On the job rotation problem



We can similarly define kmin . It is easily seen that the problem of finding kmin
is equivalent to finding a shortest cycle in a digraph and is therefore polynomially
solvable [19]. In Example 1, we have
kmin = 2 and kmax = 3.

As usual a real sequence g0, g1 , . . . is called convex [concave] if

gr-1 + gr+1 2gr

[gr-1 + gr+1 2gr]

for all r = 1, 2, ....

One class of solvable cases of the JRP is related to the fact that δk (A) for
k = 1, 2, . . . , n are the coefficients of the characteristic polynomial of A in max-



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