I have to construct an LP problem based on the ff scenario that might be similar to a scenario in another question (in the sense that I felt the need to use $mod$):
The productivities are minutes per day (as in 8 hrs). For example, if you want to produce 5 part 1's, it could be using 5 milling machines for 20 minutes or it could be using just 1 milling machine for 100 minutes.
Note:
$8 \ hrs = 480 \ min$
$\frac{3 \ min}{drill \ press}(1 \ drill \ press) + \frac{20 \ min}{machine}(5 \ machines) = 103 \ min$
$\frac{5 \ min}{drill \ press}(1 \ drill \ press) + \frac{15 \ min}{machine}(5 \ machines) = 80 \ min$
So if we let $x_i$ be # of completed assembly $i$'s for $i = 1,2$, then I think we must maximise
$$z = x_1 + x_2$$
where
$$x_1 = (103) [(3x_{1,1} + 20*5x_{1,2}) \ mod \ 103]$$
$$x_2 = (80) [(5x_{2,1} + 15*5x_{2,2}) \ mod \ 80]$$
s.t.
$$3x_{1,1} + 20*5x_{1,2} + 5x_{2,1} + 15*5x_{2,2} \le 480$$
$$|x_{i,j} - x_{k,l}| \le 30$$
where $i,j,k,l = 1,2$ and $(i,j) \ne (k,l)$
Okay, what did I do wrong?
From Chapter 2 here.
Let $x$ be the total number processed units, and $y_i$ the number of processed units on mill $i=1..5$.
So you want want to maximize $Z=x$, subject to availability constraints:
$$ \underbrace{\;3x\;}_{\mbox{drill part I}}+ \underbrace{\;5x\;}_{\mbox{drill part II}}\le 8\cdot 60\quad \mbox{(8 hour working day)}$$ $$ \underbrace{\;\frac{20x}{5}\;}_{\mbox{5 mills part I}}+ \underbrace{\;\frac{15x}{5}\;}_{\mbox{5 mills part II}}\le 8\cdot 60\quad \mbox{(8 hour working day)}$$
$$ x=y_1+y_2+y_3+y_4+y_5\quad \mbox{(all $x$ units must be processed in one mill)}$$
$$ |(20+15)y_i-(20+15)y_j|\le 30 \quad \forall i\neq j \quad \mbox{(no mill runs more than 30 minutes than another mill )}\\ x\ge 0 $$
You need to linearize constraints (3), for example: $$ |y_i-y_j|\le 10\quad \Leftrightarrow \quad-10 \le y_i - y_j \le 10 $$
Also, it is unclear if these last constraints must also hold for the drill. I assumed not, but if you need to you need to add $$ |3x+5x-20y_i-15y_i|\le 30 \quad \forall i=1,\cdots,5 $$