Project Grade 9 Math
Whitmoor Creek Dam: A Repair, Retrofit, or Removal Study
A grade 9 project where teams model twenty-year costs for three options at a fictional mill dam and recommend one to a river district board.
Objectives
- Students will build a twenty-year cost function for each of three infrastructure options and graph them on common axes.
- Students will express budget, instream-flow, and supply requirements as inequalities and test each option against them.
- Students will write and present a recommendation that names the constraint that decided the choice.
Vocabulary
Key words: impoundment, fish passage, instream flow, viable solution, life-cycle cost
Teacher overview
Thousands of small mill dams built a century ago no longer generate power, and the communities that inherited them face a choice among repair, retrofit for fish passage, and removal. The decision turns on constraints rather than on preference: in this study one option is too expensive to build, another is affordable to build but too expensive to run for twenty years, and only a comparison of each option's year-one and twenty-year costs shows which one remains available. In this project, teams study one fictional dam, build a twenty-year cost model for each option, test all three against a first-year budget cap, a twenty-year budget cap, a minimum stream flow, and the town's August water demand, then deliver a written proposal and a short presentation recommending one option to a river district board.
Project Brief
Driving question. The Whitmoor River District owns a fictional 1920s mill dam on Whitmoor Creek. The dam no longer generates power, its spillway is deteriorating, and the impoundment behind it supplies part of the town's summer water. Which option should the district choose, and which constraint decides the answer?
Your client. You are advising the board of the fictional Whitmoor River District, which will vote at its next meeting.
The three options.
- Option A, repair. Rebuild the spillway and keep the impoundment. Cost: $1.15 million in year one, plus $28,000 each year for inspection and maintenance for twenty years. Fish still cannot pass the dam.
- Option B, retrofit. Rebuild the spillway and add a fish ladder. Cost: $1.62 million in year one, plus $52,000 each year for twenty years for inspection, maintenance, and running the fish ladder (clearing debris from it and counting fish each spring).
- Option C, removal. Remove the dam and restore the channel. Cost: $1.42 million in year one, plus $6,000 each year for stream monitoring for the first ten years only. Removing the impoundment ends a supply of 0.35 MGD, so the district must also drill a replacement well at $0.90 million in year one. The new well would supply 0.35 MGD.
Water supply today. In August the town draws 0.35 MGD from the impoundment and 0.75 MGD from its existing wells. The wells, including the replacement well in Option C, draw from a deep aquifer, so pumping them does not take water out of the creek. Assume the only water removed from the creek at the dam is the town's withdrawal from the impoundment.
How to count costs. Each one-time cost is paid when the work begins, and each annual cost is paid at the end of its year. Write C(t) for the total the district has spent by the end of year t. Year-one spending is C(1), which includes the one-time costs and the first annual payment. The twenty-year total is C(20).
The constraints.
- The district may spend at most $1.8 million in year one.
- The district may spend at most $2.6 million across the full twenty years.
- At least 4 cfs must remain in Whitmoor Creek below the dam through August. The creek carries 11 cfs at the dam in an average August.
- The town's average August demand is 1.1 MGD, and it must be met by some combination of the impoundment and wells.
Conversion to use. 1 million gallons per day = 1.547 cubic feet per second.
What you turn in. A written proposal of three to four pages containing a labeled graph of all three cost functions, a completed viability table, the August water balance, and a recommendation of one option in a paragraph naming the constraint that decided it. Each team also gives a six-minute presentation to the board.
Milestones
- Session 1, site data sheet and constraints (50 min). Build a one-page data sheet listing every given quantity with its units. Define your variables, then write all four constraints as inequalities: the year-one budget cap, the twenty-year budget cap, the August instream-flow requirement, and the August supply requirement. Output: the data sheet and the four inequalities, checked by the teacher before Session 2.
- Session 2, twenty-year cost model (50 min). Write a cost function C(t) in dollars for each option, where t is the number of years since the work begins, remembering that Option C's monitoring stops after year ten, so its function has two pieces. Compute the year-one cost C(1) and the twenty-year total C(20) for each option. Graph all three functions on one set of labeled, scaled axes for 0 ≤ t ≤ 20, and draw both budget caps as horizontal lines. For any option whose graph crosses the $2.6 million line, solve an inequality to find the first year in which its running total exceeds the cap. Output: three functions, three year-one costs, three totals, the graph, and the solved inequality.
