(see Chapter 11). Consider a Bertrand, price-competition model between two firms with capacity constraints. Suppose the market consists of 20 consumers, each of which will purchase one unit of the good. Suppose that each consumer is willing to pay at most $2.00 for the good. Suppose that each unit can be produced at a cost of $0.50 (fifty cents) and this cost is the same for both firms. Suppose that each firm can produce at most 15 units of the good (so the capacity constraint is 15 units for each firm).

(i) Is there a pure-strategy Nash equilibrium?

The answer is yes there is a pure strategy Nash equilibrium.

According to Bertrand pricing competitive model, the companies will choose prices rather than choosing quantities and then the competitive outcome would be at price=marginal cost.

If any of the two firms increase its prices, all customers will buy from the lower priced firm. And if state that one firm undercuts the other by charging lower prices, it will not satisfy the additional demand because the total demand is 20 whereas the capacity constraint is 15 units. Because both firms know that one can gain the whole market by reducing the price, each firm will always strive to undercut the other until the profit equals zero. Here the pure Nash equilibrium comes in.

(ii) Find the Nash equilibria in this game.

The Nash equilibria in this game:

Capacity constraint=15       (k₁+k₂)=20

Equilibrium is given at p₁=p₂=p (k₁+k₂) where k is capacity. This is the price where no unused capacity is available.

At Nash equilibrium, usually price equals production since there is no profit. In this case it would be;

P=c=$0.5

2. For each of the following games:

(i) Determine whether they are strictly competitive games.

(ii) Find all the Nash equilibria (in pure and mixed strategies).

(iii) Find the security strategies for each player.

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(a)

(b)

3. For each of these games:

(i)

A two-player, strictly competitive game is a two-player game with the property that, for every two strategy proÖles s and sí, u1(s) > u1(s’) and u2(s) < u2(s’). An implication of such games is – if u1(s) = u1(s’), then u2(s) = u2(s’).

For (A)

u1(H,H) > u1(H,T), i.e., 1 > 1
u2(H,H) < u2(H,T), i.e., -1 < 1

u1(H,T) < u1(T,T), i.e., -1 < 1
u2(H,T) > u2(T,T), i.e., 1 > -1

u1(H,H) = u1(T,T), i.e., 1 = 1
u2(H,H) = u2(T,T), i.e., -1 = -1

u1(H,T) = u1(T,H), i.e., -1 = -1
u2(H,T) = u2(T,H), i.e., 1 = 1

Therefore, (A) is a strictly competitive game.

For (b), u1(C,C) > u1(C,D) i.e., 2 > 0 but also u2(C,C) > u2(C,D) i.e., 2 > 0 so, it’s not perfectly competiitve.

For (c)

u1(H,H) > u1(H,D), i.e., 2 < 3
u2(H,H) < u2(H,D), i.e., 2 > 1

u1(H,D) < u1(D,D), i.e., 3 > 2
u2(H,D) > u2(D,D), i.e., 1 > 2

u1(H,H) = u1(D,D), i.e., 2 = 2
u2(H,H) = u2(D,D), i.e., 2 = 2

u1(H,D) = u1(D,H), i.e., 3 = 3
u2(H,D) = u2(D,H), i.e., 1 = 1

Therefore, (c) is a strictly competitive game.

For (d), u1(A,A) < u1(A,B) i.e., 1 < 2 but also u2(A,A) < u2(A,B) i.e., 4 < 8 so, it’s not perfectly competitive.

For (e), u1(A,A) > u1(A,B) i.e., 2 > 0 but also u2(A,A) > u2(A,B) i.e., 2 > 0 so, it’s not perfectly competitive.

For (f), the strategy combinations are themselves different for the two players, so, it’s not perfectly competitive.

(ii)

The pure strategy Nash equilibrium are marked (yellow box) as shown. There is no Nash equilibrium for a pure strategy for (a) and (c). So, we have to go for the mixed strategy for both of them.

For (a)

-pH + pT = pH – pT
qH – qT = -qH + qT
pH + pT = 1
qH + qT = 1

Solving, we will get pH = pT = qH = qT = 0.5, so, Nash equilibrium will exist when both the players will play H and T with a probability of 50%.

For (b)

2pH + pD = pH + 2PD
2qH + 3qD = 3qH + 2qD
pH + pD = 1
qH + qD = 1

Solving them, we get pH = pD = 0.5, and qH = qD = 0.50. So, Nash equilibrium will exist when both the players will play H and T with a probability of 50%.

4. For each of these games:

(i) Determine whether they are strictly competitive games.

A strictly competitive game is a two-player strategic game (S1, S2, p1, p2) in which for i = 1, 2 and any two joint strategies s and s ′ pi(s) ≥ pi(s ′) iff p−i(s) ≤ p−i(s ′)

1. i) a) This is not competitive game
2. b) This is competitive game
3. c) This is also competitive game
4. d) This is competive game

(ii) Find the security strategies for each player and determine whether they are rationalizable.

1. a) There is no security strategy as there is no strategy that is dominant for any player.
2. b) For player 2 strategies X and Y are dominant over Z but for player 1 there is no strategy that is dominant. So there is no security strategy
3. c) For player 1, A is dominant

For player 2, X is dominant

w1(A) = 1

W1(B) = 0

W2(X)= 1

W2(Y) = 0

Therefore, security strategy for player 1 is A and for player 2 is X. This is also rationalizable strategy.

1. d) For player 1, dominant strategy is D

w2(X) = 4

W2(Y) = 3

Player 2’s security strategy is X.

W1(U) =1

W1(D) = 4

Player1’s security strategy is D

It’s not rationalizable strategy.

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