1a) (i). Use the pricing formula (P-MC)/P = -1/elasticity of demand. Here1b) (i). SR elasticity is price coefficient = -.40.1c) (i). Bart has no dominant strategy. Milhaus has a dominant strategy (go to class).1d) (i). Qd = Qs implies 12,000 = 300 P, so P = $ 40. This implies Q = 11,000.PMRTennis Club profits are ProfitT = p Q - 500 = pSloan School of Management 15.010/011 - Economic Analysis for Business Decisions Massachusetts Institute of Technology Professors McAdams, Montero, Stoker and van den Steen 2001 Final Exam Answers: Prepared for TA Grading Purposes 1. Short Questions 1a) 1a) (i). Use the pricing formula (P-MC)/P = -1/elasticity of demand. Here 70.2$325.119.0=⇒=−−=−PPP 1a) (ii) ElasticityiceElasticityAdSalesAdsPr$$−= in profit maximizing equilibrium. Here = – (0.5/– 1.5) = 1/3. If sales = $ 225 million, then optimal $advertising is $ 75 1b) 1b) (i). SR elasticity is price coefficient = -.40. 1b) (ii) From partial adjustment formulation of demand, LR elasticity is . 5.8.4.2.14.−=−=−− (1c) 1c) (i). Bart has no dominant strategy. Milhaus has a dominant strategy (go to class). 1c) (ii) Yes. Both go to class. 1d) 1d) (i). Qd = Qs implies 12,000 = 300 P, so P = $ 40. This implies Q = 11,000. 1d) (ii) At price = $30, Qd = 12,250, Qs = 9,250, so Qd - Qs = 3,000 shortage results.2001 Exam Answers p. 2 2. Rolling Stones 2a) a) The total cost is $1,000,000 independent of the number of seats sold (up to capacity 60,000). The marginal cost is $0 up to capacity and infinite thereafter. The inverse demand is P = 200 – 1.25 Q, which gives marginal revenue MR=200-2.5 Q. The optimal solution is found by setting MR = MC = 0, which gives Q = 200/2.5= 80, which is greater than the capacity of 60. So, we set the price to fill the 60,000 seats, which implies P = 200 – 1.25 (60) = $ 125. Profits are 125 (60,000) – 1,000,000 = 6,500,000. A picture: P MC MR 60,000 80,000 Q 2b) The MC curve becomes as depicted below (since MR = 50 at Q = 60). Beyond 60,000 seats, the MC is thus $20. Find the optimal quantity by setting MR = MC = 20. This gives 200-2.5 Q = 20 or Q = 180/2.5 = 72. To sell these 72,000 seats, we need to set price at 200-1.25*72= 110. We will pay $240,000 for the extra seats, so profits become 72,000 * 110- 1,000,000 - 240,000 = 6,680,000. P MR MC 60,000 80,000 Q DD2001 Exam Answers p. 3 3. Scholar Rocks 3a) We have MC = 0. Since the groups are identifiable, we set MR = MC for each type of person to solve for Q, and then solve for P. For each C, MR = 10 – Q, so MR = MC implies Q = 10, and P = 5. For each S, MR = 10 – 2Q, so MR = MC implies Q = 5 and P = 5. (??!!!). So it is optimal (3rd degree p.d.) to charge P = 5 for everyone. Profits are 5 (10*2,000 + 5*8000) – 50,000 = $ 250,000. 3b) We solve for the optimal two part tariff for each group separately. Since MC = 0, it is optimal to set the daily admission fee to zero for each group. The optimal membership is the consumer surplus for each group at P = 0. For C’s this is T = .5 (20)(10) = $100. For S’s this is T = .5 (10)(10) = $50. Profits are 100*2,000+ 50*8,000 – 50,000 = $ 550,000. 3c) Since the daily admission fee is 0, we can sell memberships to C’s and S’s for the S value ($50) or we can sell memberships only to C’s for the C value above ($100), and we have to check which is more profitable. For C’s only, profits are 100*2,000 – 5,000 = $ 150,000. For C’s and S’s, profits are 50*10,000 – 50,000 = $ 450,000. Therefore, it is optimal to charge $50 for a membership, with C’s and S’s becoming members, and profits are $450,000. 4. Ale and Cournot. This is the classic Cournot problem; to solve, we must compute reaction functions for A and B and solve them simultaneously. Revenue for A is RA = (12 - (QA + QB )) QA so that marginal revenue is MRA = 12 - QB - 2 QA . MC = 0, so MRA = MC implies QA = 6 - .5 QB which is A’s reaction function. By symmetry, B’s reaction function is QB = 6 - .5 QA . Solving these simultaneously gives QA = 4, QB = 4, so total supply is 8 thousand pints, and the price is P = 4, Profits for each firm are 4* 4 = $16,000. 5. Bolts 5a) Equilibrium is A “goes first,” then “Honor” and B “Trust”. If A plays “simultaneous”, then game box is relevant --- A has a dominant strategy of “Skimp,” given that B chooses “No Trust” and we end with the (box’s) Nash Equilibrium of (5,-5). If A plays “goes first” and then “Skimp”, B chooses “No Trust” and we again end up with (5, -5). If A plays “goes first” and the “Honor”, then B safely plays “Trust” and we end up with payoffs (8,8). This is the highest payoff for A (and B) and so is the chosen equilibrium. 5b) By going first, A can get 8 thousand dollars, and by playing “simultaneous,” A gets 5 thousand dollars. Therefore A would be willing to pay up to 3 thousand dollars to go first.2001 Exam Answers p. 4 6. Tennis and Racquets. Let p be the price of a membership and r be the price of a racquet. We have P = p + r in all parts of this problem. Also, for memberships, we have MCT = 0, and for racquets, we have MCR = Q. 6a) For p = 30, demand faced by Racquets R’ Us is r = P-30 = 150 – Q. So, MRR = 150 – 2Q, and MRR = MCR implies 150 - 2Q = Q or Q = 50. From demand curve, price is r = 100. Profits for Racquets R’ Us is 100*50 – TC(50) = 5,000-1,350 = $3,650. Profit to the Tennis Club is 30*50 – 500 = $ 1,000. 6b) This is the most involved question of the test. One must first solve for what Racquets R’ Us will do when the Tennis Club sets a price of p. Then we must solve for the p that maximizes Tennis Club profits. Given a price of p, Racquets R’ Us faces demand r = P-p = 180 – p - Q. Therefore, MRR = 180 – p – 2Q, and MRR = MCR implies 180 - p - 2Q = Q or Q = 60 – p/3. This summarizes Racquet’s R’ Us’s reaction and sets Q given p. Tennis Club profits are ProfitT = p Q - 500 = p (60 – p/3) – 500 = 60p – p2/3 – 500. Maximize by setting d Profit/dp = 0, or 60 – 2p/3 = …