Download Optimal Pumping Strategy for Oil Reserves: Trinomial Lattice and Bellman Equation Solution and more Assignments Systems Engineering in PDF only on Docsity!
Table 1: Possible states of oil reserves. 0 1 2 3 100000 100000 100000 100000 80000 80000 80000 64000 64000 64000 51200 51200 40960 40 960 32768 26214
Table 2: Optimal pumping strategy. 0 1 2 366744 [Enhance] 340364 [Enhance] 240000 [Enhance] 226473 [Enhance] 168000 [Enhance] 139418 [Enhance] 110400 [Enhance] 64320 [Enhance] 31920 [Normal]
Solution to Problem 1. The “trinomial lattice” that represents the possible states of oil reserves is given in Table 1. The value of the oil well at each sate and the optimal pumping strategy is given in Table 2. Consider, for example, the state of having 80000 barrels at time t = 1. Let d = 1/ 1 .1. The value at this state is the maximum of the three choices
max [
Enhance ︷ ︸︸ ︷ 0 .36(80000) − 120000 + 64320d,
N ormal ︷ ︸︸ ︷ 0 .20(80000) − 50000 + 110400d,
N o pump ︷ ︸︸ ︷ 168000 d ],
which is achieved by enhancing for a value of 226473.
Solution to Problem 2.
Part a. The Bellman equation for this problem is
V (x) = max 0 ≤z≤x {(x − z)z + dV (x − z)}. (1)
Part b. Suppose that V (x) = Kx for some K > 0. Then
Kx = max 0 ≤z≤x {(x − z)z + dK(x − z)} (2)
= max 0 ≤z≤x {(x − dK)z − z^2 + dKx}. (3)
The optimal z(x) is found by taking the derivative of the quadratic function of z inside the braces and setting the resulting equation to zero. You can verify that
z(x) := (x − dK)/ 2. (4)
Substituting z(x) back into the quadratic function of z inside the braces, and performing a bit of algebra, we have Kx = (x − dK)^2 /4 + dKx. (5)
The right-hand side of (5) is a quadratic nonlinear function of x, whereas the left-hand side is a linear function of x. The two sides “do not match,” and so the functional form for the optimal value function V (·) cannot be linear.
Part c. Here, one simply maximizes the quadratic function of z given by (x − z)z. The optimal solution is z(x) = x/2 and the objective function is (x − x/2)(x/2) = x^2 /4.
Part d. The solution to part (c) suggests that V (x) = Kx^2 for some positive K and that the optimal policy is z(x) = Cx for some positive C. You can verify that if you set V (x) = Kx^2 on the right-hand side of the Bellman equation, the right-hand side is still a quadratic function of z. The optimal value z(x) will be linear in x, and the optimal objective function value will remain quadratic upon substitution. The solution to Part (e) below will verify this and provide an exact solution.
Part e. Consider the right-hand side of the Bellman equation when V (x) = Kx^2. Taking the derivative of the function (x − z)z + dK(x − z)^2 with respect to z and setting it to zero yields 0 = x − 2 z − 2 dK(x − z), and thus
z(x) =
[ (^1) − 2 dK 2(1 − dK)
] x. (6)
Define
C(K) :=
1 − 2 dK 2(1 − dK)
