HW5,Sp2000sols - EEP101/ECON125 Spring 00 Prof.: D....

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EEP101/ECON125 Spring 00 Prof.: D. Zilberman GSIs: Malick/McGregor/St-Pierre PROBLEM SET 5 Solutions 1. Suppose r=0.1, price (P) of timber is $10 per boardfoot, and the volume of trees in a stand obeys the function Q(T) = 12 T 2 - 1/3 T 3 . a) Consider a single rotation. Set up the profit maximization problem and derive the equilibrium condition. Solve for the optimal rotation length (T*). Be sure to check to see if your answer makes sense. The problem assumed no harvesting costs. The grower’s objective is to choose the rotation length to maximize profits (or revenues in this case). Of course, since this profit is realized T years from now, we need to discount this future profit to the present. This objective is expressed as rT T e T PQ V P MAX - = Π ) ( . . Maximizing this objective with respect to T gives one first order condition which implicitly defines T*: * * * * *) ( *) ( ' 0 *) ( *) ( ' 0 : rT rT rT rT e T rPQ e T PQ e T rPQ e T PQ dT d FOC - - - - = ? = - ? = Π Since e -rt and P cancel, this leaves us with or, ( 29 r T Q T Q = ) ( ' (1) This says that the profit maximizing rotation length is such that the growth of the tree volume is equal to the interest rate. Notice that Q’(T) refers to change in growth from one period to the next and represents the change in volume of trees. The growth rate of trees multiplied by 100 is the percentage growth of trees from one period to the next. Using the information given in the description of the problem, and Q’(T) = 24T-T 2 , equation 1 is rewritten as 1 . 0 T 1/3 - T 12 24 3 2 2 = - T T (2) To find T*, we need to solve equation (2) for T. After factoring out a T and rearranging terms, equation (2) can be rewritten as *) ( *) ( ' T rQ T Q =
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0 720 66 0 24 2 . 2 0 ) 12 ( 1 . 0 24 2 2 30 1 2 3 1 = + - ? = + - ? = - - - T T T T T T T We can solve this expression for T using the quadratic formula (Refresher: a ac b b 2 4 2 - ± - where a,b, and c correspond to the coefficient of the first, second and third term of the quadratic function respectively). For our case, the value(s) of T for which equation (3) holds are found to be: 2 42 . 38 66 2 ) 720 ( 4 66 66 2 ± = - ± which reduces to T=13.785 or 52.215. This doesn’t necessarily mean that rotation lengths of either 13.785 years or 52.215 years will maximize profits. Check both answers by plugging each choice back into the profit function P*Q(T)e -- rT T=13.785: P*Q(T) e -rT = $10 [12(13.785) 2 – 1/3 (13.785) 3 ]e -- (0.1)(13.785) = $10(2280.3147 – 873.171)(0.252) = $3546.00 T=52.215: P*Q(T) e -rT = $10 [12(52.215) 2 – 1/3 (52.215) 3 ]e -- (0.1)(52.215) = $10(32716.8747 – 47453.1)(0.0054) = - $795.76 Actually what the T=52.215 is saying is that by waiting for about 52 years until harvesting these trees will result yield in a negative yield (Q(T=52.215)=
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HW5,Sp2000sols - EEP101/ECON125 Spring 00 Prof.: D....

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