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Straight Line

Engineering Economics · FE Reference Handbook section

Engineering Economics
1 formulas
10 exam-style examples
~47 min
All Engineering Economics lectures

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=8yrn = 8 yr
  • Year=4Year = 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

  1. Annual depreciation

    D=(C−S)/nD = (C - S)/n

  2. Substituting — D = ($329,000 − $14,000)/8 = $39,375 per year

  3. Book value

    BVj=C−jDBV_j = C - jD

  4. 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=5yrn = 5 yr
  • Year=3Year = 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

  1. Annual depreciation

    D=(C−S)/nD = (C - S)/n

  2. Substituting — D = ($229,000 − $22,000)/5 = $41,400 per year

  3. Book value

    BVj=C−jDBV_j = C - jD

  4. 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=5yrn = 5 yr
  • Year=4Year = 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

  1. Annual depreciation

    D=(C−S)/nD = (C - S)/n

  2. Substituting — D = ($274,000 − $10,000)/5 = $52,800 per year

  3. Book value

    BVj=C−jDBV_j = C - jD

  4. 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=10yrn = 10 yr
  • Year=3Year = 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

  1. Annual depreciation

    D=(C−S)/nD = (C - S)/n

  2. Substituting — D = ($181,000 − $10,000)/10 = $17,100 per year

  3. Book value

    BVj=C−jDBV_j = C - jD

  4. 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=14yrn = 14 yr
  • Year=2Year = 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

  1. Annual depreciation

    D=(C−S)/nD = (C - S)/n

  2. Substituting — D = ($85,000 − $8,000)/14 = $5,500 per year

  3. Book value

    BVj=C−jDBV_j = C - jD

  4. 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=8yrn = 8 yr
  • Year=4Year = 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

  1. Annual depreciation

    D=(C−S)/nD = (C - S)/n

  2. Substituting — D = ($363,000 − $19,000)/8 = $43,000 per year

  3. Book value

    BVj=C−jDBV_j = C - jD

  4. 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=10yrn = 10 yr
  • Year=4Year = 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

  1. Annual depreciation

    D=(C−S)/nD = (C - S)/n

  2. Substituting — D = ($240,000 − $34,000)/10 = $20,600 per year

  3. Book value

    BVj=C−jDBV_j = C - jD

  4. 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=14yrn = 14 yr
  • Year=3Year = 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

  1. Annual depreciation

    D=(C−S)/nD = (C - S)/n

  2. Substituting — D = ($285,000 − $40,000)/14 = $17,500 per year

  3. Book value

    BVj=C−jDBV_j = C - jD

  4. 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=5yrn = 5 yr
  • Year=4Year = 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

  1. Annual depreciation

    D=(C−S)/nD = (C - S)/n

  2. Substituting — D = ($319,000 − $19,000)/5 = $60,000 per year

  3. Book value

    BVj=C−jDBV_j = C - jD

  4. 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=12yrn = 12 yr
  • Year=2Year = 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

  1. Annual depreciation

    D=(C−S)/nD = (C - S)/n

  2. Substituting — D = ($397,000 − $5,000)/12 = $32,667 per year

  3. Book value

    BVj=C−jDBV_j = C - jD

  4. 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

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