On the job rotation problem



Remark. Matrices that are not pyramidal, may become such after simultane-
ously permuting rows and columns. It follows from (1) that the diagonal entries
of the matrix must be in descending order for (1) to be satisfied. Once rows and
columns have been simultaneously permuted in this way, additional simultane-
ous row and column permutations may be needed between rows and columns
which have a diagonal entry equal to another diagonal entry.

4.2 Monge and Hankel matrices

A matrix A will be called diagonally dominant if id ap(A). (Note that through-
out the paper
id stands for the identity permutation.) A matrix A = (aij )
R is called Monge if aij + ars ≥ ais + arj for all i, j, r, s N, i ≤ r, j ≤ s.
It is well known [6] that every Monge matrix A is diagonally dominant. It is
also easily seen that a principal submatrix of a Monge matrix is also Monge.
Hence
JRP (A, k) for Monge matrices is readily solved by finding the k biggest
diagonal entries of
A.

For a given sequence {gr R; r = 1,..., 2n — 1}, the Hankel matrix is the
matrix
H = (hij) Rn×n where hij = gi+j-1. Hankel matrices generated by
convex sequences are Monge [9]. Therefore, for these matrices, JRP is readily
solved. However, no efficient method seems to exist for Hankel matrices in
general.

In this subsection we show that finiteness of δk (H) can be easily decided for
any Hankel matrix
H. Since Hankel matrices are symmetric, we can use some
of the results of Section 3.

Theorem 15. If {gr R; r = 1,..., 2n — 1} is the sequence generating Hankel
matrix
H = (hij) Rn×n and gr = -∞ for some odd r, then δk(H) = -∞ for

all k ≤ k


max .


Proof. Let C = (cij) be defined by cj = 0 if hij = -∞ and cj = -∞ otherwise.
Assume
gr = -∞ for some odd r. So (i) cii = -∞, i.e. (i) cii = 0. We now
use Theorem 7 to give us
δk (C) = 0 for all k ≤ kmax . Then as δk (C) = 0 if and
only if
δk(H) = -∞, the theorem follows.                                 □

Theorem 16. If a matrix A = (aij) Rn×n is any matrix such that aij = -∞
if i + j is even, then δk (A) = -∞ for all odd k.

Proof. Assume A = (aij) is a matrix such that aij = -∞ if i + j is even. If aij
is finite then i + j is odd. So i and j must be of different parities.

13



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