optimal solution for the 0 1 knapsack problem

... 1, 1, 0, 0, ) x2 = (1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, ) x3 = (1, 1, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, ) x4 = (1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, ) x5 = (1, ... Therefore # sn = Hm 1 # sk+2 + [ m ]# sk+2 m ( Hm 1 + [ ])# sk+2 2 (6) The claim is proved 3.7 Proposition Let x0 = (1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, ) x1 = (1, 1, 1, 1, 1, 0, 0, 1, 1, 1, ... 0, 0, 0, , 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, , 0, ) (17 ) k1 =6 k2 k3 =18 k4 Note that, for i ∈ Oj , we have # ski = F6i = 9i For s ∈ supp µ is defined by x in (16 )...

Ngày tải lên: 13/02/2014, 04:20

... G3k +1 F3(k +1) 2.2.3 If (y3k , y3k +1 , 0, 1) is a generator, then (y3k , y3k +1 , 0, 1) ∈ { (0, 7, 0, 1) , (1, 0, 0, 1) , (1, 7, 0, 1) } 2F3k a) If (y3k , y3k +1 , 0, 1) = (0, 7, 0, 1) or (1, 0, 0, 1) , ... (1, 0, 1) 1. 1 If y3k = 0, then t(3k + 3) = (t(3k − 1) , 0, 1, 0, 1) ∪ (t(3k − 1) , 0, 0, 1, 7) Hence 2H3(k 1) +2 + G3k +1 F3(k +1) # t3k+3 1. 2 If y3k = or y3k = 1, then t(3k + 3) = (t(3k), 1, 0, 1) ... 1) , then # t3k+3 b) If (y3k , y3k +1 , 0, 1) = (1, 7, 0, 1) We consider two cases b1) If y3k 1 = or y3k 1 = 1, then # t3k+3 2F3k F3(k +1) b2) If y3k 1 = 0, then (0, 1, 7, 0, 1) ≈ (1, 0, 1, 0, 1) ,...

Ngày tải lên: 05/03/2014, 14:20

... sk +1 = sk + 3k +1 for all sk ∈ supp µk (3) Then we have # sk +1 = # sk Write sk +1 = sk + xk +1 , xk +1 ∈ D 3k +1 Since sk +1 = sk + 3k +1 ∈ supp µk and xk +1 = 0, xk +1 = or xk +1 = Which implies # sk +1 ... points in supp µn−2 , sn 1 = sn−2 + = sn−2 + 3n 1 , then sn +1 = sn−2 + 3n 1 + 0 + n +1 = sn−2 + n 1 + n + n +1 n 3 3 6 = sn−2 + n 1 + n + n +1 3 Since (1, 0, 0) ≈ (0, 1, 6), sn−2 = sn−2 , and so ... log a1 =1 2(n) log log (15 ) Combinating (14 ) and (15 ) we get α 1 Therefore log a1 log √ log (1 + 5) − log log a1 α =1 =1 log log The claim is proved To complete the proof of our Main Theorem...

Ngày tải lên: 14/03/2014, 13:20

... Pure Appl Math 12 , 623–727 (19 59) doi: 10 . 10 0 2/cpa. 31 6 01 204 05 Hung et al Boundary Value Problems 2 01 1, 2 01 1: 17 http://www.boundaryvalueproblems.com/content/ 2 01 1/ 1 /17 Page 18 of 18 13 Kozlov, VA, ... λ)), λ ∈ C, t ∈ (0, +∞) Hung et al Boundary Value Problems 2 01 1, 2 01 1: 17 http://www.boundaryvalueproblems.com/content/ 2 01 1/ 1 /17 Page 11 of 18 For every fixed l Î ℂ and t Î (0, ∞), the operator U ... Boundary Value Problems 2 01 1, 2 01 1: 17 http://www.boundaryvalueproblems.com/content/ 2 01 1/ 1 /17 Page 10 of 18 and L2 (Q) Ce2(h 1 +σ )τ |2Re(fth , uN )Q | ≤ ε|uN |2 τ + C fth t h +1 t h +1 0, Q τ ≤ ε|uN...

Ngày tải lên: 21/06/2014, 00:20

... 3u1 Du1 2 ∂ω A ∂x dx dτ ≤ K5 θ t , where 3 .11 Du1 , 1/ 2 12 u1 A1/2 E1 48K1 θ t θ t E1 E1 K θ t D , 1/ 2 24u1 E1 K4 θ t K5 θ t Du1 K3 θ t u1 u1 3 .12 u1 u1 2Kθ t K1 θ t 9θ t 3 .13 3θ t 3AE1 D E1 θ ... 2 R R 1 − ln − dx 0 0 v02 ω A ρ dx 0 R 02 dx v0 ln E2 , dx ≤ E3 , 0 sup 0 x < M1 |x|