- Session 3, August water balance (50 min). Convert the town's 1.1 MGD August demand, the 0.35 MGD impoundment supply, and the 0.75 MGD from existing wells into cubic feet per second, then build a table showing, for each option, the water supplied by each source, the water taken from the creek, and the flow left below the dam. Check each option against the 4 cfs requirement and the 1.1 MGD demand. Then write and solve an inequality for the lowest August creek flow at the dam at which each option would still leave 4 cfs below it. Output: the completed water balance table with units shown in every cell, and the three minimum creek flows.
- Session 4, viability screen and draft (50 min). Build a table with the three options down the side and the four constraints across the top, marking each cell viable or not viable and citing the number that decides it. Draft your recommendation paragraph, and name explicitly which single constraint eliminated any option you rejected. Output: the viability table and a first draft of the proposal.
- Session 5, board presentation (50 min). Present for six minutes to the class acting as the fictional district board, then turn in the finished proposal. Expect one question from the board about an option you did not recommend. Output: the presentation and the final written proposal.
From Water for Schools (waterforschools.org). This material is in the public domain (CC0) and may be copied, edited, and shared freely.
Rubric
| Criterion | 4, exemplary | 3, proficient | 2, developing | 1, beginning |
|---|---|---|---|---|
| Cost model | All three cost functions are correct, Option C's ten-year monitoring is handled correctly, and the graph is labeled and scaled so the three options can be compared at any year in the twenty-year window. | All three functions and totals are correct, and the graph is labeled, with minor scaling problems. | One function or total contains an arithmetic error, or the graph lacks units on an axis. | Functions are missing or the graph cannot be read. |
| Constraints and viability | Every constraint is written as a correct inequality, and each viability cell cites the specific number that decides it. | All four constraints are written correctly and the viability table is complete, with some cells lacking a cited number. | Constraints are stated in words rather than as inequalities, or the table has gaps. | Constraints are missing or the viability judgments do not follow from the numbers. |
| Water balance | The MGD and cfs conversions are correct, units appear in every cell, and the August flow check is right for all three options. | Conversions are correct and the table is complete, with units missing from a few cells. | One conversion is wrong or one option's flow check is missing. | The table does not track units, so the flow check cannot be verified. |
| Proposal and presentation | The recommendation names the deciding constraint, addresses the strongest objection to it, and stays within six minutes. | The recommendation names the deciding constraint and is delivered clearly within the time. | The recommendation is stated but rests on preference rather than on a constraint. | No clear recommendation, or the proposal omits required sections. |
Teacher reference
The values below follow from the Project Brief, with t in years and costs in dollars. Teams that round conversions to two decimal places will match these figures.
Cost functions (Session 2).
- Option A: C(t) = 1,150,000 + 28,000t for 0 ≤ t ≤ 20.
- Option B: C(t) = 1,620,000 + 52,000t for 0 ≤ t ≤ 20.
- Option C: C(t) = 2,320,000 + 6,000t for 0 ≤ t ≤ 10, and C(t) = 2,380,000 for 10 < t ≤ 20. The one-time cost of $2,320,000 is the removal ($1,420,000) plus the replacement well ($900,000).
| Option | Year-one cost C(1) | Twenty-year total C(20) |
|---|---|---|
| A, repair | $1,178,000 | $1,710,000 |
| B, retrofit | $1,672,000 | $2,660,000 |
| C, removal | $2,326,000 | $2,380,000 |
The $1.8 million line applies only at t = 1, so Option B's graph rising above it after year three does not count against Option B; a team that marks Option B not viable for that reason has misread the year-one cap. Only Option B's graph crosses the $2.6 million line. Solving 1,620,000 + 52,000t > 2,600,000 gives t > 18.85, so Option B's running total first exceeds the cap in year 19 (C(18) = $2,556,000 and C(19) = $2,608,000). Teams may also notice that Option B's line passes Option C's flat line during year 15 (C(14) = $2,348,000 and C(15) = $2,400,000 for Option B, against $2,380,000 for Option C), so the option with the highest first-year cost is no longer the most expensive by the end of the study.
