[動態規劃] 淺談 離散時間動態規劃 (2) -Minimax Dynamic programming

考慮系統狀態方程：
$$x(k+1) = f(x(k), u(k))$$ 現在引入一個 監測函數 (monitor function): $g(x(k))$ 此函數用來擷取我們所關心的系統狀態。

現在我們定義 cost function
$$J(u) := \displaystyle \min_{k=1,2,...,N} g(x(k))$$
我們的目標：
藉由選取最佳的 $u(k) \in \Omega_k$ 使得 $\max J(u)$，亦即
$$\max \displaystyle \min_{k=1,2,...,N} g(x(k))$$

此時我們的 Bellman equation 需要進行修正：
$$I(x(l), N-l) = \max_{u(l) \in \Omega_l} \left \{ \min \left \{ g\left( x(l+1), I(x(l+1), N-(l+1)) \right ) \right \} \right\}$$

Example
考慮如下狀態方程：
$$x\left( {k + 1} \right) = \left[ {\begin{array}{*{20}{c}} {\frac{1}{2}}&{\frac{1}{4}}\\ { - \frac{1}{2}}&{\frac{1}{4}} \end{array}} \right]x\left( k \right) + \left[ {\begin{array}{*{20}{c}} 1\\ 1 \end{array}} \right]u\left( k \right)$$ 其中 $x(k) := [x_1(k) \ x_2(k)]^T$；
現在，僅考慮 $N=2$ (只有兩步)的情況，另外控制力 $u(0), u(1) \geq 0$ 且需滿足如下拘束： $u(0) + u(1) \leq 1$，定義監測函數
$$g(x(k)) = x_1(k)$$
試求一組 $u^*(0), u^*(1)$ maximize the Floor of $g((k))$。

Solution
首先考慮 Optimal cost of one-step-to-go:  ($l=N-1 =1$)
$$\begin{array}{l} I\left( {x\left( 1 \right),1} \right) = \mathop {\max }\limits_{u\left( 1 \right) \in {\Omega _1}} \left\{ {g\left( {x\left( 2 \right)} \right)} \right\} = \mathop {\max }\limits_{u(1) \in \left[ {0,1 - u(0)} \right]} \left\{ {{x_1}\left( 2 \right)} \right\}\\ \begin{array}{*{20}{c}} {}&{}&{}&{}&{}&{} \end{array} = \mathop {\max }\limits_{u(1) \in \left[ {0,1 - u(0)} \right]} \left\{ {\frac{1}{2}{x_1}\left( 1 \right) + \frac{1}{4}{x_2}\left( 1 \right) + u\left( 1 \right)} \right\} \end{array}$$ 觀察上式可知欲得到 Maximum，我們需選
$${u^*}(1) = 1 - u(0)$$ 故將上述 $u^*(1)$ 代回，我們可得對應的 Optimal cost to go :
$$I\left( {x\left( 1 \right),1} \right) = \frac{1}{2}{x_1}\left( 1 \right) + \frac{1}{4}{x_2}\left( 1 \right) + 1 - u(0)$$ 現在我們計算 Two-steps-to-go: ($l=N-2 =0$)
$$\begin{array}{l} I\left( {x\left( l \right),N - l} \right) = \mathop {\max }\limits_{u\left( l \right) \in {\Omega _l}} \left\{ {\min \left\{ {g\left( {x\left( {l + 1} \right)} \right),I\left( {x\left( {l + 1} \right),N - l - 1} \right)} \right\}} \right\}\\  \Rightarrow I\left( {x\left( 0 \right),2} \right) = \mathop {\max }\limits_{u\left( 0 \right) \in {\Omega _0}} \left\{ {\min \left\{ {{x_1}\left( 0 \right),I\left( {x\left( 1 \right),1} \right)} \right\}} \right\}\\  \Rightarrow I\left( {x\left( 0 \right),2} \right) = \mathop {\max }\limits_{u\left( 0 \right) \in \left[ {0,1} \right]} \left\{ {\min \left\{ {{x_1}\left( 0 \right),\frac{1}{2}{x_1}\left( 1 \right) + \frac{1}{4}{x_2}\left( 1 \right) + 1 - u(0)} \right\}} \right\} \ \ (*) \end{array}$$ 由系統狀態方程可知
$$\left\{ \begin{array}{l} {x_1}\left( 1 \right) = \frac{1}{2}{x_1}\left( 0 \right) + \frac{1}{4}{x_2}\left( 0 \right) + u\left( 0 \right)\\ {x_2}\left( 1 \right) =  - \frac{1}{2}{x_1}\left( 0 \right) + \frac{1}{4}{x_2}\left( 0 \right) + u\left( 0 \right) \end{array} \right.$$ 故將上述結果代回 $(*)$ 可得
$$\Rightarrow I\left( {x\left( 0 \right),2} \right) = \mathop {\max }\limits_{u\left( 0 \right) \in \left[ {0,1} \right]} \left\{ {\min \left\{ {A + u\left( 0 \right),B - \frac{1}{4}u\left( 0 \right)} \right\}} \right\}$$ 其中
$$\left\{ \begin{array}{l} A = \frac{1}{2}{x_1}\left( 0 \right) + \frac{1}{4}{x_2}\left( 0 \right)\\ B = 1 + \frac{1}{8}{x_1}\left( 0 \right) + \frac{3}{{16}}{x_2}\left( 0 \right) \end{array} \right.$$ 為了找出 Maximum，我們定義 $F(u(0)) :={\min \left\{ {A + u\left( 0 \right),B - \frac{1}{4}u\left( 0 \right)} \right\}}$ 那麼考慮以下兩種情況:
CASE 1: $A \geq B$：則 $u^*(0)=0, u^*(1) = 1$
CASE 2: $A <B$ ：則
$${u^*}\left( 0 \right) = \left\{ \begin{array}{l} \frac{4}{5}\left( {B - A} \right),\begin{array}{*{20}{c}} {} \end{array}if\begin{array}{*{20}{c}} {} \end{array}\frac{4}{5}\left( {B - A} \right) \in \left[ {0,1} \right]\\ 1,\begin{array}{*{20}{c}} {} \end{array}otherwise \end{array} \right.$$ 且 $u^*(1) = 1 - u^*(0)$