Exercise set

Civil Infrastructure Rehabilitation Prioritization and Life-Cycle Planning Exercises

Solved civil infrastructure rehabilitation exercises for RPN, benefit-cost ranking, backlog funding, lifecycle cost, handover, residual risk and closeout.

These exercises focus on civil infrastructure rehabilitation prioritization and life-cycle planning: risk scores, benefit-cost ranking, backlog clearance, annual funding, timing, life-cycle cost, post-repair acceptance, handover records, residual risk and closeout gates. Field inspection and condition-assessment calculations are handled in a separate specialist exercise set.

Use these calculations as decision screens. Real rehabilitation programs require owner objectives, consequence classes, traffic or occupancy effects, constructability, permit constraints, procurement rules, stakeholder impacts, uncertainty budgets and responsible engineering approval.

Release Evidence Notes

Rehabilitation evidence should connect the selected intervention to the original failure mode. A completed work order is not enough; closeout should show repaired element, residual risk, acceptance measurements, record updates, future trigger, cost basis, schedule effect and accountable reviewer.

Engineering Boundary Notes

The simplified economic and risk screens below do not replace formal asset-management policy, code-based design, probabilistic risk assessment or public-sector funding governance. They teach how to make prioritization assumptions visible.

Common Release Mistakes

Common mistakes include ranking only by visible damage, ignoring consequence, double-counting benefits, accepting repairs without mechanism-specific measurements, clearing backlog on paper while deferring high-risk assets, and treating life-cycle cost as a single deterministic number.

Scenario Map

ScenarioExercisesPrimary checkEngineering decision
Risk and priority1, 2, 10, 15, 18RPN, weighted priority and risk reduction per costRank the work and identify release blockers.
Funding and timing3, 4, 5, 6, 11, 14Backlog clearance, NPV, LCC, disruption and contingencyChoose a fundable intervention plan.
Repair acceptance and closeout7, 8, 9, 12, 13, 16, 17Readings, records, residual risk, schedule and post-repair intervalAccept, hold or extend monitoring.

Exercise 1: Risk Priority Number

An asset defect has severity 8, occurrence 5 and detection score 4. Compute the RPN.

Solution

RPN=SOD=8(5)(4)=160

Engineering Comment

RPN is a ranking screen. It does not prove absolute risk because the scales are ordinal.

Plausibility Check

The maximum on a 1-to-10 scale is 1000, so 160 is a moderate-high screen.

Exercise 2: Risk Reduction After Rehabilitation

After repair, occurrence drops from 5 to 2 and detection improves from 4 to 3 while severity remains 8. Find the RPN reduction.

Solution

RPN_{new}=8(2)(3)=48,\qquad \Delta RPN=160-48=112

Engineering Comment

Severity often remains high because consequence does not disappear even when probability and detectability improve.

Plausibility Check

Both occurrence and detection scores improve, so RPN must decrease.

Exercise 3: Rehabilitation Benefit-Cost Ratio

A repair avoids expected annual losses of 85,000 dollars for 6 years. The repair costs 310,000 dollars. Ignore discounting. Find benefit-cost ratio.

Solution

BCR=\dfrac{85000(6)}{310000}=1.65

Engineering Comment

A ratio above 1 supports the repair economically, but safety-critical work may be mandatory regardless of BCR.

Plausibility Check

Total avoided loss is 510,000 dollars, larger than cost, so BCR is above 1.

Exercise 4: Backlog Funding Clearance

The rehabilitation backlog is 4.8 million dollars. Annual funding is 0.9 million dollars and new deterioration adds 0.3 million dollars per year. Estimate net clearance time.

Solution

F_{net}=0.9-0.3=0.6\ \text{million dollars/yr}
t=\dfrac{4.8}{0.6}=8\ \text{yr}

Engineering Comment

The program is reducing backlog, but eight years may be too slow for high-consequence assets.

