Boğaziçi University
Department of Economics
Spring 2017
EC 206 MICROECONOMICS II
Problem Set 3 - Solutions
1. There are four grilled chestnut sellers who are choosing where to locate on Istiklal Avenue. Each has only
three options to locate their vendor: At the Taksim entrance, at the Tunel entrance and in front of the
Galatasaray Lisesi. That is, if you think about the Istiklal Avenue as a unit length interval, [0,1], they
can locate their vendors at point 0, 1/2 or 1. There is already another vendor located at point 1/2
(This vendor is fixed and not a player in the game). They all take the price of the grilled chestnut as given
and they all have the same constant marginal cost. There are 100 buyers who are uniformly distributed
over the Istiklal Avenue. Each buyer buys from the vendor that is closest to him/her and buys only one
unit. If there is more than one vendor that is closest to a buyer, then the buyer buys from one of these
vendors with equal probability. Each seller is maximizing own sales. Find the set of pure-strategy Nash
equilibria, if any.
ANSWER: To make it easier to look at all possible cases, consider the possible distributions of the
sellers over the three points. Denote the case where there are x sellers at 0, y sellers at 1/2 and z sellers
at 1 with (x,y,z). Note that (x,y,z) is not a strategy profile!
case1: 4 sellers choose the same point:
(4,1,0), (0,5,0), (0,1,4): One seller deviates to an empty point. (25 > 25/4, 25 > 100/5 and 25 > 25/4,
respectively.)
case2: 3 sellers choose the same point:
(3,2,0), (0,2,3): One of the 3 sellers located at the same spot deviates to the empty point. (25 > 25/3)
(3,1,1), (1,1,3): One of the 3 sellers located at the same spot deviates to the other corner. (25/2 > 25/3)
(1,4,0), (0,4,1): One of the sellers at point 1/2 deviates to the empty point. (25 > 75/4)
case3: 2 sellers choose the same point:
(2,3,0), (0,3,2): A seller at point 0 or 1 deviates to the empty point. (25 > 25/2)
(2,2,1): One of the sellers at point 0 deviates to point 1/2. (50/3 > 25/2)
(1,2,2): One of the sellers at point 1 deviates to point 1/2. (50/3 > 25/2)
(2,1,2): One of the sellers at point 0 or 1 deviates to the point 1/2. (50/2 > 25/2)
(1,3,1): No deviation! Sellers at point 0 gets 25. If he deviates to 1/2 then receives 75/4, not profitable.
If he deviates to point 1 receives 12,5, again not profitable. Same is true for seller at point 1. A seller
at point 1/2 receives 50/3. If he deviates to either of the other two points, he receives 12,5, which is not
profitable. Thus, any distribution of this form gives a pure-strategy Nash equilibrium.
The cases are exhaustive. Therefore N E = {(s1 , s2 , s3 , s4 ) : exactly one si is 1, exactly one si is 0, exactly
two si is 1/2 }
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2. Two candidates, A and B, compete in an election where there are n many voters and n is even. Of the n
voters, n/2 support candidate A and n/2 support candidate B. Each citizen decides whether to vote, at
a cost, for the candidate she supports, or to abstain. A citizen who abstains receives the payoff of 2 if the
candidate she supports wins, 1 if this candidate ties for first place, and 0 if this candidate loses. A citizen
who votes receives the payoffs 2 − c, 1 − c, and −c in these three cases, respectively, where 0 < c < 1.
Find the set of all pure-strategy Nash equilibria.
ANSWER: There are four possible cases:
case1: Everyone votes: In this case A and B are tied and each voter gets a payoff of 1 − c. If a voter does
not vote then this voter gets 0. Thus there is no profitable deviation.
case2: Not everyone votes and candidates are tied: In this case, someone who does not vote and gets 1,
has a profitable deviation to voting to get 2 − c which is strictly larger than 1 since c < 1.
case3: Not everyone votes and one candidate wins by one vote: In this case, a voter who does not vote
for the losing candidate would profitably deviate to voting, since 0 < 1 − c.
case4: Not everyone votes and one candidate wins by more than one vote: In this case, a voter who votes
for the wining candidate would profitably deviate to not voting, since 2 − c < 2.
Thus, the only Nash equilibrium is the one where everyone votes.
3. Two players are playing a game in which each player requests an amount of money, simultaneously. The
amount must be an integer between 11 and 20, inclusive. Each player will receive the amount she requests
in TL. A player will receive an additional amount of 20 TL if she asks an amount that is exactly 1 less
than the other player’s amount. All of the above is common knowledge.
(a) Find the set of all pure-strategy Nash equilibria.
ANSWER: N Epure = ∅. To see this analyze case by case. Let n1 and n2 be the integers picked by
player 1 and 2, respectively. Then,
case1: 11 = n1 = n2 : player 1 deviates to 20.
case2: 11 < n1 = n2 ≤ 20: player 1 deviates to n2 − 1.
case3: 11 < n1 < n2 ≤ 20: player 2 deviates to n1 − 1.
case4: 11 = n1 < n2 < 20: player 2 deviates to 20.
case5: 11 = n1 < n2 = 20: player 1 deviates to 19.
(b) Now suppose we add one more rule to the game above: If the players pick the same number, then
they both get an additional 10 TL. Find the set of all pure-strategy Nash equilibria of this new
game.
