Break-Even, Benefit-Cost and Life-Cycle Analysis

NCEES FE Civil spec: Engineering Economics, C. Analyses (e.g., break-even, benefit-cost, life cycle, sustainability, renewable energy). FE Reference Handbook: Engineering Economics chapter, break-even and benefit-cost analysis.

Section C pulls the time value of money into decision tools: break-even analysis, benefit-cost ratios for public projects, and life-cycle cost comparisons. Civil engineers use these constantly, which is why public-works style questions (roads, pump stations, bridges) are common on the FE Civil exam.

What the NCEES spec covers

The spec lists break-even, benefit-cost, life cycle, sustainability and renewable energy analyses. Sustainability and renewable energy questions usually reduce to the same math: a higher first cost traded for lower operating costs, evaluated by PW, AW, payback or B/C.

Key concepts

Break-even is the quantity (or other variable) at which two options cost the same, or revenue equals cost.
Benefit-cost ratio compares the equivalent worth of benefits to users with the equivalent worth of costs to the owner. B/C ≥ 1 means the project is justified at that interest rate.
Conventional vs modified B/C. Conventional B/C puts O&M in the denominator with first cost. Modified B/C subtracts O&M from benefits. They give different ratios but always agree on whether B/C is above or below 1.
Incremental analysis. For mutually exclusive projects, compare the increment between the cheaper and more expensive option. The highest B/C ratio is not automatically the best choice.
Life-cycle cost includes first cost, operation, maintenance, rehabilitation and salvage over the whole service life, usually compared as PW or AW.
Simple payback is first cost divided by annual savings. It ignores interest and cash flows after payback.

Core formulas

Q_{B E} = \frac{F C}{p - v} (price p, variable cost v per unit)

Two options: F C_{1} + v_{1} Q = F C_{2} + v_{2} Q \Rightarrow Q = \frac{F C _{2} - F C _{1}}{v _{1} - v _{2}}

(\frac{B}{C})_{conv} = \frac{P W _{benefits}}{P W _{first cost} + P W _{O & M}} (\frac{B}{C})_{mod} = \frac{P W _{benefits} - P W _{O & M}}{P W _{first cost}}

Where it is in the FE Reference Handbook

The Engineering Economics chapter defines break-even analysis and benefit-cost analysis in its nomenclature section, and the interest factor table and interest tables supply the factors you need to convert annual amounts.

Worked example

A highway improvement costs $1,500,000 now. It provides user benefits of $220,000 per year and needs $40,000 per year of maintenance for 25 years. At 6%, find the conventional and modified B/C ratios.

(P / A, 6%, 25) = \frac{1.0 6 ^{25} - 1}{0.06 ( 1.06 ) ^{25}} = 12.7834

P W_{B} = 220, 000 (12.7834) = $2, 812, 300 P W_{O & M} = 40, 000 (12.7834) = $511, 300

(\frac{B}{C})_{conv} = \frac{2 , 812 , 300}{1 , 500 , 000 + 511 , 300} = 1.40

(\frac{B}{C})_{mod} = \frac{2 , 812 , 300 - 511 , 300}{1 , 500 , 000} = 1.53

Both ratios exceed 1, so the project is justified. The answer choices on the exam often include both values, so read whether O&M is meant to be a cost or a reduction in benefits.

Common mistakes

Picking the alternative with the highest B/C instead of doing incremental analysis.
Counting a disbenefit to users (such as construction delays) as a cost to the owner, or the reverse.
Mixing PW for benefits and AW for costs in one ratio.
Using contribution margin incorrectly in break-even: divide fixed cost by price minus variable cost, not by price.
Treating simple payback as a measure of profitability.

Try the break-even analysis example for the other half of this section.

Practice Problems

Try these from the free PolyCalc FE Civil problem bank. Each one has a full worked solution.

Fixed cost is $30,000, selling price is $80 per unit, and variable cost is $50 per unit. What is the break-even quantity? See the worked solution or practice the Engineering Economics deck
Machine A has fixed costs of $40,000 per year and variable costs of $12 per unit. Machine B has fixed costs of $25,000 per year and $18 per unit. The annual output at which both cost the same is most nearly: See the worked solution or practice the Engineering Economics deck
A road project costs $1,200,000 now. It produces user benefits of $150,000 per year and needs $30,000 per year of maintenance, both for 20 years. At 5%, the conventional benefit-cost ratio is most nearly: See the worked solution or practice the Engineering Economics deck
A loader costs $200,000, lasts 20 years and has a $20,000 salvage value. At 7% interest, its equivalent annual cost of ownership is most nearly: See the worked solution or practice the Engineering Economics deck

Related Calculators and Examples

Frequently Asked Questions

What is the difference between conventional and modified benefit-cost ratio?

The conventional ratio divides benefits by first cost plus operating and maintenance cost. The modified ratio subtracts operating and maintenance cost from benefits and divides by first cost. The values differ, but both are greater than 1 for the same projects.

How do you find a break-even quantity?

Set the two cost (or revenue and cost) expressions equal and solve for the quantity. For revenue versus cost, the break-even quantity is fixed cost divided by price minus variable cost per unit.

Why use incremental analysis for mutually exclusive projects?

A higher-cost alternative may have a lower overall ratio but still be worth its extra cost. Comparing the increment checks whether the additional benefits justify the additional cost.

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