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Therefore, the sole equilibrium in the Bertrand model emerges when both firms establish a price equal to unit cost, known as the competitive price. [9] It is to highlight that the Bertrand equilibrium is a weak Nash-equilibrium. The firms lose nothing by deviating from the competitive price: it is an equilibrium simply because each firm can ...
Repeated interaction or repeated price competition can lead to the price above MC in equilibrium. [7] More money for higher price. It follows from repeated interaction: If one company sets their price slightly higher, then they will still get about the same amount of buys but more profit for each buy, so the other company will raise their price ...
As a solution to the Bertrand paradox in economics, it has been suggested that each firm produces a somewhat differentiated product, and consequently faces a demand curve that is downward-sloping for all levels of the firm's price.
The Bertrand equilibrium is the same as the competitive result. [ 53 ] [ clarification needed ] Each firm produces where P = MC {\displaystyle P={\text{MC}}} , resulting in zero profits. [ 49 ] A generalization of the Bertrand model is the Bertrand–Edgeworth model , which allows for capacity constraints and a more general cost function.
In microeconomics, the Bertrand–Edgeworth model of price-setting oligopoly looks at what happens when there is a homogeneous product (i.e. consumers want to buy from the cheapest seller) where there is a limit to the output of firms which are willing and able to sell at a particular price. This differs from the Bertrand competition model ...
Market equilibrium computation (also called competitive equilibrium computation or clearing-prices computation) is a computational problem in the intersection of economics and computer science. The input to this problem is a market , consisting of a set of resources and a set of agents .
Bertrand's box paradox: the three equally probable outcomes after the first gold coin draw. The probability of drawing another gold coin from the same box is 0 in (a), and 1 in (b) and (c). Thus, the overall probability of drawing a gold coin in the second draw is 0 / 3 + 1 / 3 + 1 / 3 = 2 / 3 .
If the matrix is invertible then this is a linear system of equations with a unique solution, and so given some final demand vector the required output can be found. Furthermore, if the principal minors of the matrix I − A {\displaystyle I-A} are all positive (known as the Hawkins–Simon condition ), [ 6 ] the required output vector x ...