Create the freest arrangement of white chess pieces on board by consequently moving your pieces

Question

You start with this board. You're the White. Freedom Index of initial arrangment of your chess pieces is equal to 20 (each pawn has two moves, each knight has two moves)

Your goal is to have as much freedom of movement for your chess pieces as possible, i.e. do your best to get high Freedom Index. By this I mean that you must arrange your chess pieces in such way, that when you will count number of moves each of your chess pieces has and then sum said numbers, then resulting sum (also known as Freedom Index) must be as high as you can get.You can create such arrangment only by consequently moving your chess pieces in accordance with chess rules (i.e. each chess pieces can move only as it can move in chess), with exception that castling is forbidden and consequently doesn't count when Freedom Index is calculated. You can make as many moves as you want.Your answer must contain picture of your final arrangment of chess pieces. Pawn promotion is allowed. Use of computer for finding an arrangmenet is allowed, assuming that it follows the rules of the puzzle (in particular, no castling and bishops must be walking on different colors). Also, you should give Freedom Index of your arrangmenet in your post.

I will select answer with the highest Freedom Index, compared to other answers.

P.S. The upper bound for Freedom Index is 321. The lower bound for Freedom Index is 0 (i.e.you stalemated yourself)

Answer 1

222: \begin{matrix}R_{11} & . & . & . & . & . & . & R_{10} \\. & K_7 & . & . & . & Q_{18} & . & . \\. & . & . & Q_{24} & . & . & . & . \\. & Q_{18} & . & . & . & . & Q_{20} & . \\. & . & . & . & Q_{24} & . & . & . \\. & . & Q_{21} & . & . & . & . & Q_{16} \\Q_{15} & . & . & . & . & Q_{18} & . & . \\. & . & B_5 & B_7 & N_4 & N_4 & . & . \\\end{matrix}

I used integer linear programming as follows. Let $P$ be the set of pieces, with number $n_p$ of pieces available: $n_\text{king}=1, n_\text{bishop}=n_\text{knight}=n_\text{rook}=2, n_\text{queen}=9$. Let $C=\{1,\dots,8\}^2$ be the set of cells. For each piece $p\in P$ and cell $(i,j)\in C$, let $N_{p,i,j} \subseteq C$ be the set of neighboring cells with respect to possible moves of $p$. For $p\in P$, $(i,j)\in C$, and $(i_2,j_2)\in N_{p,i,j}$, let $B_{p,i,j,i_2,j_2} \subseteq C$ be the set of cells strictly between $(i,j)$ and $(i_2,j_2)$. Let binary decision variable $x_{p,i,j}$ indicate whether piece $p$ occupies cell $(i,j)$. Let binary decision variable $m_{p,i,j,i_2,j_2}$ indicate whether piece $p$ occupies cell $(i,j)$ and can move to cell $(i_2,j_2)$. The problem is to maximize $$\sum_{p\in P} \sum_{(i,j)\in C} \sum_{(i_2,j_2)\in N_{p,i,j}} m_{p,i,j,i_2,j_2}$$ subject to \begin{align} \sum_{p\in P} x_{p,i,j} &\le 1 &&\text{for $(i,j)\in C$} \tag1\\ \sum_{(i,j)\in C} x_{p,i,j} &\le n_p &&\text{for $p\in P$} \tag2\\ m_{p,i,j,i_2,j_2} &\le x_{p,i,j} &&\text{for $p\in P, (i,j)\in C, (i_2,j_2)\in N_{p,i,j}$} \tag3\\ m_{p,i,j,i_2,j_2} &\le 1-\sum_{p_2} x_{p_2,i_2,j_2} &&\text{for $p\in P, (i,j)\in C, (i_2,j_2)\in N_{p,i,j}$} \tag4\\ m_{p,i,j,i_2,j_2} &\le 1-\sum_{p_2} x_{p_2,i_3,j_3} &&\text{for $p\in P, (i,j)\in C, (i_2,j_2)\in N_{p,i,j}, (i_3,j_3)\in B_{p,i,j,i_2,j_2}$} \tag5 \\ \sum_{\substack{(i,j) \in C:\\ \mod(i+j,2) = r}} x_{\text{bishop},i,j} &\le 1 &&\text{for $r \in \{0,1\}$} \tag6 \end{align} Constraint $(1)$ places at most one piece per cell. Constraint $(2)$ places at most $n_p$ copies of piece $p$. Constraint $(3)$ enforces $m_{p,i,j,i_2,j_2} = 1 \implies x_{p,i,j} = 1$. Constraint $(4)$ enforces $m_{p,i,j,i_2,j_2} = 1 \implies x_{p_2,i_2,j_2} = 0$. Constraint $(5)$ enforces $m_{p,i,j,i_2,j_2} = 1 \implies x_{p_2,i_3,j_3} = 0$. Constraint $(6)$ enforces at most one bishop per color.