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Unformatted text preview: University of Toronto, Department of Economics, ECO 204 2008‐2009 S. Ajaz Hussain ECO 204 2008‐2009 Ajaz Hussain HW 18 Solutions Question 1 In this question, you will practice 3rd degree price discrimination. Suppose you are pricing manager of parking Toronto’s Pearson airport. You’ve identified two market segments: “short” and “long” term parkers. The demands for each segment are: Short term: Ps = 3 – Qs/200 Long term: PL = 2 – QL/200 P is hourly rate and Q is number of cars. There is a single garage with 600 spaces and you have to allocate spaces between the two segments. Let the MC of parking be negligible. (a) What is the garage owner’s objective? Answer: Given that MC = 0, maximizing profits is the same as maximizing revenues since: MR = MC, MC = 0, MR = 0 (b) Suppose the garage charges a single price for both segments. What is the optimal price? Hint: the short and long term parkers are being “aggregated”. Answer: If the garage “lumps” the segments and treats it as one segment, it will have charge all customers a single uniform price P. The total market demand is: Q = QS + QL where: 1 University of Toronto, Department of Economics, ECO 204 2008‐2009 S. Ajaz Hussain PS = 3 – Qs/200 → Qs = 200(3 – P) And: PL = 2 – QL/200 → QL = 200(2 – P) Thus: aggregate demand is: Q = QS + QL → Q = 200(3 – P + 2 – P) → Q = 200(5 – 2P) → Q = 1000 – 400P → 400P = 1000 – Q → P = 1000/400 – Q/400 Thus, MR (using same intercept, twice the slope short cut) is: MR = 1000/400 – Q/200 = 0 → Q/200 = 1000/400 → Q = 500 Note how this answer is feasible since it is below the capacity of 600. The answer is therefore Q = 500 and the price is: P = 1000/400 – 500/400 → P = 500/400 =$1.25 an hour. (c) Suppose the garage charges each segment a different price‐ what are each segment’s optimal prices? Hint: first try the problem as if there is no constraint and then if necessary as if there is a constraint (you did the same thing for the single output case with capacity in Lecture 13). What is the price elasticity in each segment? 2 University of Toronto, Department of Economics, ECO 204 2008‐2009 S. Ajaz Hussain Answer In Lecture 13 when we discussed output under capacity, the algorithm was to first solve the problem as if there is no constraint. If the answer was feasible, then we should keep the answer. If the answer is infeasible (i.e. it exceeds capacity) we should re‐solve the problem with the constraint that output = capacity. In the same way, in Lecture 20 when we discussed 3rd degree price discrimination we should: • With unlimited capacity set each segment’s M∏ = 0 (which means each segment’s M∏ are equal to each other) • With limited capacity set each segment’s M∏ equal to each other and then add the constraint that total output is = capacity. o Now, how we do know the capacity matters? You should do exactly what you did in the single output case: first solve the problem without the capacity constraint and if feasible keep the answer; otherwise re‐solve with the constraint outputs = capacity. o Optional: For students with advanced math skills here is how to solve the problem without doing the “two‐step” process above. You maximize: ∏Total = ∏A + ∏B subject to: QA + QB ≤ Capacity The reason we don’t do this in this course is that we only know how to handle problems with the constraint QA + QB = Capacity not QA + QB ≤ Capacity. To do the latter see Martin Osborne’s Math Tutorial for Kuhn‐Tucker conditions. This is why ‐‐ in the undergraduate course ‐‐ we must do the two‐step process. Accordingly, first solve the problem as if the constraint does not exist. Profit maximization for each segment requires: Setting M∏S = 0 → MRS = MCS → MRS = 0 Setting M∏L = 0 →MRL = MCL → MRL = 0 3 University of Toronto, Department of Economics, ECO 204 2008‐2009 S. Ajaz Hussain Using short cut to derive the MR: MRs = 3 – Qs/100 = 0 → Qs = 300 → Ps = 3 ‐ 300/200 = $1.50 MRL = 2 – QL/100 = 0 → Qs = 200 → PL = 2 ‐ 200/200 = $1.00 The solutions are feasible as Qs + QL < 600 Note how, compared to the uniform pricing case ‐‐ the short term parkers pay a higher price and the long term parkers pay a lower price. Finally: Es = 200(1.50/300) = ‐1 EL = 200(1/200) = ‐1 