A culvert costs $85,000 now and $4,000 per year to maintain for 20 years. At i = 6%, what is the present worth of the total cost?

Start with the thinking

• Uniform series converts with (P/A, i, n).
• Costs are all negative — keep one sign convention.

Step-by-step solution

1. Series factor

   $$(P/A) = [(1 + i)^n - 1]/[i(1 + i)^n]$$

2. Growth term

   $$(1.06)^{20} = 3.2071$$

3. Factor

   $$(P/A) = (3.2071 - 1)/[0.06(3.2071)] = 2.2071/0.19243 = 11.47$$

4. Maintenance PW — 4,000(11.47) = $45,880

5. Total — PW = 85,000 + 45,880 = $130,880

Why the other options are there

• $165,000 (annual costs summed undiscounted)
• $45,900 (initial cost omitted)

Reference: FE Reference Handbook — Engineering Economics — Uniform series

$26,000 is invested at 9.5% compounded annually for 10 years. What is the future worth?

Start with the thinking

• Single sum moved forward — use (F/P, i, n).
• i and n must be in the same period.

Step-by-step solution

1. Factor

   $$F = P(1 + i)ⁿ$$

2. Substituting — F = $26,000(1 + 0.095)^10

3. Factor value

   $$(1 + i)ⁿ = 2.4782$$

4. Result — F = $64,434

Why the other options are there

• $50,700 (simple interest)
• $10,491 (factor inverted)

Reference: FE Reference Handbook — Engineering Economics → G�������������Uniform gradient amount per interest period

Maintenance is $13,000 in year 1 and grows by $7,000 each year for 10 years. At 7.5%, what is the 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

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

2. Evaluate

   $$(P/A) = 6.8641$$

3. Gradient part

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

4. Evaluate

   $$(P/G) = 26.8286$$

5. Combine — PW = $13,000(6.8641) + $7,000(26.8286) = $277,033

Why the other options are there

• $89,233 (gradient omitted)
• $569,719 (gradient treated as uniform)

Reference: FE Reference Handbook — Engineering Economics → G�������������Uniform gradient amount per interest period

$89,000 is invested at 8.0% compounded annually for 18 years. What is the future worth?

Start with the thinking

• Single sum moved forward — use (F/P, i, n).
• i and n must be in the same period.

Step-by-step solution

1. Factor

   $$F = P(1 + i)ⁿ$$

2. Substituting — F = $89,000(1 + 0.080)^18

3. Factor value

   $$(1 + i)ⁿ = 3.9960$$

4. Result — F = $355,646

Why the other options are there

• $217,160 (simple interest)
• $22,272 (factor inverted)

Reference: FE Reference Handbook — Engineering Economics → G�������������Uniform gradient amount per interest period

Maintenance is $18,000 in year 1 and grows by $5,000 each year for 14 years. At 7.0%, what is the 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

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

2. Evaluate

   $$(P/A) = 8.7455$$

3. Gradient part

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

4. Evaluate

   $$(P/G) = 47.3718$$

5. Combine — PW = $18,000(8.7455) + $5,000(47.3718) = $394,277

Why the other options are there

• $157,418 (gradient omitted)
• $769,601 (gradient treated as uniform)

Reference: FE Reference Handbook — Engineering Economics → G�������������Uniform gradient amount per interest period

$122,000 is invested at 8.0% compounded annually for 17 years. What is the future worth?

Start with the thinking

• Single sum moved forward — use (F/P, i, n).
• i and n must be in the same period.

Step-by-step solution

1. Factor

   $$F = P(1 + i)ⁿ$$

2. Substituting — F = $122,000(1 + 0.080)^17

3. Factor value

   $$(1 + i)ⁿ = 3.7000$$

4. Result — F = $451,402

Why the other options are there

• $287,920 (simple interest)
• $32,973 (factor inverted)

Reference: FE Reference Handbook — Engineering Economics → G�������������Uniform gradient amount per interest period

Maintenance is $14,000 in year 1 and grows by $2,000 each year for 10 years. At 6.5%, what is the 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

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

2. Evaluate

   $$(P/A) = 7.1888$$

3. Gradient part

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

4. Evaluate

   $$(P/G) = 28.6395$$

5. Combine — PW = $14,000(7.1888) + $2,000(28.6395) = $157,923

Why the other options are there

• $100,644 (gradient omitted)
• $244,420 (gradient treated as uniform)

Reference: FE Reference Handbook — Engineering Economics → G�������������Uniform gradient amount per interest period

$60,000 is invested at 7.0% compounded annually for 19 years. What is the future worth?

Start with the thinking

• Single sum moved forward — use (F/P, i, n).
• i and n must be in the same period.

Step-by-step solution

1. Factor

   $$F = P(1 + i)ⁿ$$

2. Substituting — F = $60,000(1 + 0.070)^19

3. Factor value

   $$(1 + i)ⁿ = 3.6165$$

4. Result — F = $216,992

Why the other options are there

• $139,800 (simple interest)
• $16,590 (factor inverted)

Reference: FE Reference Handbook — Engineering Economics → G�������������Uniform gradient amount per interest period

Maintenance is $28,000 in year 1 and grows by $5,000 each year for 6 years. At 8.5%, what is the 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

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

2. Evaluate

   $$(P/A) = 4.5536$$

3. Gradient part

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

4. Evaluate

   $$(P/G) = 10.3049$$

5. Combine — PW = $28,000(4.5536) + $5,000(10.3049) = $179,025

Why the other options are there

• $127,500 (gradient omitted)
• $264,108 (gradient treated as uniform)

Reference: FE Reference Handbook — Engineering Economics → G�������������Uniform gradient amount per interest period

$131,000 is invested at 10.5% compounded annually for 8 years. What is the future worth?

Start with the thinking

• Single sum moved forward — use (F/P, i, n).
• i and n must be in the same period.

Step-by-step solution

1. Factor

   $$F = P(1 + i)ⁿ$$

2. Substituting — F = $131,000(1 + 0.105)^8

3. Factor value

   $$(1 + i)ⁿ = 2.2228$$

4. Result — F = $291,185

Why the other options are there

• $241,040 (simple interest)
• $58,935 (factor inverted)

Reference: FE Reference Handbook — Engineering Economics → G�������������Uniform gradient amount per interest period