#01 Language Impact Score

Explanation

The master formula: a single number between 0 and 1 for overall language health at time t. Every variable feeds in through its normalized value and importance weight.

Data & measurement

Each Kifuliiru variable v_i(t) is observed on its native scale, then norm_i maps it to [0,1]. Weights α_i reflect importance for Kifuliiru documentation and adoption and should sum to a documented total; only variables in 𝒞 ∪ 𝒜 with known values enter each run. Version the weight table whenever stakeholders rebalance Kifuliiru corpus vs adoption emphasis.

Solution & proof

LIS(t) is a weighted average of normalized Kifuliiru scores: numerator ∑ α_i norm_i(v_i) and denominator ∑ α_i over the active set. Hence LIS lies in [0,1] when every norm_i does. Missing variables are excluded from both sums so remaining weights renormalize—this is the score-with-missing-data policy applied to the full Kifuliiru index.

Examples

1. 1. Two active variables

   Word problem

   Only two Kifuliiru impact variables are known this quarter: variable 1 has weight α_1 = 0.6 and normalized value 0.8; variable 2 has α_2 = 0.4 and norm 0.5. Compute LIS.

   Kifuliiru weights and normalized values

   Variable	α_i	norm_i
   i = 1	0.6	0.8
   i = 2	0.4	0.5

   Solution

   Step 1 — Weighted sum in the numerator for Kifuliiru: 0.6×0.8 + 0.4×0.5 = 0.48 + 0.20 = 0.68.

   Step 2 — Sum of weights: 0.6 + 0.4 = 1.0.

   Step 3 — LIS = 0.68 / 1.0 = 0.68.

   $$\mathrm{LIS}=\frac{0.6(0.8)+0.4(0.5)}{0.6+0.4}=\frac{0.68}{1}=0.68$$

2. 2. Missing third variable

   Word problem

   Suppose a third Kifuliiru variable would have α_3 = 0.2 but is unknown this month. Only variables 1 and 2 (same table as before) are used. Recompute LIS with renormalized weights over known variables only.

   Known Kifuliiru subset

   Variable	α_i	norm_i	Note
   i = 1	0.6	0.8	known
   i = 2	0.4	0.5	known
   i = 3	0.2	—	missing

   Solution

   Step 1 — Omit i = 3 from numerator and denominator.

   Step 2 — Numerator stays 0.6×0.8 + 0.4×0.5 = 0.68; denominator of active weights is still 0.6 + 0.4 = 1.0.

   Step 3 — LIS = 0.68 again here because the unknown Kifuliiru variable's weight was not in the active set. If the known set were only part of the full weight budget, you would divide by the sum of active α_i only (see "Score with missing data" in the library).

   $\mathrm{LIS}=\frac{{\alpha }_1{\mathrm{norm}}_1+{\alpha }_2{\mathrm{norm}}_2}{{\alpha }_1+{\alpha }_2}=\frac{0.68}{1}=0.68$

Source

Impact measurement framework — Core Architecture

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