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Inflation

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.

  • To account for inflation, the dollars are deflated by the general inflation rate per interest period f, and then they are shifted
  • over the time scale using the interest rate per interest period i. Use an inflation adjusted interest rate per interest period d for
  • computing present worth values P.

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
Present worth of an arithmetic gradient — Inflation

Maintenance is $60,000 in year 1 and grows by $2,000 each year for 15 years. At 8.5%, what is the present worth?

Given

  • A₁ = $60,000

  • G = $2,000/yr

  • n=15n = 15
  • i=8.5i = 8.5%

Find

Total present worth

Start with the thinking

  • Split the cash flow into a uniform base plus a gradient.
  • The gradient factor starts in year 2 by definition.

Step-by-step solution

  1. Uniform part

    (P/A,i,n)=[(1+i)n−1]/[i(1+i)n](P/A, i, n) = [(1 + i)ⁿ - 1]/[i(1 + i)ⁿ]

  2. Evaluate

    (P/A)=8.3042(P/A) = 8.3042

  3. Gradient part

    (P/G,i,n)=(1/i)[(P/A)−n/(1+i)n](P/G, i, n) = (1/i)[(P/A) - n/(1 + i)ⁿ]

  4. Evaluate

    (P/G)=45.7899(P/G) = 45.7899

  5. Combine — PW = $60,000(8.3042) + $2,000(45.7899) = $589,834

Answer:

PW ≈ $589,834

Why the other options are there

  • $498,254 (gradient omitted)
  • $747,381 (gradient treated as uniform)

Reference: FE Reference Handbook — Engineering Economics → Inflation

Example 2
Present worth of an arithmetic gradient — Inflation (2)

Maintenance is $57,000 in year 1 and grows by $6,000 each year for 9 years. At 7.0%, what is the present worth?

Given

  • A₁ = $57,000

  • G = $6,000/yr

  • n=9n = 9
  • i=7.0i = 7.0%

Find

Total present worth

Start with the thinking

  • Split the cash flow into a uniform base plus a gradient.
  • The gradient factor starts in year 2 by definition.

Step-by-step solution

  1. Uniform part

    (P/A,i,n)=[(1+i)n−1]/[i(1+i)n](P/A, i, n) = [(1 + i)ⁿ - 1]/[i(1 + i)ⁿ]

  2. Evaluate

    (P/A)=6.5152(P/A) = 6.5152

  3. Gradient part

    (P/G,i,n)=(1/i)[(P/A)−n/(1+i)n](P/G, i, n) = (1/i)[(P/A) - n/(1 + i)ⁿ]

  4. Evaluate

    (P/G)=23.1404(P/G) = 23.1404

  5. Combine — PW = $57,000(6.5152) + $6,000(23.1404) = $510,211

Answer:

PW ≈ $510,211

Why the other options are there

  • $371,368 (gradient omitted)
  • $723,191 (gradient treated as uniform)

Reference: FE Reference Handbook — Engineering Economics → Inflation

Example 3
Present worth of an arithmetic gradient — Inflation (3)

Maintenance is $56,000 in year 1 and grows by $3,000 each year for 13 years. At 9.0%, what is the present worth?

Given

  • A₁ = $56,000

  • G = $3,000/yr

  • n=13n = 13
  • i=9.0i = 9.0%

Find

Total present worth

Start with the thinking

  • Split the cash flow into a uniform base plus a gradient.
  • The gradient factor starts in year 2 by definition.

Step-by-step solution

  1. Uniform part

    (P/A,i,n)=[(1+i)n−1]/[i(1+i)n](P/A, i, n) = [(1 + i)ⁿ - 1]/[i(1 + i)ⁿ]

  2. Evaluate

    (P/A)=7.4869(P/A) = 7.4869

  3. Gradient part

    (P/G,i,n)=(1/i)[(P/A)−n/(1+i)n](P/G, i, n) = (1/i)[(P/A) - n/(1 + i)ⁿ]

  4. Evaluate

    (P/G)=36.0731(P/G) = 36.0731

  5. Combine — PW = $56,000(7.4869) + $3,000(36.0731) = $527,486

Answer:

PW ≈ $527,486

Why the other options are there

  • $419,267 (gradient omitted)
  • $711,256 (gradient treated as uniform)

Reference: FE Reference Handbook — Engineering Economics → Inflation

Example 4
Present worth of an arithmetic gradient — Inflation (4)

Maintenance is $11,000 in year 1 and grows by $7,000 each year for 11 years. At 6.5%, what is the present worth?

