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

A widening saves $210,000 per year in delay, costs $1.6 M to build (20-yr life, i = 7%, no salvage) and $30,000 per year to maintain. Compute B/C.

Start with the thinking

• Convert everything to the same time basis — annual is easiest here.
• Maintenance belongs in the denominator, not as a negative benefit.

Step-by-step solution

1. Capital recovery

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

2. Growth term

   $$(1.07)^{20} = 3.8697$$

3. Factor

   $$(A/P) = 0.07(3.8697)/2.8697 = 0.09439$$

4. Annual cost — 1,600,000(0.09439) + 30,000 = 151,024 + 30,000 = $181,024

5. Ratio

   $$B/C = 210,000/181,024 = 1.16$$

Why the other options are there

• 1.31 (O&M left out)
• 0.86 (ratio inverted)

Reference: FE Reference Handbook — Engineering Economics — Benefit-cost analysis

A public improvement costs $784,000 with $17,000 annual maintenance and yields $84,000 of annual benefits over 35 years at 4.5%. What is B/C?

Start with the thinking

• Convert the first cost to an equivalent uniform annual cost.
• Maintenance belongs in the denominator, not netted from benefits.

Step-by-step solution

1. Capital recovery

   $$CRF = i(1 + i)ⁿ/[(1 + i)ⁿ - 1]$$

2. Evaluate

   $$CRF = 0.05727$$

3. Annual cost of capital — EUAC = $784,000 × 0.05727 = $44,900/yr

4. Ratio — B/C = $84,000/($44,900 + $17,000) = 1.357

5. Decision — B/C ≥ 1, project justified

Why the other options are there

• 4.94 (capital cost omitted)
• 0.107 (first cost not annualised)

Reference: FE Reference Handbook — Engineering Economics → Benefit-Cost Analysis

A public improvement costs $569,000 with $18,000 annual maintenance and yields $148,000 of annual benefits over 22 years at 5.0%. What is B/C?

Start with the thinking

• Convert the first cost to an equivalent uniform annual cost.
• Maintenance belongs in the denominator, not netted from benefits.

Step-by-step solution

1. Capital recovery

   $$CRF = i(1 + i)ⁿ/[(1 + i)ⁿ - 1]$$

2. Evaluate

   $$CRF = 0.07597$$

3. Annual cost of capital — EUAC = $569,000 × 0.07597 = $43,227/yr

4. Ratio — B/C = $148,000/($43,227 + $18,000) = 2.417

5. Decision — B/C ≥ 1, project justified

Why the other options are there

• 8.22 (capital cost omitted)
• 0.260 (first cost not annualised)

Reference: FE Reference Handbook — Engineering Economics → Benefit-Cost Analysis

A public improvement costs $1,164,000 with $24,000 annual maintenance and yields $123,000 of annual benefits over 22 years at 5.0%. What is B/C?

Start with the thinking

• Convert the first cost to an equivalent uniform annual cost.
• Maintenance belongs in the denominator, not netted from benefits.

Step-by-step solution

1. Capital recovery

   $$CRF = i(1 + i)ⁿ/[(1 + i)ⁿ - 1]$$

2. Evaluate

   $$CRF = 0.07597$$

3. Annual cost of capital — EUAC = $1,164,000 × 0.07597 = $88,430/yr

4. Ratio — B/C = $123,000/($88,430 + $24,000) = 1.094

5. Decision — B/C ≥ 1, project justified

Why the other options are there

• 5.13 (capital cost omitted)
• 0.106 (first cost not annualised)

Reference: FE Reference Handbook — Engineering Economics → Benefit-Cost Analysis

A public improvement costs $693,000 with $36,000 annual maintenance and yields $133,000 of annual benefits over 38 years at 4.0%. What is B/C?

Start with the thinking

• Convert the first cost to an equivalent uniform annual cost.
• Maintenance belongs in the denominator, not netted from benefits.

