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Given the following text description, write Python code to implement the functionality described below step by step
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Description:
 Linear programming 
Step1:  Scipy's syntax
Example for a problem of 2 dimensions
Step2:  TODO
Step3:  Hand write the model
Step4:  Automatically make the model
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Python Code:
%matplotlib inline
import matplotlib.pyplot as plt
import scipy.optimize
Explanation: Linear programming : stock problem
See https://docs.scipy.org/doc/scipy/reference/generated/scipy.optimize.linprog.html
End of explanation
plt.plot([1, 2, 3], [10, 30, 20], "o-")
plt.xlabel("Unit of time (t)")
plt.ylabel("Price of one unit of energy (c)")
plt.title("Cost of energy on the market")
plt.show();
Explanation: Scipy's syntax
Example for a problem of 2 dimensions:
$$
\begin{align}
    \min_{x_1,x_2} & \quad \color{\red}{c_1} x_1 + \color{\red}{c_2} x_2 \
    \text{s.t.}    & \quad \color{\orange}{A_{1,1}} x_1 + \color{\orange}{A_{1,2}} x_2 \leq \color{\green}{b_1} \
                   & \quad \color{\orange}{A_{2,1}} x_1 + \color{\orange}{A_{2,2}} x_2 \leq \color{\green}{b_2} \
                   & \quad \color{\purple}{B_{1,1}} \geq x_1 \geq \color{\purple}{B_{1,2}} \
                   & \quad \color{\purple}{B_{2,1}} \geq x_2 \geq \color{\purple}{B_{2,2}} \
\end{align}
$$
$$
\color{\red}{
\boldsymbol{c} = \begin{pmatrix}
    c_1 \
    c_2
\end{pmatrix}
}
\quad
\color{\orange}{
\boldsymbol{A} = \begin{pmatrix}
    A_{1,1} & A_{1,2} \
    A_{2,1} & A_{2,2}
\end{pmatrix}
}
\quad
\color{\green}{
\boldsymbol{b} = \begin{pmatrix}
    b_1 \
    b_2
\end{pmatrix}
}
\quad
\color{\purple}{
\boldsymbol{B} = \begin{pmatrix}
    B_{1,1} & B_{1,2} \
    B_{2,1} & B_{2,2}
\end{pmatrix}
}
$$
$$
\text{scipy.optimize.linprog}(\color{\red}{\boldsymbol{c}}, ~ \color{\orange}{\boldsymbol{A}}, ~ \color{\green}{\boldsymbol{b}}, ~ \color{\purple}{\boldsymbol{B}})
$$
Stock problem
End of explanation
# Price of energy on the market
price = [10, 30, 20]
plt.plot(price);
stock_max = 100      # battery capacity
Explanation: TODO: put pictures and explanations from my PhD
$x_t < 0$ = buy $|x|$ at time $t$
$x_t > 0$ = sell $|x|$ at time $t$
$s_t$ = the battery level at time $t$
$$
\begin{align}
    \max_{x_1,x_2,x_3} & \quad p_1 x_1 + p_2 x_2 + p_3 x_3 \
    \text{s.t.}        & \quad s_0 = 0 \
                       & \quad s_t = s_{t-1} - x_t \
                       & \quad x_t \geq -(s_{\max} - s_{t-1}) \quad \quad \text{can't buy more than the "free space" of the battery} \
                       & \quad x_t \leq s_{t-1}               \quad \quad \quad \quad \quad \quad \text{can't sell more than what have been stored in the battery} \
                       & \
                       & \
    \Leftrightarrow \min_{x_1,x_2,x_3} & \quad - p_1 x_1 - p_2 x_2 - p_3 x_3 \
    \text{s.t.}        & \quad s_0 = 0 \
                       & \quad s_t = s_{t-1} - x_t \
                       & \quad x_t \geq -(s_{\max} - s_{t-1}) \Leftrightarrow x_t \geq -s_{\max} + s_{t-1} \
                       & \quad x_t \leq s_{t-1} \
                       & \
                       & \
    \Leftrightarrow \min_{x_1,x_2,x_3} & \quad - p_1 x_1 - p_2 x_2 - p_3 x_3 \
    \text{s.t.}        & \quad s_1 = -x_1 \
                       & \quad x_1 \geq -s_{\max} \
                       & \quad x_1 \leq 0 \
                       & \quad s_2 = s_1 - x_2 = -x_1 - x_2 \
                       & \quad x_2 \geq -s_{\max} + s_1 \Leftrightarrow x_2 \geq -s_{\max} - x_1 \
                       & \quad x_2 \leq s_1 \Leftrightarrow x_2 \leq -x_1 \
                       & \quad s_3 = s_2 - x_3 = -x_1 - x_2 - x_3 \
                       & \quad x_3 \geq -s_{\max} + s_2 \Leftrightarrow x_3 \geq -s_{\max} - x_1 - x_2 \
