Midterm Exam - Solutions

EC 70 - Math for Economists Samson Alva Department of Economics, Boston College October 13, 011 Midterm Exam - Solutions 1 Quasilinear Preferences (a) There are a number of ways to define the Lagrangian for this problem, depending on how one chooses to treat the nonnegativity constraints; here, I will work with the formulation that explicitly incorporates them into the Lagrangian, defined as: L(x, y, λ, µ, ν) = ln(x) + y λ(px + qy I) + µx + νy (b) Given the way the Lagrangian has been defined above, the Karush-Kuhn-Tucker (KKT) theorem implies that the values x and y that solve the consumers problem, together with the associated values λ, µ, and ν of the three multipliers, must satisfy L 1 (x, y, λ, µ, ν ) = 1 x λ p + µ = 0 (1) L (x, y, λ, µ, ν ) = 1 λ q + ν = 0 () L 3 (x, y, λ, µ, ν ) = px qy + I 0 (3) L 4 (x, y, λ, µ, ν ) = x 0 (4) L 5 (x, y, λ, µ, ν ) = y 0 (5) and the nonnegativity and complementary slackness conditions λ 0 λ (px + qy I) = 0 (6) µ 0 µ x = 0 (7) ν 0 ν y = 0 (8) (c) The form of the utility function implies that the budget constraint will always bind The form of the utility function also implies that x > 0 and hence, by (7), that µ = 0 as well Suppose first that y = 0 In this case, (1)-(3) and (8) require that 1 x λ p = 0, 1 λ q + ν = 0, I px = 0, ν 0 must hold Thus, x = I, p λ = 1, and I ν = q 1 But since I ν 0, we see that for y = 0 it must be that q 1 0 ie q I I Thus, whenever q I, it is optimal for the consumer to choose x = I and p y = 0 1

(d) Now suppose y > 0 which implies ν = 0 Then, equations (1)-(3) require that 1 x λ p = 0, 1 λ q = 0, px + qy = I So, first notice that λ = 1 and so q x = q, which then yields, via the budget p constraint, y = I 1, which is strictly positive as long as q < I q Thus, whenever q < I, it is optimal for the consumer to choose x = q and p y = 0 (e) The results from parts (c) and (d) above imply that only consumers with income I larger than q will purchase good y In this sense, y is best regarded as the luxury good in this example Lemma of the person who herds wooly mammals The econometricians estimates, together with the assumptions made about the firms behavior, imply that the firms minimum cost function, defined by subject to the constraint C(w, r, y) min wl + rk k,l y f(k, l) has the form ln C(w, r, y) = 1 ln w + 1 ln r + ln y Given the estimated form, we know that C = wry Then, using Shephard s Lemma, we find that l C = 1 C C w = 1 w = l = y r w and k C = 1 C C r = 1 r = k = y w r Multiplying k and l yields kl = y 4 and so we get y = kl Thus, it must be that the production function f(k, l) = k 1 l 1 3 Inventory Problem (a) Immediately after inventory has been reordered, the level of inventory is x The rate of depletion is constant, and so the inventory level declines linearly over time from x to 0, at which point the inventory jumps instantaneously to x again (see Figure 1) Thus, if T is the amount of time between reorders, the average inventory level is 1 xt = x, where the numerator is the area under the inventory T curve (b) The Lagrangian for the problem is L = C h x + C on + λ(a xn)

Figure 1: Inventory over time and the first order conditions are L x = C h λn = 0 (9) L n = C o λx = 0 (10) L λ = A xn = 0 (11) Using L x and L n, we get (λx)(λn) = (C o ) ( Ch ) = λ = C oc h A and so we get λ Co C h = ± A As should be clear from the expressions below, the negative solution for the multiplier does not make sense because it yields negative values for x and n Assuming λ takes on a positive value, x = C o λ = ACo, C h and n = C h λ = ACh C o The multiplier is the marginal inventory cost of increasing annual supply (c) If the firm can delay fulfilling orders, its inventory cost function changes to x U U C(x, n, U) = C h + C p + C on (assuming x U; see discussion at the end of this paragraph) Keep in mind that orders still have to be fulfilled, though they may be delayed by paying the penalty cost Thus, the annual delivery requirement is still A, but now when the order amount of x is received, the highest 3

