Sample Exam Solutions May 28, 2019 Question 1 The demand and supply functions for two goods X and Y are given below. Demand: QX = 120− 2PX + 3PY , QY = 150 + 6PX − 4PY Supply: QX = −240 + 6PX , QY = −150 + 6PY (a) Determine the matrix A in the equation of the form A PX PY = 360 300 which determine market equilibrium. Answer: A = 8 −3 −6 10 (b) Show that A is invertible, by verifying that det(A) 6= 0. Answer: A is a 2× 2 matrix with det(A) = 8 · 10− (−3) · (−6) = 62 6= 0. So A is invertible. (c) Using A−1, obtain the prices of the two goods at market equilibrium. Show your working and provide answers correct to 2 decimal places. Answer: A−1 = 1 det(A) 10 3 6 8 = 1 62 10 3 6 8 So PX PY = A−1 360 300 = 1 62 10 3 6 8 360 300 = (10 ∗ 360 + 3 ∗ 300)/62 (6 ∗ 360 + 8 ∗ 300)/62 ≈ 72.58 73.55 At market equilibrium, the prices of good X and Y are, to 2 decimal place, 72.58 and 73.55 respectively. Question 2 A monopolist faces a demand function P = 173− 2Q, with P denoting the market price and Q denoting the quantity demanded. There is a fixed cost of 200, the total variable cost function is TV C(Q) = 13Q 3 − 10Q2 + 188Q. (a) Obtain an expression for the profit function f(Q) defined for Q > 0. Answer: f(Q) =Total Revenue− Total Cost =Q(173− 2Q)− ( 200 + 1 3 Q3 − 10Q2 + 188Q ) =− 1 3 Q3 + 8Q2 − 15Q− 200 (b) Determine the stationary point(s) of the profit function. Answer: The stationary point(s) is(are) the solution(s) of the equation f ′(Q) = −3 · 1 3 Q2 + 2 · 8Q− 15 = −Q2 + 16Q−15 = 0. Using abc-formula with a = −1, b = 16, c = −15, we have ∆ = b2 − 4ac = 162 − 4 · (−1) · (−15) = 196 > 0 and the equation have two roots Q1 = −b+√∆ 2a = −16 +√196 2 · (−1) = 1, Q2 = −b−√∆ 2a = −16−√196 2 · (−1) = 15. So the stationary points are Q = 1 and Q = 15. (c) Assume the maximal point of the profit function exist. What is the maximal value of the profit function? Answer: The maximal point is one of the stationary points from (b). Compare f(1) = −622 3 f(15) = 250 So the maximal value of the profit function is 250. (d) Compute the elasticity ElQf(Q) at point Q = 10, to 2 decimal places. Provide a brief interpretation of the computed value of the price elasticity, in the context of this example. Answer: ElQf(Q) = f ′(Q)Q f(Q) = (−Q2 + 16Q− 15)Q −13Q3 + 8Q2 − 15Q− 200 Plug in Q = 10 yields ElQf(10) ≈ 3.86. If the demand quantity Q increases by 1% from 10 units, the profit increases by 3.86% approximately. Question 3 Suppose the market price P can be written as a function of the market demand quantity Q ∈ [0, 5) P (Q) = 100− 10Q. The current demand quantity is Q0 = 2. Showing all steps of your working, evaluate the consumer surplus in this market CS = ∫ Q0 0 P (Q)dQ− P0Q0 where Q0 is the current market demand quantity. Answer: The current price is P0 = P (Q0) = 100− 10 · 2 = 80. CS = ∫ Q0 0 P (Q)dQ− P0Q0 = ∫ 2 0 (100− 10Q)dQ− 80 · 2 = [ 100Q− 5Q2]Q=2 Q=0 − 160 =180− 0− 160 = 20. Question 4 Energy company GreenVolt (GV) owns a property at the wind-swept, sunny location of Ocean Heads. GV is evaluating two projects: a wind farm and a solar energy plant. The wind farm requires an initial investment of $10m, and a $5m loss is expected for the first year. For the following 3 years, GV expects annual returns of $8m from electricity sales. The solar plant requires an initial investment of $15m. GV expects a loss of $2m for the first year and annual returns of $9m for the following 3 years. Assume a discount (interest) rate of 8% compounded annually. (a) Calculate, in $m to 3 decimal places, the present value of the two projects. Answer: The present value of the cash flows stream for the wind farm and the solar energy plant