IF&PA in 20 minutes

1 · Problem

One quantity was measured six times. Which value to accept?

Each result is an interval “value ± uncertainty”. Five agree with each other. The sixth is far away — and states the smallest uncertainty. Is it wrong, or are the others?

This is what interlaboratory comparisons, sensor readings and results of different methods look like.

2 · Why not average

The mean and the weighted mean are pulled by the outlier

The weighted mean is the metrology standard: a result weighs 1/u². A narrow interval gets a huge weight. Try it: drag an end of interval I₆ to make it even narrower — watch the dashed lines.

The more confident a wrong result is, the harder it pulls the weighted mean. We need a way in which confidence gives no power.

3 · The idea

Hold a vote

Social choice theory studies how to turn many opinions into one. Seven voters rank three candidates. Different counting rules give different winners — see which.

4 · Condorcet paradox

A collective opinion can be contradictory

Each of three experts reasons consistently, but together they prefer a₁ to a₂, a₂ to a₃ and a₃ again to a₁. The pairwise-win graph closes into a cycle. The Condorcet rule finds no winner here.

5 · Kemeny rule

Choose the order closest to all opinions

The Kemeny rule goes through all strict orders and picks the ones closest to the profile. The histogram shows the distances of all 120 orders for the five microclimate zones: the minimum is reached by two at once — multiple solutions.

6 · An interval is an opinion

An interval votes for the values it covers

Split the axis into discrete values — the “candidates”. An interval prefers values inside it to values outside. Such a ranking is called an interval-induced ranking. Try it: move the intervals and watch their “ballots” — the rows of the matrix — change.

7 · The algorithm step by step

Six steps of IF&PA

Press “Next” and watch the result emerge from the intervals. The intervals can be dragged at any step.

8 · The key property

One interval, one vote

An interlaboratory comparison (a teaching example from the calculator). Laboratory L7 is wrong and states the smallest uncertainty. Try it: make L7 even narrower or move it further — the PM stays, the weighted mean follows L7.

9 · Partition norm

The result depends on the partition norm — and not monotonically

The PM is one of the discrete values (or the midpoint between two), so as the number of values n changes the result moves in jumps within the norm h. Try it: move the n slider. Hence the uncertainty component proportional to h, and the idea of self-refinement — a finer partition around the result found.

10 · When the method helps

Honest data versus confident outliers

The browser generates 200 data sets in two situations and compares the errors of ten estimators. The shorter the bar, the more accurate the estimator. The preferential median is highlighted.

With honest data the weighted mean is better — it uses the precision of every result. When confident results are wrong, the PM wins. Set your own conditions →

11 · Where it is used

From the Planck constant to weld seams

12 · Norm from history

IF&PA estimates a labour norm

The norm b is how many shifts a unit of work volume takes. From the history it is found from the ratios T_k/V_k — actual duration to the volume done. The ratios are usually close together, but delays — bad weather, downtime, waiting for materials — give rare large values, always in the same direction. Each observation becomes an interval, and the norm is chosen by voting. Try it: move the delays even further — the mean grows, the PM stays with the typical values.

A norm overstated by the mean means extra months in the plan; an understated one means missed deadlines. Next — how a norm turns into a schedule.

13 · The knapsack problem

A month is a knapsack

The capacity of the knapsack is the monthly capacity R in crew-shifts. The items are work packages — parts of works on sections of the route. Package j needs r_j = b_j·V_j shifts — the norm of its work times the volume — and gives the volume V_j. Choose packages so that they fit and the volume done is the largest. Try it yourself, then compare with the optimum and with the greedy choice.

This is the classical knapsack problem: Σ r_jx_j ≤ R, where x_j = 1 if the package is taken. It is NP-hard — enumeration grows exponentially with the number of items — but problems of practical size are solved exactly by branch and bound.

14 · Multidimensional knapsack

Several capacities — the packages must fit all of them

A month has more than one capacity: crews, machinery, materials. A package uses each of them, and for every resource type l we need Σ r_j^lx_j ≤ R^l. This is the multidimensional knapsack. Try it: a set that fitted the crews may now fail on machinery.

How “profitable” a package is depends on which resource is scarce, so simple greedy rules fail here more often than in the one-dimensional problem.

15 · Schedule

A project is a chain of month-knapsacks

Each month t is a knapsack of capacity R; the decision x_jt = 1 puts package j into month t. The order is kept: earthworks on a section not before clearing, installation not before earthworks. The norm depends on the month: in winter clearing and installation take one and a half times longer, earthworks twice as long. The smallest duration H is found by increasing H until the first feasible plan. Try it: set an error of the norms — the plan is built with the wrong norms and executed with the true ones.

Understated norms give a short plan that cannot be executed; overstated ones give extra months. That is why the norm must be accurate and unbiased — and why a robust estimator is needed. The full model with several resources and an exact solver is in the Planner.

16 · Self-check

Six questions

17 · What next

Now — on your own data