ECO4145 Mathematical Economics II Assignment 3
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ECO4145 Mathematical Economics II
Assignment 3
1. Find a continuous function c : t 
ct, 0 ≤ t ≤ 1, to minimize(ct)0≤t≤1 
∫0(1)ct(4) 
t
subject to 
= xt + ct, x0 = x0, x1 = 0.
The Hamiltonian is
ℋ = 
c4 + λ (x + c)
4
4
The state equation
eq [1] = x' [t ] 
x [t ] + c [t ]
x′ [t ] 
c [t ] + x [t ]
The first-order condition for minimizing the Hamiltonian is
ℋ t = ℋ / . {x 
x [t ] , λ 
λ[t ]}
4
4
foct = D[ℋt, c ] 
0
偏导
c3 + λ[t ] 
0
The second-order condition for minimizing the Hamiltonian is satisfied.
soct = D[foct〚1〛, c ]
偏导
3 c2
The optimal control at time t
s [0] = Solve[foct, c ]
解方程
c 
- λ[t ]1/3 
, 
c 
(- 1)1/3 λ[t ]1/3 
, 
c 
- (- 1)2/3 λ[t ]1/3 
Note that the three solutions are the same. So, any one of them will do. Let us pick the first 0ne.
In[
]:= s [1] = s [0]〚1〛
Out[
]= 
c 
- λ[t ]1/3 
The optimal control at time t
In[
]:= c [t_ ] := Evaluate[c / . s [1]]
计算
In[
]:= c [t ]
Out[
]= - λ[t ]1/3
In[
]:= c [1]
Out[
]= - λ[1]1/3
The adjoint equation is
In[
]:= dℋdx = D[ℋ , x ]
偏导
Out[
]= λ
In[
]:= dℋdxt = dℋdx / . {x 
x [t ] , c 
c [t ] , λ 
λ[t ]}
Out[
]= λ[t ]
In[
]:= eq [2] = λ ' [t ] 
- dℋdxt
Out[
]= λ′ [t ] 
- λ[t ]
In[
]:= s [2] = DSolve[{eq [1], eq [2], x [0] 
x0, λ[0] 
λ0} , {x, λ}, t ] // Flatten
求解微分方程 
压平
Out[
]= 
λ 
Function {t } , 
-t λ0 , x 
Function {t } , - 
In[
]:= x [1] / . s [2]
- 4 x0 λ0 + 3 λ04/3 - 
Out[
]= -
4 λ0
The final condition for xt
In[
]:= eq [3] = (x [1] / . s [2]) 
0
0
In[
]:= s [3] = Solve[eq [3] , λ0] // Flatten
解方程 
压平
Solve : There may be values of the parameters for which some or all solutions are not valid .
Out[
]= 
λ 0 
In[
]:= s [3] = s [3] // FullSimplify
完全简化
64 
4 x03
Out[
]= 
λ 0 
In[
]:= λ0 = λ0 / . s [3]
Out[
]= 
With the value ofλ0 just found, solve the system again
In[
]:= s [3] = DSolve[{eq [1], eq [2], x [0] 
x0, λ[0] 
λ0} , {x, λ}, t ] // Flatten
求解微分方程 
压平
Out[
]= 
λ 
Function {t } , 
,
-4+t - 
4 x04 + 
16/3 x04 - 
16/3 x03 4/3 + 
4 -t x03 4/3
x 
Function {t } , 
The optimal control at time t is
In[
]:= c [t ]
Out[
]= - λ[t ]1/3
In[
]:= c [t ] / . s [3]
4 
4-t x03 1/3
Out[
]= -
3 × - 1 + 
4/3
The minimum cost is
In[
]:= v = 
01 
c [t ]4 / . s [3]
t
16 
4 x03 4/3
Out[
]= 27 - 1 + 
4/3 3
In[
]:= v = v // PowerExpand
幂展开
Out[
]= 
In[
]:= ClearAll[ℋ, x, c, λ , x0, λ0, v, s, eq, foc]
清除全部
2. Find a continuous function c : t 
ct, 0 ≤ t ≤ 2, to
maximize(ct)0≤t≤2 ∫0(2)
2 xt - 3 ct - c 
tt(2)
subject to 
= xt + ct, x0 = 5, x2 = 0.
4 
ECO4145_Assignment 3_04_04_2022_Solution.nb
In[
]:= ℋ = 2 x - 3 c - c2 + λ (x + c)
Out[
]= - 3 c - c2 + 2 x + (c + x ) λ
The state equation is
In[
]:= eq [1] = x' [t ] 
x [t ] + c [t ]
Out[
]= x′ [t ] 
c [t ] + x [t ]
Maximizing the Hamiltonian
In[
]:= ℋ t = ℋ / . {x 
x [t ] , λ 
λ[t ]}
Out[
]= - 3 c - c2 + 2 x [t ] + (c + x [t ]) λ[t ]
The optimal control at each instant is obtained by maximizing the Hamiltonian ℋ[xt(*), c, λt, t] at that instant. The first-order condition that characterizes ct(*) is
In[
]:= foct = D[ℋt, c ] 
0
偏导
Out[
]= - 3 - 2 c + λ[t ] 
0
The second-order condition that characterizes ct(*) is satisfied.