This should not be surprising: we are able to reach the R maximizing solution in each segment. (d) Repeat question (c) with a capacity of 400 instead of 600. Answer: Now if capacity is 400 we know that the previous answer can’t be feasible since Qs + QL = 500 < 400. In this case, the discussion in Lecture 20 tells you that: • Set M∏ of each segment equal. This is equation 1 • Use the constraint Qs + QL = 400. This is equation 2. o The astute student will notice that we’ve said Qs + QL = 400 and not Qs + QL ≤ 400. This is because the unconstrained solution has Qs + QL = 500 and so we know that the constrained solution must be Qs + QL = 400. Equation 1: Setting M∏ equal: M∏s = M∏L → MRs ‐ MC = MRL ‐ MC 4 University of Toronto, Department of Economics, ECO 204 2008‐2009 S. Ajaz Hussain → MRs = MRL → 3 – Qs/100 = 2 – QL/100 → 100 = Qs – QL Equation 2: Qs + QL = 400 Solving the two equations simultaneously: 2Qs = 500 → Qs = 250 → Ps = 3 – 250/200 = $1.75 → QL = 150 → Ps = 2 ‐ 150/200 = $1.25 These prices are higher than part (d) because we’re renting fewer spaces in each segment. But still, the more inelastic segment is being charged a higher price. Question 2 (2007‐2008 Final Exam Question) Poo‐Joe is a French car maker. It manufactures cars in France at MC = $100. Demand for cars in France is given by the equation PF = 3,000 – 5QF. Poo‐Joe also exports cars to Japan where demand is given by the equation PJ = 1,500 – 6QJ. Suppose transportation cost between countries is $200 per car. (a) How many cars should be manufactured for the French and Japanese markets? Assume that French Customs doesn’t allow cars bound for Japan to be “re‐imported” back into France. Calculate the French and Japanese prices. Show all calculations clearly. Hint: That cars cannot be re‐imported means there can be on no arbitrage between France and Japan. Moreover, think carefully what the MC in France and Japan are. Answer: Note how the question makes no mention of capacity (i.e. Poo‐Joe has unlimited capacity). Thus as we discussed in Lecture 20, Poo‐Joe should allocate cars to whichever market has the greater Marginal ∏ (“marginal analysis” see HW 17 for instance). At the optimum, the French and Japanese markets will have: 5 University of Toronto, Department of Economics, ECO 204 2008‐2009 S. Ajaz Hussain M∏F = 0 .. Equation 1 M∏J = 0 .. Equation 2 Which as we argued implies that M∏F = M∏J and therefore MRF = MRJ. Now, equation 1 implies that: MRF = MCF and equation 2 implies that: MRJ = MCJ . For the French market: PF = 3000 – 5 QF → MRF = 3000 – 10 QF MRF = MCF yields: 3000 – 10 QF = 100 → QF = 290 → PF = 1550 For the Japanese market: PJ = 1500 – 6 QJ → MRJ = 1500 – 12 QJ MRJ = MCJ yields: 1500 – 12 QJ = 300 where we have used the fact the car must be manufactured in France (MC = 100) and then transported to Japan (an extra 200). → QJ = 100 → PJ = 900 6 University of Toronto, Department of Economics, ECO 204 2008‐2009 S. Ajaz Hussain Note how the Japanese market is cheaper than the French market. Given that the price gap is larger than $300 arbitrators would like to buy cars in Japan and re‐import to France. However, in this question they can’t as French law (and honest customs) prevents such arbitrage. (b) Now suppose French Customs allows cars exported to Japan to be re‐imported back to Japan. What should be the French and Japanese prices and number of cars? Show all calculations clearly. Hint: Look at Lecture 20 discussion on the price gap needed to prevent arbitrage. Answer: In Lecture 20 (slide 22) we discussed how to maximize profits across segments when arbitrage cannot be prevented: Max ∏Total = ∏F + ∏J Subject to: Price gap ≤ cost of arbitrage = $200 To maximize profits Poo‐Joe should make the price gap as large as possible: i.e. Price gap between France and Japan = $200. The problem is therefore: Max ∏Total = ∏F + ∏J Subject to: PF ‐ PJ = 200 Now: ∏Total = ∏F + ∏J → ∏Total = RF ‐ CF + RJ ‐ CJ → ∏Total = PF QF ‐ MCF QF + PJ QJ ‐ MCJ QJ → ∏Total = (PF ‐ MCF) QF + (PJ ‐ MCJ) QJ → ∏Total = (PF ‐ 100) QF + (PJ ‐ 300) QJ Now: 7 University of Toronto, Department of Economics, ECO 204 2008‐2009 S. Ajaz Hussain QJ = 250 – 0.1667 PJ QF = 600 – 0.2 PF Substituting in profits and choosing Price in Japan as the decision