551621quiz4.dvi = 2 1 100 10 1 20 2 max f g 1 =0 ¡0 1 5 3 + 1 6 1 12 2 = 24 10 = 5 3 + 6 12 = 0 = 0 fgdfgdfgdfgdfgdfgddfgdfgdfgdfg...

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551621quiz4.dvi = 2 1 100 10 1 20 2 max f g 1 =0 ¡0 1 5 3 + 1 6 1 12 2 = 24 10 = 5 3 + 6 12 = 0 = 0 fgdfgdfgdfgdfgdfgddfgdfgdfgdfg sdgdsggsdfgdfgdsfgsddfgdfgdfgdfgdfgsdgsdgsdgsdgsdgsdgsdgsdgsddggsdgdfgdfger gsdgdgdgdgdssdgsdd dfgdfgdfgdfgdfgdfgfdgdf
Answered Same DayFeb 03, 2021

Answer To: 551621quiz4.dvi = 2 1 100 10 1 20 2 max f g 1 =0 ¡0 1 5 3 + 1 6 1 12 2 = 24 10 = 5 3 + 6 12 = 0 = 0...

Ishmeet Singh answered on Feb 03 2021
139 Votes
Q1
Objective: Find Hamiltonian and max. Principle
Given: Eq. of motionEq’n: 1
dx/dt = 2x (1-x/100
)-y
x is stock & y is control variable as a function of time.
Each periods objective: 10y-1/20y^2
Discount rate: 10%
Sol’n: Hamiltonian, H= f(x,y)
Dx/dt or x*= 2x (1-x/100)-y         From Eq’n: 1
Similarly, dy/dt or y* = 10-y/10
Therefore,
Y(0) = 10-y/10 => 10 (Given)
Y(10) = 9
Max. (x) =⌡2x(1-x/100)-y        {limit 0 to T}
Q2
Therefore taking Hamilton equation as,
H= 2x(1-x/100)-y +ƛx
Solving for dH/dx,
· 2x –x^2/50 –y +ƛx
· 2-x/25+ƛ = 0
· ƛ =(x-50)/25
· x = 25 ƛ + 50
Therefore it is verified that the Hamiltonian is st. line in y. Next in light of the fact that the control variable is dependent on the costate variable, we would need to solve for the solution for λ.
Now, ƛ = -dH/dy =-1
ƛ = 1
Since the integrand is not dependent on the state variable, y λ˙ = − ∂H ∂y = -1 ⇒ λ = 1
where...
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