Weighted scoring matrix
Score each option against weighted criteria and sum, then test whether the winner survives a small change in the weights.
- Time cost
- 45 min
- Output
- A scored matrix, a ranking, and a flip distance per criterion.
- Steps
- 6
Use when
- Three or more options, four or more criteria, and no option is best on everything.
- You need to show your reasoning to someone who was not in the room.
- The criteria are genuinely independent of one another.
Do not use when
- One criterion dominates. Then it is a single-variable comparison and the matrix is theatre.
- The scores are invented. A matrix of guesses produces a confident total from nothing.
- You already know the answer. The weights will be tuned until the matrix agrees.
Inputs required
- Two or more options
- Criteria with weights
- A score per option per criterion
Procedure
- 01
List criteria before options
Write what matters before looking at what is available. Criteria chosen afterwards describe the option you prefer.
- 02
Weight them
Distribute a fixed budget — 100 points across the criteria. Forcing a fixed sum stops everything being important.
- 03
Score each cell
Use a consistent scale, 0–10, and anchor the ends: what would a 0 look like, what would a 10 look like. Score a column at a time, not a row at a time, to reduce halo effects.
- 04
Normalise and total
Weighted score is weight × score, summed per option. Normalise weights to sum to 1 so totals are comparable across runs.
- 05
Delete flat criteria
Any criterion where every option scores the same cannot change the ranking. It adds nothing but a number. Remove it.
- 06
Test the winner
For each criterion, compute how far its weight must move before the ranking changes. If any single small move flips the winner, the matrix is not deciding this.
Characteristic failure mode
Worked example
Choosing between three rental flats.
- 01Criteria and weights: rent 35, commute 30, space 20, light 15.
- 02Totals come out 6.85, 6.60, 6.40 — the first wins.
- 03Sensitivity: moving the commute weight from 30 to 34 flips first and second.
Result
The matrix does not separate the top two. That is the useful finding: the choice rests on how much a commute is worth, which is a question to answer directly rather than through a total.
Where this disagrees with another method
The matrix optimises; satisficing says stop at good enough. For a reversible decision with many similar options, satisficing wins — the matrix will spend an hour separating options that are within noise of each other.
Where to go next