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Jun 5 at 14:01 comment added Max Alekseyev @ClaudeChaunier: I see, thanks! For some reason I was thinking about variation rather than variance. Still, what I describe provides a heuristic (now indicated in the answer) that can do well for many practical instances.
Jun 5 at 13:58 history edited Max Alekseyev CC BY-SA 4.0
indicating heuristic
Jun 5 at 9:10 comment added Claude Chaunier A perfect matching with edge weights $1, 3, 3, 3, 3, 3, 3, 3, 3, 5$ would have variance $8$ and $t - w = 4$, while a perfect matching with edge weights $2, 2, 2, 2, 2, 4, 4, 4, 4, 4$ would have variance $10$ and $t-w = 2$. As @BrendanMcKay hinted at, the minimum-range $t-w$ method would pick the second choice instead of the better first choice.
Jun 4 at 9:55 comment added Max Alekseyev @ClaudeChaunier: I do not follow your point. Could you please elaborate?
Jun 4 at 7:08 comment added Claude Chaunier I am afraid this is not going to help. The minimal variance might involve a minimal $w$ and a maximal $t$.
Jun 3 at 11:01 comment added Manfred Weis that seems to be the only reasonable method...
Jun 3 at 10:54 history answered Max Alekseyev CC BY-SA 4.0