<p> Dixit Entry Deterrence Model Analysis</p><p>Basic Assumptions Firm 1 is the incumbent monopolist Firm 2 is a potential entrant The firms produce identical goods</p><p>Demand: P = A - q1 – q2 Marginal cost of capacity for both firms = r Marginal cost of labor for both firms = w One unit of capacity and one unit of labor is required to produce one unit of output</p><p>Firm 1’s cost given output q1, capacity K1, and fixed cost F1 is Firm 2’s cost given output q2 and fixed cost F2 is. Firm 2 can avoid its fixed costs by not producing.</p><p>The game Dixit analyzes has 3 stages.</p><p>Stage 1: Firm 1 chooses a level of capacity K1 which generates a cost rK1. Stage 2: Firm 2 decides to enter firm 1’s market or to stay out.</p><p>Stage 3: If there is no entry in stage 2, Firm 1 chooses its output, q1. If there is entry in stage 2, firms 1 and 2 simultaneously and non-cooperatively choose their output levels, q1 and q2.</p><p>In this analysis, we will graphically illustrate the subgame perfect Nash equilibrium for a range of values of F1 and F2. Figure 1 reflects the case in which the level of firm 1 output above which firm 2’s profit is negative is greater than firm 1’s monopoly output, M.</p><p>Subgame perfection requires that we first analyze two types of subgames. In one type, entry has occurred. In the other type, entry has not occurred.</p><p>Stage 3 with no entry. The equilibrium will involve the monopolist firm 1 producing the larger of M and Z. </p><p>Stage 3 with entry: We need to derive the best-responses for each firm. These are illustrated in Figure 1. </p><p>Firm 1: The red curve, R1(q2), is firm 1’s best-response. For q1 ≤ K1, firm 1 has sufficient capacity so its </p><p> marginal cost of production is equal to w as the capacity cost is sunk. The upper portion of R1 corresponds to the situation in which firm 1’s profit-maximizing best-response in stage 3 requires no additional capacity. For q1 > K1, firm 1 must add extra capacity for each additional unit of output so its marginal cost now equals r + w. The lower portion of R1 corresponds to the situation in which firm 1 chooses to produce above K1. The middle, vertical segment of R1 corresponds to the situation in which firm 1 produces exactly K1. Given the values of q2 associated with this vertical segment, at any q1 < K1 firm 1’s marginal revenue is greater than w (this means firm 1 wants to produce at least q1) while at any q1 > K1 firm 1’s marginal revenue is less than r+w (this means firm 1 wants to produce no more than q1.</p><p>Firm 2: The solid blue segment of R2(q1) is firm 2’s best response but, unlike in standard Cournot games, </p><p> firm 2’s best-response output is positive only when q1 ≤ Z because when q1 > Z, the best firm 2 can do is earn negative profit. Thus, it will prefer to produce zero in order to avoid paying its fixed cost. Smaller values of F2 imply larger values of Z. </p><p>(1)</p><p>Now we can put this information together to determine the stage 3 equilibrium. First note that any equilibrium quantities in which both firms produce must fall between points B and C. It also cannot involve firm 1 producing more than Z units. If K1 lies to the left of B, the equilibrium will be at B. If</p><p>K1 ≥ Z, the equilibrium will involve only firm 1 producing at the maximum of M and Z. If K1 < Z and K1</p><p> lies to the right of B, then the equilibrium will be at point E and will imply q1 = K1. Note that because of the assumptions of linear demand and constant marginal costs, the Stackelberg equilibrium is at point S. </p><p>That is, at point S, firm 1’s profits are maximized on R2.</p><p>Stage 2 analysis.</p><p>Given the stage 3 analysis, firm 2 will enter only if K1 ≤ Z and it will stay out if K1 > Z. For K1 = Z, firm </p><p>2 is indifferent between entering and staying out. For simplicity, we will assume that firm 2 stays out at </p><p>K1 = Z.</p><p>Stage 1 analysis. Given the above analysis of stages 2 and 3, firm 1 will find itself in one of three cases. Case 1. Z lies to the left of B. Entry will never occur regardless of firm 1’s capacity choice. Any choice of capacity between 0 and M will generate the same profit for firm 1 as it will end up producing its monopoly quantity, M. In this case, entry is naturally blockaded.</p><p>Case 2. Z lies to the right of B and Z ≤ M. In this case, any value of K1 between Z and M will deter entry and still result in firm 1 producing its monopoly quantity, M.</p><p>Case 3. Z > M. This case creates a trade-off for firm 1. One option is to invest in M units of capacity. Entry will occur. The stage 3 equilibrium will be at S so that firm 1 earns its Stackelberg profit. The other option is to invest in Z units of capacity. Entry will be deterred and firm 1 will produce Z units. Firm 1 could deter entry with more capacity but because Z > M, firm 1’s profit when it deters entry is decreasing in K1. Thus, K1=Z is the profit-maximizing entry-deterring capacity. The profit firm 1 will earn be choosing K1=Z is decreasing in Z. Thus, if Z is too large (that is, if F2 is too small), firm 1 will prefer to accommodate entry and choose K1=M. However, for Z closer to M, firm 1will prefer to deter entry with K1 = Z. Homework: Assume A = 150, r = 30, w = 60, and F1 = 200. Show that entry is naturally blockaded when F2 > 400 (Case 1), that firm 1 deters entry and ends up producing its monopoly quantity when 225 < F2 < 400 (Case 2), that firm 1 deters entry and produces more than its monopoly quantity when 19.3 < F2 < 225, and that firm 1 accommodates entry when F2 < 19.3. </p>