sol_set5

# sol_set5 - Solutions to Problem Set 5 ARE 261 November 4...

This preview shows pages 1–3. Sign up to view the full content.

Solutions to Problem Set 5 ARE 261 November 4, 2001 Question 1 The Euler equation, using Calculus of Variations, is ¨ x =0 (1) If we integrate this equation twice and use the boundary conditions we get the following solution x ( t )= t +2 (2) The necessary conditions using the Maximum Principle are ∂H ∂u = x 2 u + λ =0 2 H ∂u 2 = 2 0 ∂H ∂x = ˙ λ = u +2 x λ ∂H ∂λ = ˙ x = x + u Now guess that the co-state variable is linear in the state, or that, λ = s ( t ) x . Di f erentiate this equation with respect to time. This yields ˙ λ = x ˙ s + s ˙ x (3) From the f rst necessary condition we know that u = λ + x 2 . Use this to get rid o f u in the third and fourth necessary condition and then substitute these into equation (3). This gives the following ODE in s 2 ˙ s +6 s + s 2 3=0 (4) Question 2 The f rst thing we are asked to f nd is the stock level that maximizes steady state harvest. The equation for steady state level harvest is given by setting the equation of motion for the stock of f sh to zero. Thus h ss = αβx ss α ( x ss ) 2 (5) 1

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

View Full Document
where the superscript ss denotes steady state. To determine the maximum steady state level of harvest di f erentiate equation 5 with respect to x ss and set the expression to zero. This implies that h m = αβ 2 4 x m = β 2 Next we want to f nd the stock level that maximizes the steady state F ow of net utility. Net utility at steady state is given by U ( h ss ) c ( x ss ) h ss (6) where h ss is de f n edbyequa t
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 5

sol_set5 - Solutions to Problem Set 5 ARE 261 November 4...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online