In[
]:= soct = D[foct〚1〛, c ]
偏导
Out[
]= - 2
The optimal control at time t
In[
]:= s [0] = Solve[foct, c ] // Flatten
解方程 
压平
Out[
]= 
c 
× (- 3 + λ[t ]) 
In[
]:= c [t_ ] := Evaluate[c / . s [0]]
计算
In[
]:= c [t ]
Out[
]= 
× (- 3 + λ[t ])
In[
]:= eq [1]
Out[
]= x′ [t ] 
x [t ] + 
× (- 3 + λ[t ])
The adjoint equation
In[
]:= dℋdx = D[ℋ , x ]
偏导
Out[
]= 2 + λ
In[
]:= dℋdxt = dℋdx / . {x 
x [t ] , c 
c [t ] , λ 
λ[t ]}
Out[
]= 2 + λ[t ]
In[
]:= eq [2] = λ ' [t ] 
- dℋdxt
Out[
]= λ′ [t ] 
- 2 - λ[t ]
In[
]:= eq [1]
Out[
]= x′ [t ] 
x [t ] + 
× (- 3 + λ[t ])
Solve the system ofdifferential equations {eq[1], eq[2]}
In[
]:= s [1] =
DSolve[{eq [1], eq [2], x [0] 
5, λ[0] 
λ0} , {x, λ}, t ] // Flatten // FullSimplify
求解微分方程 
压平 
完全简化
Out[
]= 
x 
Function {t } , 
-t 1 + 
t × - 2 + 12 
t - λ0 + 
t λ0
,
λ 
Function {t } , - 
-t - 2 + 2 
t - λ0
The final condition for xt
In[
]:= eq [3] = (x [2] / . s [1]) 
0
1 + 
2 × - 2 + 12 
2 - λ0 + 
2 λ0
Out[
]= 
0
4 
2
In[
]:= s [2] = Solve[eq [3] , λ0] // Flatten
解方程 
压平
Out[
]= 
λ 0 
- 
In[
]:= λ0 = λ0 / . s [2]
Out[
]= - 
Use the value ofλ0, solve the system ofdifferential equations {eq[1], eq[2]} again
In[
]:= s [3] =
DSolve[{eq [1], eq [2], x [0] 
5, λ[0] 
λ0} , {x, λ}, t ] // Flatten // FullSimplify
求解微分方程 
压平 
完全简化
Out[
]= 
x 
Function {t } , - 
,
λ 
Function {t } , - 
The optimal control at time t
In[
]:= c [t ] / . s [3]
Out[
]= 
× - 3 - 
The optimal payoff
In[
]:= v = 
2 
2 x[t ] - 3 c[t ] - c [t ]2 
/ . s [3]
t
Out[
]= - 
In[
]:= ClearAll[ℋ , ℋ t, dℋdx, dℋdxt, eq, s, λ , λ0, foct, c, x, v]
清除全部
3. Find a continuous function c : t 
ct, 0 ≤ t ≤ 2, to maximize(ct)0≤t≤2 ∫0(2)
2 xt - 3 ct - ct(2)
t
subject to 
= xt + ct, x0 = 5.
Note that the endpoint x2 is free, i.e., not constrained.
Let
(1) v[ξ , τ ] = maximize(ct) τ ≤t≤ 2 
2 xt - 3ct - ct(2)
t
subject to 
= xt + ct, xτ = ξ , x2 = 0.
As defined, v[ξ , τ ] gives the value attained by the objective function, given that the system begins at time 0 ≤ τ ≤ 2 in state ξ. For τ = 2, we have v[ξ , 2] = 0 because the problem begins at the end of the time horizon and ends immediately, with the ensuing consequence that there is no payoff: v[ξ , 2] = 0 no matter what the value of ξ is, and, thus, D1 v[ξ , 2] = 0.
Let ct(*) be the optimal control at time t, 0 ≤ t ≤ 2, for the original problem and xt(*), 0 ≤ t ≤ 2, be the state of the system at time t under the optimal control ct(*), 0 ≤ t ≤ 2. Also, let λt= D1 v[xt(*), t] denote the co-state at time t along the optimal trajectory.
The Hamiltonian is
In[
]:= ℋ = 2 x - 3 c - c2 + λ (x + c)
Out[
]= - 3 c - c2 + 2 x + (c + x ) λ
The state equation is
In[
]:= eq [1] = x' [t ] 
x [t ] + c [t ]
Out[
]= x′ [t ] 
c&nbs
2022-04-12