Lecture 10: Economic Life, Replacement, and Public Sector Analysis
1. Economic Life and Replacement Analysis
Previous lectures focused on evaluating new alternatives. This lecture addresses economic life—determining the optimal usage duration of an asset—and replacement analysis—deciding when to replace an existing asset with a new one.
Economic Life of a New Asset (Challenger)
The economic life of a new asset (the challenger) is defined as the number of years that minimizes the Equivalent Uniform Annual Cost (EUAC) of ownership. This life is often shorter than the asset’s physical or useful life.
The concept relies on balancing two opposing cost dynamics:
- Capital Costs: If an asset is kept for only a few years, the initial cost is spread over a short time, resulting in a high annual capital recovery cost. This component decreases as the asset life increases.
- Operating and Maintenance (O&M) Costs: As an asset ages, maintenance, repair, and the risk of unplanned breakdowns typically increase. This component causes the total EUAC to rise if the asset is kept for too long.
The minimum cost life is achieved at the period where the total EUAC curve bottoms out, balancing these two factors.
Replacement Analysis: Defender versus Challenger
Replacement analysis addresses whether existing equipment (the defender) should be retained or replaced now by the best available new equipment (the challenger).
A key concept is the marginal cost of the defender, which represents the year-to-year cost of keeping the existing asset for one more year. This marginal cost includes lost market value, O&M expenses, and risk costs for that specific year.
The recommended replacement rule, used when the defender’s marginal costs are increasing and service is indefinitely required, is:
Maintain the defender as long as the marginal cost of ownership for one more year is less than the challenger’s minimum EUAC.
Once the defender’s marginal cost exceeds the challenger’s minimum EUAC, it is economically beneficial to replace the defender immediately.
Solved Example 1: Calculating Minimum Cost Life
Problem: Determine the economic life (minimum EUAC) for a machine purchased for \(\$15,000\), if the interest rate is \(10\%\). Salvage values (\(S\)) and annual operating and maintenance costs (O&M) are estimated as follows:
| Year (\(n\)) | \(S_n\) | O&M Cost |
|---|---|---|
| 1 | \(\$10,000\) | \(\$2,000\) |
| 2 | \(\$6,000\) | \(\$3,000\) |
| 3 | \(\$3,000\) | \(\$4,500\) |
| 4 | \(\$1,000\) | \(\$6,500\) |
Solution:
We must calculate the EUAC for each possible life (\(n=1\) to \(4\)) using the Capital Recovery formula, \(\text{EUAC} = P(A/P, i, n) - S_n(A/F, i, n) + \text{EUAC}_{\text{O\&M}}\). Since O&M costs are irregular, we convert them to an Equivalent Uniform Annual Cost (\(\text{EUAC}_{\text{O\&M}}\)) over the specific life \(n\).
| \(n\) | Capital Recovery Cost | \(\text{EUAC}_{\text{O\&M}}\) (Over \(n\) years) | Total EUAC |
|---|---|---|---|
| 1 | \(\$15,000(A/P, 10\%, 1) - \$10,000(A/F, 10\%, 1) = \$5,500\) | \(\$2,000\) | \(\$7,500\) |
| 2 | \(\$15,000(0.5762) - \$6,000(0.4762) = \$5,860\) | \(\frac{2,000(P/F, 10\%, 1) + 3,000(P/F, 10\%, 2)}{(P/A, 10\%, 2)} \times (A/P, 10\%, 2) = \$2,476\) | \(\$8,336\) |
| 3 | \(\$15,000(0.4021) - \$3,000(0.3021) = \$5,110\) | \(\frac{2,000(P/F, 10\%, 1) + 3,000(P/F, 10\%, 2) + 4,500(P/F, 10\%, 3)}{(P/A, 10\%, 3)} \times (A/P, 10\%, 3) = \$3,109\) | \(\$8,219\) |
| 4 | \(\$15,000(0.3155) - \$1,000(0.2155) = \$4,527\) | \(\frac{\sum (\text{O\&M})(P/F, 10\%, t)}{(P/A, 10\%, 4)} \times (A/P, 10\%, 4) = \$3,788\) | \(\$8,315\) |
The minimum EUAC of \(\$7,500\) occurs at \(n=1\) year. This asset’s economic life is 1 year.
2. Economic Analysis in the Public Sector
Economic analysis in the public sector (governments, agencies, municipalities) differs significantly from private-sector analysis due to divergent objectives, viewpoints, and unique financial considerations.
