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The hyperbolic bound [7] is a tighter sufficient condition for schedulability than the one presented by Liu and Layland: = (+), where U i is the CPU utilization for each task. It is the tightest upper bound that can be found using only the individual task utilization factors.
A real number x is the least upper bound (or supremum) for S if x is an upper bound for S and x ≤ y for every upper bound y of S. The least-upper-bound property states that any non-empty set of real numbers that has an upper bound must have a least upper bound in real numbers.
If (,) is a partially ordered set, such that each pair of elements in has a meet, then indeed = if and only if , since in the latter case indeed is a lower bound of , and since is the greatest lower bound if and only if it is a lower bound. Thus, the partial order defined by the meet in the universal algebra approach coincides with the original ...
For any sets X and Y, X join Y, written X ⊕ Y, is defined to be the union of the sets {2n : n ∈ X} and {2m+1 : m ∈ Y}. The Turing degree of X ⊕ Y is the least upper bound of the degrees of X and Y. Thus is a join-semilattice. The least upper bound of degrees a and b is denoted a ∪ b.
It is at least the absolute value of the difference of the sizes of the two strings. It is at most the length of the longer string. It is zero if and only if the strings are equal. If the strings have the same size, the Hamming distance is an upper bound on the Levenshtein distance. The Hamming distance is the number of positions at which the ...
1 Proof. 2 References. 3 See also. ... Download QR code; Print/export Download as PDF; ... Because the supremum is the least upper bound, ...
For example, the least upper bound of a bounded increasing computable sequence of computable real numbers need not be a computable real number. [9] A sequence with this property is known as a Specker sequence , as the first construction is due to Ernst Specker in 1949. [ 10 ]
The rational number line Q does not have the least upper bound property. An example is the subset of rational numbers = {<}. This set has an upper bound. However, this set has no least upper bound in Q: the least upper bound as a subset of the reals would be √2, but it does not exist in Q.