Adjusted winner procedure

The adjusted winner procedure

The adjusted winner procedure is a method for dividing assets between two individuals (e.g., in the case of a divorce). It requires each person to divide 100 points among the assets, reflecting their relative values of the items. It also requires that the assets be divisible (e.g., a case of wine can be divided, as opposed to a mantle clock which cannot be divided). Each item is then tentatively assigned to the party which placed more points on that item. The person who has a lower point total (call him Louis) then acquires from the other (call her Helga) part of the asset for which the ratio of the recipient's (Louis') point assignment to the donor's (Helga's) point assignment is highest (this ratio will be less than one, since the donor has the item because she valued it more). The amount of the asset that is transferred is determined so that after it is transferred, the point totals of the two parties will be equal (it may be necessary to transfer part of another asset if Helga still has more points when all of the first asset is transferred). This is proportionate and envy free, as the cake cutting procedure was, but is also Pareto optimal (no other division could increase the satisfacton of both parties.

To illustrate this procedure, consider the division of an estate consisting of an apple orchard (A), a blueberry patch (B), a cranberry bog (C), a date grove (D), and an elderberry patch (E). The points assigned to these by Ralph (R) and Susan (S), respectively, are A:9,12; B:14,15; C:22,25; D: 22,32; and E:33,16 (9+14+22+22+33=100=12+15+25+32+16). The preliminary assignment gives A,B,C,D to Susan because she valued them more, and E to Ralph, But Susan's point total is 84, while Ralph's is only 33, therefore Ralph needs more. The highest ratio (Ralph/Susan) on Susan's items is 14/15=.93, therefore we transfer B to Ralph, because that item increases Ralph's points most relative to the loss of points for Susan. The new point totals are Ralph:47 (BE), Susan:69 (ACD). Susan still has more points, so based on the ratio 22/25=.88 being the highest among ACD, C should be transferred to Ralph, but that would result in new totals Ralph:69, Susan:44, which is transferring too many points. Fortunately, the cranberry bog can be divided. Giving x of it to Ralph and leaving 1-x of it with Susan, we need 47+x×22 = 44+(1-x)×25 (47 is Ralph's valuation of the blueberry and elderberry patches, 22 is his value for the cranberry bog; 44 is Susan's valuation of the apple orchard and date grove, 25 is her value for the cranberry bog). This equation is readily solved yielding x=22/47=.47 and Ralph (with the blueberry and elderberry patches, and .47 of the cranberry bog) and Susan (with the apple orchard, date grove, and .53 of the cranberry bog) each get 57.3 points.

Exercise: What is the adjusted winner division between Ahmed and Beauregard for Frankincense, Gold, Hemlock, Indigo, and Jasmine with the respective point assignments F:5,10; G:25,10; H:35,20; I:15,35; J:20,25?

If you know your opponents weightings

One maximizes what one gets by being charged as few points as possible for getting it. Therefore, Ralph could increase his share if he knew Susan's point schedule. In particular, if Ralph knew he was going to end up with B and E, he should barely go over Susan's values, e.g., give B 16 and E 17 points. That leaves him with 14 points to redistribute, he should distribut them to items he does not get, but making sure they do not raise those items in his preference list. The ratio .88 gave him a share of the cranberry bog. If Ralph gives 3 points to the cranberry bog to make its ratio 1, then he can give 2 to the apple orchard and 9 to the date grove. The new points are: A:11,12; B:16,15; C:25,25; D: 31,32; and E:17,16. Ralph gets B and E costing 33 points, Susan gets A and D costing 44 points. The cranberry bog must be divided between them yielding 33+x×25 = 44+(1-x)×25 or x=.72. The new point totals of 51 are not of interest, what is of interest is that Ralph will get .72 instead of .47 of the cranberry bog. Careful adjustment of the point values would result in even more of the cranberry bog going to Ralph.

The problem of discrete objects

What would happen if the cranberry bog could not be divided? Then the apple orchard or the blueberry patch would need to be used to equal the point values. If Ralph took the cranberry bog, Sarah would only keep (25/29) of the blueberry patch, resulting in a point total of 56.93 for each. If Susan took the cranberry bog, Ralph would take the entire apple orchard and (1/54) of the elderberry patch, resulting in a point total of 56.41. If some assets cannot be divided, the level of equal compensation will be lower, or perhaps equal compensation will not be attainable.

Competencies: If Vera's and Walter's preference weightings for the Wedding flowers are respectively: Astors 12,20; Begonias 18,10; Chrysanthmums 20,15; daffodils 25,15; edelweiss 13,25; and ferns 12,15 ((it was a short marriage), how should the flowers be distributed?
How many additional flowers could Vera get by adjusting her preference weightings?
If all of each type of flower must go to one person, what is the fairest possible division?



May 2003

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