# Dynamic Cash Flow Homework Solutions

Table 1: Possible states of oil reserves.

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 re presents 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

which is achieved by enhancing for a value of 226473.

Solution to Problem 2.

Part a. The Bellman equation for this problem is

Part b. Suppose that V (x) = Kx for some K > 0. Then

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

Substituting z (x) back into the quadratic function of z inside the braces, and performing a bit
of algebra , we have

The right-hand side of (5) is a quadratic non linear 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) be low 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 - 2z- 2dK(x - z), and thus

Define

Given the value for K, we can express the optimal policy as

The optimal objective function value of the right-hand side of the Bellman equation is therefore

The right-hand side function must be identical to the left-hand side function, which is, by
assumption, Kx^2. If our as sumption is correct , then a value of K exists such that

Assume d = 1/1.1. We can solve for the value of K and thus C(K) using successive
approximations. That is, we can define

where C(Ki) is defined in (7). Assuming V0(•) = 0 is identical to setting K0 = 0. This yields
C(0) = 0.5 from (7). Substituting C(0) = 0.5 into the right-hand side of (11) yields K1 = 0.25.
(We're just repeating the myopic solution so far.) The value of C(0.25) is 0.3529. Substituting
C(0.25) = 0.3529 into the right-hand side of (11) yields K2 = 0.3235. The value of C(0.3235)
is 0.2917. Substituting C(0.3235) = 0.2917 into the right-hand side of (11) yields K3 = 0.3402.
The process continues. The Ki converge to := 0.38416876 and C() = 0.231662479.

Remark. The identity (11) can be directly solved for K, but it requires a bit of lengthy algebra.
It turns out that the right-hand side simplifies to the reciprocal of 4(1 - dK). Thus, the K
value we seek satisfies the identity 1 = 4(1-dK)K or 4dK^2 -4K +1 = 0. The solution to this

(The larger root will be larger than 1/2d, which will yield a negative value for C(K).) Since

it follows that

Note that C = 0.5 when d = 0, as it should.

 Prev Next

Start solving your Algebra Problems in next 5 minutes!

2Checkout.com is an authorized reseller
of goods provided by Sofmath

Attention: We are currently running a special promotional offer for Algebra-Answer.com visitors -- if you order Algebra Helper by midnight of February 19th you will pay only \$39.99 instead of our regular price of \$74.99 -- this is \$35 in savings ! In order to take advantage of this offer, you need to order by clicking on one of the buttons on the left, not through our regular order page.

If you order now you will also receive 30 minute live session from tutor.com for a 1\$!

You Will Learn Algebra Better - Guaranteed!

Just take a look how incredibly simple Algebra Helper is:

Step 1 : Enter your homework problem in an easy WYSIWYG (What you see is what you get) algebra editor:

Step 2 : Let Algebra Helper solve it:

Step 3 : Ask for an explanation for the steps you don't understand:

Algebra Helper can solve problems in all the following areas:

• simplification of algebraic expressions (operations with polynomials (simplifying, degree, synthetic division...), exponential expressions, fractions and roots (radicals), absolute values)
• factoring and expanding expressions
• finding LCM and GCF
• (simplifying, rationalizing complex denominators...)
• solving linear, quadratic and many other equations and inequalities (including basic logarithmic and exponential equations)
• solving a system of two and three linear equations (including Cramer's rule)
• graphing curves (lines, parabolas, hyperbolas, circles, ellipses, equation and inequality solutions)
• graphing general functions
• operations with functions (composition, inverse, range, domain...)
• simplifying logarithms
• basic geometry and trigonometry (similarity, calculating trig functions, right triangle...)
• arithmetic and other pre-algebra topics (ratios, proportions, measurements...)

ORDER NOW!