project, respectively, are PV1 = −10 + −5 1.08 + 8 (1 + 0.08)2 + 8 (1 + 0.08)3 + 8 (1 + 0.08)4 ≈ $4.460m PV2 = −15 + −2 1.08 + 9 (1 + 0.08)2 + 9 (1 + 0.08)3 + 9 (1 + 0.08)4 ≈ $4.624m (b) Based on the present values that you have carried out in (a), explain which project you think is preferable. Answer: The Solar energy plant project is preferable because it has a higher present value. (c) Calculate, in percentage to 3 decimal places, the internal rate of return of the two projects. Answer: Solve the equation −10 + −5 1 + r + 8 (1 + r)2 + 8 (1 + r)3 + 8 (1 + r)4 = 0 yields the internal rate of return for the wind farm project IRR1 ≈ 19.594% Solving the equation −15 + −2 1 + r + 9 (1 + r)2 + 9 (1 + r)3 + 9 (1 + r)4 = 0 yields the internal rate of return for the solar energy plant project IRR2 ≈ 17.717% (d) On the basis of the internal rate of return of the two projects, which project do you think is preferable? Answer: The wind farm project is preferable because it has a higher internal rate of return. Question 5 Given the Cobb-Douglas production function Q = 100L0.3K, L,K > 0. (a) Write down the equation of the isoquant for Q = 800 in the form K = f(L) Answer: Rewrite 100L0.3K = 800 to get K = 8L−0.3 (b) Show by differentiation that f(L) is convex. Answer: f ′(L) = 8L−0.3−1 · (−0.3) = −2.4L−1.3 f ′′(L) = −2.4L−1.3−1 · (−1.3) = 3.12L−2/3 So f ′′(L) > 0 for all L > 0, and therefore f is a convex function. (c) Find the values of L and K, to 2 decimal places, for which the production is maximised under the budget restriction L+ 2K = 30 using Lagrange method. Answer: Construct the Lagrange function g(L,K, λ) = Q(L,K) + λ(30− L− 2K) = 100L0.3K + λ(30− L− 2K) The stationary points are the solutions to the system of equations g′L(L,K, λ) = 100 · 0.3 · L0.3−1K − λ = 30L−0.7K − λ = 0 g′K(L,K, λ) = 100L 0.3 · 1− 2λ = 100L0.3 − 2λ = 0 g′λ(L,K, λ) = 30− L− 2K = 0 From the first two equations, we have 100L0.3 = 2λ = 60L−0.7K that is L = 60 100 K = 0.6K. Substituting back to the last equation (the budget constraint) in the system yields 30− 0.6K − 2K = 0⇒ K = 30 2.6 = 300 26 ≈ 11.54 Hence, from above, L = 0.6K = 18026 ≈ 6.92. (Alternatively, L = 0.6K ≈ 0.6 · 11.54 ≈ 6.92 or L = 30− 2K ≈ 30− 2 · 11.54 = 6.92) The production is maximised with L ≈ 6.92 and K ≈ 11.54 under the required budget constraint. (d) If the budget increases by 1 (that is, increases from 30 to 31), compute the resulting change (rounded off to 2 decimal places) in the maximal level of production, using the Lagrange multiplier method. Answer (preferred): The corresponding value of Lagrange multiplier at the optimal point in question (c) is λ∗ = 30 · (L∗)−0.7 · K∗ = 30 · (18026 )−0.7 · 30026 ≈ 89.3428. If the budget increases by 1 (from 30), the maximal level of production will increase by approximately 89.3428 units. Remark 1. You may solve the question with the constraint L + 2K = 31, as in question (c), and calculate the true change in the maximal level of production. Note that the final outcome will be different. Please try on your own, and here I do not provide the details. Question 6 The following difference equation models the salary scale for part-time staff Yt = 20 + 1.2Yt−1 where Yt denotes the salary (in dollars) in year t = 0, 1, 2, . . .. (a) If Y0 = 2, 300, deduce Y1, Y2 and Y3 directly from the difference equation. Answer: Y1 = 20 + 1.2Y0 = 20 + 1.2 · 2300 = 2780 Y2 = 20 + 1.2Y1 = 20 + 1.2 · 2780 = 3356 Y3 = 20 + 1.2Y2 = 20 + 1.2 · 3356 = 4047.2 (b) Solve the difference equation. In other words, determine Yt for all t = 0, 1, . . . . Answer: The equilibrium state is x∗ = 201−1.2 = −100, and we can rewrite the difference equation as Yt − x∗ = 1.2 · (Yt−1 − x∗) , that is Yt + 100 = 1.2 · (Yt−1 + 100) So Y0 + 100, Y1 + 100, Y2 + 100, . . . is a geometric sequence with common ratio r = 1.2. Hence, Yt + 100 = r t(Y0 + 100) = 2400 · 1.2t and it follows Yt = 2400 · 1.2t − 100, for all t ≥ 1. (c) Calculate the number of year until the salary first exceeds $15,000. Answer: Solve Yt = 2400 · 1.2t − 100 > 15000, that is, 1.2t > 151002400 , or eln(1.2)·t > 15100 2400 Using the fact that f(x) = ex is an increasing function on (−∞,∞), we have ln(1.2) · t > ln ( 15100 2400 ) ⇔ t > ln ( 15100 2400 ) ln(1.2) ≈ 10.1, where the inequality sign did not change since ln(1.2) ≈ 0.18 > 0. Noting that the time index t is an integer, so the number of year until the salary first exceeds $15,000 is 11. Question 7 Suppose that a firm’s capital stock K(t) satisfies the differential equation dK dt = I − δK(t) where investment I is constant, and δK(t) denotes depreciation, with δ a positive constant (a) Find the solution of the equation if the capital stock at time t = 0 is K0. Answer: We have K ′(t) = I − δK(t) so( K(t)− I δ )′ = −δ ( K(t)− I δ ) . In other words, defining K˜(t) := K(t)− Iδ , then it satisfies the differential equation K˜ ′(t) = −δK˜(t). It follows that K˜(t) = Ae−δt for some constant A, and therefore K(t) = K˜(t) + I δ = Ae−δt + I δ Plug-in K(0) = K0 we have A+ I δ = K0 ⇔ A = K0 − I δ So the firm’s capital stock K(t) = ( K0 − I δ ) e−δt + I δ (b) Let δ = 0.05 and I = 10. Explain what happens as t → ∞ when: (i) K0 = 150; (ii) K0 = 250. Answer: Plugging in δ = 0.05 and I = 10 we have K(t) = (K0 − 200) e−0.05t + 200 As t→∞, for both the cases of (i) K0 = 150; (ii) K0 = 250 we have K(t)→ 0 + 200 = 200 since e−0.05t = ( e−0.05 )t → 0 where e−0.05 ≈ 0.95 < 1. Question 8 A daily diet mix requires a minimum of: 160 mg of Vitamin K and 1000 mg of Vitamin D. Two foods A and B contain these vitamins: Vitamin K Vitamin D Cost per 1 kg Food A 10 mg 100 mg 40 Food B 8 mg 40 mg 20 (a) Write down the inequality constraints for each vitamin. Denote the consumption (in kg) of food A and food B as x and y. Answer: • Vitamin K: 10x+ 8y ≥ 160; • Vitamin D: 100x+ 40y ≥ 1000; (b) Using extreme point theorem, determine the number of units of food A and B which fulfil the daily requirements at a minimum cost. You may assume the optimal point exist. Answer: The question is a linear programming problem as follows: Min 40x+ 20y, subject to the following (five) constraints 10x+ 8y ≥ 160 100x+ 40y ≥ 1000 x ≥ 0, y ≥ 0 Setting any two of the constraint active, we have candidates of the corner points as follows: (1) 10x+ 8y = 160 and 100x+ 40y = 1000: by Cramer’s rule we have x = ∣∣∣∣∣∣ 160 81000 40 ∣∣∣∣∣∣∣∣∣∣∣∣ 10 8100 40 ∣∣∣∣∣∣ = 160 · 40− 8 · 1000 10 · 40− 8 · 100 = −1600 −400 = 4 y = ∣∣∣∣∣∣ 10 160100 1000 ∣∣∣∣∣∣∣∣∣∣∣∣ 10 8100 40 ∣∣∣∣∣∣ = 10 · 10000− 160 · 100 −400 = −6000 −400 = 15 It satisfies the other constraints x ≥ 0 and y ≥ 0 (2) 10x+ 8y = 160, x = 0: y = 20. It does not satisfy 100x+ 40y ≥ 1000. (3) 10x+ 8y = 160, y = 0: x = 16. It satisfies 100x+ 40y ≥ 1000 and x ≥ 0. (4) 100x+ 40y = 1000, x = 0: y = 25. It satisfies 10x+ 8y ≥ 160 and x ≥ 0. (5) 100x+ 40y = 1000, y = 0: x = 10. It does not satisfy 10x+ 8y ≥ 160. (6) x = 0, y = 0. It does not satisfy 10x+ 8y ≥ 160, nor 100x+ 40y ≥ 1000. We have found in total 6 candidate points, but only 3 corner points (4, 15), (16, 0), (0, 25) corresponding to the costs 40x+ 20y equal to, respectively 460, 640, 500. Hence, the minimal cost is 460, which can be obtained by consuming 4 units of food A and 15 units of food B. END OF EXAMINATION Formulae are provided in the next two pages.
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