solution to 2&4

solution to 2&4 - maximize the sum of utilities d...

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1 2. a. Claim: n=1 is PE. Proof: Suppose that there is another allocation that Pareto dominates n=1. Clearly 1 is a admitted in this allocation, otherwise he would have been worse o/. But since other students are also admitted in this new allocation, 1 is worse o/. Thus, n=1 cannot be Pareto dominated. b. Let us denote an allocation as a vector x = ( x 1 ; :::; x 15 ) , where each coordinate takes the value 1 if i not. Also let us denote n ( x ) = P x i : Note that n ( x ) is the number of admitted students under allocation x: The PE allocations are the vectors that satisfy the following two conditions: (1) x 6 = (0 ; :::; 0) and (2) x i = 1 implies v i n ( x ) > 0 : c. It&s optimal to admit the ±rst 5 students; once letting mr.6 into the class, each one of the ±rst ±ve students losses a unit of utility - gross loss of 5- while mr.6 contributes only additional 4 units. So we should stop at 5, if we want to
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Unformatted text preview: maximize the sum of utilities. d. The following dominates n=1: admit also student 2, and let him pay one unit to student 1. Then 1 is exactly as happy as before, and 2 got a positive utility instead of zero. All the others are indi/erent. 4. Think of this as a game: each one has two strategies- to stick to the old path or to take the new path. Suppose that Z takes the old path: if I took the old path she would spent 30 minutes, while in the new path it would take only 29. Suppose that Z takes the new path: if I took the old path she would spent 35 minutes, while in the new path it would take only 34. So no matter what Z does it is optimal for I to take the new path. Similarly, no matter what I does , it is optimal for Z to take the new path. 1...
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