Water Main Rehabilitation: EMV, Exposure Limits, and Break-Even
A municipal utility must rehabilitate 6.4 km of 60-year-old cast iron water main. The capital committee has three shortlisted options. Option A, open-cut replacement, costs $8.2M; historical data for this corridor indicate a 12% chance of a permit delay adding $1.5M and an 8% chance of contaminated soil adding $2.4M. Option B, trenchless lining, costs $10.6M with a 25% chance of failing the integrity test and requiring $3.1M of remediation, plus a 10% chance of a $1.0M overrun. Option C, phased repairs, costs $6.4M now and has a 40% chance of requiring full replacement in five years; the present value of that future replacement is $6.0M. The utility's board has stated it will not accept a worst-case exposure above $13M and prioritizes uninterrupted service.
1. Calculate the expected monetary value (EMV) of each option using the stated probabilities and costs, and state which option has the lowest EMV. Show your arithmetic.
Read the worked answer
EMV is the probability-weighted sum of the cost outcomes. Treat the listed events as independent and additive, and ignore the future replacement's timing beyond the given present value.
- A: $8.2M + (0.12 × $1.5M) + (0.08 × $2.4M) = $8.2M + $0.18M + $0.192M = $8.572M.
- B: $10.6M + (0.25 × $3.1M) + (0.10 × $1.0M) = $10.6M + $0.775M + $0.10M = $11.475M.
- C: $6.4M + (0.40 × $6.0M) = $6.4M + $2.4M = $8.8M.
Option A has the lowest EMV at $8.572M, about $0.228M below C and $2.903M below B. However, EMV is not a decision rule by itself; it does not capture the board's exposure limit or service-continuity objective. Also the independence assumption matters: if the permit delay and contaminated soil events are correlated, A's EMV would rise.
2. Apply the board's $13M worst-case exposure constraint. Which options remain feasible, and does the constraint change the ranking? Explain any judgement you must make.
Read the worked answer
Worst-case exposure is the sum of the base cost and all adverse events, assuming they can occur together:
- A: $8.2M + $1.5M + $2.4M = $12.1M.
- B: $10.6M + $3.1M + $1.0M = $14.7M.
- C: $6.4M + $6.0M = $12.4M.
B exceeds the $13M limit and is eliminated unless the limit can be waived or the overrun and integrity failure cannot co-occur. A and C remain feasible on exposure. The constraint does not change the EMV ranking between A and C: A is still lower by $0.228M. But C's exposure is close to the limit, and its future replacement is conditional on a 40% event; if the board treats that as a near-certain liability in its planning, C becomes less attractive. The judgement is whether the board's limit applies to the nominal worst case or to a high percentile. If the former, A is preferred; if the latter, a probabilistic exposure analysis should be run before selecting.
3. The contaminated-soil probability for Option A is the least certain input. At what value would Option A cease to have the lowest EMV among the feasible options? Show the break-even and interpret it.
Read the worked answer
Hold all other inputs constant. A's EMV is $8.2M + $0.18M + p × $2.4M, where p is the contaminated-soil probability. C's EMV is $8.8M. Set:
$8.38M + $2.4M × p = $8.8M → $2.4M × p = $0.42M → p = 0.175, or 17.5%.
So A remains cheaper than C until the contaminated-soil probability exceeds 17.5%, more than double the 8% estimate. The break-even is useful because it shows the decision is robust to plausible estimation error. If the geotechnical data are sparse, the team should test whether 17.5% is within the credible range; if it is, the choice between A and C is sensitive and further sampling may be justified. The same logic can be applied to the permit-delay probability: A's EMV equals C's when q × $1.5M = $0.408M, or q = 27.2%, versus the 12% estimate. The break-even analysis does not override the board's $13M exposure constraint, which A satisfies.