# Minimise Total $x$, Maximum $x$ or $|x|$ in integer/linear programming

Suppose we have a linear program (it may be integer, it probably doesn't matter).

Suppose the variables $t_j$ give the tardiness for job $j$, and we want to minimise something to do with this tardiness.

Now, there are at least two options that I know of:

1) Minimise total tardiness: $min \sum_j t_j$

2) Minimise maximum tardiness: $min_z\{z \geq t_j \forall j\}$

So here are the problems with these two: (1) will give a small number of total tardiness, however two solutions with the same total tardiness could look very different (one solution has many slightly tardy jobs, and another solution has only 1 tardy job, but it's extremely late). (2) prevents one exceptionally tardy job, however if only one job is tardy by $1$ unit, then all other jobs are now permitted to be tardy for free. (If a solution with $n$ jobs can be constructed with zero tardiness, then it's possible $n+1$ jobs will lead to one job being tardy by $1$ unit, but since we're only minimising the maximum tardiness, all other jobs can now be tardy by up to $1$ unit without any affect on the objective value).

A third approach which is a bit different would be to minimise the total number of tardy jobs. So if $\hat{t}_j$ is $1$ if job $t_j>0$, or $0$ otherwise: $min \sum_j \hat{t}_j$.

All of these can give wildly different solutions. I understand they are minimising different things (obviously), and I understand mathematics is a very precise science; so technically we are getting exactly what we've asked for. However, somehow, asking for (1), (2), or (3) alone seem to lack something "human". No human would produce a timetable with one exceptionally late jobs in favour of spreading lateness out a little and saying "it's okay - the total tardiness is low". Nor would a human let all jobs be late because one job has to be late and say "it's okay, one job is bit late so all jobs can be a bit late". Nor would they allow one job to be arbitrarily late and say "it's okay. Only one of the jobs is arbitrarily late".

So what is the solution? I don't know.

Perhaps multiply (1) and (2) together? Or some linear combination?