Ngày tải lên: 21/06/2014, 16:20

... ) 1, 600 , 200 732,638 35,268 203 ,7 10 10 , 477 76, 606 16 , 805 12 ,753 78 ,06 6 65,244 9, 807 2 ,18 4 68 ,14 8 4, 703 283,7 91 1 40, 0 20 69 ,09 5 10 , 936 5,369 23 ,09 0 22,5 10 5 ,02 0 10 0 , 01 0 56, 914 8,5 70 8,469 26 ,05 7 ... phu/hethongvanban?class_id=2&_page =1& mode= detail&document_id= 10 1 900 / Shannon, K ( 200 9) “Greater Hanoi - Megacity in the Making” In Topos, p.p 98 - 10 3 Thanh Thủy ( 2 01 1) Điều chỉnh quy hoạch khu đô thị Tây Nam hồ Linh Đàm 21 ... use pattern of Linh Dam (Mapping - Quyen Phuong, 2 01 1) Table Land use planning of Linh Dam (Total area: 1, 8 40, 2 30 m2) Types of use I 10 11 12 13 14 II III Linh Dam Penisula Water body Street network...

Ngày tải lên: 29/08/2013, 08:15

... 3. 8 01 3. 700 2.7 90 Hài lòng % 16 ,19 6,67 1, 9 Bình thường % 17 ,14 22,86 19 ,05 Không hài lòng % 3, 81 12,38 Thu nhập bình quân Mức độ hài lòng với công việc Nguồn: Tổng hợp số liệu điều tra năm 2 01 1 ... 87 ,14 879 ,14 6 41 Sức khỏe Nghìn đồng/người/tháng 63, 71 43 ,14 24,57 Mua đồ dùng thiết yếu Nghìn đồng/người/tháng 99,86 13 9, 71 10 9 Vui chơi giải trí/tháng Nghìn đồng/người/tháng 42,86 19 , 71 15, 71 ... làm việc 10 11 tháng/năm (Bảng 1) 3 .1. 2 Môi trường làm việc Khi đánh giá mức độ đảm bảo sức khỏe an toàn lao động môi trường làm việc 10 0 % lao động giúp việc cho đảm bảo Trái lại, 10 0 % lao động...

Ngày tải lên: 29/08/2013, 08:15

... + βn 1 λn 1 + · · · + 1 1 + 0 , then p(M ) = M n + βn 1 M n 1 + · · · + 1 M + 0 I = Therefore M n = −βn 1 M n 1 − βn−2 M n−2 − · · · − 1 M − 0 I, and so bT M n N = bT (−βn 1 M n 1 − ... x1 , x2 ∈ K(t, x0 ) Then there exists 1 , α2 ∈ A such that t x1 = X(t)x0 + X(t) X 1 (s)N 1 (s) ds t x2 = X(t)x0 + X(t) X 1 (s)N α2 (s) ds Let ≤ λ ≤ Then t λx1 + (1 − λ)x2 = X(t)x0 + X(t) X 1 ...  1  sgn x =   1 36 x >0 x =0 x < = 0; Therefore the optimal control α∗ switches at most once; and if h1 = 0, then α∗ is constant Since the optimal control switches at most once, then the...

Ngày tải lên: 24/03/2014, 12:20

... 1 − 1 n +1 n +1 n +1 + B( 1 , 1 ; un ) , 1 M t φ n n +1 ≥ ( τ n +1 − 1 ) + |D1/2 (un )∇h 1 |2 0 t φ n +1 + [[ 1 ]] 0, h −C n +1 φ 1 (7:28) Luo et al Boundary Value Problems 2 01 1, 2 01 1: 48 http://www.boundaryvalueproblems.com/content/ 2 01 1/ 1/48 ... using the boundedness of u M and n +1 n +1 n +1 (un · ∇h 1 , 1 ) ≤ ε |D1/2 (un )∇h 1 | M n +1 φ 1 : +C n +1 n +1 n +1 (un · ∇h 1 , 1 ) ≤ ε |D1/2 (un )∇h 1 | M +C n +1 φ 1 (7:27) Thus n +1 n 1 − 1 ... (φct , 1 ) , 1 , 1 M t t + φ n +1 n τ2 − τ2 n +1 n +1 n +1 n +1 + B(τ2 , 1 ; un ) + B(cn +1 , 1 ; un ) , 1 M M t (7:25) n +1 n +1 n +1 −B(cn +1 , 1 ; un ) + L( 1 ; un , cn +1 ) − L( 1 ; un , cn +1 ) M...

