Practice_midterm_2_solutions.pdf - 1 For the general sum games below nd safety strategies(for both players and all mixed and pure Nash Equilibrium and

Practice_midterm_2_solutions.pdf - 1 For the general sum...

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Unformatted text preview: 1. For the general sum games below nd safety strategies (for both players), and all mixed and pure Nash Equilibrium, and the respective expected payos for both players in each case. (a)   (1, −2) (2, 1) (b)  (2, 2) (0, −1) (2, 0) . (−1, −1)  (−1, 0) . (1, 1) Solution First we need to nd safety strategies. For convenience let's write out the payo matrices of both players separately. Since the matrix B is a payo to the second player, it is more convenient to look at its transpose. a)     1 2 −2 1 T A= , B = . 2 −1 0 −1     2 −1 2 −1 T A= , B = . 0 1 0 1 b) We are to nd safety strategies for the 2 × 2 games. First, check that there are no saddle points. Highlite the maximal elements in every column:     1 2 −2 1 T a) A = 2 −1 , B = 0 −1 . b) A =   2 −1 , 0 1 T  B =  2 −1 . 0 1 Note that the highlited elements are never minimal elements in their row. Thus, there is no pure saddle point, and we just need to nd equalizing strategies of the rst player: a) x1 + 2(1 − x1 ) = 2x1 − (1 − x1 ) =⇒ x1 = 0.75, −2y1 = y1 − (1 − y1 ) =⇒ y1 = 0.25, b) 2x1 = −x1 + (1 − x1 ) =⇒ x1 = 0.25, 2y1 = −y1 + (1 − y1 ) =⇒ y1 = 0.25. Thus the pair of safety strategies for the part a is (0.75, 0.25)T and (0.25, 0.75)T . For the part b the safetystrategies are (0.25, 0.75)T and (0.25, 0.75)T . Th corresponding pairs of expected payos are given by xT Ay, xT By which is equal to a)(1.25, −0.5) b)(0.5, 0.5). Now we search for all the Nash equilibria. For any equilibrium (x∗ , y∗ ) there are 3 possibilities: • Both (x∗ , y ∗ ) are pure. • Both (x∗ , y ∗ ) are fully mixed. • One of (x∗ , y ∗ ) is pure and another is fully mixed. We know that the fully mixed strategy is an optimal response i the another players strategy is equalizing. Since there are no pure equalizing strategies in this game, the third case never holds. Thus, we only need to look for pure or fully mixed equilibria. Pure equilibria: leave only the greatest elements in each column for the matrix of the rst player, and in a row for the second player:   (∗, ∗) (2, 0) . (2, 1) (∗, ∗)   (2, 2) (∗, ∗) . (∗, ∗) (1, 1) We've found 2 pure equilibria in each game. Fully mixed equilibria are exactly the pairs of equalizing strategies: a) − 2x1 + (1 − x1 ) = −(1 − x1 ) =⇒ x1 = 0.5, y1 + 2(1 − y1 ) = 2y1 − (1 − y1 ) =⇒ y1 = 0.75, b) 2x1 − (1 − x1 ) = (1 − x1 ) =⇒ x1 = 0.5, 2y1 − (1 − y1 ) = (1 − y1 ) =⇒ y1 = 0.5, Thus, each game gas one fully mixed equilibrium. Finally, we compute the average payos to the players, which gives us the following answer:   a) The pair ofsafety strategies is (0.75, 0.25)T, (0.25,0.75)T with the corresponding payos (1.25, −0.5). The Nash  equilibria are (0, 1)T , (1, 0)T , (1, 0)T , (0, 1)T , and (0.5, 0.5)T , (0.75, 0.25)T with the corresponding payos (2, 1), (2, 0) and (1.25, −0.5).   b) The pair of safety strategies is (0.25, 0.75)T , (0.25, 0.75)T with the corresponding payos (0.5, 0.5). The Nash      equilibria are (1, 0)T , (1, 0)T , (0, 1)T , (0, 1)T , and (0.5, 0.5)T , (0.5, 0.5)T with the corresponding payos (2, 2), (1, 1) and (0.5, 0.5). 2. A simultaneous congestion game: There are two drivers, one who will travel from A to C, the other from B to D. Each road is labelled (x, y), where x is the cost to any driver who travels the road alone, and y is the cost to each driver if both drivers use this road. Write the game in matrix form, and nd all of the pure Nash equilibria. (1,5) A D (3,6) (2,4) (1,2) B C Solution. The rst driver can pick one of two possible routes: ADC or ABC . The second driver also has two choices: BAD and . Suppose the rst driver chooses ADC and the second  BCD. Then the road CD is used by two drivers with cost for both, the rst player also uses the road AD for the price 1, and the second driver  BC for price 1. Thus, their payments are (5, 5). Analogously we can compute the payments for all the other their choices: denote XY [k]  the price of the road XY when used by k drivers and write the matrix of the game (in terms of costs): BCD 4 ABC ADC BAD BCD   AB[2] + BC[1], AB[2] + AD[1]  BC[2] + AB[1], BC[2] + CD[1]  AD[2] + DC[1], AD[2] + AB[1] CD[2] + AD[1], CD[2] + BC[1]   (7, 7) (5, 4) . (7, 8) (5, 5) To nd Nash equilibria leave only the smallest (as we are dealing with costs) elements in each column for the matrix of the rst player, and in a row for the second player:  (7, ∗) (7, ∗)  (5, 4) . (5, 5) Thus, there are two pure Nash equilibria (ABC, BCD) and (ADC, BCD). 