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Dominance, the Pareto frontier, and why a weight is a value judgement written as a coefficient.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to decide whether one plan dominates another, compute a Pareto frontier from a set of candidates, and say exactly what the frontier settles and what it deliberately leaves open. You will also be able to form a weighted objective, find the weight at which a choice flips, and explain what a weight asserts about the units it combines — so that a value judgement is reported as one rather than presented as a calculation.
From lesson 2: a model has one objective, and "maximise profit and minimise risk" is two models and a decision nobody has made. This lesson is that decision, treated properly — including what can be settled without making it, and what cannot.
With several objectives there is usually no single best plan. What there is, is a way to eliminate the plans nobody would choose.
Dominance. Plan A dominates plan B when A is at least as good as B on every objective and strictly better on at least one. A dominated plan can be discarded without any statement of preference: whatever you care about, A is at least as good.
The Pareto frontier is the set of plans nothing dominates. Every plan on it is better than every other on something, so choosing among them requires a preference the analysis does not contain.
That division is the honest structure of the problem:
| Question | Who answers it |
|---|---|
| Which plans are dominated? | the mathematics, undisputably |
| Which point of the frontier do we want? | a person, by stating a preference |
Scalarisation. The usual way to pick a point is to combine the objectives into one:
$$\min\ w_1 f_1(x) + w_2 f_2(x), \qquad w_i > 0.$$
Every choice of weights gives a point on the frontier, and every weight is an exchange rate between the objectives' units. "Cost plus $50 \times$ delay hours" asserts that an hour is worth $50$. That is a value judgement written as a coefficient, and the coefficient makes it look like a measurement.
The epsilon-constraint method is often more honest: optimise one objective and put the others in as constraints — minimise cost subject to delay at most two hours. Nobody has to price an hour; somebody has to say what delay is tolerable, which is a question people can usually answer.
Another way: picture
Points scattered on a pair of axes, one per plan. The ones on the upper-right boundary are the frontier; everything inside is dominated by something on it. A weighted objective is a straight line swept across the picture, and it stops on whichever frontier point its slope happens to touch first.
Another way: steps
With more than one objective:
Minimise $\text{cost} + 50 \times \text{delay hours}$. Cost is in pounds and delay is in hours, so the sum only makes sense if one hour is worth fifty pounds. That number is not in the data; somebody chose it.
And it decides the answer. Two plans, one costing $1{,}000$ with $4$ hours of delay and one costing $1{,}200$ with $0$: at $50$ per hour they score $1{,}200$ and $1{,}200$ — a tie. At $40$ the first wins; at $60$ the second. A number nobody defended is selecting the plan.
This is not an argument against weights. It is an argument for reporting them: say what the exchange rate is, say who chose it, and show how the answer moves as it varies. A model that hides a value judgement inside a coefficient is presenting a decision as a calculation, and that is the failure this lesson exists to prevent.
Averaging first. Combining objectives before computing the frontier throws away the picture of what is actually available, and leaves nobody able to see what was traded.
Weights on incomparable units. Money and lives, cost and safety. Sometimes such a rate has to be chosen; it should never be chosen silently.
Normalising away the meaning. Scaling both objectives to $[0,1]$ and weighting equally feels neutral and is not — it makes the exchange rate depend on the ranges of the data, which is an arbitrary rate chosen by accident rather than on purpose.
Reporting one frontier point as the answer. It is an answer to a question about preferences that somebody supplied. Report which question was asked.
Assigning weights feels like the technical way to handle several objectives, and it does produce a single number and a single answer. But the weights are the trade-off — every exchange rate corresponds to a point on the frontier, and choosing the rate chooses the point. Nothing has been computed that was not decided. The practical consequence is a reporting obligation rather than a mathematical one: state the weights, state that they are a judgement and whose, and show how far the answer moves when they change. A recommendation that is stable across a wide range of weights is a strong one; a recommendation that flips at a weight nobody defended is a decision in disguise.
Four plans, both objectives maximised: A $(8,5)$, B $(6,4)$, C $(7,3)$, D $(5,8)$.
The candidates.
B is dominated by A ($8 \ge 6$ and $5 \ge 4$, strictly better on both). C is dominated by A too ($8 \ge 7$, $5 \ge 3$). D is not: it beats A on the second objective.
Two eliminated with no preference stated.
Frontier: A and D. The decision is now between two plans rather than four, and it is a genuine trade — A for the first objective, D for the second — which is exactly the choice that needed a person. That is what the analysis is for.
The real choice, isolated.
Instead of minimise cost + 50 × delay, ask: minimise cost subject to delay ≤ 2 hours.
One objective, one constraint.
Nobody has to price an hour. Somebody has to say what delay is acceptable — a question an operations manager can answer from experience, where "what is an hour worth" is not.
A question people can actually answer.
And varying the tolerance traces the frontier: run it at $1$, $2$, $3$ hours and the costs come back as a table of real alternatives. That table is usually a better deliverable than any single optimum, because it shows the shape of the trade rather than one point on it.
Sweeping the constraint draws the frontier.
Minimising both cost and time: P $(100, 8)$, Q $(120, 5)$, R $(130, 9)$, S $(90, 12)$.
Both minimised, so lower is better on each.
R: cost $130$ and time $9$, against P's $100$ and $8$ — P is better on both, so R is dominated and out.
P, Q and S: P is cheaper than Q and slower; S is cheaper than P and slower still. Each beats the others on something, so all three are on the frontier. Three real alternatives, one eliminated, and the choice among the three needs someone to say what an hour of time is worth to them — which is the conversation the analysis has now made possible rather than pre-empted.
Both objectives are to be maximised. A scores $8$ and $5$; B scores $6$ and $4$. Does A dominate B?
The two objectives are combined as $f_1 + 2 f_2$. What is A's combined score, given $7$ and $3$?
Answer:
Of $19$ candidate plans, $4$ are on the Pareto frontier. What has been achieved?
Plan A scores $(7, 4)$ and plan B scores $(4, 7)$ on two objectives, both maximised, combined as $f_1 + t f_2$. At what $t$ do they tie?
Answer:
Both objectives are to be maximised. A scores $5$ and $8$; B scores $9$ and $2$. Does A dominate B?
A model minimises $\text{cost} + 90 \times \text{delay hours}$. What has the $90$ asserted?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
The two objectives are combined as $f_1 + 4 f_2$. What is A's combined score, given $7$ and $3$?
Answer:
You can compute a frontier, use dominance to eliminate, and say what a weight asserts. Next: what happens when the numbers in the model are not known.
9. Your turn: which of these are on the frontier?, step 3