variable (which from PF ‐ PJ = 200 implies that PF ‐ 200 = PJ) yields: ∏ = (PJ + 200 –100) (600 – 0.2 PJ – 40) + (PJ – 300) (250 – 0.16667 PJ) → ∏ = (PJ +100) (560 – 0.2 PJ) + (PJ – 300)(250 – 0.1667 PJ) → ∏ = 560PJ – 20PJ + 56000 – 0.2PJ 2 + 250PJ – 0.1667PJ 2 – 75000 + 50PJ → ∏ = – 19000 + 840 PJ – 0.3667PJ 2 Note that you could just as well have chosen the Price in France as the decision variable: Maximizing profit: d∏/dPJ =0 → 840 – 0.7333PJ = 0 → PJ = 1145.45 → PF = 1345.45 → QJ = 59.09 → QF = 330.90 Note: You may be wondering why this problem can’t be solved by setting M∏F = M∏J. To see why look at the figure below which depicts the profit functions in France and Japan. Because Poo‐Joe has to prevent arbitrage by setting the price gap equal to $200, there has to be a corresponding gap between the French and Japanese outputs. 8 University of Toronto, Department of Economics, ECO 204 2008‐2009 S. Ajaz Hussain In this figure, the top panel shows the French and Japanese demands while the bottom panel shows the profits. Note that the profit maximizing output for Japan is less than that of France. Preventing arbitrage requires that we create a “wedge” of $200 between the French and Japanese prices. The price gap is equivalent to an output gap between France and Japan. This output gap does not result in equal Marginal profits across France and Japan. In fact the M∏ in Japan is positive (you can show it is 490.91) and the M∏ in France is negative (you can show that it is ‐409.09). Ideally, to maximize profits we’d equate the Marginal profits but can’ because to prevent arbitrage we must create a price gap of $200 which can only happen if the marginal profits are not equal 1 You can also see this point another way: suppose you wanted a price gap = $200 and had set M∏F = M∏J .Then: M∏F = MRF ‐ MCF = 3000 ‐ 10 QF ‐ 100 = 2,900 ‐ 10 QF M∏J = MRJ ‐ MCJ = 1500 ‐ 12 QJ ‐ 300 = 1,200 ‐ 12 QJ Setting M∏F = M∏J : → 2,900 ‐ 10 QF = 1,200 ‐ 12 QJ → 1,700 = 10 QF ‐ 12 QJ .. Equation 1 Now: we know that PF ‐ PJ = 200: 1 Optional: If you know “solver” in Excel here is the spreadsheet solution: bring up solver and solve. 9 University of Toronto, Department of Economics, ECO 204 2008‐2009 S. Ajaz Hussain → 3,000 ‐ 5QF ‐ (1,500 ‐ 6QJ) = 200 → 2,800 ‐ 5QF ‐ 1,500 + 6 QJ = 0 → 1,300 ‐ 5QF + 6 QJ = 0 → 1,300 = 5QF ‐ 6 QJ .. Equation 2 → 2,600 = 10QF ‐ 12 QJ .. Equation 2 Thus: 1,700 = 10 QF ‐ 12 QJ .. Equation 1 2,600 = 10QF ‐ 12 QJ .. Equation 2 These equations cannot be solved as these are parallel. Hence in this question equating M∏ does not work. Question 3 (2007‐2008 Test 3 Question) Alma Construction Company (ACC) has a manufacturing plant in London, Ontario with cost function: C(Q) = 500 + 5Q2 and demand function Q = 120 – 0.2P. Suppose ACC builds an identical manufacturing plant adjacent to the first manufacturing plant. Argue that ACC should divide output equally between the two plants. Show all calculations and steps clearly. Hint: Use marginal analysis technique in Lecture 20. Answer: This problem is really easy, if you understand the logic (as we did in Lecture 20) of multi‐plant output. If the two plants are identical, they have the same cost functions. Optimal Multi‐plant allocation of inputs requires: MCA = MCB → 10QA = 10QB 10 University of Toronto, Department of Economics, ECO 204 2008‐2009 S. Ajaz Hussain → QA = QB That is, it is optimal to divide output equally between the two plants. Question 4 Ajax produces output from two factories located in Buffalo, NY and Toronto, ON. The Buffalo factory has cost equation: C = 20QBuffalo + QBuffalo2 and the Toronto factory has cost equation: C = 10QToronto + (5/2)QToronto2. If Ajax is currently minimizing cost its costs and produces 5 units of output in Buffalo how many units is it producing in Toronto? Answer: If Ajax sells into one market using output from two factories, it must be that: MCBuffalo = MCToronto → 20 + 2QBuffalo =10 + 5QToronto → 20 + 2(5) =10 + 5QToronto → 20 = 5QToronto → QToronto = 4 11 ...
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This note was uploaded on 05/02/2011 for the course ECO 204 taught by Professor Hussein during the Fall '08 term at University of Toronto- Toronto.

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