Plausibility Check

If only 0.6 million dollars is cleared each year, several years are required for a 4.8 million dollar backlog.

Exercise 5: Annual Funding Gap

To clear the same 4.8 million dollar backlog in 5 years while new deterioration adds 0.3 million dollars per year, find required annual funding.

Solution

F=\dfrac{4.8}{5}+0.3=1.26\ \text{million dollars/yr}

Engineering Comment

Funding targets should include new deterioration, otherwise the backlog plan is structurally underfunded.

Plausibility Check

The required amount is greater than 0.96 million dollars per year because new deterioration is added.

Exercise 6: Life-Cycle Cost Comparison

Option A costs 500,000 dollars now and 40,000 dollars per year for 10 years. Option B costs 720,000 dollars now and 15,000 dollars per year for 10 years. Ignore discounting. Which is cheaper?

Solution

LCC_A=500000+10(40000)=900000
LCC_B=720000+10(15000)=870000

Option B is cheaper by 30,000 dollars.

Engineering Comment

A higher capital repair can be justified if it reduces recurring maintenance enough.

Plausibility Check

Option B costs 220,000 dollars more initially but saves 250,000 dollars in maintenance.

Exercise 7: Post-Repair Crack Acceptance

Acceptance requires average crack width below 0.20 mm. Readings after repair are 0.16, 0.19, 0.21 and 0.18 mm. Find the average and decision.

Solution

\bar{w}=\dfrac{0.16+0.19+0.21+0.18}{4}=0.185\ \text{mm}

The average passes, but the 0.21 mm local reading should be reviewed against the local limit.

Engineering Comment

Average acceptance can hide a localized active crack. Closeout should state both average and maximum.

Plausibility Check

Most readings are below 0.20 mm, so an average below 0.20 mm is plausible.

Exercise 8: Handover Record Hold

A rehabilitation package requires 18 closeout records. The contractor submits 15, and 2 of those are incomplete. Find usable record completion.

Solution

C=\dfrac{15-2}{18}\times100=72.2\%

Engineering Comment

The asset should remain on handover hold if missing records affect inspection intervals, warranties or future load decisions.

Plausibility Check

Only 13 usable records exist out of 18, so completion near 70 percent is reasonable.

Exercise 9: Residual Risk Screen

Initial annual failure probability is 0.020. Rehabilitation reduces it by 70 percent. Consequence is 2.5 million dollars. Find residual expected annual loss.

Solution

p_r=0.020(1-0.70)=0.006
EAL=0.006(2500000)=15000\ \text{dollars/yr}

Engineering Comment

Residual risk is not zero. It should be linked to monitoring and future inspection triggers.

Plausibility Check

The probability is less than one percent after repair, so expected annual loss is much less than consequence.

Exercise 10: Weighted Priority Score

Three projects have condition, consequence and cost scores. Project A: 80, 60, 50. Project B: 60, 90, 40. Project C: 70, 70, 80. Weights are 0.4, 0.4 and 0.2. Rank by weighted score.

Solution

P_A=0.4(80)+0.4(60)+0.2(50)=66
P_B=0.4(60)+0.4(90)+0.2(40)=68
P_C=0.4(70)+0.4(70)+0.2(80)=72

Rank: C, B, A.

Engineering Comment

The cost score must be defined consistently. If higher cost should reduce priority, the score must already be inverted.

Plausibility Check

Project C is balanced and has the highest cost score, so it leads under these weights.

Exercise 11: User Disruption Cost

A lane closure affects 18,000 vehicles per day, adds 4 minutes per vehicle and lasts 12 days. Value of time is 18 dollars per hour. Estimate disruption cost.

Solution

C=18000(12)\left(\dfrac{4}{60}\right)(18)=259200\ \text{dollars}

Engineering Comment

User cost can change the preferred construction window or staging strategy.

Plausibility Check

The total delay is 14,400 vehicle-hours, and 14,400 times 18 dollars is 259,200 dollars.

