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| future_fabulators:formalised_decision_making [2013-10-17 19:42] – timbo | future_fabulators:formalised_decision_making [2014-02-11 07:54] – nik | ||
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| ==== Formalised Decision Making ==== | ==== Formalised Decision Making ==== | ||
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| + | 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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| //(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 | + | Some calculations show that the number of ways of doing this is $(n-1)! |
| 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)! |
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| + | but this needs some work to check it. (**note:** feel free to use inline $\LaTeX$ formatting) | ||
| - | but this needs some work to check it. | ||
maja