BuildIt Construction uses steel beams with a uniform cross section but of different

lengths in its various projects. For its projects over the next month, it will require beams

of lengths $4,7,9,10,14,18,23,$ and $30$ ; the number of beams needed are $40,30,60,110,$

$100,75,90,$ and $125,$ respectively. While it can hold all required beams in its storage

center, BuildIt typically will store only some of these beams lengths and use them to

satisfy the demand for smaller lengths, with the scrap pieces being discarded. For

example, an $18ft$ beam can be trimmed to satisfy all required lengths except for the $23$

and $30ft$ beams. However, for example, if it is used to satisfy the demand for $7ft$ beam,

then each time it is used there will be a scrap of $11ft$ which cannot be used to satisfy the

demand of any other beams. This extra $11ft$ should be discarded. It costs BuildIt $$100$ to

prepare the storage center for holding beams of any particular length, and a per$-$beam

amount of which is given below for each beam length.

Thus, if it were told to hold $200$ beams of length $18ft,$ the cost would be

$$100+200($$$25)=$$$5100.$ BuildIt wants to minimize the total cost of storing the required

beams. Formulate this problem as a shortest path problem. Drawing the network$/$graph$,$

putting down the costs$/$lengths of each arc$/$edge and explaining the objective $($such as we

are trying to find the shortest path from node $x$ to node $y)$ on this graph is sufficient. You

do not have to write down the formulation explicitly. $($Hint: If we were to store beams of

lengths $10$ and $30ft,$ would we use a beam of length $30ft$ to satisfy the demand for a

beam of length $7ft$ in an optimal solution?$)$