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20 Sep 2011: Kenyon, J.B. Orlin, P.W. Shor and R.R. Weber. , in Proc. ... In addition, we present a ran- domized O(nB log B)-time on-line algorithm SS , based on SS, whose expected behavior is essentially optimal for all discrete distributions.

26 Oct 2011: B(x,r δ) π(C(π1(x),r)) B(x,r δ), for all R > R0. (2.5). ... Also for all A π1(B(0,K)) we have that. µ(A) vol(π(A)) 0, as R.

22 Aug 2011: 3. adjust the divisor d in such a way that the sum of the seatnumbers of all Member States equals the given Parliament-size. ... 2. Suppose, at some stage, that State i has been allocated ai seatsin all.

17 Feb 2011: Wenext prove fk B(s, L) for all k and suitable c′, η. ... 3.5. Proof of Theorem 4. For simplicity of notation, we suppress the supremaover B(s, L) in most of what follows—uniformity of the bound follows from track-ing all

30 Nov 2011: Trans-mission is assumed to occur simultaneously through all resources on route i. ... By Proposition 4.3,there exists a compact set B(R) in RI+ such that n(t) B(R) for all t 0 for anyfluid model solution n satisfying |n(0)|

8 Jun 2011: pt =1011 for all t. {c(0),c(1),c(2),c(3),c(4),. } = {0, 316,12,1,1,. }. Note that c(i)2 c(i 1)c(i 1) ... Adv Appl Probab 42(3):795–815. Bartroff J, Goldstein L, Samuel-Cahn E (2010b) The spend-it-all regionand small time results for the continuous

26 Oct 2011: 1 P(Ψ(B) > 0)). 1 exp(BΛ. P(Ψ(B) > 0)). Taking sup over all Λ Γ of type 1, we deduce that P(Ψ(X) > 0) = 1. • ... Pu(B hits A) > 0,for all starting points u. By neighborhood recurrence of Brownian motion, we get that thisprobability is indeed

20 Jan 2011: This is the same as requiring an equal sizeof resource contribution from all participants. ... So N B(n 1, 1 αq). Theexpected net benefit of all the agents is.

25 Oct 2011: µqA(B) := q. Rd. Ex[eqTA 1{XTAB}. ]dx for all B B(Rd). ... 8.2& 8.4], provided we replace Gq by. G0(y,B) :=. 0. Py(Xt B) dt for all B B(Rd), (1.21).

9 May 2011: We write. g(x) z = b, z 0. It will frequently be mathematically useful to turn all our constraints into equalities. ... You can confirm for all b thatprimal and dual here have the same optimal values.

15 Sep 2011: In this case any bidder can obtain greater profits by bidding bA E on A, b E on B, and 0 on all other items. ... for all c and N. For c > 2 and p > 1.5, 1 - I/p seems very close to B.

21 Oct 2011: The Invincible Fighter Problem corresponds to the investment problem of Derman et al andso (A), (B) and (C) all hold. ... We summarise in Table 1 the status of (A), (B) and (C) for the set of problems discussed above,all of which are defined by cases of

15 Sep 2011: As ν , all states are eventually in S2. The solution tothe ν-wage problem can be succinctly described in terms of the 2n indices producedby the algorithm. ... Also, bi < b for all states i in which theactive action is taken.

15 Sep 2011: Using this heuristic limfl-0o( 1/S)log P(B) < - 5 if and only if I{c a)/a > 5for all a. ... because it is unlikely that any theoretical model is rich enough to adequatelymodel all traffic classes.

18 Sep 2011: fa i 1u r era t ere pair s t rat e gy rna x i rn i z e s for all t the pro b a biIi t y ... 352 P. NASH AND R. R. VEBER. B)O, and the result follows by taking c as a vector which is in the component corresponding to all jobs complete and 1

15 Sep 2011: This is the same as requiring an equal sizeof resource contribution from all participants. ... So N B(n 1, 1 αq). Theexpected net benefit of all the agents is.

15 Sep 2011: We desire to stabilize t > l e system by ensuring t h a t for all i a,nd t we have E ( z i ( t ) ) < B for some R. ... Similarly, r = {x : Bx 2 0) for sonie B with coefficients t h a t a,re all -1, 0 or 1.

15 Sep 2011: We desire to stabilize the system by ensuring that for all i and t we have E(xi(t)) < B for some B. ... Similarly, = f x Bx? O} for some B with coefficients that are all -1, 0, or 1.

15 Sep 2011: Thus, we need to establish the optimality of 7r, amongst all dispatching functions. ... That is, B, is the set of states for which the confining of all future Cj job assignments to P,,. "? ",

15 Sep 2011: So for all m_. mo we have [b m(c6 - p,, (6))] > b and. ... Thus I(c, b)? bH(c) for all b. Moreover, for those b for which b =. m(cp'(H(c))-c), m= 1, 2,--., we have I(c, b)=bH(c).