Solving dynamic optimization with non-binding inequality constraint
I want to solve a problem similar to the following discrete and finite time horizon dynamic optimization problem :
begin{equation}
begin{split}
&max_{{d_t}} sum_{t=0}^{T} - left [ f(s_t) + k(d_t) right ] + F(s_{T+1})\
& s.t. left { begin{array}{ll}
s_{t+1} = s_t + d_t \
s_t < A \
s_0 in mathbb{R}_+ \
s_T = A \
end{array}
right.
end{split}
end{equation}
I am able to solve the problem without the inequality constraint $s_t < A$ using Lagrangian multiplier which leads to a recursive equation on $s_t$ which I am able to solve numerically afterwards.
I wish to find a similar recursive equation, but this time when considering $s_t < A$. How should I take this inequality constraint into account when computing the Lagrangian first order conditions ?
inequality optimization lagrange-multiplier constraints
|
show 6 more comments
I want to solve a problem similar to the following discrete and finite time horizon dynamic optimization problem :
begin{equation}
begin{split}
&max_{{d_t}} sum_{t=0}^{T} - left [ f(s_t) + k(d_t) right ] + F(s_{T+1})\
& s.t. left { begin{array}{ll}
s_{t+1} = s_t + d_t \
s_t < A \
s_0 in mathbb{R}_+ \
s_T = A \
end{array}
right.
end{split}
end{equation}
I am able to solve the problem without the inequality constraint $s_t < A$ using Lagrangian multiplier which leads to a recursive equation on $s_t$ which I am able to solve numerically afterwards.
I wish to find a similar recursive equation, but this time when considering $s_t < A$. How should I take this inequality constraint into account when computing the Lagrangian first order conditions ?
inequality optimization lagrange-multiplier constraints
1
Introduce a slack variable $epsilon$ such that $s_t-A+epsilon^2 = 0$
– Cesareo
Dec 26 at 14:06
1
Thank you @Cesareo for your answer. Then how should I deal with the slack variable when deriving the first order condition ? Is there a direct method regarding slack variables ?
– AlexC75
Dec 26 at 14:11
1
Handle it as more a variable. The Lagrangian now reads $L(d_t,lambda,epsilon)$ so regarding $epsilon$ we have $L_{epsilon} = 2lambda_iepsilon = 0$ and $L_{lambda_i} = s_t -A+epsilon^2=0$ one additional condition.
– Cesareo
Dec 26 at 14:18
1
But there is no new co-state variable ? I am not sure on how to write the Lagrangian is this framework.
– AlexC75
Dec 26 at 14:26
$L(d_t,lambda,epsilon) = sum_{t=0}^{T} - left [ f(s_t) + k(d_t) right ] + F(s_{T+1})+lambda_1(s_{t+1} - s_t - d_t)+lambda_2(s_t-A+epsilon_1^2)+lambda_3(S_T-A)+lambda_4(s_0-epsilon_2^2)$
– Cesareo
Dec 26 at 14:31
|
show 6 more comments
I want to solve a problem similar to the following discrete and finite time horizon dynamic optimization problem :
begin{equation}
begin{split}
&max_{{d_t}} sum_{t=0}^{T} - left [ f(s_t) + k(d_t) right ] + F(s_{T+1})\
& s.t. left { begin{array}{ll}
s_{t+1} = s_t + d_t \
s_t < A \
s_0 in mathbb{R}_+ \
s_T = A \
end{array}
right.
end{split}
end{equation}
I am able to solve the problem without the inequality constraint $s_t < A$ using Lagrangian multiplier which leads to a recursive equation on $s_t$ which I am able to solve numerically afterwards.
I wish to find a similar recursive equation, but this time when considering $s_t < A$. How should I take this inequality constraint into account when computing the Lagrangian first order conditions ?
inequality optimization lagrange-multiplier constraints
I want to solve a problem similar to the following discrete and finite time horizon dynamic optimization problem :
begin{equation}
begin{split}
&max_{{d_t}} sum_{t=0}^{T} - left [ f(s_t) + k(d_t) right ] + F(s_{T+1})\
& s.t. left { begin{array}{ll}
s_{t+1} = s_t + d_t \
s_t < A \
s_0 in mathbb{R}_+ \
s_T = A \
end{array}
right.
end{split}
end{equation}
I am able to solve the problem without the inequality constraint $s_t < A$ using Lagrangian multiplier which leads to a recursive equation on $s_t$ which I am able to solve numerically afterwards.
