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future_fabulators:formalised_decision_making

Formalised Decision Making

By Tim Boykett

In the process of creating scenarios, forecasts and otherwise moving onwards, we are left with a problem of decision making and selecting a small amount of elements from larger clusters, for example in selecting two critical uncertainties in the 2×2 scenario matrix (see scenario methods). On this page we look briefly at some ideas and thoughts.

Some language that we use here:

  • A factor is one of the things we are selecting for future work
  • An evaluation is what the participants in a workshop do
  • A selection is the factors we select as being the most relevant ones to keep working with afterwards.

Evaluations

Some of the techniques that we have seen used that will be relevant:

  • dots: each participant get a number of dots to allocate to factors. More dots should indicate more of whatever it is that participants seek, whether that be relevance, importance, etc.
  • numbers: giving factors a numerical value, whether from 1 (uninteresting) to 5 (thrilling) or with a middle level from which two extremes vary i.e. plus and minus points, or having e.g. 5 as neutral, 10 as love and 0 as hate.
  • ordering: selecting the factors in a list from highest to lowest in the evaluation.

Partially ordered sets and Coverings

Note: the following is based upon some formal mathematical ideas: we endeavour to be accurate but not overload the following text with technicalities.

One way we have of proceeding is to rank various choices by a number of parameters. Whether these are importance, variability, causality, unpredictability or many others, we are left with the problem of selecting some of these to move forwards with. A set if called totally ordered, or linearly ordered, if for any pair of elements in that set, call them A and B, one is above or below the other in the order. So either A=B, A<B or A>B. It is a basic thesis of rational economics that a person, given the choice of two packages, can select one of them. This is their preference. Since any two distinct packages can be ordered by this preference, the set of packages is totally ordered.

If the choice of things we had to select from were to be totally ordered, then there would be a maximal element, so we could select that. But this is not the case, in the formats we have experimented with, we have various ways of measuring things and we want to select the maximum.

A partial order, on the other hand, does not have the requirement that every pair of elements has an ordering. We require only the following axioms (note that we write A<B for “A is less than or equal to B” because we cannot find the symbol right now):

  • Transitivity: if A<B and B<C then A<C. If love is more important than sex and sex is more important than food, then love is more important than food.
  • Antisymmetry: if A<B and B<A then A=B: If time is more important than money and money is more important than time, then time is money.
  • Reflexivity: A <A, a sunset is at least as good as a sunset.

Given two total orderings, we can define a new partial order from them by saying that A>B if A is above B in both orders. This is called the product of the two orders.

In our process of selecting factors to carry on with, we want to select ones that are maximal (nothing is higher than them in the scale) and are better than others. In general, we would select all the maximal elements in the partial oder, then we have a collection of factors than are the most relevant. If we are using the product order, this collection of maximal elements is the factors that are the top of one of the total orders that we selected.

Numerical Techniques

One modality has been to give each factor a numerical value, a score, and to use this for ordering. Each factor gets a value from 0 to 10, possibly with decimal points. A natural way to select which are the most important is to add these numbers together to get a total order.

This has one significant problem: how do we know that the (subjective) difference between 3 and 6 in one evaluation is the same as the difference between 5 and 8 in another evaluation?

Downsets

Each element in a partially ordered set (poset) has a downset, the set of elements that are below it in the order. The downset of a set of elements is the union of the downsets of each member of that set.

We might say that a set dominates its downset. We will call a 2-selection a selection of size 2 that dominates the whole set.

So a possible selection technique is to take a set of factors S so that the number of factors not in the downset of S is as small as possible.

Some calculations

In Scenario planning, we want to select 2 (or perhaps 3) factors that are highest in our ordering of importance and uncertainty. So we have two orders, giving a product partial order, from which we want to select 2 factors. A mathematical question arises: how often will we have the situation that this is (not) possible?

Example: suppose we have the two orders that are precisely opposite, in one we have A>B>C>D>… and in the other A<B<C<D<… so there are no orderings in the product. Then no selection other than everything will dominate. This is highly unlikely, but it is still there.

Formalities. We have n factors. Let us take one of our orders to be 1<2<3<…<n, then the other order will be a permutation of this one. we will write it as a sequence of numbers xxxx. With some thought, it becomes clear that to have a single selection, the second order (permutation) must be xxxxxn so that n is at the top of both orders. This occurs with a chance of 1 in n. In order to have a 2 factor selection we need the permutation to be xxxnyyyz where z is the largest number not to occur in the set xxxn. Then the selection n and z will be enough to dominate.

(need to check the following calculations - done on paper in a shaky plane and then train!)

Some calculations show that the number of ways of doing this is $(n-1)! \sum\limits_{i=1}^{n-1} (1/i)$, where k! means the factorial of k. As the total number of permutations is n!, the proportion of permutations that have a 2-selection is $1/n \sum_{i=1}^{n-1} 1/i$. This means that with 6 factors, the likelihood of having a 2-selection is around 5/12, less than 50%. The likelihoof of a 1-selection is 1/6, so there is around 7/12 chance of having a 1 or 2 selection. For 14 factors, the likelihood goes down to one quarter.

To Do: work out the corect formula for counting 3-selections. These would be xxxnyyyYzzzZ where Y is the largest factor not in xxxn and Z is the largest factor not in xxxnyyyY. My current conjecture is:

$(n-1)! \sum\limits_{i=1}^{n-1} ( (1/i) \sum\limits_{j=1}^{i-1} (1/j))$

but this needs some work to check it. (note: feel free to use inline $\LaTeX$ formatting)

future_fabulators/formalised_decision_making.txt · Last modified: 2014/03/04 07:01 by maja