Estimate a cost model from observations
Before using a cost estimate, decide whether the data refer to comparable output levels, technologies, and price dates.
| Observation | Output | Cost ($) |
|---|---|---|
| 1 | 100 | 3,000.00 |
| 2 | 500 | 9,000.00 |
This extrapolates beyond the observations; the linear model may stop being valid.
Use the two-point slope v = (C₂ − C₁)/(q₂ − q₁).
Cash-flow dates and signs are part of the answer. Model values retain precision; displays round money to cents. Illustrative assumptions exclude taxes and inflation unless stated.
What the model means
A two-point line determines its slope and intercept exactly but does not measure uncertainty. With more observations, least squares minimizes the sum of squared residuals. Residual patterns can reveal curvature; R² measures fit within the sample, not proof that extrapolation is safe.
Build the calculation
For observations (q₁,C₁),(q₂,C₂), find v=(C₂−C₁)/(q₂−q₁), then F=C₁−vq₁. For n observations use v=[nΣqC−ΣqΣC]/[nΣq²−(Σq)²], F=mean(C)−v mean(q). Estimation can also use expert judgment, analogy, or bottom-up component costs; explain the data and assumptions.
Set up. Calculate. Explain.
Rebuild the starting scenario
Cost at 100 units: 3000 $; Cost at 500 units: 9000 $; Target output: 700 units. Calculate the estimated cost at the target output.
Setup hint
Use the two-point slope v = (C₂ − C₁)/(q₂ − q₁).
Open the visual solution
- Use the two-point slope v = (C₂ − C₁)/(q₂ − q₁).
- Recover the intercept F = C₁ − vq₁ = $1,500.00.
- Predict F + vq; with more observations use least squares and inspect residuals, not only R².
| Observation | Output | Cost ($) |
|---|---|---|
| 1 | 100 | 3,000.00 |
| 2 | 500 | 9,000.00 |
This extrapolates beyond the observations; the linear model may stop being valid.
Use the two-point slope v = (C₂ − C₁)/(q₂ − q₁).
Cash-flow dates and signs are part of the answer. Model values retain precision; displays round money to cents. Illustrative assumptions exclude taxes and inflation unless stated.
Checks allow 0.1% numerical rounding (minimum $0.05 or 0.001 in the stated nonmonetary unit). Interpolated factors and rates are approximations; keep table precision consistent.
Transfer the idea to a new scenario
Costs are $4,000 at 100 units and $10,000 at 500 units. Fit the line and predict cost at 800 units.
Setup hint
Use the two-point slope v = (C₂ − C₁)/(q₂ − q₁).
Open the visual solution
- Use the two-point slope v = (C₂ − C₁)/(q₂ − q₁).
- Recover the intercept F = C₁ − vq₁ = $2,500.00.
- Predict F + vq; with more observations use least squares and inspect residuals, not only R².
| Observation | Output | Cost ($) |
|---|---|---|
| 1 | 100 | 4,000.00 |
| 2 | 500 | 10,000.00 |
This extrapolates beyond the observations; the linear model may stop being valid.
Use the two-point slope v = (C₂ − C₁)/(q₂ − q₁).
Cash-flow dates and signs are part of the answer. Model values retain precision; displays round money to cents. Illustrative assumptions exclude taxes and inflation unless stated.
Checks allow 0.1% numerical rounding (minimum $0.05 or 0.001 in the stated nonmonetary unit). Interpolated factors and rates are approximations; keep table precision consistent.
What is the key risk in predicting far beyond the measured output range?
Can you defend your setup?
Can you place each amount on the timeline, match the interest period, calculate the equivalent quantity, and state the decision or limitation?
Completion is saved on this browser.