The scenario is that we own a chemical company with 2 warehouses. We have had the following quantities of chemical A sent to 4 different ports ready for transportation to our warehouses:
We require the following amounts of chemical A at our warehouses:
The cost in £ to ship 1L of chemical A from a port to a warehouse is given below
We want to find a linear program to transport the chemical as cheaply as possible and fulfill the requirements at each warehouse (this is the simple part). The program I have formulated is as follows
Let $x_{i,j}$ denote the amount of chemical A we ship from port $i$ to warehouse $j$, where $i\in \{1,2\}$ and $j \in \{1,2,3,4\}$. $$\text{Minimize } 20x_{1,1} +13x_{1,2} +10x_{2,1}+28x_{2,2}+45x_{3,1}+18x_{3,2}+15x_{4,1}+11x_{4,2}$$ $$\text{Subject to } x_{1,1}+x_{2,1}+x_{3,1}+x_{4,1} \ge 280$$ $$x_{1,2} +x_{2,2}+x_{3,2}+x_{4,2} \ge 330$$ $$x_{1,1}+x_{1,2} \le 300$$ $$x_{2,1}+x_{2,2} \le 70$$ $$x_{3,1}+x_{3,2} \le 90$$ $$x_{4,1}+x_{4,2} \le 150$$ $$x_{i,j} \ge 0 \in \{1,2\} \text{ and } j \in \{1,2,3,4\}$$
Now, the tricky part is the following: "Suppose now that all is as in the previous question but that only 220L of Important Chemical A are now required at Warehouse 1. Any excess chemical can be transported to either Warehouse 1 or 2 for storage, in which case the company must pay only the relevant transportation costs, or can be disposed of at the port in which case the company pays no transportation costs but pays a disposal fee of £16 per L. You want to find out how to ship or dispose of all 610L of chemical imported at a minimum cost, while still ensuring that the required amounts (220L and 330L, respectively) delivered to Warehouse 1 and 2. Describe how to modify your linear program from the previous question to model this problem."
By listing the costs of shipping or disposing I have found that it is cheapest to send any excess from: port 1 to warehouse 2., port 2 to warehouse 1, port 3 to be disposed, port 4 to warehouse 2.
I think the only thing we would need to change is the objective function and the first constraint to be $\ge 220$.
We can model the amount of excess and cost to deal with it, say from port 1, as follows: $13(300-(x_{1,1} +x_{1,2})$. We pay £13 for every 1L of chemical excess in port 1 to transport it (to the cheapest destination) warehouse 2.
So would my objective function look like this then:
$$\text{Minimize } 20x_{1,1} +13x_{1,2}+10x_{2,1}+28x_{2,2}+45x_{3,1}+18x_{3,2}+15x_{4,1}+11x_{4,2} +13(300-(x_{1,1} +x_{1,2}))+10(70-(x_{2,1}+x_{2,2})) + 16(90-(x_{3,1}+x_{3,2}))+11(150-(x_{4,1}+x_{4,2}))$$
What you have is correct, but a simpler and less error-prone approach is to introduce a third "dummy" warehouse to represent disposal. Modify the $\ge 280$ constraint to $\ge 220$ as you indicated, and introduce new variables $x_{i,3} \ge 0$. Modify each supply constraint to include exactly one $x_{i,3}$, and change $\le$ to $=$. Finally, add $16\sum_i x_{i,3}$ to the objective function.