Overview

Engineering Economics · FE Reference Handbook section

Engineering Economics

0 formulas

10 exam-style examples

~45 min

All Engineering Economics lectures

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

Factor Name Converts Symbol Formula

Core formulas for this FE topic

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

This section is conceptual; there are no equations to memorise.

Worked exam-style examples

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

Example 1

Capitalized cost of a perpetual public works asset — Overview

A bridge deck costs $644,000 to build and $32,000 per year to maintain forever. At 5.5% interest, determine the capitalized cost.

Given

First cost = $644,000
A = $32,000/yr
$i = 5.5%$

Find

Capitalized cost of the asset

Start with the thinking

Capitalized cost is the present worth of a cash flow that continues indefinitely.
For a perpetuity P = A/i, so a lower interest rate raises the capitalized cost sharply.

Step-by-step solution

Formula
$CC = First cost + A/i$
Perpetuity term — A/i = $32,000/0.055 = $581,818
Substituting — CC = $644,000 + $581,818
Evaluate — CC = $1,225,818

Answer:

Capitalized cost ≈ $1,225,818

Why the other options are there

$645,760 (multiplied instead of divided)
$581,818 (omitted the first cost)

Reference: FE Reference Handbook — Engineering Economics → Overview

Example 2

Capitalized cost of a perpetual public works asset — Overview (2)

A bridge deck costs $505,000 to build and $9,000 per year to maintain forever. At 6.0% interest, determine the capitalized cost.

Given

First cost = $505,000
A = $9,000/yr
$i = 6.0%$

Find

Capitalized cost of the asset

Start with the thinking

Capitalized cost is the present worth of a cash flow that continues indefinitely.
For a perpetuity P = A/i, so a lower interest rate raises the capitalized cost sharply.

Step-by-step solution

Formula
$CC = First cost + A/i$
Perpetuity term — A/i = $9,000/0.060 = $150,000
Substituting — CC = $505,000 + $150,000
Evaluate — CC = $655,000

Answer:

Capitalized cost ≈ $655,000

Why the other options are there

$505,540 (multiplied instead of divided)
$150,000 (omitted the first cost)

Reference: FE Reference Handbook — Engineering Economics → Overview

Example 3

Capitalized cost of a perpetual public works asset — Overview (3)

A bridge deck costs $624,000 to build and $33,000 per year to maintain forever. At 7.0% interest, determine the capitalized cost.

Given

First cost = $624,000
A = $33,000/yr
$i = 7.0%$

Find

Capitalized cost of the asset

Start with the thinking

Capitalized cost is the present worth of a cash flow that continues indefinitely.
For a perpetuity P = A/i, so a lower interest rate raises the capitalized cost sharply.

Step-by-step solution

Formula
$CC = First cost + A/i$
Perpetuity term — A/i = $33,000/0.070 = $471,429
Substituting — CC = $624,000 + $471,429
Evaluate — CC = $1,095,429

Answer:

Capitalized cost ≈ $1,095,429

Why the other options are there

$626,310 (multiplied instead of divided)
$471,429 (omitted the first cost)

Reference: FE Reference Handbook — Engineering Economics → Overview

Example 4

Capitalized cost of a perpetual public works asset — Overview (4)

A bridge deck costs $837,000 to build and $32,000 per year to maintain forever. At 7.0% interest, determine the capitalized cost.

Given

First cost = $837,000
A = $32,000/yr
$i = 7.0%$

Find

Capitalized cost of the asset

Start with the thinking

Capitalized cost is the present worth of a cash flow that continues indefinitely.
For a perpetuity P = A/i, so a lower interest rate raises the capitalized cost sharply.

Step-by-step solution

Formula
$CC = First cost + A/i$
Perpetuity term — A/i = $32,000/0.070 = $457,143
Substituting — CC = $837,000 + $457,143
Evaluate — CC = $1,294,143

Answer:

Capitalized cost ≈ $1,294,143

Why the other options are there

$839,240 (multiplied instead of divided)
$457,143 (omitted the first cost)

Reference: FE Reference Handbook — Engineering Economics → Overview

Example 5

Capitalized cost of a perpetual public works asset — Overview (5)

A bridge deck costs $566,000 to build and $44,000 per year to maintain forever. At 4.0% interest, determine the capitalized cost.

Given

First cost = $566,000
A = $44,000/yr
$i = 4.0%$

Find

Capitalized cost of the asset

Start with the thinking

Capitalized cost is the present worth of a cash flow that continues indefinitely.
For a perpetuity P = A/i, so a lower interest rate raises the capitalized cost sharply.