I wish to find a similar recursive equation, but this time when considering $s_t < A$. How should I take this inequality constraint into account when computing the Lagrangian first order conditions ?
inequality optimization lagrange-multiplier constraints
inequality optimization lagrange-multiplier constraints
asked Dec 26 at 13:58
AlexC75
215
215
1
Introduce a slack variable $epsilon$ such that $s_t-A+epsilon^2 = 0$
– Cesareo
Dec 26 at 14:06
1
Thank you @Cesareo for your answer. Then how should I deal with the slack variable when deriving the first order condition ? Is there a direct method regarding slack variables ?
– AlexC75
Dec 26 at 14:11
1
Handle it as more a variable. The Lagrangian now reads $L(d_t,lambda,epsilon)$ so regarding $epsilon$ we have $L_{epsilon} = 2lambda_iepsilon = 0$ and $L_{lambda_i} = s_t -A+epsilon^2=0$ one additional condition.
– Cesareo
Dec 26 at 14:18
1
But there is no new co-state variable ? I am not sure on how to write the Lagrangian is this framework.
– AlexC75
Dec 26 at 14:26
$L(d_t,lambda,epsilon) = sum_{t=0}^{T} - left [ f(s_t) + k(d_t) right ] + F(s_{T+1})+lambda_1(s_{t+1} - s_t - d_t)+lambda_2(s_t-A+epsilon_1^2)+lambda_3(S_T-A)+lambda_4(s_0-epsilon_2^2)$
– Cesareo
Dec 26 at 14:31
|
show 6 more comments
1
Introduce a slack variable $epsilon$ such that $s_t-A+epsilon^2 = 0$
– Cesareo
Dec 26 at 14:06
1
Thank you @Cesareo for your answer. Then how should I deal with the slack variable when deriving the first order condition ? Is there a direct method regarding slack variables ?
– AlexC75
Dec 26 at 14:11
1
Handle it as more a variable. The Lagrangian now reads $L(d_t,lambda,epsilon)$ so regarding $epsilon$ we have $L_{epsilon} = 2lambda_iepsilon = 0$ and $L_{lambda_i} = s_t -A+epsilon^2=0$ one additional condition.
– Cesareo
Dec 26 at 14:18
1
But there is no new co-state variable ? I am not sure on how to write the Lagrangian is this framework.
– AlexC75
Dec 26 at 14:26
$L(d_t,lambda,epsilon) = sum_{t=0}^{T} - left [ f(s_t) + k(d_t) right ] + F(s_{T+1})+lambda_1(s_{t+1} - s_t - d_t)+lambda_2(s_t-A+epsilon_1^2)+lambda_3(S_T-A)+lambda_4(s_0-epsilon_2^2)$
– Cesareo
Dec 26 at 14:31
1
1
Introduce a slack variable $epsilon$ such that $s_t-A+epsilon^2 = 0$
– Cesareo
Dec 26 at 14:06
Introduce a slack variable $epsilon$ such that $s_t-A+epsilon^2 = 0$
– Cesareo
Dec 26 at 14:06
1
1
Thank you @Cesareo for your answer. Then how should I deal with the slack variable when deriving the first order condition ? Is there a direct method regarding slack variables ?
– AlexC75
Dec 26 at 14:11
Thank you @Cesareo for your answer. Then how should I deal with the slack variable when deriving the first order condition ? Is there a direct method regarding slack variables ?
– AlexC75
Dec 26 at 14:11
1
1
Handle it as more a variable. The Lagrangian now reads $L(d_t,lambda,epsilon)$ so regarding $epsilon$ we have $L_{epsilon} = 2lambda_iepsilon = 0$ and $L_{lambda_i} = s_t -A+epsilon^2=0$ one additional condition.
– Cesareo
Dec 26 at 14:18
Handle it as more a variable. The Lagrangian now reads $L(d_t,lambda,epsilon)$ so regarding $epsilon$ we have $L_{epsilon} = 2lambda_iepsilon = 0$ and $L_{lambda_i} = s_t -A+epsilon^2=0$ one additional condition.
