{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

sol8 - Solutions to Problem Set 8 ARE 261 Question 1 1 The...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Solutions to Problem Set 8 ARE 261 December 15, 2004 Question 1 1. The dynamic programming equation (DPE) for the control problem is J t ( x t , t ) = max u t ½ 1 2 e rt ( u 2 t + x 2 t ) + J x ( x t , t )( x t + u t ) ¾ (1) where J ( x t , t ) is the value function, and J t ( x t , t ) and J x ( x t , t ) are partial derivatives of the value function. 2. J ( x T , T ) = e rT ax 2 T 2 . 3. Substitute the guess for the value function, that is J ( x t , t ) = s ( t ) x 2 t 2 , into the DPE, di ff erentiate with respect to u t , and set the expression to zero. Note that J t ( x t , t ) = ˙ s x 2 t 2 and J x ( x t , t ) = sx . This gives the following policy function u t = e rt s ( t ) x t (2) Substitute the optimal policy function into the DPE. This in turns gives the di ff erential equation that solves s ( t ) ˙ s = e rt s 2 + e rt 2 s t (3) With the boundary condition that s ( T ) = ae rT . It is easier to work with a transformation of s ( t ) . Let z ( t ) = e rt s ( t ) . Then dz dt = z ( t ) 2 (2 r ) z ( t ) + 1 (4) With the boundary condition that z T = a . 4. λ t from the Maximum Principle is equal to J x ( x t , t ) from the DPE. 5. We already have an autonomous di ff erential equation in z ( t ) . By solv- ing this equation “backwards” we convert the boundary condition at the terminal period into a boundary condition at the initial time. Rede fi ne 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
the di ff erential equation for z ( t ) in terms of τ where τ = T t , the time remaining. In terms of τ the di ff erential equation is dz = z ( τ ) 2 + (2 r ) z ( τ ) 1 (5) This equation has two steady states, one positive and one negative. These are solved for by setting dz = 0 and then solving the quadratic in z . The positive root is equal to z 1 = (2 r )+ (2 r ) 2 +4 2 and the negative root is equal to z 2 = (2 r ) (2 r ) 2 +4 2 . The negative root is stable while the positive root is unstable. The fi gure shows the graph of dz = 0 for r = 1 -1 0 1 2 3 4 5 dz/d(tau) -3 -2 -1 1 2 z In the steady state, we have s = z 2 e rt , so ds dt = rz 2 e rt = rs -4 -2 0 2 4 ds/dt -4 -2 2 4 s 2
Background image of page 2
6. So long as the initial point a lies to the left of the positive root z 1 , the di ff erential equation for s will converge to the steady state as T → ∞ .
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}