Step-by-step solution

Formula
$CC = First cost + A/i$
Perpetuity term — A/i = $44,000/0.040 = $1,100,000
Substituting — CC = $566,000 + $1,100,000
Evaluate — CC = $1,666,000

Answer:

Capitalized cost ≈ $1,666,000

Why the other options are there

$567,760 (multiplied instead of divided)
$1,100,000 (omitted the first cost)

Reference: FE Reference Handbook — Engineering Economics → Overview

Example 6

Capitalized cost of a perpetual public works asset — Overview (6)

A bridge deck costs $681,000 to build and $8,000 per year to maintain forever. At 5.5% interest, determine the capitalized cost.

Given

First cost = $681,000
A = $8,000/yr
$i = 5.5%$

Find

Capitalized cost of the asset

Start with the thinking

Capitalized cost is the present worth of a cash flow that continues indefinitely.
For a perpetuity P = A/i, so a lower interest rate raises the capitalized cost sharply.

Step-by-step solution

Formula
$CC = First cost + A/i$
Perpetuity term — A/i = $8,000/0.055 = $145,455
Substituting — CC = $681,000 + $145,455
Evaluate — CC = $826,455

Answer:

Capitalized cost ≈ $826,455

Why the other options are there

$681,440 (multiplied instead of divided)
$145,455 (omitted the first cost)

Reference: FE Reference Handbook — Engineering Economics → Overview

Example 7

Capitalized cost of a perpetual public works asset — Overview (7)

A bridge deck costs $477,000 to build and $20,000 per year to maintain forever. At 5.5% interest, determine the capitalized cost.

Given

First cost = $477,000
A = $20,000/yr
$i = 5.5%$

Find

Capitalized cost of the asset

Start with the thinking

Capitalized cost is the present worth of a cash flow that continues indefinitely.
For a perpetuity P = A/i, so a lower interest rate raises the capitalized cost sharply.

Step-by-step solution

Formula
$CC = First cost + A/i$
Perpetuity term — A/i = $20,000/0.055 = $363,636
Substituting — CC = $477,000 + $363,636
Evaluate — CC = $840,636

Answer:

Capitalized cost ≈ $840,636

Why the other options are there

$478,100 (multiplied instead of divided)
$363,636 (omitted the first cost)

Reference: FE Reference Handbook — Engineering Economics → Overview

Example 8

Capitalized cost of a perpetual public works asset — Overview (8)

A bridge deck costs $610,000 to build and $25,000 per year to maintain forever. At 6.0% interest, determine the capitalized cost.

Given

First cost = $610,000
A = $25,000/yr
$i = 6.0%$

Find

Capitalized cost of the asset

Start with the thinking

Capitalized cost is the present worth of a cash flow that continues indefinitely.
For a perpetuity P = A/i, so a lower interest rate raises the capitalized cost sharply.

Step-by-step solution

Formula
$CC = First cost + A/i$
Perpetuity term — A/i = $25,000/0.060 = $416,667
Substituting — CC = $610,000 + $416,667
Evaluate — CC = $1,026,667

Answer:

Capitalized cost ≈ $1,026,667

Why the other options are there

$611,500 (multiplied instead of divided)
$416,667 (omitted the first cost)

Reference: FE Reference Handbook — Engineering Economics → Overview

Example 9

Capitalized cost of a perpetual public works asset — Overview (9)

A bridge deck costs $357,000 to build and $39,000 per year to maintain forever. At 9.0% interest, determine the capitalized cost.

Given

First cost = $357,000
A = $39,000/yr
$i = 9.0%$

Find

Capitalized cost of the asset

Start with the thinking

Capitalized cost is the present worth of a cash flow that continues indefinitely.
For a perpetuity P = A/i, so a lower interest rate raises the capitalized cost sharply.

Step-by-step solution

Formula
$CC = First cost + A/i$
Perpetuity term — A/i = $39,000/0.090 = $433,333
Substituting — CC = $357,000 + $433,333
Evaluate — CC = $790,333

Answer:

Capitalized cost ≈ $790,333

Why the other options are there

$360,510 (multiplied instead of divided)
$433,333 (omitted the first cost)

Reference: FE Reference Handbook — Engineering Economics → Overview

Example 10

Capitalized cost of a perpetual public works asset — Overview (10)

A bridge deck costs $731,000 to build and $24,000 per year to maintain forever. At 9.5% interest, determine the capitalized cost.

Given

First cost = $731,000
A = $24,000/yr
$i = 9.5%$

Find

Capitalized cost of the asset

Start with the thinking

Capitalized cost is the present worth of a cash flow that continues indefinitely.
For a perpetuity P = A/i, so a lower interest rate raises the capitalized cost sharply.

Step-by-step solution

Formula
$CC = First cost + A/i$
Perpetuity term — A/i = $24,000/0.095 = $252,632
Substituting — CC = $731,000 + $252,632
Evaluate — CC = $983,632

Answer:

Capitalized cost ≈ $983,632

Why the other options are there

$733,280 (multiplied instead of divided)
$252,632 (omitted the first cost)

Reference: FE Reference Handbook — Engineering Economics → Overview