Talent scheduling

Consider a film shoot composed of ${\displaystyle n}$ shooting days and involving a total of ${\displaystyle m}$ actors. Then we use the day out of days matrix (DODM) ${\displaystyle T^{0}\in \{0,1\}_{m\times n}}$ to represent the requirements for the various shooting days. The matrix with the ${\displaystyle (i,j)}$ entry given by:

${\displaystyle t_{m\times n}^{0}={\begin{cases}1,&{\mbox{if actor i is required in shooting day j,}}\\0,&{\mbox{otherwise.}}\end{cases}}}$

Then we define the pay vector ${\displaystyle {\mathfrak {R}}^{m}}$, with the ${\displaystyle i}$th element given by ${\displaystyle c_{i}}$ which means rate of pay per day of the ${\displaystyle i}$th actor. Let v denote any permutation of the n columns of ${\displaystyle T^{0}}$, we have:

${\displaystyle \sigma$ :\{1,2,...,n\}\rightarrow \{1,2,...,n\}}

${\displaystyle \sigma _{n}}$ is the permutation set of the n shooting days. Then define ${\displaystyle T(\sigma )}$ to be the matrix ${\displaystyle T^{0}}$ with its columns permuted according to ${\displaystyle \sigma }$, we have:

${\displaystyle t_{i,j}(\sigma )=t_{i,\sigma (j)}^{0}}$ for ${\displaystyle i\in \{1,2,...,n\},j\in \{1,2,...,n\}}$

Then we use ${\displaystyle l_{i}(\sigma )}$ and ${\displaystyle e_{i}(\sigma )}$ to represent denote respectively the earliest and latest days in the schedule ${\displaystyle S}$ determined by a which require actor ${\displaystyle i}$. So we can find actor ${\displaystyle i}$ will be hired for ${\displaystyle l_{i}(\sigma )-e_{i}(\sigma )+1}$ days. But in these days, only ${\displaystyle r_{i}=\sum _{j=1}^{n}t_{ij}^{0}}$ days are actually required, which means ${\displaystyle h_{i}(S)}$ days are unnecessary, we have:

${\displaystyle h_{i}(S)=h_{i}(\sigma )=l_{i}(\sigma )-e_{i}(\sigma )+1-r_{i}=l_{i}(\sigma )-e_{i}(\sigma )+1-\sum _{j=1}^{n}t_{i,j}^{0}}$

The total cost of unnecessary days is:

${\displaystyle K(\sigma )=\sum _{j=1}^{m}c_{i}h_{i}(\sigma )=\sum _{j=1}^{m}c_{i}[l_{i}(\sigma )-e_{i}(\sigma )+1-\sum _{j=1}^{n}t_{i,j}^{0}]}$

${\displaystyle K(\sigma )}$ will be the objective function we should minimize.[1]

In talent scheduling problem, we can prove that is NP-hard by a reduction from the optimal linear arrangement(OLA) problem.[2] And in this problem, even we restrict each actor is needed for just two days and all actors' salaries are 1, it's still polynomially reducible to the OLA problem. Thus, this problem is unlikely to have pseudo-polynomial algorithm.[3]

The integer programming model is given by:[4]

Minimize	${\displaystyle \sum _{i=1}^{m}c_{i}(l_{i}-e_{i}+1)}$
subject to	${\displaystyle \sum _{j=1}^{n}x_{j,k}=\sum _{k=1}^{n}x_{j,k}=1,}$		${\displaystyle 1\leq j,k\leq n}$
	${\displaystyle t_{i,j}e_{i}\leq \sum _{k=1}^{n}t_{i,j}kx_{j,k}\leq l_{i},}$		${\displaystyle 1\leq i\leq m,1\leq j\leq n}$
	${\displaystyle l_{i},e_{i}\geq 1,}$		${\displaystyle 1\leq i\leq m}$
	${\displaystyle x_{j,k}\in \{0,1\},}$		${\displaystyle 1\leq j,k\leq n}$
	${\displaystyle l_{i},e_{i}\in \mathbb {Z} ,}$		${\displaystyle 1\leq i\leq m}$

In this model, ${\displaystyle e_{i}}$ means the earliest shooting day for talent ${\displaystyle i}$, ${\displaystyle l_{i}}$ is the latest shooting day for talent ${\displaystyle i}$, ${\displaystyle x_{j,k}}$ is the scheduling for the project, i.e.

${\displaystyle x_{j,k}={\begin{cases}1&{\text{if scene }}j{\text{ is scheduled in day }}k{\text{ of shooting }}\\0&{\text{otherwise}}\end{cases}}}$