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future_fabulators:formalised_decision_making [2013-10-17 19:10] – timbo | future_fabulators:formalised_decision_making [2014-03-04 07:01] (current) – maja | ||
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==== Formalised Decision Making ==== | ==== Formalised Decision Making ==== | ||
- | In the process of creating scenarios, forecasts and otherwise moving onwards, we are left with a problem of decision making. On this page we look briefly at some ideas and thoughts. | + | By Tim Boykett |
+ | |||
+ | In the process of creating scenarios, forecasts and otherwise moving onwards, we are left with a problem of decision making | ||
Some language that we use here: | Some language that we use here: | ||
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Some of the techniques that we have seen used that will be relevant: | 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. | + | * 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. | + | * ordering: selecting the factors in a list from highest to lowest in the evaluation. |
===Partially ordered sets and Coverings== | ===Partially ordered sets and Coverings== | ||
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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): | 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: | + | |
- | * Antisymmetry: | + | * Antisymmetry: |
- | * Reflexivity: | + | * Reflexivity: |
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. | 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. | ||
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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. | 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. | ||
- | 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. | + | We might say that a 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> | ||
+ | |||
+ | Formalities. We have n factors. Let us take one of our orders to be 1< | ||
+ | |||
+ | //(need to check the following calculations - done on paper in a shaky plane and then train!)// | ||
- | =Some calculations= | + | Some calculations |
- | In Scenario planning, we want to select 2 (or perhaps 3) factors that are highest in our ordering | + | |
- | Example: suppose we have the two orders that are precisely opposite, | + | 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: |
- | Formalities. We have n factors. Let us take one of our orders to be 1< | + | $(n-1)! \sum\limits_{i=1}^{n-1} ( (1/i) \sum\limits_{j=1}^{i-1} (1/j))$ |
- | // | + | but this needs some work to check it. (**note:** feel free to use inline $\LaTeX$ formatting) |
- | Some calculations show that the number of ways of doing this is (n-1)! sum(i=1, | ||