Figure : Inventory over time inventory level will be x U ie x is the amount by which inventory will jump up after a reorder, and U is the level from which it will jump up It is still the case that A = nx Moreover, it should be apparent, upon reflection, that the reorder amount x can be assumed to be at least as large in magnitude as U, so that the post reorder level x U is at least 0 If this were not true, then the firm would always have a level of inventory deficit, and so have a higher steady state average penalty cost This argument is analogous to the one for part a), where we recognized that the firm shouldn t reorder until I falls to 0 See Figure for the time-path of inventory The Lagrangian becomes (keeping in mind we are minimizing not maximizing) L = C h x U and the KKT conditions are + C p U + C on + λ(a nx) + µ x (U x) µ n n µ U U L x = C h λn µ x = 0 (1) L n = C o λx µ n = 0 (13) L U = C h + C p + µ x µ U = 0 (14) L λ = A nx = 0 (15) x U, µ x 0, µ x (U x) = 0 (16) n 0, µ n 0, µ n n = 0 (17) U 0, µ U 0, µ U U = 0 (18) Now, suppose x > U, so that µ x = 0 Then λ n = C h, and µ U = Cp C h 0, implying C p C h Thus, we conclude that a necessary condition for x > U is that C p C h Next, suppose U > 0, so that µ U = 0 Then µ x = C h Cp 0, implying C h C p Thus, we conclude that a necessary condition for U > 0 is that C h C p Thus, we can conclude that the only way x > U and U > 0 is if C h = C p Otherwise, C h > C p implies x = U 0 and C p > C h implies x U = 0 4

So now suppose C p > C h, so that x U = 0 Assume A > 0, so that n x = A > 0 implying x > 0 and n > 0 Thus, µ x = 0, µ n = 0, and µ U = Cp C h, implying that λ n = C h and λ x = C o These yield the same solutions as part b) Thus, if C p > C h it is optimal to reorder inventory when the level falls to 0, to avoid penalty costs Next, suppose C p < C h, so that x = U 0 With A > 0, we have that n x = n U = A > 0 implying n > 0 and U > 0 Then, µ U = 0, µ n = 0 and µ x = C h C p Next, λ n = Cp and λ x = C o It s easy to see that the solutions here resemble those of part b), with C p taking the place of C h Thus, and x = U = n = AC o C p ACp C o Thus, if C p < C h, it is optimal to bear the penalty cost and reorder inventory solely to fulfill the backlog of orders, taking the level of inventory from U to 0 Basically, holding inventory and bearing penalty costs are perfect substitutes Finally, if C h = C p, it should be clear that the solution for n is still the same ie n = AC h C o, and so the annual order constraint implies x stays the same at AC o C p AC o C h, but U can be any amount from 0 to 4 Multimarket Monopolist and Arbitrageurs (a) Defining elasticity as ε = p q q, we have p ε 1 = p 1 q 1 = p 1 a p 1, ε = p q = p b p (b) Profit is π(p 1, p ) = p 1 q 1 + p q First order conditions are π p 1 = a p 1 = 0, π p = b p = 0, which implies that p 1 = q 1 = a and p = q = b Maximum profit is π = a +b 4 (c) We need q 1 0 and q 0 for the demand functions to make sense, and so we have the constraints p 1 a and p b Incorporating these as well as the arbitrage constraints into the Lagrangian yields L = p 1 (a p 1 ) + p (a p ) λ 1 (p 1 τp ) λ (p τp 1 ) µ 1 (p 1 a) µ (p b) (d) The constraint functions are linear in the choice variables p 1 and p, so the nondegenerate constraint qualification is satisfied automatically, as long as the constraint functions are linearly independent, which is clearly the case given τ > 1 5

