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future_fabulators:formalised_decision_making [2013-10-17 19:20] – timbofuture_fabulators:formalised_decision_making [2014-02-11 07:54] – nik
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 ==== 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. On this page we look briefly at some ideas and thoughts. 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.
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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. +  * 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: 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.  +  * 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. +  * 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.+  * 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. 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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 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.  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=+==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? 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?
  
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 //(need to check the following calculations - done on paper in a shaky plane and then train!)// //(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(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.+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: 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(i=1,n-1)((1/i) sum(j=1,i-1)(1/j))+$(n-1)! \sum\limits_{i=1}^{n-1} ( (1/i) \sum\limits_{j=1}^{i-1} (1/j))$ 
 + 
 +but this needs some work to check it. (**note:** feel free to use inline $\LaTeX$ formatting)
  
-but this needs some work to check it. 
  
  • future_fabulators/formalised_decision_making.txt
  • Last modified: 2014-03-04 07:01
  • by maja