ANSWER: N Epure = (11, 11). To see this first note that 11 is a best response to 11. Thus (11,11)
is a NE. Now look at other profiles.
case1: 11 < n1 = n2 ≤ 20: player 1 deviates to n2 − 1.
case2: 11 < n1 < n2 ≤ 20: player 2 deviates to n1 − 1.
case3: 11 = n1 < n2 ≤ 20: player 2 deviates to 11.
4. There are n players. Each player has one unit of a discrete private good and can either consume her
private good, or contribute it to the production of a public good. The public good will be produced if and
only if k or more players contribute. Each player’s payoff is the sum of a private good payoff and a public
good payoff: her total payoff if the public good is produced and she consumes her private good is v + c; it
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is v if the public good is produced and she does not consume her private good; it is 0 if the public good
is not produced and she does not consume her private good. Assume 2 ≤ k ≤ n, and v > c > 0. The
players make their contribution decisions simultaneously and irrevocably. Find all pure strategy Nash
equilibria of the game. Are they efficient?
ANSWER: Consider a strategy profile in which exactly k players contribute. Any player not contributing
is receiving v + c, the maximum feasible payoff, and therefore cannot possibly profit from deviating.
Consider a player who is contributing. This player receives v under the strategy profile we are considering.
By deviating to not contributing, he would cause the public good to not be produced and his payoff would
become c < v. Therefore, any strategy profile in which exactly k players contribute is a pure strategy
equilibrium. All these equilibria result in an efficient allocation.
Now consider the strategy profile in which zero players contribute, and as a result all receive a payoff of
c. Any player who deviates (to contributing) will receive 0 < c, since the public good will still not be
produced. Therefore, the strategy profile in which no player contributes is also a pure strategy equilibrium.
These are not efficient.
There are no other pure strategy equilibria. To see why, consider any strategy profile in which m players,
where 0 < m < k players contribute. Contributing players receive 0 and can profit by deviating to not
contributing and receive c > 0. Finally, consider any strategy profile in which m > k players contribute.
Now, contributing players receive v and can profit by deviating to not contributing and receive v + c > v,
since the public good will still be produced.
5. In a college there are n students. They are simultaneously sending data over the college’s data network.
Let xi ≥ 0 be the size of data sent by student i. Each student i chooses xi , simultaneously. The speed
of network is inversely proportional to the total size of the data, so that it takes x1 + ... + xn minutes to
send the message. The payoff of student i is ui (xi , x−i ) = xi − xi (x1 + ... + xn ).
(a) Find the pure-strategy Nash equilibrium of this game, and compute the equilibrium payoffs.
ANSWER: Given x−i , the best response of player i is given by
1 − (x1 + ... + xi + ... + xn ) − xi = 0
that is,
1 − (x1 + ... + xi−1 + xi+1 + ... + xn ) − 2xi = 0
Applying symmetry, xi = x for all i, we get
1 − (n + 1)x = 0, that is, x∗ =
Thus the unique Nash equilibrium is xi =
1
,
(n+1)2
1
n+1 ,
1
n+1
for all i. And the equilibrium payoffs are ui (x∗ , ..., x∗ ) =
for each i.
(b) Now, suppose we introduce a new program so that we pay each student a fixed amount M and
charge p ≤ 1 for each unit of data the student sends. Student i’s payoff will be ui (xi , x−i ) =
M + xi − xi (x1 + ... + xn ) − pxi . Find the pure-strategy Nash equilibrium under this program, and
the equilibrium payoffs.
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ANSWER: Given x−i , the best response of player i is given by
1 − (x1 + ... + xi + ... + xn ) − xi − p = 0
that is,
1 − (x1 + ... + xi−1 + xi+1 + ... + xn ) − 2xi = p
Applying symmetry, xi = x for all i, we get
1 − (n + 1)x = p, that is, x∗ =
Thus the unique Nash equilibrium is xi =
1−p
n+1 ,
1−p
n+1
for all i. And the equilibrium payoffs are ui (x∗ , ..., x∗ ) =
1−p 2
M + ( n+1
) , for each i.
Remark: Here, for a given n, one can choose p and M and increase social welfare: If M =
p(1 − p)/(n + 1) and p such that 1 + n > (2 − p)/(1 − p), then each student is better off in the
equilibrium of part (b) than the one in part (a). (Why?)
6. In the following normal-form games below, find all pure-strategy Undominated Nash equilibria, if any,
resulting from iterated elimination of Nash equilibria with weakly dominated strategies.
Game 1:
(1)/(2)
N
E
W
S
U
(2,1)
(1,6)
(2,3)
(4,6)
M
(4,2)
(6,5)
(1,5)
(3,2)
D
(5,6)
(8,7)
(5,3)
(4,3)
ANSWER: N E = {(D, E), (U, S)}. S is weakly dominated, so U N E = {(D, E)}.
Game 2:
(1)/(2)
N
E
W
S
U
(6,3)
(7,3)
(3,2)
(1,3)
M
(3,2)
(11,4)
(5,4)
(1,1)
D
(3,10)
(9,9)
(2,1)
(1,10)
ANSWER: N E = {(U, N ), (M, E), (M, W ), (U, S), (D, S)}. For player 1, D is weakly dominated, and
for player 2, both S and W are weakly dominated, so {(U, N ), (M, E)} are the ones that survive the
first round of iterated elimination of Nash equilibria with weakly dominated strategies. Now, in the
reduced game, N E = {(U, N ), (M, E)}, but N is weakly dominated for player 1 in this reduced game, so
U N E = {(M, E)}.
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