Lesson 11 Flashcards — Decision Matrices

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1. What problem does a decision matrix solve?

Choosing between options when no option wins on every measure. It does not make the decision — it makes the reasoning explicit and auditable.

2. What makes something a valid measure of merit?

You must be able to put a number on it, for every option. Efficiency, mass, cost, and MTBF qualify; 'quality' and 'reliability' as stated do not.

3. How many measures of merit should you use?

Three to six. Fewer and the decision is trivial; more and the weights become noise.

4. How do you normalize a bigger-is-better measure?

\(x_\text{norm} = x/x_\text{max}\). The best option scores 1.0 and everything else is a fraction of it.

5. How do you normalize a smaller-is-better measure?

\(x_\text{norm} = x_\text{min}/x\). Cost and mass work this way — getting the direction backward silently produces a confident, wrong recommendation.

6. What must the weights sum to, and what do they represent?

They sum to 1.0, and they encode the customer's priorities, not yours. A satellite program weights mass heavily; a ground installation barely weights it at all. On Project 1 the weights are given by the solicitation — apply them exactly. In practice you choose them, and that is the first thing a reviewer challenges.

7. A customer states preferences as ratios, not weights. How do you convert?

Write the stated preference as a ratio of whole numbers, add the parts, then divide each part by that sum — the weights now total 1.0. Project 1: cost weighted twice efficiency, efficiency twice volume, so cost : efficiency : volume \(= 4:2:1\), sum 7, and \(w = 4/7,\ 2/7,\ 1/7 = 0.571,\ 0.286,\ 0.143\). PP10: payload and programming ease are equal and each worth twice cost or flight time, so \(1:2:2:1\), sum 6, and \(w = 1/6,\ 1/3,\ 1/3,\ 1/6 = 0.167,\ 0.333,\ 0.333,\ 0.167\).

8. How is the final score computed?

\(\text{Score}_i = \sum_j w_j\,x_{ij,\text{norm}}\) — multiply each normalized value by its weight and sum across the row.

9. Can an option win overall while losing on the highest-weighted measure?

Yes. Losing a little on one measure and winning outright on the others can produce a higher total. That is not a bug in the method. In the lesson's three power supplies (96/80/88% efficient, 2.4/2.0/3.2 kg, $1500/1000/800) at weights \(0.5/0.2/0.3\), A is best on efficiency — the most heavily weighted measure — and still finishes last: A \(0.827\), B \(0.857\), C \(\mathbf{0.883}\). C wins, and C is also the heaviest of the three.

10. What should you do after computing the winner?

Report the margin, and test the weights — change them by a reasonable amount and see whether the winner flips. A recommendation that survives that is worth defending.

11. Where is this objective assessed?

Not on any quiz or the midterm — only through Project 1, Part 2, where the decision matrix is the deliverable.