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CORNELL ECON 2040 - practice-final solutions

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Networks: Fall 2012 Practice Final Exam: solutionsDavid Easley and Eva Tardos December 1, 2013Details of the final exam are listed on handout with the questions. Here you get solutions,and list of other topics to think about.(1)(a) Yes, U is dominant strategy for player A(b) The unique Nash is (U,M)(c) No mixed Nash, as Player A has a dominant strategy, and hence cannot be neutralbetween the two options, and Player B has a single best response.(2)(a) If we write a for the PageRank of A, then the PageRank of B and C is a/4, thePageRank of D is a/2, the PageRank of E is 3a/8, and the PageRank of F is a/2.(b) Adding up all these PageRank values, we get 23a/8 = 1, and a = 8/23, and then wecan fill in the other values from that.(3)(a) Currently X has 2 points, Y has 3 points, and Z has 1. So for X to come out firstwithout a tie, it has to be ranked first and Y has to be ranked last; this puts Z in the middle.Thus we can tell what the ranking has to be.(b) We can’t tell, because this will happen in either of the two rankings where Y is first.(4) Two (of many) possible examples: Put three politicians at mutual 120-degree angles,creating a triangle with three − signs; or put one politician at the point (1,0), another 50degrees above this pint, and a third 50 degrees below this point. Then we get a triangle withtwo + signs and one − sign.(5)(a) No. The expected value of a car for a buyer is 15 × 1/3 + 10 × 1/3 + 5 × 1/3 = 10.This is a highest a car will be bought. Price is too low for the good car owners to sell.(b) Yes. In this case the expected value of a car for a buyer is 10 × 1/2 + 5 × 1/2 = 7.5.The price will have to be between 7 and 7.5.(c) Yes, and any price between 2 and 5 is possible.(d) Not possible, any equilibrium must sell bad cars, and if it sells good cars than also sellsmedium cars. Those equilibria are listed above.(6) (a) Yes: A becomes stronger because they’d now have C as an extra option in additionto B; as a result B is getting weaker, as his neighbor A got stronger, and C becomes strongerbecause their alternate option B is now weaker. So they both want this edge.(b) No: A is currently getting more than $0, and after the edge, A would expect to get$5 by the symmetry of its exchange with B. So it’s not worth $5 to add the edge.(7)(a) You’d expect B to be the closer friend. We know that more triangles have formedaround the A-B edge than the A-C edge. From the part of the course on strong and weakties, we know that triadic closure happens more effectively on strong ties, so this suggeststhat the A-B edge is a stronger tie.(b) The natural conjecture is that y > x: a close friend is more likely to comment ona photo than an arbitrary friend, and so y is an average of a sample that’s enriched withstrong ties. By (a), we know that these tend to have higher scores.(8)(a) Since it hasn’t spread into Ontario, we know the threshold is greater than 2/6 = 1/3.Also, since it no one in New York is ceasing to use (i.e. deciding to drop the feature) it’sreasonable to assume that the threshold is at most 4/6 = 2/3.(b) After using Strategy A, Ontario users have 4 friends in Ontario and 3 friends in NewYork. After using Strategy B, Ontario users have 3 friends in Ontario and 3 friends in NewYork.(c) Strategy B is a better choice, because it creates a graph where the feature will spreadas long as its threshold q satisfies q ≤ 1/2. With Strategy A, we would need q ≤ 3/7.(9)(a) Three equilibria: 0, and two solutions to the f(z)r(z) = p equation, that is (1 − z)(1 −2z)z = p. See the figure 1. Note that the maximum of the function (1 − z)(1 − 2z)z isactually a little less than 0.1 so there is a small range with p very close to 0.1 wherethere is only one equilibria at 0. We do not require students to notice this fact.011/2pFigure 1: Sketch of the curve f(z)r(z) and the line at p.(b) 0 and the higher equilibrium z2is stable, the middle one z1is not. The reason is thatin the range [0, z1] and [z2, 1] the fraction using the product is shrinking, while in therange [z1, z2] it is growing.(c) There are now three equilibria, consisting of the solutions to f(z)r(z) = (1 − z)(1 −2z)z = 0: these are z = 0, z = 1/2, and z = 1.(d) The middle equilibrium z is stable, 0 is not, and the equilibrium at z = 1 also not.The reason is that in the interval [0, z] the fraction using the product is growing, whilein the range [z, 1] it is shrinking.(10)(a) It is never useful to bid above your value. You can only win an auction with a bid babove your value, if you wouldn’t win with your value if the price is above the value.(b) Good idea. See the example above, A had to pay 1, 200, and could have paid less if hebids in the range (1000, 1200).Other topics not mentioned on this test• traffic equilibrium in networks• matching markets (market equilibrium)• web structure• sponsored search• information cascades• betting and beliefs• evolutionary game theory• epidemics: SIR model•


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