– Cesareo
Dec 26 at 14:18
1
1
But there is no new co-state variable ? I am not sure on how to write the Lagrangian is this framework.
– AlexC75
Dec 26 at 14:26
But there is no new co-state variable ? I am not sure on how to write the Lagrangian is this framework.
– AlexC75
Dec 26 at 14:26
$L(d_t,lambda,epsilon) = sum_{t=0}^{T} - left [ f(s_t) + k(d_t) right ] + F(s_{T+1})+lambda_1(s_{t+1} - s_t - d_t)+lambda_2(s_t-A+epsilon_1^2)+lambda_3(S_T-A)+lambda_4(s_0-epsilon_2^2)$
– Cesareo
Dec 26 at 14:31
$L(d_t,lambda,epsilon) = sum_{t=0}^{T} - left [ f(s_t) + k(d_t) right ] + F(s_{T+1})+lambda_1(s_{t+1} - s_t - d_t)+lambda_2(s_t-A+epsilon_1^2)+lambda_3(S_T-A)+lambda_4(s_0-epsilon_2^2)$
– Cesareo
Dec 26 at 14:31
|
show 6 more comments
1 Answer
1
active
oldest
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I will present a reduced example shoving how to solve this kind of problems using the classical procedure of Lagrange multipliers. To apply this technique, previously we transform the inequalities into equalities with the help of the $epsilon_k$ slack variables.
First form the Lagrangian
$$
L(x,u,lambda,epsilon) = phi(x_{n+1})-f(x,u)+sum_{k=1}^nlambda_k(x_{k+1}-x_k-u_k) + sum_{k=1}^nlambda_{n+k}(x_k-A+epsilon_k^2)+lambda_{2n+1}(x_n-A)
$$
where
$$
phi(x_{n+1}) = x_{n+1}^2\
f(x,u) = min(max)sum_{k=1}^na x_k^2+b u_k^2
$$
Now considering $n = 5$ the stationary conditions are
$$
nabla L = left{
begin{array}{rcl}
2 epsilon _1 lambda _6&=&0 \
2 epsilon _2 lambda _7&=&0 \
2 epsilon _3 lambda _8&=&0 \
2 epsilon _4 lambda _9&=&0 \
2 epsilon _5 lambda _{10}&=&0 \
-u_1-x_1+x_2&=&0 \
-u_2-x_2+x_3&=&0 \
-u_3-x_3+x_4&=&0 \
-u_4-x_4+x_5&=&0 \
-u_5-x_5+x_6&=&0 \
epsilon _1^2-A+x_1&=&0 \
epsilon _2^2-A+x_2&=&0 \
epsilon _3^2-A+x_3&=&0 \
epsilon _4^2-A+x_4&=&0 \
epsilon _5^2-A+x_5&=&0 \
x_5-A&=&0 \
2 b u_1+lambda _1&=&0 \
2 b u_2+lambda _2&=&0 \
2 b u_3+lambda _3&=&0 \
2 b u_4+lambda _4&=&0 \
2 b u_5+lambda _5&=&0 \
-lambda _1+lambda _6-2 a x_1&=&0 \
lambda _1-lambda _2+lambda _7-2 a x_2&=&0 \
lambda _2-lambda _3+lambda _8-2 a x_3&=&0 \
lambda _3-lambda _4+lambda _9-2 a x_4&=&0 \
lambda _4-lambda _5+lambda _{10}+lambda _{11}-2 a x_5&=&0 \
lambda _5+2 x_6&=&0 \
end{array}
right.
$$
Solving this system of equations we obtain for the parametric values $a = 2, b = 2, A = 1, n = 5$
follows the $min$ strategy (in red $u$ and in blue $x$)
and the $max$ strategy
Any way, this kind of problem can be satisfactorily solved with a quadratic programming package, without the need for the slack variables introduction. See https://en.wikipedia.org/wiki/Quadratic_programming
I hope this helps.
This definitely help. Thank's a lot.
– AlexC75
Dec 27 at 15:26
add a comment |
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I will present a reduced example shoving how to solve this kind of problems using the classical procedure of Lagrange multipliers. To apply this technique, previously we transform the inequalities into equalities with the help of the $epsilon_k$ slack variables.
