Suppose that we have $M$ production stations $A_1, \dots, A_M$ of a product and $N$ destination stations $B_1, \dots, B_N$ of the product.
We suppose that $x_{ij}$ units of the product are transferred from $A_i$ to $B_j$ where $i=1, \dots, M, j=1, \dots, N$. The cost of transportation of $x_{ij}$ units of the product is $c_{ij} x_{ij}$ where $c_{ij}>0$.
We want to see what quantities the stations $A_1, \dots, A_M$ have to produce and simultaneously how these quantities have to be distributed to the stations $B_1, \dots, B_N$ so that the cost of transport gets minimized.
Solution of the problem of transport using techniques of dynamic programming
We suppose that $M=2$.
Then $a_1, a_2$ are the available quantities of the product at the stations $A_1, A_2$ respectively.
We define $f_n^{\star}(x_{1n}, x_{2n})$ the cost of the optimal strategy for the transport of the product from the two production stations with quantities $x_{1n}, x_{2n}$ respectively to the $n$ destination stations.
Then if we subtract $y_1$ units from the production station $A_1$ and $y_2$ units from $A_2$, for the equipment of the $n$-th destination station $B_n$, then there will be $(x_{1n}-y_1, x_{2n}-y_2)$ quantities left for the equipment of the remaining $n-1$ destination stations.
The mathematical formulation of the above is:
$$f_n^{\star} (x_{1n}, x_{2n})= \min_{y \in S_n} \left( c_{1n} y+ c_{2n}(b_n-y)+f_{n-1}^{\star}(x_{1n}-y, x_{2n}-(b_n-y))\right)\\ S_n=\{y: y \geq \max \{ 0, b_n-x_{2n}\} \text{ and } y\leq \min \{x_{1n}, b_n \} \} \text{ and } x_{1n} \leq a_1, x_{2n} \leq a_2$$
Could you explain to me how we get this formula?
Notice that the notation $x_{ij}$ is used in two different contexts in this question. Initially it refers to the quantity that is sent from production station $i$ to destination $j$. Later, it refers to the quantity that is sent from production station $i$ to $\textbf{all}$ the first $j$ destinations.
The formula uses $x_{ij}$ in the second context, the quantity that is sent from production station $i$ to all the first $j$ destinations.
$b_n$ refers to the quantity that the $n$-th denstination requires.
We are told that production station 1 is to provide $x_{1n}$ products and production station 2 is to provide $x_{2n}$ products to the first $n$ stations, the goal if to minimize the cost given this setting. Clearly, $x_{1n}\leq a_1$ as that is the available products at production station 1. Similarly, $x_{2n} \leq a_2$.
If we are further told that production station 1 has to send $y$ unit of items to destination $n$, then production station 2 has to send $b_n-y$ unit of items to destination $n$ to satisfy the demand from destination $n$. By doing so, production station 1 has to send $x_{1n}-y$ items to the first $n-1$ destinations and production station 2 has to send $x_{2n}-(b_n-y)$ items to the first $n-1$ destinations. The minimal cost of doing that would be $$c_{1n}y+c_{2n}(b_n-y)+f_{n-1}^{\star}(x_{1n}-y,x_{2n}-(b_n-y)).$$
We have to choose $y$ carefully to minimize the cost. Recall that $y$ is the quantity of products to be sent from production station 1 to destination station $n$ and $b_n-y$ is the quantity of products to be sent from production station 2 to destination station $n$. Such quantities has to be nonnegative and they cannot be more than what their corresponding production station provides. Hence,
$$0 \leq y \leq x_{1n}$$
$$0 \leq b_n-y \leq x_{2n}$$
Solving for y in this inequalities system gives us the domain of $y$ and we call it $S_n=\left\{y:y \in [max(0,b_n-x_{2n}),min(x_{1n},b_n)]\right\}.$
Hence, $$f_n^{\star}(x_{1n},x_{2n})=\min_{y \in S_n}\left(c_{1n}y+c_{2n}(b_n-y)+f_{n-1}^{\star}(x_{1n}-y,x_{2n}-(b_n-y))\right).$$