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CORNELL ECON 2040 - ps3_sol(1)

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Homework 3 Solutions(1) We know that in a second-price, sealed-bid auction that bidding your true value is adominant strategy. That is, no matter what the others do, you are better biddingtruthfully. So the fact that bidders 2 and 3 overbid does not affect bidder 1’s optimalbid. He should bid v1. It is enough to just state that bidding one’s true valueis a dominant strategy in a second-price sealed-bid auction, you did notneed to prove the assertion.(2a) Similar to question 1, it is a dominant strategy in a second-price, sealed bid auctionfor bidders 1 and 2 to bid their true values, v1and v2, respectively.(2b) Since bidder 3 does not know his true value, he cannot bid it. So our strategy from theprevious question does not apply in this situation. Instead we need to consider possibleoutcomes of the auction and their corresponding payoff to bidder 3 for various bids bybidder 3. Let the bid amount of bidder 3 be b3. There are three possible outcomes ofinterest. The first is that bidder 3 does not win the auction (in other words b3< v1orb3< v2). In this case, the net pay off to bidder 3 is zero (he doesn’t win the auction,but doesn’t have to pay anything). In the second case, bidder 3 wins the auction andbidder 2 is the second highest bid. In this case, bidder 3’s net pay off is again 0 sincehe gets v2(we are given v3= v2in the problem) of value from the item, but also hasto pay v2(since bidder 2 bids v2) for it . In the third case, bidder 3 wins the auction,but bidder 1 is the second highest bidder. In this case bidder 3’s net pay off is v2− v1,but since bidder 1 bid more than bidder 2, we know v1> v2which implies that bidder3’s net pay off is negative. In each of the cases, the best bidder 3 can receive is 0. If hechooses to bid any positive amount, it is possible he will end up in case 3, which hasa negative payoff. Therefore he should submit a bid of 0 or equivalently should choosenot to bid. When solving this question, it was also important to remember that bidder3 does not know v1or v2.(3) The optimal bid is less than vi. There were two possible routes for explaining this:• The first is to note that by cheating, the seller has actually turned the auction intoa first price auction. The winning bidder now pays the amount they bid for the item(instead of the second highest bid). As we know, it is not optimal to bid your truevalue in a first price auction. Instead the bidder should bid an amount less than theirtrue value.• The second is to reason through potential strategies for bidding. Bidding more than viis clearly a bad idea. If i wins then i would be offered the opportunity to buy the itemat a price above viwhich he would not accept. Bidding viis also a bad idea since if iwins he is offered the chance to buy the item at viwhich would yield 0 profit. Only bybidding less than vican i have a chance of making a positive profit. Therefore, bidderi should bid less than vi.Please note that it is not possible to actually derive the optimal value of viin this question,as you were not given the distribution of values. Instead, you just needed to determine thatthe bidder should bid less than viand provide justification for that fact.1(4) Let us label the residents A through I, going from top to bottom of the diagram and thehospitals 1 through 4 (likewise top to bottom). To begin this question, we need to lookfor either a perfect matching or a constricted set. We can find two such constricted sets.First, consider the residents. The four residents B, D, G, and H are qualified for twohospitals (1 and 3) that need a total of only three residents. The second constricted setis found among hospitals 2 and 4. These two hospitals need six total residents betweenthem, but only five residents are actually qualified for positions at these hospitals (A,C, E, F, and I). We can explain this problem to the hospital administrator by showinghim either 1) the two hospitals who need six residents but only have five qualifiedcandidates or 2) the four residents who are competing for only three eligible positions.This explanation should avoid any explicit mention of network related terminology.(5) With the given prices and values we can construct a table of net values for all threebuyers:Buyer Value for a Value for b Value for cprice 6 2 0x r−6 s−2 ty 3 1 1z 1 1 2(5a) For the prices to be market clearing, we know each buyer must get one of the itemsthat gives him the highest net value. Therefore, from the table above we know thata’s item must go to y and that c’s item must go to z. This implies that x must buyb’s item for the market to clear.(5b+c) We begin by answering part c. As in part a, we know for the market to clear thatx must value b’s item at least as much as a or c’s item at the given prices. We canuse this information and the net value table above to restrict the possible values for r,s, and t. Since x must value b’s item at least as much as a’s we know from the tableabove that r −6 ≤ s − 2 which implies r − 4 ≤ s. Also, we know that x must also valueb’s item at least as much as c’s. Similarly, from the table above we know that at theseprices: t ≤ s − 2 which implies t + 2 ≤ s. Therefore, the two inequalities that definethe range of possible values of r, s, and t are t + 2 ≤ s and r − 4 ≤ s (which is theanswer for part c). Any three values of r, s, and t for which those two inequalities holdcan be chosen as the answer for part b. Please note that the reason the inequalitiesare not strict (i.e. we use ≤ instead of <) is that market clearing can hold when twoor more items both give the buyer the same highest net value, as long as he gets oneof the items, he is indifferent as to which he gets and market clearing is satisfied.(6a) The prices of 10 for a and 12 for b are not market clearing. To see this, we constructanother net value table:2Buyer Value for a Value for bprice 10 12x 2 0y −2 −4z 4 2From the table, we see that both x and z have the highest net value for item a. Giventhey both value the two items equally, but a has the lower price, they will prefer thegood with the cheaper price. Therefore the set {x, z} is constricted and the marketwill not clear.(6b) As we saw in part a, if the items do not have the same price then the buyers willprefer the cheaper item and markets will not clear. Therefore, we must give the twoitems identical prices for markets to clear. Next, we must determine the range ofpossible prices. To do


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