First form the Lagrangian
$$
L(x,u,lambda,epsilon) = phi(x_{n+1})-f(x,u)+sum_{k=1}^nlambda_k(x_{k+1}-x_k-u_k) + sum_{k=1}^nlambda_{n+k}(x_k-A+epsilon_k^2)+lambda_{2n+1}(x_n-A)
$$
where
$$
phi(x_{n+1}) = x_{n+1}^2\
f(x,u) = min(max)sum_{k=1}^na x_k^2+b u_k^2
$$
Now considering $n = 5$ the stationary conditions are
$$
nabla L = left{
begin{array}{rcl}
2 epsilon _1 lambda _6&=&0 \
2 epsilon _2 lambda _7&=&0 \
2 epsilon _3 lambda _8&=&0 \
2 epsilon _4 lambda _9&=&0 \
2 epsilon _5 lambda _{10}&=&0 \
-u_1-x_1+x_2&=&0 \
-u_2-x_2+x_3&=&0 \
-u_3-x_3+x_4&=&0 \
-u_4-x_4+x_5&=&0 \
-u_5-x_5+x_6&=&0 \
epsilon _1^2-A+x_1&=&0 \
epsilon _2^2-A+x_2&=&0 \
epsilon _3^2-A+x_3&=&0 \
epsilon _4^2-A+x_4&=&0 \
epsilon _5^2-A+x_5&=&0 \
x_5-A&=&0 \
2 b u_1+lambda _1&=&0 \
2 b u_2+lambda _2&=&0 \
2 b u_3+lambda _3&=&0 \
2 b u_4+lambda _4&=&0 \
2 b u_5+lambda _5&=&0 \
-lambda _1+lambda _6-2 a x_1&=&0 \
lambda _1-lambda _2+lambda _7-2 a x_2&=&0 \
lambda _2-lambda _3+lambda _8-2 a x_3&=&0 \
lambda _3-lambda _4+lambda _9-2 a x_4&=&0 \
lambda _4-lambda _5+lambda _{10}+lambda _{11}-2 a x_5&=&0 \
lambda _5+2 x_6&=&0 \
end{array}
right.
$$
Solving this system of equations we obtain for the parametric values $a = 2, b = 2, A = 1, n = 5$
follows the $min$ strategy (in red $u$ and in blue $x$)
and the $max$ strategy
Any way, this kind of problem can be satisfactorily solved with a quadratic programming package, without the need for the slack variables introduction. See https://en.wikipedia.org/wiki/Quadratic_programming
I hope this helps.
This definitely help. Thank's a lot.
– AlexC75
Dec 27 at 15:26
add a comment |
I will present a reduced example shoving how to solve this kind of problems using the classical procedure of Lagrange multipliers. To apply this technique, previously we transform the inequalities into equalities with the help of the $epsilon_k$ slack variables.
First form the Lagrangian
$$
L(x,u,lambda,epsilon) = phi(x_{n+1})-f(x,u)+sum_{k=1}^nlambda_k(x_{k+1}-x_k-u_k) + sum_{k=1}^nlambda_{n+k}(x_k-A+epsilon_k^2)+lambda_{2n+1}(x_n-A)
$$
where
$$
phi(x_{n+1}) = x_{n+1}^2\
f(x,u) = min(max)sum_{k=1}^na x_k^2+b u_k^2
$$
Now considering $n = 5$ the stationary conditions are
$$
nabla L = left{
begin{array}{rcl}
2 epsilon _1 lambda _6&=&0 \
2 epsilon _2 lambda _7&=&0 \
2 epsilon _3 lambda _8&=&0 \
2 epsilon _4 lambda _9&=&0 \
2 epsilon _5 lambda _{10}&=&0 \
-u_1-x_1+x_2&=&0 \
-u_2-x_2+x_3&=&0 \
-u_3-x_3+x_4&=&0 \
-u_4-x_4+x_5&=&0 \
-u_5-x_5+x_6&=&0 \
epsilon _1^2-A+x_1&=&0 \
epsilon _2^2-A+x_2&=&0 \
epsilon _3^2-A+x_3&=&0 \
epsilon _4^2-A+x_4&=&0 \
epsilon _5^2-A+x_5&=&0 \
x_5-A&=&0 \
2 b u_1+lambda _1&=&0 \
2 b u_2+lambda _2&=&0 \
2 b u_3+lambda _3&=&0 \
2 b u_4+lambda _4&=&0 \
2 b u_5+lambda _5&=&0 \
-lambda _1+lambda _6-2 a x_1&=&0 \
lambda _1-lambda _2+lambda _7-2 a x_2&=&0 \
lambda _2-lambda _3+lambda _8-2 a x_3&=&0 \
lambda _3-lambda _4+lambda _9-2 a x_4&=&0 \
lambda _4-lambda _5+lambda _{10}+lambda _{11}-2 a x_5&=&0 \
lambda _5+2 x_6&=&0 \
end{array}
right.