                       & \quad x_3 \leq s_2 \Leftrightarrow x_3 \leq -x_1 - x_2 \
                       & \
                       & \
    \Leftrightarrow \min_{x_1,x_2,x_3} & \quad - p_1 x_1 - p_2 x_2 - p_3 x_3 \
    \text{s.t.}        & \quad x_1 \geq -s_{\max} \
                       & \quad x_1 \leq 0 \
                       & \quad x_2 \geq -s_{\max} - x_1 \
                       & \quad x_2 \leq -x_1 \
                       & \quad x_3 \geq -s_{\max} - x_1 - x_2 \
                       & \quad x_3 \leq -x_1 - x_2 \
                       & \
                       & \
    \Leftrightarrow \min_{x_1,x_2,x_3} & \quad - p_1 x_1 - p_2 x_2 - p_3 x_3 \
    \text{s.t.}        & \quad x_1 \geq -s_{\max} \
                       & \quad x_1 \leq 0 \
                       & \quad x_1 + x_2 \geq -s_{\max} \
                       & \quad x_1 + x_2 \leq 0 \
                       & \quad x_1 + x_2 + x_3 \geq -s_{\max} \
                       & \quad x_1 + x_2 + x_3 \leq 0 \
                       & \
                       & \
    \Leftrightarrow \min_{x_1,x_2,x_3} & \quad - p_1 x_1 - p_2 x_2 - p_3 x_3 \
    \text{s.t.}        & \quad -x_1 \leq s_{\max} \
                       & \quad  x_1 \leq 0 \
                       & \quad -x_1 - x_2 \leq s_{\max} \
                       & \quad  x_1 + x_2 \leq 0 \
                       & \quad -x_1 - x_2 - x_3 \leq s_{\max} \
                       & \quad  x_1 + x_2 + x_3 \leq 0 \
                       & \
                       & \
\end{align}
$$
$$
\color{\red}{
\boldsymbol{c} = \begin{pmatrix}
    -p_1 \
    -p_2 \
    -p_3
\end{pmatrix}
}
\quad
\color{\orange}{
\boldsymbol{A} = \begin{pmatrix}
    -1 &  0 &  0 \
     1 &  0 &  0 \
    -1 & -1 &  0 \
     1 &  1 &  0 \
    -1 & -1 & -1 \
     1 &  1 &  1
\end{pmatrix}
}
\quad
\color{\green}{
\boldsymbol{b} = \begin{pmatrix}
    s_{\max} \
    0 \
    s_{\max} \
    0 \
    s_{\max} \
    0
\end{pmatrix}
}
\quad
\color{\purple}{
\boldsymbol{B} = \begin{pmatrix}
    - & - \
    - & - \
    - & -
\end{pmatrix}
}
$$
End of explanation
# Coefficients of the linear objective function to be minimized
p = -np.array(price)
# 2-D array which, when matrix-multiplied by x, gives the values of the upper-bound inequality constraints at x.
A = [[-1,  0,  0],
     [ 1,  0,  0],
     [-1, -1,  0],
     [ 1,  1,  0],
     [-1, -1, -1],
     [ 1,  1,  1]]
# 1-D array of values representing the upper-bound of each inequality constraint (row) in A.
b = [stock_max, 0, stock_max, 0, stock_max, 0]
# Sequence of (min, max) pairs for each element in x, defining the bounds on that parameter.
# Use None for one of min or max when there is no bound in that direction.
# By default bounds are (0, None) (non-negative).
# If a sequence containing a single tuple is provided, then min and max will be applied to all variables in the problem.
x0_bounds = (None, None)
x1_bounds = (None, None)
x2_bounds = (None, None)
bounds = (x0_bounds, x1_bounds, x2_bounds)
scipy.optimize.linprog(p, A_ub=A, b_ub=b, bounds=bounds)
Explanation: Hand write the model
End of explanation
# Cost of energy on the market
#price = [10, 30, 20]          # -> -100, 100, 0
#price = [10, 30, 10, 30]      # -> [-100.,  100., -100.,  100.]
#price = [10, 30, 10, 30, 30]  # -> [-100.,  100., -100.,  100.,    0.]
#price = [10, 20, 30, 40]      # -> [-100.,  0., 0.,  100.]
price = [10, 30, 20, 50]
price = [10, 30, 20, 50]
plt.plot(price);
p = -np.array(price)
A = np.repeat(np.tril(np.ones(len(price))), 2, axis=0)
A[::2, :] *= -1
A
b = np.zeros(A.shape[0])
b[::2] = stock_max
b
bounds = tuple((None, None) for _p in price)
bounds
%%time
res = scipy.optimize.linprog(p, A_ub=A, b_ub=b, bounds=bounds)   # , method='revised simplex'
res
res.x.round(decimals=2)
Explanation: Automatically make the model
End of explanation
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