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

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
$BV_j = C - jD$
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
$BV_j = C - jD$
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
$BV_j = C - jD$
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
$BV_j = C - jD$
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
$BV_j = C - jD$
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
$BV_j = C - jD$
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
$BV_j = C - jD$
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
$BV_j = C - jD$
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
$BV_j = C - jD$
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
$BV_j = C - jD$
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)