$$
Solving this system of equations we obtain for the parametric values $a = 2, b = 2, A = 1, n = 5$
follows the $min$ strategy (in red $u$ and in blue $x$)
and the $max$ strategy
Any way, this kind of problem can be satisfactorily solved with a quadratic programming package, without the need for the slack variables introduction. See https://en.wikipedia.org/wiki/Quadratic_programming
I hope this helps.
This definitely help. Thank's a lot.
– AlexC75
Dec 27 at 15:26
add a comment |
I will present a reduced example shoving how to solve this kind of problems using the classical procedure of Lagrange multipliers. To apply this technique, previously we transform the inequalities into equalities with the help of the $epsilon_k$ slack variables.
First form the Lagrangian
$$
L(x,u,lambda,epsilon) = phi(x_{n+1})-f(x,u)+sum_{k=1}^nlambda_k(x_{k+1}-x_k-u_k) + sum_{k=1}^nlambda_{n+k}(x_k-A+epsilon_k^2)+lambda_{2n+1}(x_n-A)
$$
where
$$
phi(x_{n+1}) = x_{n+1}^2\
f(x,u) = min(max)sum_{k=1}^na x_k^2+b u_k^2
$$
Now considering $n = 5$ the stationary conditions are
$$
nabla L = left{
begin{array}{rcl}
2 epsilon _1 lambda _6&=&0 \
2 epsilon _2 lambda _7&=&0 \
2 epsilon _3 lambda _8&=&0 \
2 epsilon _4 lambda _9&=&0 \
2 epsilon _5 lambda _{10}&=&0 \
-u_1-x_1+x_2&=&0 \
-u_2-x_2+x_3&=&0 \
-u_3-x_3+x_4&=&0 \
-u_4-x_4+x_5&=&0 \
-u_5-x_5+x_6&=&0 \
epsilon _1^2-A+x_1&=&0 \
epsilon _2^2-A+x_2&=&0 \
epsilon _3^2-A+x_3&=&0 \
epsilon _4^2-A+x_4&=&0 \
epsilon _5^2-A+x_5&=&0 \
x_5-A&=&0 \
2 b u_1+lambda _1&=&0 \
2 b u_2+lambda _2&=&0 \
2 b u_3+lambda _3&=&0 \
2 b u_4+lambda _4&=&0 \
2 b u_5+lambda _5&=&0 \
-lambda _1+lambda _6-2 a x_1&=&0 \
lambda _1-lambda _2+lambda _7-2 a x_2&=&0 \
lambda _2-lambda _3+lambda _8-2 a x_3&=&0 \
lambda _3-lambda _4+lambda _9-2 a x_4&=&0 \
lambda _4-lambda _5+lambda _{10}+lambda _{11}-2 a x_5&=&0 \
lambda _5+2 x_6&=&0 \
end{array}
right.
$$
Solving this system of equations we obtain for the parametric values $a = 2, b = 2, A = 1, n = 5$
follows the $min$ strategy (in red $u$ and in blue $x$)
and the $max$ strategy
Any way, this kind of problem can be satisfactorily solved with a quadratic programming package, without the need for the slack variables introduction. See https://en.wikipedia.org/wiki/Quadratic_programming
I hope this helps.
I will present a reduced example shoving how to solve this kind of problems using the classical procedure of Lagrange multipliers. To apply this technique, previously we transform the inequalities into equalities with the help of the $epsilon_k$ slack variables.
