isi question - spcial planning

[RichaS]

[shiva]

a) The location could be the average of the positions of individuals, such as y = (x1+x2+........xn)/N. This will maximize the utility of the worst off individual also.

b) Same location as in (a) because it maximizes the utilities of all the individuals, and hence also the sum of utilities.

c) If the average of individual locations is (<1/2), ideal location is at y=1

if the average of individual locations is (>1/2), ideal location is at y=0
if the average = 1/2, the location will be either at y = 0 or y =1, depends on the skewed locations of the individuals.

I think this should be correct, even I am not so sure. Please check this and let me know if there are any corrections.

[Abhitesh]

Let x1<x2<x3<...<xN.
a) The location will be (x1+xN)/2. In case of average suppose that |x1-<x>|>|xN-<x>| then you'll have to move the location closer to x1 and vice-versa. Ultimately you'll settle for average of x1 and xN.

b)The average absolute deviation is minimum for median. So the answer for this part will be median of xi's. Try proving this.

c)The location of garbage can be 0 or 1 as explained by Shiva. Also if you choose either 0 or 1 then the utility will be N<x> or N(1-<x>). There is no dependence on skewed distribution. If <x>=1/2 then y=0 or 1 will be equally desirable.

[RichaS]

Thanks!