Investment Objective and Viewpoint
- Objective: Public investment seeks to promote the general welfare of citizens and to ensure that the value of benefits exceeds the costs to whomsoever they accrue. This contrasts with the private sector’s focus on maximizing stockholder wealth.
- Viewpoint: The viewpoint for analysis must be chosen carefully to avoid suboptimal decisions. The proper viewpoint should be at least as broad as those who pay the costs and those who receive the benefits. For public-sector problems, this is typically the combined viewpoint of the government and the citizenry.
Selecting an Interest Rate
Determining the appropriate interest rate for public projects is complex because tax-funded projects do not always rely on conventional borrowing. Possibilities for setting the discount rate include:
- No Time-Value-of-Money Concept (0%): Arguing that tax money is collected and spent quickly, minimizing time delay. Often advocated by proponents of marginal projects.
- Cost of Capital Concept: Using the rate at which the government borrows money (e.g., municipal bond rates).
- Opportunity Cost Concept: Setting the rate based on the best investment opportunity foregone, either by the government agency or by the taxpayer.
The general recommendation is to select the largest of the cost of capital, government opportunity cost, or taxpayer opportunity cost interest rates.
The Benefit-Cost Ratio
The primary economic measure used in the public sector is the Benefit-Cost (B/C) Ratio.
\[\text{B/C ratio} = \frac{\text{Equivalent Worth of Benefits}}{\text{Equivalent Worth of Costs}}\]
The equivalent worth can be calculated using Present Worth (PW), Future Worth (FW), or Annual Worth (AW), all yielding the same ratio.
Decision Rule: A project is acceptable if the B/C ratio is \(\mathbf{\ge 1.0}\).
Variations of the B/C Ratio
- Conventional B/C Ratio: Operating and Maintenance (O&M) costs are included in the denominator (Costs).
- Modified B/C Ratio: O&M costs are subtracted from the benefits in the numerator, leaving only initial investment costs in the denominator.
Both versions produce consistent recommendations for acceptance or rejection, although their numerical values will differ.
Incremental B/C Analysis
When comparing mutually exclusive public projects, incremental B/C analysis must be used.
\[\frac{\Delta B}{\Delta C} = \frac{\text{Equivalent Worth of Incremental Benefits}}{\text{Equivalent Worth of Incremental Costs}}\]
Decision Rule: If \(\Delta B/\Delta C \ge 1.0\), the incremental investment is justified, and the higher-cost alternative is selected.
Solved Example 2: Benefit-Cost Ratio Analysis
Problem: A city is evaluating a park renovation project requiring an initial cost of \(\$500,000\). Annual maintenance costs are estimated at \(\$50,000\). Annual user benefits are estimated at \(\$120,000\). The project life is 10 years, and the selected public discount rate is \(7\%\). Determine whether the project is justified using both the conventional and modified B/C ratios.
(Factors at 7% interest, \(n=10\): \((P/A, 7\%, 10) \approx 7.024)\)
Solution:
First, calculate the Present Worth of the Annual Benefits and Costs:
- \(\text{PW}_{\text{Benefits}} = \$120,000 \times 7.024 = \$842,880\)
- \(\text{PW}_{\text{O\&M Costs}} = \$50,000 \times 7.024 = \$351,200\)
- \(\text{Initial Cost} = \$500,000\)
1. Conventional B/C Ratio: (O&M costs are in the denominator)
\[\text{B/C}_{\text{Conv}} = \frac{\text{PW of Benefits}}{\text{PW of Initial Costs} + \text{PW of O\&M Costs}}\]
\[\text{B/C}_{\text{Conv}} = \frac{\$842,880}{\$500,000 + \$351,200} = \frac{\$842,880}{\$851,200} \approx \mathbf{0.990}\]
2. Modified B/C Ratio: (O&M costs are subtracted from benefits in the numerator)
\[\text{B/C}_{\text{Mod}} = \frac{\text{PW of Benefits} - \text{PW of O\&M Costs}}{\text{PW of Initial Costs}}\]
\[\text{B/C}_{\text{Mod}} = \frac{\$842,880 - \$351,200}{\$500,000} = \frac{\$491,680}{\$500,000} \approx \mathbf{0.983}\]
Conclusion:
Since both the conventional B/C ratio (\(\approx 0.990\)) and the modified B/C ratio (\(\approx 0.983\)) are less than 1.0, the project is not economically justified at the \(7\%\) interest rate.