First form the Lagrangian
$$
L(x,u,lambda,epsilon) = phi(x_{n+1})-f(x,u)+sum_{k=1}^nlambda_k(x_{k+1}-x_k-u_k) + sum_{k=1}^nlambda_{n+k}(x_k-A+epsilon_k^2)+lambda_{2n+1}(x_n-A)
$$
where
$$
phi(x_{n+1}) = x_{n+1}^2\
f(x,u) = min(max)sum_{k=1}^na x_k^2+b u_k^2
$$
Now considering $n = 5$ the stationary conditions are
$$
nabla L = left{
begin{array}{rcl}
2 epsilon _1 lambda _6&=&0 \
2 epsilon _2 lambda _7&=&0 \
2 epsilon _3 lambda _8&=&0 \
2 epsilon _4 lambda _9&=&0 \
2 epsilon _5 lambda _{10}&=&0 \
-u_1-x_1+x_2&=&0 \
-u_2-x_2+x_3&=&0 \
-u_3-x_3+x_4&=&0 \
-u_4-x_4+x_5&=&0 \
-u_5-x_5+x_6&=&0 \
epsilon _1^2-A+x_1&=&0 \
epsilon _2^2-A+x_2&=&0 \
epsilon _3^2-A+x_3&=&0 \
epsilon _4^2-A+x_4&=&0 \
epsilon _5^2-A+x_5&=&0 \
x_5-A&=&0 \
2 b u_1+lambda _1&=&0 \
2 b u_2+lambda _2&=&0 \
2 b u_3+lambda _3&=&0 \
2 b u_4+lambda _4&=&0 \
2 b u_5+lambda _5&=&0 \
-lambda _1+lambda _6-2 a x_1&=&0 \
lambda _1-lambda _2+lambda _7-2 a x_2&=&0 \
lambda _2-lambda _3+lambda _8-2 a x_3&=&0 \
lambda _3-lambda _4+lambda _9-2 a x_4&=&0 \
lambda _4-lambda _5+lambda _{10}+lambda _{11}-2 a x_5&=&0 \
lambda _5+2 x_6&=&0 \
end{array}
right.
$$
Solving this system of equations we obtain for the parametric values $a = 2, b = 2, A = 1, n = 5$
follows the $min$ strategy (in red $u$ and in blue $x$)
and the $max$ strategy
Any way, this kind of problem can be satisfactorily solved with a quadratic programming package, without the need for the slack variables introduction. See https://en.wikipedia.org/wiki/Quadratic_programming
I hope this helps.
edited Dec 27 at 10:10
answered Dec 27 at 1:34
Cesareo
8,2133516
8,2133516
This definitely help. Thank's a lot.
– AlexC75
Dec 27 at 15:26
add a comment |
This definitely help. Thank's a lot.
– AlexC75
Dec 27 at 15:26
This definitely help. Thank's a lot.
– AlexC75
Dec 27 at 15:26
This definitely help. Thank's a lot.
– AlexC75
Dec 27 at 15:26
add a comment |
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1
Introduce a slack variable $epsilon$ such that $s_t-A+epsilon^2 = 0$
– Cesareo
Dec 26 at 14:06
1
Thank you @Cesareo for your answer. Then how should I deal with the slack variable when deriving the first order condition ? Is there a direct method regarding slack variables ?
– AlexC75
Dec 26 at 14:11
1
Handle it as more a variable. The Lagrangian now reads $L(d_t,lambda,epsilon)$ so regarding $epsilon$ we have $L_{epsilon} = 2lambda_iepsilon = 0$ and $L_{lambda_i} = s_t -A+epsilon^2=0$ one additional condition.
– Cesareo
Dec 26 at 14:18
1
But there is no new co-state variable ? I am not sure on how to write the Lagrangian is this framework.
– AlexC75
Dec 26 at 14:26
$L(d_t,lambda,epsilon) = sum_{t=0}^{T} - left [ f(s_t) + k(d_t) right ] + F(s_{T+1})+lambda_1(s_{t+1} - s_t - d_t)+lambda_2(s_t-A+epsilon_1^2)+lambda_3(S_T-A)+lambda_4(s_0-epsilon_2^2)$
– Cesareo
Dec 26 at 14:31