solnt9 - Partial solution to Problem set 9 Answer to part...

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

View Full Document Right Arrow Icon
Partial solution to Problem set 9 Answer to part a) y = F + γN The DPE for the autonomous problem is: rJ ( N )=max F μ ay b 2 y 2 + J 0 ( N )( AN + BF ) I am going to "guess" that the value function is quadratic in the state: J = λ + µN + ν N 2 2 Substitute this guess into the DPE to obtain r μ λ + µN + ν N 2 2 =max F μ ay b 2 y 2 +( µ + νN )( AN + BF ) Now I’m going to ask Scienti f cWorkp laceto f nd the max ay b 2 y 2 +( µ + νN )( AN + BF ) Candidate(s) for extrema: 1 2 b ¡ PN 2 + KN + L ¢ P B 2 ν 2 +2 νAb 2 νBbγ K 2 aBν +2 B 2 µν +2 µAb 2 µBbγ L a 2 +2 aBµ + B 2 µ 2 The optimal control rule is F = a bγN + + BνN b
Background image of page 1

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

View Full DocumentRight Arrow Icon
The maximized DPE is r μ λ + µN + ν N 2 2 = 1 2 b ¡ PN 2 + KN + L ¢ Equating coe cients in N 2 implies = P b or = B 2 ν 2 +2 νAb 2 νBbγ b or 1 b ¡ B 2 ν 2 +2 νAb 2 νBbγ ¢ ν =0 , Solution is: { ν =0 } , © ν = b 2 A +2 +1 B 2
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 3

solnt9 - Partial solution to Problem set 9 Answer to part...

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

View Full Document Right Arrow Icon
Ask a homework question - tutors are online