A library must build shelving to shelve $200$ 4-inch high books, $100$ 8-inch high books, and $80$ 12-inch high books. Each book is 0.5-inch thick.
The library has several ways to store books. For example, an 8-inch high shelf may be built to store all books of height no more than $8$ inches, and a 12-inch high shelf may be built for the 12-inch high books. Alternatively, a 12-inch high shelf may be built to store all books. The library believes it costs ${\it£}2300$ to build a shelf and that a cost of ${\it£}5$ per square inch is incurred for book storage. (Assume that the area required to store a book is given by the height of storage area times book thickness.)
Formulate and solve a shortest path problem that could be used to help the library determine how to shelve the books at minimum cost. (Hint: Have nodes $0$, $4$, $8$, and $12$, with being the total cost of shelving all books of height $> i$ and $\le j$ on a single shelf.)
I went on with the hint, and I discovered that the cost is minimised when we build only one shelf which is for all books (a 12-inch shelf). So the cost would be the cost of shelving all the 4-inch, 8-inch and 12-inch books) plus that of building the shelve which is $2000+2000+2400+2300$, or ${\it£}8700$.
Would this be the correct answer? I'm confused since this is a pretty obvious answer, which I think is not correct.
There is an answer on Chegg solutions which gives me the following table of costs:
\begin{array}{rr}\hline\text{Shelves}&\text{Books}&\text{Building}&\text{Storage Costs}&\text{Unshelved}\\&&\text{Costs}&&\\\hline4''&4''&{\it£}2300&200\times0.5\times5\times4={\it£}2000&8'',12''\\8''&8''&{\it£}2300&100\times0.5\times5\times8={\it£}2000&4'',12''\\8''&4'',4''&{\it£}2300&2\times200\times0.5\times5\times4={\it£}4000&8'',12''\\12''&12''&{\it£}2300&80\times0.5\times5\times12={\it£}2400&4'',8''\\12''&4'',4'',4''&{\it£}2300&\small{\it£}2000+{\it£}2000+{\it£}2000+{\it£}2400={\it£}8400&8''\\12''&\small4'',8'',12''&{\it£}2300&{\it£}2000+{\it£}2000+{\it£}2400={\it£}6400&-\\\hline\end{array}
For reference, this is the shortest path problem they're trying to optimize:
In the above table, the building cost is given to be ${\it£}2300$. From the above table, it is clear that four nodes are to be taken, that is $0$, $4$, $8$ and $12$ with $C_{i,j}$ being the total cost of shelving all books of height $>i$ and $\le j$ on a single shelf. The network will have four nodes ($0$, $4$, $8$ and $12$). The network for minimizing cost is given below.
and they calculate the cost of building a shelf for the 4-inch books only with the following formula:
Now $X_{i,j}=$ one unit of flow from node $i$ to node $j$ through $x(i,j)$. It is now seen that the length of any path from node $0$ to node $12$ is the net cost incurred. The cost from node $0$ to node $4$ is calculated as below. \begin{align}C_{0,4}=\frac{\text{Building cost}+\text{Storage cost}}{\text{Number of books}\times\text{per square inch cost}}=\frac{{\it£}2300+{\it£}2000}{200\times5}=\frac{{\it£}4300}{1000}={\it£}4.3.\end{align}
Why, in the above way of calculating $C_{0,4}$ , do they divide by (number of books per square inch x cost per square inch) . What is the logic behind this? If someone could explain if this is the right path to the solution and if not, explain how would I solve this.