Water balance (Session 3). The town's demand is 1.1 × 1.547 = 1.70 cfs. The impoundment supplies 0.35 × 1.547 = 0.54 cfs and the existing wells supply 0.75 × 1.547 = 1.16 cfs.
| Option | Supply from impoundment | Supply from wells | Taken from creek | Flow left below dam | Lowest creek flow that still leaves 4 cfs |
|---|---|---|---|---|---|
| A, repair | 0.35 MGD (0.54 cfs) | 0.75 MGD (1.16 cfs) | 0.54 cfs | 11 − 0.54 = 10.46 cfs | 4 + 0.54 = 4.54 cfs |
| B, retrofit | 0.35 MGD (0.54 cfs) | 0.75 MGD (1.16 cfs) | 0.54 cfs | 10.46 cfs | 4.54 cfs |
| C, removal | none | 1.1 MGD (1.70 cfs) | 0 cfs | 11 cfs | 4 cfs |
Viability screen (Session 4).
| Option | Year one ≤ $1.8 million | Twenty years ≤ $2.6 million | Flow below dam ≥ 4 cfs | Supply ≥ 1.1 MGD |
|---|---|---|---|---|
| A, repair | Viable, $1,178,000 | Viable, $1,710,000 | Viable, 10.46 cfs | Viable, 0.35 + 0.75 = 1.1 MGD |
| B, retrofit | Viable, $1,672,000 | Not viable, $2,660,000 is $60,000 over | Viable, 10.46 cfs | Viable, 0.35 + 0.75 = 1.1 MGD |
| C, removal | Not viable, $2,326,000 is $526,000 over | Viable, $2,380,000 | Viable, 11 cfs | Viable, 0.75 + 0.35 = 1.1 MGD |
Option A is the only option that passes all four constraints. The year-one cap eliminates Option C and the twenty-year cap eliminates Option B, so a team that checks only first-year cost will wrongly treat Option B as available. The flow and supply requirements are met by every option and do not decide anything in an average August. Even in a dry August, Options A and B fail the flow requirement only if the creek falls below 4.54 cfs, which is a reasonable point for a team to raise when the board asks about drought.
The strongest objection to Option A is that fish still cannot pass the dam. A level 4 proposal answers it with numbers: Option B misses the twenty-year cap by $60,000, about 2.3 percent of the cap, and it would become viable if its annual cost were $49,000 or less, because 1,620,000 + 20a ≤ 2,600,000 gives a ≤ 49,000. A team may reasonably recommend Option A while noting that the board could reconsider Option B if the district raised its twenty-year cap or found a partner to share the fish-ladder costs. A team that recommends Option B without addressing the cap scores at level 2 on the last criterion, because the recommendation does not follow from the stated constraints.
Differentiation
Support
Give teams a partly completed cost table for this fictional dam with the year-one figures already entered, so the first task is only the twenty-year extension. Provide a worked conversion showing 1 MGD expressed in cubic feet per second, and allow teams to use it directly in Session 3 rather than deriving it.
Extension
Ask teams to add a fourth option to the fictional study: repair now and remove the dam at the start of year 11, using the same Option A and Option C costs with no price increase. Have them build its cost function as a piecewise expression, test it against both budget caps, and explain why a piecewise model is needed when a project changes character partway through its life. For reference, C(t) = 1,150,000 + 28,000t for 0 ≤ t ≤ 10, then C(t) = 3,750,000 + 6,000(t − 10) for 10 < t ≤ 20, because the removal and the well add $2,320,000 when year 11 begins. The option passes the year-one cap at $1,178,000 and fails the twenty-year cap at $3,810,000.
ELL supports
Provide a labeled diagram of the fictional dam showing the impoundment, the spillway, the fish ladder, and the reach below the dam, with each vocabulary term placed on the feature it names. Give a presentation frame: "We recommend Option ___ because ___ meets the ___ constraint, while Option ___ fails it by ___."
Extensions
Have teams research a dam removal or fish-passage project completed in their own state, then compare its reported cost structure with the pattern in this fictional study: a large one-time cost against a small recurring cost. Ask them to write one paragraph on what their model left out, such as permitting time, sediment behind the dam, or the value of restored fish runs.