Ngày tải lên: 20/06/2014, 22:20

... Processing 2 01 1, 2 01 1: 119 http://asp.eurasipjournals.com/content/ 2 01 1/ 1 /11 9 1. 2 1. 18 0. 9 1. 16 0. 8 1. 14 0. 7 1. 12 0. 6 1. 1 L(σ )0. 5 1 .08 0. 4 1 .06 0. 3 1 .04 0. 2 1 .02 I(σ) Page of 11 0. 1 1 σ 10 (a) 0 σ 10 (b) ... Rev 55(3), 2 71 297 ( 200 2) doi: 10 . 11 15 /1. 145 8 01 3 doi: 10 . 11 86 /16 87- 61 80- 2 01 1- 119 Cite this article as: Martínez-Rodríguez et al.: A general solution to the continuous-time estimation problem under ... ), sin(4π e−t ), } 0. 14 0. 12 0. 1 0. 08 WL errors 0. 06 0. 04 0. 02 0 0 .1 0. 2 0. 3 0. 4 0. 5 0. 6 0. 7 0. 8 0. 9 t Figure MS errors of the WL estimator (5) (solid line) and the estimator calculated in...

Ngày tải lên: 20/06/2014, 22:20

... the obstacle problem Indiana Univ Math J 200 1, 50( 3): 10 7 7 -11 12 10 Evans LC, Gariepy R: Measure Theory and Fine Properties of Functions CRC Press, Boca Raton; 19 92 doi: 10 . 11 86 /16 87 -18 47- 2 01 1- 19 ... Equations 2 01 1, 2 01 1: 19 http://www.advancesindifferenceequations.com/content/ 2 01 1/ 1 /19 Uraltseva N: Two-phase obstacle problem, function theory and phase transitions J Math Sci 200 1, 10 6 : 307 3- 307 7 Bozorgnia ... Difference Equations 2 01 1, 2 01 1: 19 http://www.advancesindifferenceequations.com/content/ 2 01 1/ 1 /19 Page of 16 The regularity of the solution, the Hausdorff dimension and the regularity of the free boundary...

Ngày tải lên: 21/06/2014, 02:20

... 1. 4 If a function φ satisfies 1. 10 and 1. 11 , then α φ 1 s ≤ K 1 φ s , α φ θ2 s ≤ 2aθ2 φ s , for every s > and < θ2 ≤ ≤ 1 < ∞, where 1 log2 K and α2 1. 12 loga Remark 1. 5 Under condition 1. 12 ... × 0, ∞ φ D2 u dz Rn × 0, ∞ φ |ut | dz ≤ C Rn × 0, ∞ φ f dz, 1. 17 One has φ ∈ Δ2 ∩ ∇2 1. 18 Theorem 1. 9 Assume that φ ∈ Δ2 ∩ ∇2 If u is the solution of 1. 1 - 1. 2 with f ∈ Lφ Rn × 0, ∞ , then 1. 8 ... × 0, ∞ , in Rn × {t 3 .19 0} with the estimate D2m v Lp Rn × 0, ∞ ≤ C fλ Lp Rn × 0, ∞ 3. 20 10 Boundary Value Problems Therefore we see that D2m v Lp Q 10 i zi ≤ D2m v ≤ C fλ C fλ Set w Lp Rn × 0, ∞...

Ngày tải lên: 21/06/2014, 16:20

... method for nonexpansive mappings,” Bulletin of the Australian Mathematical Society, vol 65, no 1, pp 10 9 11 3, 200 2 11 H.-K Xu, “Iterative algorithms for nonlinear operators,” Journal of the London ... conditions C1 , C2 , and C4 lim n→∞ αn αn 1. 7 Actually, Xu 10 , 11 and Wittmann proved the following approximate fixed points theorem Also see 12 , 13 Fixed Point Theory and Applications Theorem 1. 3 Let ... ∞ C1 lim αn n→∞ 0, C2 αn ∞, ∞ C3 n |αn − αn | < ∞ 1. 6 n In 200 2, Xu 10 , 11 extended wittmann’s result to a uniformly smooth Banach space, and gained the strong convergence of {xn } under the...

Ngày tải lên: 21/06/2014, 20:20

... p 1 p = p 1 p = p 1 × p 1 × Ω l1 σ p/ (1 p) l1 σ p/ (1 p) ∇ 1 p = = Ω l1 σ p/ (1 p) p p 1 Ω p−2 1 ∇ 1 p 1 ∇ 1 · ∇w ∇ 1 | p−2 ∇ 1 ∇ 1 w − w ∇ 1 Ω p 1 ∇ 1 Ω p−2 p ∇ 1 ∇ 1 w − p 1 l1 σ p/ (1 ... p 1 w l1 σ p/ (1 p) |∇ 1 | p w p p 1 l1 σ p/ (1 p) p 1 Ω 1 1 p−2 p 1 1 w − p 1 l1 σ p/ (1 p) p 1 by (1. 2) p 1 Ω 1 1 p − ∇ 1 p w ∀w ∈ W (2 .1) Thus ψ is a subsolution if p p 1 l1 σ p/ (1 ... Theorem 1. 1 in Section 2 Proof of Theorem 1. 1 p/(p 1) First we construct a positive subsolution of (1. 1) For this, we let ψ = l1 σ p/ (1 p) 1 1/(p 1) Since ∇ψ = p/(p − 1) l1 σ p/ (1 p) 1 ∇φ1...