The objective function is concave and continuously differentiable, so the necessary and sufficient conditions of the KKT theorem hold Any solutions we find to the KKT equations must be a constrained maximizer of our problem as defined by the Lagrangian (e) The KKT conditions are: L 1 = a p 1 λ 1 + λ τ µ 1 = 0 (19) L = b p + λ 1 τ λ µ = 0 (0) p 1 τp, λ 1 0, λ 1 (p 1 τp ) = 0 (1) p τp 1, λ 0, λ (p τp 1 ) = 0 () p 1 a, µ 1 0, µ 1 (p 1 a) = 0 (3) p b, µ 0, µ (p b) = 0 (4) (f) Both arbitrage constraints cannot bind is this would imply that p 1 > p and p > p 1, given that τ > 1, a contradiction (g) If p = τp 1 then p > p 1 Given that a > b, this is not consistent with our solution to part b), where we find that p 1 > p when the arbitrage constraints don t bind, where the inconsistency is that the optimal prices p 1 and p would could only reverse magnitude ordering if they were discontinuous functions of the exogenous parameter τ But intuitively, and mathematically from the Berge maximum theorem, the continuity of the objective function over the domain and the continuity of the domain as a function of the parameter τ rules out discontinuity of the optimal solutions Thus, we should not expect the constraint p τp 1 to bind at a maximum (h) We know both constraints can t bind, and we know intuitively that the second one couldn t bind if the first one didn t, so we really only have to consider the two cases where the first arbitrage constraint either does or does not bind Suppose it doesn t bind Then, it is immediately clear that, with µ 1 = 0 and µ = 0, we recover the arbitrage-free solution of p 1 = a and p = b (i) In Figure 3, the brown region containing the unconstrained maximizer pricing scheme is the feasible region Since there is a unique unconstrained maximum, the isoprofit curve through the maximum is just the point itself there are no other price combinations that will yield as high a profit Isoprofit curves are circles of increasing radius centered on this point (j) Assuming positive quantities, we know that µ 1 = µ = 0 We have that p 1 = τp and p < τp 1, implying λ = 0 Replace p 1 in the first order conditions to get a τp λ 1 = 0 and b p + τλ 1 = 0 Combining these, we get τ(a τp ) + b p = 0, so p = 1 and p 1 = τp = τ This give us π = p 1(a p 1) + p (b p ) = p (τ(a τp )+b p ) Notice that the combined first order condition above can be rewritten as τ(a τp ) + b p = (1 + τ )p Substituting this into 1+τ (τa+b) 6 1+τ (τa+b)

Figure 3: p 1 on horizontal axis, p on vertical axis, circular isoprofit curves, brown central region is feasible the maximum profit expression yields π = (1 + τ )p = 1 (τa + b) 1 + τ 4 (k) Simple substitution of τ = a into the expressions in the previous part produce the b unconstrained price and maximum profit Quite clearly, if τ > a, then the first b (and also second) arbitrage constraint do not bind, since the costs of arbitrage are much too high Therefore, the arbitrage constraints are binding if 1 τ a b (l) Even though we assumed for the problem above that τ > 1, we can take the limit of our solutions as τ approaches 1 to find π = (a+b), which is less that the 8 unconstrained profit a +b The difference between these profit levels is equal to 4 (a b) > 0 since a > b 8 (m) Since p = b, q = 0, and so the problem is to maximize p 1 (a p 1 ), which immediately yields the optimal price of p 1 = a a, and profit of Now, we computed in 4 the previous part that the profit when τ = 1 is (a+b) Therefore, for the monopolist not to find it optimal to choke market, it must be that (a+b) a, which 8 4 8 immediately yields (a + b) a Keep in mind that we did not verify that the nonnegativity constraints on quantities were satisfied by our constrained optimal solutions What conditions on the parameters are necessary for these quantity nonnegativity constraints not to be violated? Well, we need p τ (τa+b) 1 a implying a and p 1+τ b implying 7

5 Topology 1 τa+b b Then, τ b ( + τ )a and τa (1 + τ )b so 1+τ τ + τ a b (1 + τ ) τ Since a > b, the left-hand side inequality is trivially satisfied So, it is the righthand side inequality that matters, and our conclusions above are valid if it is satisfied Solving the problem for the case where the right-hand side inequality does not hold is more challenging because the inverse demand functions are not differentiable at p 1 = a and p = b, requiring case-by-case analysis Topologists attribute to sets the properties of being open or being closed However, unlike the everyday notion of openness and closedness being mutually exclusive attributes, topologists do not make this either-or distinction, and hence the confusion of the customer, and the mirth of the topologist, stems from the customer s assumption that a store being not closed implies that it is open The subset (0, 1] R is neither closed nor open 8