I'm having difficulties with the logic with the last part of the reformulation part of the problem below.
Let $x_i$ be the the number of ships of type $i$ to purchase.
For $4a:$ (the formulation)
$$\text{Min} \ z = 20,000x_1 + 1,000x_2$$
s.t.
- 7500 soldiers
$$2,000x_1 + 1,000x_2 \le 7,500$$
- 55,000 gallons
$$12,000x_1 + 7,000x_2 \le 55,000$$
- 900 crew people (crewmen?)
$$250x_1 + 100x_2 \le 900$$
- integer constraint for ships
$$x_1, x_2 \in \{0,1,2,...\}$$
For $4b:$ (the reformulation)
$$\text{Min} \ z = 20,000x_1 + 1,000x_2 \color{red}{+ 2,000y_1 + 1,000y_2}$$
s.t.
same
same
same
same
$$x_2 \le My_1, y_1 \in \{0,1\}$$
$$x_2 - 2 \le My_2, y_2 \in \{0,1\}$$
7.
I think we have to say, linearly, that:
if $x_2 \ge 4$, then $y_3 = 0$.
if $y_3 = 0$, then $x_1 \le 3$.
then say exactly one of the following (also linearly):
converse of 'if $x_2 \ge 4$, then $y_3 = 0$'
converse of 'if $y_3 = 0$, then $x_1 \le 3$'
and finally say
- $y_3$ is binary.
I believe these translate to:
$$x_2 - 3 \le M(1-y_3)$$
$$x_1 - 3 \le M(y_3)$$
$$4 - x_2 \le My_3$$
$$4 - x_1 \le M(1-y_3)$$
$$y_3 \in \{0,1\}$$
Is that right? I think if I use $4$ instead of $3$, I'm going to get the contrapositive of $(iii)$.