Step-by-step solution

1. Capital recovery

   $$CRF = i(1 + i)ⁿ/[(1 + i)ⁿ - 1]$$

2. Evaluate

   $$CRF = 0.05163$$

3. Annual cost of capital — EUAC = $693,000 × 0.05163 = $35,781/yr

4. Ratio — B/C = $133,000/($35,781 + $36,000) = 1.853

5. Decision — B/C ≥ 1, project justified

Why the other options are there

• 3.69 (capital cost omitted)
• 0.192 (first cost not annualised)

Reference: FE Reference Handbook — Engineering Economics → Benefit-Cost Analysis

A public improvement costs $999,000 with $28,000 annual maintenance and yields $137,000 of annual benefits over 36 years at 6.0%. What is B/C?

Start with the thinking

• Convert the first cost to an equivalent uniform annual cost.
• Maintenance belongs in the denominator, not netted from benefits.

Step-by-step solution

1. Capital recovery

   $$CRF = i(1 + i)ⁿ/[(1 + i)ⁿ - 1]$$

2. Evaluate

   $$CRF = 0.06839$$

3. Annual cost of capital — EUAC = $999,000 × 0.06839 = $68,326/yr

4. Ratio — B/C = $137,000/($68,326 + $28,000) = 1.422

5. Decision — B/C ≥ 1, project justified

Why the other options are there

• 4.89 (capital cost omitted)
• 0.137 (first cost not annualised)

Reference: FE Reference Handbook — Engineering Economics → Benefit-Cost Analysis

A public improvement costs $1,149,000 with $30,000 annual maintenance and yields $180,000 of annual benefits over 31 years at 7.5%. What is B/C?

Start with the thinking

• Convert the first cost to an equivalent uniform annual cost.
• Maintenance belongs in the denominator, not netted from benefits.

Step-by-step solution

1. Capital recovery

   $$CRF = i(1 + i)ⁿ/[(1 + i)ⁿ - 1]$$

2. Evaluate

   $$CRF = 0.08392$$

3. Annual cost of capital — EUAC = $1,149,000 × 0.08392 = $96,420/yr

4. Ratio — B/C = $180,000/($96,420 + $30,000) = 1.424

5. Decision — B/C ≥ 1, project justified

Why the other options are there

• 6.00 (capital cost omitted)
• 0.157 (first cost not annualised)

Reference: FE Reference Handbook — Engineering Economics → Benefit-Cost Analysis

A public improvement costs $605,000 with $34,000 annual maintenance and yields $73,000 of annual benefits over 34 years at 4.5%. What is B/C?

Start with the thinking

• Convert the first cost to an equivalent uniform annual cost.
• Maintenance belongs in the denominator, not netted from benefits.

Step-by-step solution

1. Capital recovery

   $$CRF = i(1 + i)ⁿ/[(1 + i)ⁿ - 1]$$

2. Evaluate

   $$CRF = 0.05798$$

3. Annual cost of capital — EUAC = $605,000 × 0.05798 = $35,079/yr

4. Ratio — B/C = $73,000/($35,079 + $34,000) = 1.057

5. Decision — B/C ≥ 1, project justified

Why the other options are there

• 2.15 (capital cost omitted)
• 0.121 (first cost not annualised)

Reference: FE Reference Handbook — Engineering Economics → Benefit-Cost Analysis

A public improvement costs $872,000 with $17,000 annual maintenance and yields $120,000 of annual benefits over 27 years at 5.0%. What is B/C?

Start with the thinking

• Convert the first cost to an equivalent uniform annual cost.
• Maintenance belongs in the denominator, not netted from benefits.

Step-by-step solution

1. Capital recovery

   $$CRF = i(1 + i)ⁿ/[(1 + i)ⁿ - 1]$$

2. Evaluate

   $$CRF = 0.06829$$

3. Annual cost of capital — EUAC = $872,000 × 0.06829 = $59,551/yr

4. Ratio — B/C = $120,000/($59,551 + $17,000) = 1.568

5. Decision — B/C ≥ 1, project justified

Why the other options are there

• 7.06 (capital cost omitted)
• 0.138 (first cost not annualised)

Reference: FE Reference Handbook — Engineering Economics → Benefit-Cost Analysis