  In this paper, it is shown that all programmes of all television channels can
be modelled as an interval graph. The programme slots are taken as the vertices
of the graph and if the time duration of two {programme slots} have non-empty
intersection, the corresponding vertices are considered to be connected by an
edge. The number of viewers of a programme is taken as the weight of the
vertex. A set of programmes that are mutually exclusive in respect of time
scheduling is called a session. We assume that a company sets the objective of
selecting the popular programmes in $k$ parallel sessions among different
channels so as to make its commercial advertisement reach the maximum number of
viewers, that is, a company selects $k$ suitable programme slots simultaneously
for advertisement. The aim of the paper is, therefore, to {help} the companies
to select the programme slots, which are mutually exclusive with respect to the
time schedule of telecasting time, in such a way that the total number of
viewers of the selected programme in $k$ parallel slots rises to the optimum
level. It is shown that the solution of this problem is obtained by solving the
maximum weight $k$-colouring problem on an interval {graph}. An algorithm is
designed to solve this just-in-time optimization problem using $O(kMn^2)$ time,
where $n$ and $M$ represent the total number of programmes of all channels and
the upper bound of the viewers of all programmes of all channels respectively.
The problem considered in this paper is a daily life problem which is modeled
by $k$-colouring problem on interval graph.