Given

  • A₁ = $11,000

  • G = $7,000/yr

  • n=11n = 11
  • i=6.5i = 6.5%

Find

Total present worth

Start with the thinking

  • Split the cash flow into a uniform base plus a gradient.
  • The gradient factor starts in year 2 by definition.

Step-by-step solution

  1. Uniform part

    (P/A,i,n)=[(1+i)n−1]/[i(1+i)n](P/A, i, n) = [(1 + i)ⁿ - 1]/[i(1 + i)ⁿ]

  2. Evaluate

    (P/A)=7.6890(P/A) = 7.6890

  3. Gradient part

    (P/G,i,n)=(1/i)[(P/A)−n/(1+i)n](P/G, i, n) = (1/i)[(P/A) - n/(1 + i)ⁿ]

  4. Evaluate

    (P/G)=33.6417(P/G) = 33.6417

  5. Combine — PW = $11,000(7.6890) + $7,000(33.6417) = $320,071

Answer:

PW ≈ $320,071

Why the other options are there

  • $84,579 (gradient omitted)
  • $676,636 (gradient treated as uniform)

Reference: FE Reference Handbook — Engineering Economics → Inflation

Example 5
Present worth of an arithmetic gradient — Inflation (5)

Maintenance is $13,000 in year 1 and grows by $4,000 each year for 8 years. At 7.0%, what is the present worth?

Given

  • A₁ = $13,000

  • G = $4,000/yr

  • n=8n = 8
  • i=7.0i = 7.0%

Find

Total present worth

Start with the thinking

  • Split the cash flow into a uniform base plus a gradient.
  • The gradient factor starts in year 2 by definition.

Step-by-step solution

  1. Uniform part

    (P/A,i,n)=[(1+i)n−1]/[i(1+i)n](P/A, i, n) = [(1 + i)ⁿ - 1]/[i(1 + i)ⁿ]

  2. Evaluate

    (P/A)=5.9713(P/A) = 5.9713

  3. Gradient part

    (P/G,i,n)=(1/i)[(P/A)−n/(1+i)n](P/G, i, n) = (1/i)[(P/A) - n/(1 + i)ⁿ]

  4. Evaluate

    (P/G)=18.7889(P/G) = 18.7889

  5. Combine — PW = $13,000(5.9713) + $4,000(18.7889) = $152,783

Answer:

PW ≈ $152,783

Why the other options are there

  • $77,627 (gradient omitted)
  • $268,708 (gradient treated as uniform)

Reference: FE Reference Handbook — Engineering Economics → Inflation

Example 6
Present worth of an arithmetic gradient — Inflation (6)

Maintenance is $20,000 in year 1 and grows by $8,000 each year for 7 years. At 5.5%, what is the present worth?

Given

  • A₁ = $20,000

  • G = $8,000/yr

  • n=7n = 7
  • i=5.5i = 5.5%

Find

Total present worth

Start with the thinking

  • Split the cash flow into a uniform base plus a gradient.
  • The gradient factor starts in year 2 by definition.

Step-by-step solution

  1. Uniform part

    (P/A,i,n)=[(1+i)n−1]/[i(1+i)n](P/A, i, n) = [(1 + i)ⁿ - 1]/[i(1 + i)ⁿ]

  2. Evaluate

    (P/A)=5.6830(P/A) = 5.6830

  3. Gradient part

    (P/G,i,n)=(1/i)[(P/A)−n/(1+i)n](P/G, i, n) = (1/i)[(P/A) - n/(1 + i)ⁿ]

  4. Evaluate

    (P/G)=15.8347(P/G) = 15.8347

  5. Combine — PW = $20,000(5.6830) + $8,000(15.8347) = $240,337

Answer:

PW ≈ $240,337

Why the other options are there

  • $113,659 (gradient omitted)
  • $431,906 (gradient treated as uniform)

Reference: FE Reference Handbook — Engineering Economics → Inflation

Example 7
Present worth of an arithmetic gradient — Inflation (7)

Maintenance is $21,000 in year 1 and grows by $2,000 each year for 12 years. At 5.5%, what is the present worth?

Given

  • A₁ = $21,000

  • G = $2,000/yr

  • n=12n = 12
  • i=5.5i = 5.5%

Find

Total present worth

Start with the thinking

  • Split the cash flow into a uniform base plus a gradient.
  • The gradient factor starts in year 2 by definition.