Exercise 12: Work Package Critical Path

Design review takes 3 weeks, procurement 6 weeks, repair 4 weeks and acceptance testing 2 weeks in sequence. Find total duration.

Solution

T=3+6+4+2=15\ \text{weeks}

Engineering Comment

If tasks are truly sequential, procurement often controls the release date more than field repair duration.

Plausibility Check

All four durations are positive, so the total must exceed the largest single task.

Exercise 13: Post-Repair Inspection Interval

Baseline inspection interval is 24 months. Residual risk factor is 1.5. Use interval equal to baseline divided by risk factor.

Solution

I=\dfrac{24}{1.5}=16\ \text{months}

Engineering Comment

Shortening the interval is appropriate when residual uncertainty remains after repair.

Plausibility Check

A factor greater than 1 reduces the interval below 24 months.

Exercise 14: Contingency Allowance

Base rehabilitation cost is 1.2 million dollars. Design uncertainty contingency is 12 percent and access contingency is 8 percent. Estimate total with additive contingencies.

Solution

C=1.2(1+0.12+0.08)=1.44\ \text{million dollars}

Engineering Comment

Separate contingencies help explain whether the uncertainty is technical, access-related or market-related.

Plausibility Check

The total contingency is 20 percent, and 20 percent of 1.2 million is 0.24 million.

Exercise 15: Risk Reduction per Dollar

Project A reduces RPN by 112 at a cost of 310,000 dollars. Project B reduces RPN by 150 at a cost of 600,000 dollars. Compare RPN reduction per 100,000 dollars.

Solution

E_A=\dfrac{112}{3.1}=36.1,\qquad E_B=\dfrac{150}{6.0}=25.0

Project A gives more RPN reduction per 100,000 dollars.

Engineering Comment

Efficiency is useful for portfolios, but high-consequence mandatory work may still outrank efficient low-cost work.

Plausibility Check

Project B reduces more total RPN but costs almost twice as much, so lower efficiency is plausible.

Exercise 16: Pavement Treatment Timing

If IRI grows from 2.6 to a trigger of 3.4 m/km at 0.20 m/km per year, estimate years to treatment.

Solution

t=\dfrac{3.4-2.6}{0.20}=4\ \text{yr}

Engineering Comment

Treatment can be planned before the trigger if mobilization, traffic control or budget cycles require lead time.

Plausibility Check

The required increase is 0.8 m/km, and four increments of 0.20 reach it.

Exercise 17: Drainage Retrofit Benefit

A drainage retrofit costs 180,000 dollars and reduces expected flood-damage loss from 70,000 dollars per year to 25,000 dollars per year. Estimate simple payback.

Solution

S=70000-25000=45000\ \text{dollars/yr}
t=\dfrac{180000}{45000}=4\ \text{yr}

Engineering Comment

Payback is useful, but hydraulic safety and road closure risk may justify action even with longer payback.

Plausibility Check

Four years of 45,000 dollar savings equals the 180,000 dollar cost.

Exercise 18: Rehabilitation Closeout Gate

A closeout gate requires acceptance testing pass, records at least 95 percent complete and residual high-risk items equal to zero. Results are pass, 92 percent and one high-risk item. Decide the gate.

Solution

\text{testing}=\text{pass},\qquad 92\%<95\%,\qquad 1>0

The gate fails because records are incomplete and one high-risk residual item remains.

Engineering Comment

Passing field tests cannot close the rehabilitation if records and residual risk do not support future asset management.

Plausibility Check

Two of the three criteria fail, so the decision must be hold.

Validation Package Checklist

  • intervention is linked to the original failure mode and consequence class;
  • risk score, benefit basis, cost basis and uncertainty are stated;
  • user disruption, funding constraints and schedule drivers are visible;
  • post-repair measurements include acceptance threshold and maximum value where relevant;
  • handover records are complete enough for future inspection and maintenance;
  • residual risk, future trigger and accountable owner decision are documented.
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See also