost Recovery System (MACRS) (4)

Equipment costing $165,000 has a salvage value of $31,000 after 11 years. Find the annual depreciation and the book value at the end of year 2.

Given

C = $165,000
S = $31,000
n = 11 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 = ($165,000 − $31,000)/11 = $12,182 per year
Book value
$B V_{j} = C - j D$
Substituting — BV = $165,000 − 2($12,182) = $140,636

Answer: D = $12,182/yr; BV(2) = $140,636

Why the other options are there

D = $15,000 (salvage not deducted)
BV = $31,000 (final-year value reported)

Reference: FE Reference Handbook — Engineering Economics → Modified Accelerated Cost Recovery System (MACRS)

Example 5

Straight-line depreciation and book value — Modified Accelerated Cost Recovery System (MACRS) (5)

Equipment costing $284,000 has a salvage value of $39,000 after 11 years. Find the annual depreciation and the book value at the end of year 3.

Given

C = $284,000
S = $39,000
n = 11 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 = ($284,000 − $39,000)/11 = $22,273 per year
Book value
$B V_{j} = C - j D$
Substituting — BV = $284,000 − 3($22,273) = $217,182

Answer: D = $22,273/yr; BV(3) = $217,182

Why the other options are there

D = $25,818 (salvage not deducted)
BV = $39,000 (final-year value reported)

Reference: FE Reference Handbook — Engineering Economics → Modified Accelerated Cost Recovery System (MACRS)

Example 6

Straight-line depreciation and book value — Modified Accelerated Cost Recovery System (MACRS) (6)

Equipment costing $395,000 has a salvage value of $30,000 after 8 years. Find the annual depreciation and the book value at the end of year 2.

Given

C = $395,000
S = $30,000
n = 8 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 = ($395,000 − $30,000)/8 = $45,625 per year
Book value
$B V_{j} = C - j D$
Substituting — BV = $395,000 − 2($45,625) = $303,750

Answer: D = $45,625/yr; BV(2) = $303,750

Why the other options are there

D = $49,375 (salvage not deducted)
BV = $30,000 (final-year value reported)

Reference: FE Reference Handbook — Engineering Economics → Modified Accelerated Cost Recovery System (MACRS)

Example 7

Straight-line depreciation and book value — Modified Accelerated Cost Recovery System (MACRS) (7)

Equipment costing $120,000 has a salvage value of $36,000 after 12 years. Find the annual depreciation and the book value at the end of year 4.

Given

C = $120,000
S = $36,000
n = 12 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 = ($120,000 − $36,000)/12 = $7,000 per year
Book value
$B V_{j} = C - j D$
Substituting — BV = $120,000 − 4($7,000) = $92,000

Answer: D = $7,000/yr; BV(4) = $92,000

Why the other options are there

D = $10,000 (salvage not deducted)
BV = $36,000 (final-year value reported)

Reference: FE Reference Handbook — Engineering Economics → Modified Accelerated Cost Recovery System (MACRS)

Example 8

Straight-line depreciation and book value — Modified Accelerated Cost Recovery System (MACRS) (8)

Equipment costing $127,000 has a salvage value of $7,000 after 11 years. Find the annual depreciation and the book value at the end of year 4.

Given

C = $127,000
S = $7,000
n = 11 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 = ($127,000 − $7,000)/11 = $10,909 per year
Book value
$B V_{j} = C - j D$
Substituting — BV = $127,000 − 4($10,909) = $83,364

Answer: D = $10,909/yr; BV(4) = $83,364

Why the other options are there

D = $11,545 (salvage not deducted)
BV = $7,000 (final-year value reported)

Reference: FE Reference Handbook — Engineering Economics → Modified Accelerated Cost Recovery System (MACRS)

Example 9

Straight-line depreciation and book value — Modified Accelerated Cost Recovery System (MACRS) (9)

Equipment costing $192,000 has a salvage value of $21,000 after 13 years. Find the annual depreciation and the book value at the end of year 3.

Given

C = $192,000
S = $21,000
n = 13 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 = ($192,000 − $21,000)/13 = $13,154 per year
Book value
$B V_{j} = C - j D$
Substituting — BV = $192,000 − 3($13,154) = $152,538

Answer: D = $13,154/yr; BV(3) = $152,538

Why the other options are there

D = $14,769 (salvage not deducted)
BV = $21,000 (final-year value reported)

Reference: FE Reference Handbook — Engineering Economics → Modified Accelerated Cost Recovery System (MACRS)

Example 10

Straight-line depreciation and book value — Modified Accelerated Cost Recovery System (MACRS) (10)

Equipment costing $106,000 has a salvage value of $25,000 after 8 years. Find the annual depreciation and the book value at the end of year 4.

Given

C = $106,000
S = $25,000
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 = ($106,000 − $25,000)/8 = $10,125 per year
Book value
$B V_{j} = C - j D$
Substituting — BV = $106,000 − 4($10,125) = $65,500

Answer: D = $10,125/yr; BV(4) = $65,500

Why the other options are there

D = $13,250 (salvage not deducted)
BV = $25,000 (final-year value reported)

Reference: FE Reference Handbook — Engineering Economics → Modified Accelerated Cost Recovery System (MACRS)

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

Modified Accelerated Cost Recovery System (MACRS) 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.