Step-by-step solution

  1. Uniform part

    (P/A,i,n)=[(1+i)n−1]/[i(1+i)n](P/A, i, n) = [(1 + i)ⁿ - 1]/[i(1 + i)ⁿ]

  2. Evaluate

    (P/A)=8.6185(P/A) = 8.6185

  3. Gradient part

    (P/G,i,n)=(1/i)[(P/A)−n/(1+i)n](P/G, i, n) = (1/i)[(P/A) - n/(1 + i)ⁿ]

  4. Evaluate

    (P/G)=41.9407(P/G) = 41.9407

  5. Combine — PW = $21,000(8.6185) + $2,000(41.9407) = $264,870

Answer:

PW ≈ $264,870

Why the other options are there

  • $180,989 (gradient omitted)
  • $387,833 (gradient treated as uniform)

Reference: FE Reference Handbook — Engineering Economics → Inflation

Example 8
Present worth of an arithmetic gradient — Inflation (8)

Maintenance is $14,000 in year 1 and grows by $8,000 each year for 8 years. At 7.0%, what is the present worth?

Given

  • A₁ = $14,000

  • G = $8,000/yr

  • n=8n = 8
  • i=7.0i = 7.0%

Find

Total present worth

Start with the thinking

  • Split the cash flow into a uniform base plus a gradient.
  • The gradient factor starts in year 2 by definition.

Step-by-step solution

  1. Uniform part

    (P/A,i,n)=[(1+i)n−1]/[i(1+i)n](P/A, i, n) = [(1 + i)ⁿ - 1]/[i(1 + i)ⁿ]

  2. Evaluate

    (P/A)=5.9713(P/A) = 5.9713

  3. Gradient part

    (P/G,i,n)=(1/i)[(P/A)−n/(1+i)n](P/G, i, n) = (1/i)[(P/A) - n/(1 + i)ⁿ]

  4. Evaluate

    (P/G)=18.7889(P/G) = 18.7889

  5. Combine — PW = $14,000(5.9713) + $8,000(18.7889) = $233,910

Answer:

PW ≈ $233,910

Why the other options are there

  • $83,598 (gradient omitted)
  • $465,761 (gradient treated as uniform)

Reference: FE Reference Handbook — Engineering Economics → Inflation

Example 9
Present worth of an arithmetic gradient — Inflation (9)

Maintenance is $45,000 in year 1 and grows by $1,000 each year for 13 years. At 5.5%, what is the present worth?

Given

  • A₁ = $45,000

  • G = $1,000/yr

  • n=13n = 13
  • i=5.5i = 5.5%

Find

Total present worth

Start with the thinking

  • Split the cash flow into a uniform base plus a gradient.
  • The gradient factor starts in year 2 by definition.

Step-by-step solution

  1. Uniform part

    (P/A,i,n)=[(1+i)n−1]/[i(1+i)n](P/A, i, n) = [(1 + i)ⁿ - 1]/[i(1 + i)ⁿ]

  2. Evaluate

    (P/A)=9.1171(P/A) = 9.1171

  3. Gradient part

    (P/G,i,n)=(1/i)[(P/A)−n/(1+i)n](P/G, i, n) = (1/i)[(P/A) - n/(1 + i)ⁿ]

  4. Evaluate

    (P/G)=47.9234(P/G) = 47.9234

  5. Combine — PW = $45,000(9.1171) + $1,000(47.9234) = $458,192

Answer:

PW ≈ $458,192

Why the other options are there

  • $410,269 (gradient omitted)
  • $528,791 (gradient treated as uniform)

Reference: FE Reference Handbook — Engineering Economics → Inflation

Example 10
Present worth of an arithmetic gradient — Inflation (10)

Maintenance is $39,000 in year 1 and grows by $1,000 each year for 9 years. At 9.5%, what is the present worth?

Given

  • A₁ = $39,000

  • G = $1,000/yr

  • n=9n = 9
  • i=9.5i = 9.5%

Find

Total present worth

Start with the thinking

  • Split the cash flow into a uniform base plus a gradient.
  • The gradient factor starts in year 2 by definition.

Step-by-step solution

  1. Uniform part

    (P/A,i,n)=[(1+i)n−1]/[i(1+i)n](P/A, i, n) = [(1 + i)ⁿ - 1]/[i(1 + i)ⁿ]

  2. Evaluate

    (P/A)=5.8753(P/A) = 5.8753

  3. Gradient part

    (P/G,i,n)=(1/i)[(P/A)−n/(1+i)n](P/G, i, n) = (1/i)[(P/A) - n/(1 + i)ⁿ]

  4. Evaluate

    (P/G)=19.9858(P/G) = 19.9858

  5. Combine — PW = $39,000(5.8753) + $1,000(19.9858) = $249,122

Answer:

PW ≈ $249,122

Why the other options are there

  • $229,136 (gradient omitted)
  • $282,014 (gradient treated as uniform)

Reference: FE Reference Handbook — Engineering Economics → Inflation

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