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4 Nov 1999: where:. Stage 1 is a simultaneous, multiple-round auction, conducted in three substages,with progressive eligibility requirements and an improvement margin requirement,with bidders submitting bids on individual properties; and. ... Stage 2 is a

28 Apr 2010: n, which can be reached in just one pivotfrom x1 = = xn1 = xn = 0 by pivoting in thecolumn with the smallest entry in the bottom row! ... requirements byskill categories to meet the overall force structure (anLP with 21,000 constraints and 43,000

24 Feb 2010: remaining work requirement of not more than x. Then, N sx/Ns G(x). ... also when this would cause an upstream entry point to obtain more than.

21 Oct 2009: 5. of the matrix A, corresponding to resource requirements at pooled resources,may be non-integral. ... jk}. Once a customer has completed its service requirement at allqueues it leaves the network.

28 Apr 2010: n, which can be reached in just one pivotfrom x1 = = xn1 = xn = 0 by pivoting in thecolumn with the smallest entry in the bottom row! ... requirements byskill categories to meet the overall force structure (anLP with 21,000 constraints and 43,000

19 Nov 2010: Formally, this can be doneby adding the requirement. suptR. t. pn(y, j ) dy Fn(t). ... It follows that A1 is also Toeplitzand we denote its entries by bkl = g(|k l|) for some function g.

15 Mar 2016: All entries in the last row are non-positive, so this solution is optimal. ... An IP is a linear program with the additional requirement thatvariables should be integral.

4 Apr 2007: The bold entries indicate downloads from the server; un-bold entriesindicate downloads from a peer who has the corresponding part. ... to less than 1; for example, the bottom two entriesin column 2 and the bottom entry in column 3 are 0 for n = M 2.

15 Aug 2012: Above (1.7) we required that J1, J2 > 0, but actually this requirement may be weakenedto obtain an interesting conclusion. ... may contain multiple entries, since there may exist (e1, z1), (e2, z2) F suchthat e1 z1.τ (e1) = e2 z2.τ (e2).

16 Sep 2019: To obtain a consistent theory we must place requirements on F :. F1: F and F. F2: A F = Ac F. F3: A1,A2,. ... We also place requirements on P : it is to be a real-valued function defined on F whichsatisfies three axioms (known as the Kolmogorov axioms):.