3. Find all the ESS and ESMS for the game (1, 1) (1, 1) (0, 0) (1, 1) (0, 0) (0, 1) (0, 0) (1, 0) . (0, 0) Solution First, we need to nd all the symmetric Nash equilibria. Let's start with the pure equilibria  we check the diagonal elements of the matrix for equilibrium. It is easy to see, that only the top left corner element of the matrix is a symmetric pure Nash equilibrium. Next we nd mixed equilibria. Write x = (x1 , x2 , x3 )T Note that xT B = (x1 + x2 , x1 + x3 , 0), which means that optimal response to x can never have a positive third coordinate (as the third strategy always gives zero payo, which is strictly less then x1 + x2 or x1 + x3 ). Thus the non-pure symmetric Nash equilibrium x should have . Let's nd such equilibria from the equalizing principle: the optimal response to has two rst coordinates positive i x1 6= 0, x2 6= 0, x3 = 0 (x1 + x2 , x1 + x3 , 0) x1 + x2 = x1 + x3 ≥ 0, x3 = 0 =⇒ x2 = 0. We met a contradiction, so there are no mixed symmetric Nash equilibria in this game. Now we need to check if the equilibrium x = (1, 0, 0)T is ESS or ESMS. Let's show that it is ESMS. Then it will also follow that x is ESS. We need to check that if for some mixed z 6= x holds zT Ax = xT Ax, then xT Az > zT Az. However, we have xT AT = (1, 1, 0), z T Ax = xT AT z = (1, 1, 0) · z = z1 + z2 , xT Ax = 1, z T Ax = xT Ax ⇐⇒ z1 + z2 = 1 ⇐⇒ z3 = 0. So if zT Ax = xT Ax then z3 = 0. Furtermore, xT A = (1, 1, 0), xT Az = (1, 1, 0) · z = z1 + z2 = 1,   1 1 z1 z T Az = (z1 , z2 ) = z12 + 2z1 z2 = (z1 + z2 )2 − z22 = 1 − z22 , 1 0 z2  xT Az ≤ z T Az ⇐⇒ 1 ≤ 1 − z22 ⇐⇒ z2 = 0 ⇐⇒ z = (1, 0, 0)T = x. So we got that for any mixed z 6= x if zT Ax = xT Ax, then xT Az > zT Az. Thus, (1, 0, 0)T is a unique ESMS (and ESS) in this game. 4. Show that Theorem 8.1.8 (in the book) holds in the presence of multiple trac ows. Specically, let G be a network where ri > 0 units of trac are routed from source si to destination ti , for each i = 1 . . . k. Suppose that the latency function on each edge e is ane; that is, e(x) = ae x + be , for constants ae , be > 0. Show that the price of anarchy is at most 4/3; that is, the total latency in equilibrium is at most 4/3 that of the optimal ow. In fact, we just repeat the proof of the Theorem 8.1.8. First, let's proove that if f is an equilibrium ow, and f 0 is some other path ow, then Solution X (Fe0 − Fe )l(Fe ) ≥ 0, e where are edge ows, corresponding to f, f 0 respectively. (This is Lemma 8.1.7 from the book, with the only dierence that now we have multiple trac ows.) Indeed, x any i and let Pi denote the set of all paths from si to ti . Since f is a Nash equilibrium, for any p ∈ Pi if fp > 0 then Lp (f ) = minπ∈P Lπ (f ), where Lp (f ) is a latency of the path p, corresponding to the path ow f . Denote Fe , Fe0 i Li = min Lp . p∈Pi Then we can write X Fe l(Fe ) = e XX i fp Lp (f ) = X Li i p∈Pi X fp = X Li ri . i p∈Pi Next, " X e Fe0 l(Fe ) = X e l(Fe ) # X p:e∈p fp0 = " X fp0 p # X l(Fe ) = e:e∈p X fp0 Lp (f ) = p XX i p∈Pi Combining the results above we get what we wanted: X Fe0 l(Fe ) ≥ e X i X e Li ri = X Fe l(Fe ), e (Fe0 − Fe )l(Fe ) ≥ 0. fp0 Lp (f ) ≥ X i Li X p∈Pi fp0 = X i Li ri . Now it is only left to repeat the proof of the theorem 8.1.8 verbatim: consider f ∗  an optimal ow, f  an equilibrium ow. Denote the average latency of the ow φ as L(φ) and write L(f ) = ∗ L(f ) − L(f ) ≤ X X Fe l(Fe ) ≤ X e e Fe∗ l(Fe∗ ) l(Fe ) − e  Fe∗ l(Fe ), = X Fe∗ ae (Fe − Fe∗ ). e Using the inequality x(y − x) ≤ y2 /4 we obtain L(f ) − L(f ∗ ) ≤ X e Fe∗ ae (Fe − Fe∗ ) ≤ 1 1X 1X ae Fe2 ≤ Fe (ae Fe + be ) = L(f ), 4 e 4 e 4 L(f ) 4 < . L(f ∗ ) 3 ...
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