Ngày tải lên: 23/06/2014, 00:20

... Structures of the Template First Trial Run [ viii ] 19 8 19 9 2 01 2 01 202 203 204 205 205 206 208 2 10 2 10 211 212 214 216 219 219 2 20 2 20 2 20 2 20 2 21 2 21 223 224 224 224 225 226 226 227 2 30 232 Simpo ... Settings 99 10 0 10 0 10 1 10 2 Logout Summary 10 3 10 3 Chapter 7: The Menus Menu 10 5 Menus Customizing an Existing Menu Menus Icon Default Icon Publish/Unpublish Icon Move Icon 10 6 10 6 10 7 10 7 10 8 10 8 [ ... Popular Search Administrator Module 16 7 16 7 16 8 16 9 17 1 17 1 17 2 17 2 17 3 17 4 17 5 17 6 17 6 17 6 17 6 17 7 17 8 17 8 17 8 17 9 17 9 17 9 1 80 18 1 18 1 18 2 Logged in Users Popular Recent added Articles Menu...

Ngày tải lên: 27/06/2014, 00:20

... 1 1 11 00 11 00 11 00 11 00 11 00 11 00 11 00 11 00 11 00 1 1 11 00 11 00 11 00 11 00 11 00 1 1 1 1 11 00 11 00 11 00 11 00 11 00 11 00 11 00 11 00 11 00 11 00 11 00 11 00 11 00 1 11 00 11 00 ... 00 11 00 11 00 11 00 1 1 11 00 11 00 11 00 11 00 11 00 11 00 11 00 1 11 00 11 00 11 00 1 1 1 1 1 1 11 00 11 00 11 00 11 00 11 00 1 11 00 11 00 11 00 1 1 1 1 1 1 11 00 11 00 11 00 11 00 11 00 11 ... 11 00 11 00 11 00 1 1 1 11 00 11 00 11 00 11 00 1 11 00 11 00 11 00 11 00 11 00 11 00 11 00 11 00 a 1- galaxy 11 00 11 00 1 1 11 00 11 00 11 00 11 00 11 00 11 00 2-galaxies 1 11 00 11 00 11 00 ...

Ngày tải lên: 07/08/2014, 08:22

... were measured ( 10 0 0 individuals) but a 10 0 % power when 500 Fls were l added to these 10 0 0 F2s From our simulations, the further inclusion of parental P1 and P2 performances in the analyses appears ... from the 10 % empirical o quantiles of the test statistic distribution, for each population structure studied, was defined by the group sizes n The power at the 10 % level was simply estimated for ... differs between the F2 (a and the non-segregating subpopulations ) Simulations were performed considering for the structures C 10 to C19 and !FZ and cr! /1. 25 or their equivalent with the total number...

Ngày tải lên: 09/08/2014, 18:21

... J Cardiothorac Surg 2 01 0, 5:58 doi: 10 . 11 86 /17 49- 809 0-5- 10 0 Cite this article as: Tagarakis et al.: An alternate solution for the treatment of ascending aortic aneurysms: the wrapping technique ... major comorbidities are combined with the patient’s wish to avoid major aortic surgery Received: September 2 01 0 Accepted: November 2 01 0 Published: November 2 01 0 Reference Ang KL, Raheel F, Bajaj ... Tagarakis et al Journal of Cardiothoracic Surgery 2 01 0, 5: 10 0 http://www.cardiothoracicsurgery.org/content/5 /1/ 10 0 Page of Table Patients’ baseline characteristics and comorbidities...

Ngày tải lên: 10/08/2014, 09:22

... brief analysis of the literature Anaesthesist 200 5, 54:2 20- 224 13 Weisfeldt ML, Becker LB: Resuscitation after cardiac arrest: a 3-phase time-sensitive model JAMA 200 2, 288: 303 5- 303 8 14 Wik L, Kramer-Johansen ... 200 5, 293: 305 -3 10 16 Nolan JP, Deakin CD, Soar J, Bottiger BW, Smith G: European Resuscitation Council guidelines for resuscitation 200 5 Section Adult advanced life support Resuscitation 200 5, ... Care 200 6, 10 : R13 10 Krismer AC, Wenzel V, Stadlbauer KH, Mayr VD, Lienhart HG, Arntz HR, Lindner KH: Vasopressin during cardiopulmonary resuscitation: a progress report Crit Care Med 200 4, 32:...

Ngày tải lên: 12/08/2014, 23:22

Ngày tải lên: 26/03/2015, 09:41