k litres of Milk delivery with minimum cost Note:
This is the exact same as a question asked 12 days ago but I do not understand the answer to that and I could not comment on the question so here we are. This is the link to that question. I think the answer is checking all the routes and the costs and then choosing the best one. Is that the best algorithm?
Assume that you are the CEO of a milk delivery company and want to deliver $k$ litres of milk to one of your customers. You can buy milk from another company. Then you should use an arbitrary path to reach your customer.
These paths are given to you as a graph $G=(V,E)$. At first, you are at vertex $p$. The objective is to reach vertex $q$. But there are some rules.  
Assume that $u$ and $v$ are two arbitrarily chosen vertices. When you go from vertex $u$ to $v$, You should pay $C_{uv} \gt 0$ per litre of milk. Also, for example if before going from $u$ to $v$, the amount of milk was $m$ litres, after going from $u$ to $v$, this amount becomes $m \times Y_{uv}$ such that $0 \lt y_{uv} \le 1$.  
Assume that the cost of buying $1$ litre of milk is $\alpha$.  
Notice that at the end, there should be $k$ litres of milk left that can be delivered to the customer.  
Question:  

Find an algorithm to deliver $k$ litres of milk to the customer, with
  the minimum cost.

Note1 (The meaning of cost):  
Assume that you want to go from vertex $a$ to vertex $b$, $\alpha=2$ and the amount of milk you want to deliver is $1$ litre. There are $2$ paths from $a$ to $b$:
1. Going directly from $a$ to $b$:
   You should buy $1$ litre of milk, which costs $2$ units. Then you should go from $a$ to $b$, which costs $C_{ab}=10$. Also, if $Y_{ab}=1$, the amount of milk doesn't change. So, the  total cost becomes $2+10=12$. 
2. Going from $a$ to $c$, and then from $c$ to $b$:
   You should buy $3$ litres of milk, which costs $6$ units. Then, you should go from $a$ to $c$. If we assume that $Y_{ac}=0.5$, then the amount of milk becomes $1.5$ litres. Also, if $C_{ac}=5$, then we should pay $5 \times 3 = 15$ units for going from $a$ to $c$. Then we go from $c$ to $b$. When we reach $c$, there are $1.5$ litres of milk left. Also, If $C_{cb}=6$, then the price of going from $c$ to $b$ will be $6 \times 1.5 = 9$ units. Also, $Y_{cb}= \frac{2}{3}$. Thus, $1$ litre of milk remains at the end. So, the total cost will be $6+15+9=30$ units. 
 A: No, that algorithm suggested by @RossMillikan does not enumerate all possible paths.  On the other hand, it is not a straightforward dynamic programming algorithm because the minimum-cost solution to supply milk to node $v$ from node $u$ may not be used in the minimum-cost solution to supply $v$ from $w$, even though milk is sent from $w$ to $v$ via $u$.
Consider the following example: The cost of milk, $\alpha$, is $2$.  We have three nodes, $w$, $u$ and $v$.  We want to send one liter of milk from $w$ to $v$.  The graph has five edges: two between $w$ and $u$, and three between $u$ and $v$.  The three edges between $u$ and $v$ represent different modes of transportation:


*

*Cost $10$ per liter and scaling factor $0.2$.

*Cost $30$ per liter and scaling factor $0.5$.

*Cost $20$ per liter and scaling factor $0.2$.


If milk were bought directly at $u$, the cheapest option would be to buy $5$ liters for $10$ and ship them at a cost of $10$ per liter, for a total cost of $60$.  That would beat the alternatives of buying $2$ liters for $4$ and shipping them at a cost of $30$ per liter for a total cost of $64$, or buying $5$ liters and shipping them at $20$ per liter.
However, the milk has to come from $w$.  The cost of buying additional milk and moving it from $w$ to $u$ so as to get $3$ extra liters to $u$ may offset the advantage of solution (1) over solution (2).  Solution (3), on the other hand, is inferior to solution (1) and needs not be considered any further.
For each of the two edges from $w$ to $u$, we need to consider two possibilities to find the cheapest way to send milk to $v$ using that edge.  We end up considering four combinations, though there are six distinct paths.  If there are enough "inferior solutions," the advantage may be substantial. 
