Lovable App

1. Objective

Estimate, per model, the additional cost the company would absorb by extending the factory warranty from 12 to 24 months, and propose a commercial premium that covers this risk with a defined margin.

2. Data sources

Claims (last 5 years): 26.684 approved warranty claims, with failure date (CTL_FAILURE_DATE) and delivery date (CTL_DELIVERY_DATE).
Fleet (last 10 years): 13.798 delivered machines — used to measure the exposed fleet in each window.
Model: grouped by the prefix before the '-' (e.g., D51EX-22 → D51EX, PC200-8M0 → PC200).

3. Core assumption — time-based window

A failure is allocated to Year 1 if it occurs within 12 months of delivery; to Year 2 if it occurs between 12 and 24 months. Only machines that have actually completed the window enter the denominator (≥12m for Y1, ≥24m for Y2). This avoids overstating cost using new fleet that has not yet had the chance to fail (censoring treatment).

4. Formulas

Frequency Yn (claims/machine) = Claims Yn / Fleet exposed ≥n×12m

Avg cost per claim Yn = Total cost Yn / Claims Yn

Cost / machine Yn = Frequency Yn × Avg cost per claim Yn

Weibull bathtub model — projection of years 3 and 4

Weibull reliability: R(t) = exp[−(t/η)^β] · cumulative hazard H(t) = (t/η)^β · hazard h(t) = (β/η)(t/η)^(β−1)

β < 1 → infant mortality (decreasing hazard) · β = 1 → constant · β > 1 → wear-out (increasing hazard)

r21 = (Cost / machine Y2) / (Cost / machine Y1) → β = log2(1 + r21)

Weibull bathtub: H(t) = (t/η)^β → increment of year k ∝ k^β − (k−1)^β

Cost / machine Y3 = Y2 × (3^β−2^β)/(2^β−1) · Y4 = Y3 × (4^β−3^β)/(3^β−2^β)

Δ Cost (extension 12 → 48m) = Y2 + Y3 + Y4

Suggested premium = Δ Cost × (1 + margin%)

Important: Frequency is not failure probability — it is the average number of claims per machine in the window. Values above 100% (e.g., 230%) mean that, on average, each machine generated more than one claim in the period (different components, multiple interventions). This is normal in heavy-equipment warranty and is the correct metric to project cost (Frequency × Avg ticket).

Year 1 is already covered today, so the extension cost starts at Year 2. Years 3 and 4 are not observed yet: they are projected from the Weibull bathtub shape β fitted on the measured Y2/Y1 ratio, keeping the same claims base and 12-month windows.

5. Numeric example — WA200

Fleet exposed ≥12 months	2.059
Fleet exposed ≥24 months	1.747
Claims Y1 / Y2	4.719 / 957
Frequency Y1 / Y2 (claims/machine)	2.29 / 0.55
Avg cost per claim Y1 / Y2	$2,735 / $4,474
Cost / machine Y1	$6,269
Cost / machine Y2 (observed)	$2,451
r21 → β (Weibull bathtub)	39.1% → 0.476
Cost / machine Y3 (projected)	$1,856
Cost / machine Y4 (projected)	$1,552
Δ Cost (extension 12 → 48m)	$5,859
Suggested premium (20% margin)	$7,031

6. Decision read

Rate Y2 < Y1 in most models: failures concentrate in the early months (infant mortality). Good signal to extend warranty.
Avg cost Y2 ≥ Y1: when a failure occurs in Year 2, it usually involves larger components — that's why Δ is not proportional to the rate.
Models with high Δ and high Rate Y2 require a higher premium or exclusion from the standard offer; models with low Δ are natural candidates for the extension.

7. Limitations and mitigations

Time-based window (not operated hours). Models with highly variable usage profiles may have under/overestimated rates.
5-year claims history — recently launched models have smaller samples; the minimum-fleet filter mitigates this noise.
Does not include parts/labor cost inflation; annual review is recommended.
The premium margin (dashboard slider) absorbs variability, admin expenses, and profit — it does not replace formal actuarial analysis for large contracts.