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future_fabulators:formalised_decision_making [2013-10-17 19:42] timbofuture_fabulators:formalised_decision_making [2014-02-11 07:40] maja
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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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 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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 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(i=1,n-1) ( (1/i) sum(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