... connection 7 /19 /2008 11 Networking devices (cont.) 7 /19 /2008 12 Networking devices (cont.) Cisco 15 03 Micro Hub 7 /19 /2008 13 Networking devices (cont.) Cisco Catalyst 19 24 Switch 7 /19 /2008 14 Networking ... user networks over large geographic areas 7 /19 /2008 Network history 7 /19 /2008 Network history (cont.) 7 /19 /2008 Network history (cont.) 7 /19 /2008 Networking devices Equipment that connects directly ... manage a network networking technology could increase productivity while saving money In the mid -19 80s, each company that created network hardware and software used its own 7 /19 /2008 Data networks...
... using cables that are only a couple of meters long 0945_01f.book Page 11 Wednesday, July 2, 2003 3:53 PM Perspectives on Networking 11 Bam Bam happens by and sees that Pebbles is stressed Pebbles ... well—you are going to be finished soon, right?” 0945_01f.book Page 10 Wednesday, July 2, 2003 3:53 PM 10 Chapter 1: Introduction to Computer Networking Concepts So much for using first names for ... computer is typically connected to a network via some cable Figure 1- 1 shows the basic end-user perspective of networking Figure 1- 1 End-User Perspective on Networks Home User PC with Modem Office...
... Destination: 1.1 .1. 1 Source: 2.2.2.2 Bob sends the packet to R2, which makes a routing decision R2 chooses to send the packet to R1 because the destination address of the packet is 1.1 .1. 1, and R1 knows ... topology to know that 1.1 .1. 1 (Larry) is on the other side of R1 Similarly, when R1 gets the packet, it forwards the packet over the Ethernet to Larry And if the link between R2 and R1 fails, IP allows ... source 0945_01f.book Page 27 Wednesday, July 2, 2003 3:53 PM The TCP/IP Protocol Architecture Figure 2-4 27 IP Services Provided to TCP Bob - 2.2.2.2 Larry - 1.1 .1. 1 HTTP GET R2 TCP R1 IP R3 HTTP...
... and a trailer? 10 If a Fast Ethernet NIC currently is receiving a frame, can it begin sending a frame? 11 What are the two key differences between a 10 -Mbps NIC and a 10 /10 0 NIC? 12 What is the ... collide All devices on a 10 BASE2, 10 BASE5, or 10 BASE-T network using a hub risk collisions between 0945_01f.book Page 61 Wednesday, July 2, 2003 3:53 PM Early Ethernet Standards 61 the frames that they ... 3 -11 , which shows the full-duplex circuitry used with a single PC cabled to a LAN switch 0945_01f.book Page 63 Wednesday, July 2, 2003 3:53 PM Ethernet Data-Link Protocols Figure 3 -11 63 10 BASE-T...
... Cú pháp file: IP addressFully.Qualified.Name[host_alias]* 19 2 .16 8 .1. 10 centos -1. nhatnghe.com centos -1 Các ứng dụng trước tiên sử dụng file cần truy vấn máy tính tên File /etc/sysconfig/network ... ifconfig dùng để cấu hình địa IP, netmask, địa broadcast tham số cấu hình khác ifconfig eth0 19 2 .16 8 .1. 10 netmask 255.255.255.0 man ifconfig Lệnh ifconfig cấu hình cho card mạng (từng interface) ... file, dùng ethereal để phân tích gói tin, xác định loại traffic, tìm kiếm dấu hiệu mong muốn 10 Hỏi & Đáp 11 ...
... conditions ]} Dik ≤ Dik -1 for all k ≤ h so that h +1 h h 1 h Di = D j + d ij ≤ D j + d ij = Di j [ ] Dih ≤ Di1 = di1 = di1 + D1h j [ ] shortest (≤ h + 1) walk length = min{ , min[D + d ]} D Combining ... cycles without node 1, and it yields the shortest path lengths to node 18 h j Proof of Bellman-Ford (1) Proof is by induction on hop count h For h = we have Di1 = di1 for all i ≠ 1, so the result ... B Walk: sequence of nodes (n1, n2, …, nk), where (n1, n2), (n2, n3), …, (nk -1, nk) are arcs Path: a walk with no repeated nodes Cycle: a walk (n1, n2, …, nk), with n1=nk and no other repeated...
... averages of these times: 1 W2 = R + Q1 + Q2 + λ1W2 1 µ2 11 λ1W1 + λ 2W2 + λ1W2 1 µ2 1 Solving for W2 and using the expression for W1 : R W2 = (1 − 1 ) (1 − 1 − ρ2 ) =R+ 10 -30 Non-Preemptive ... p)n np np (1 ¡ p) (1 ¡ p)k 1 p k = 1; 2; : : : pet 1 (1 p)et p 1 p p2 r p r (1 p) p2 ¸ ¸ ¡n¢ k Geometric p Negative Bin (r; p) Moment Gen Fun MX (t) ³ k 1 r 1 ´ r (1 k¡r p ¡ p) k = r; r + 1; : : : ... terms: Tk = Rk + + Tk ( 1 + µ k − 1 − − ρ k Final solution: Tk = + ρ k 1 ) (1/ µ k ) (1 − 1 − − ρk ) + Rk (1 − 1 − − ρk 1 ) (1 − 1 − − ρ k ) where: k Rk = ∑ λ i X i2 i =1 ...
... ∑ n n 11 ρ22 ρ nk k ∑ n n 11 ρ22 ρ nk + k n1 + + nk = M = n1 + + nk = m nk = = ∑ n1 + + nk 1 = m n n 11 ρ22 ∑ n1 + + nk = m nk > ρnk− 1 + k n n 11 ρ 22 ∑ n1 + + nk = m nk > n n 11 ρ 22 ... mK = M mi′ +1> = n 11 ∑ ρini 1 ρnK K m1 1 ρimi′ m1 +…+ mi′ +…+ mK = M 1 mi′≥ ρ mK K ρ mK K = n 11 ρini 1 ρ nK K G ( M − 1) This is the probability of state ( n1 ,…, ni − 1, … nK ) in an identical ... 11 ∑ m∈F ( M ) mi > ρini m 1 ρ nK K ρimi ρ mK K Changing index mi = mi′ + 1, mi′ ≥ in the sum in the denominator: αij ( n ) = ∑ n 11 ρini ρnK K m1 1 ρimi′ +1 m1 +…+ mi′ +1+ …+ mK = M mi′ +1> ...
... leaves with probability 1- p Composite arrival rate and steady-state distribution: λ′ = λ + λ′r 11 = λ + λ′p ⇒ λ′ = λ / (1 − p ) p( n ) = (1 − ρ)ρ n , n ≥ 0; ρ = λ ′ / µ = λ / (1 − p )µ Probability ... 1{ ni > 0} + ∑ µi ri 01{ ni > 0} ij m ≠n i i, j i i i i i = ∑ γ i + ∑ µi [∑ rij + ri ] ⋅ 1{ ni > 0} j = ∑ γ i + ∑ µi 1{ ni > 0} ∑ q* (n, m) = ∑ γ* + ∑ µi1{ni > 0} = ∑ λ i ri + ∑ µi1{ni > 0} i m ≠n i ... distribution of the network is K p( n ) = ∏ pi ( ni ), n1 ,…, nK ≥ i =1 where for every node i =1, 2,…,K pi ( ni ) = (1 − ρi )ρin , ni ≥ k 1 rik j rij i i γi ri 8-6 Jackson’s Theorem (proof) Guess...
... distribution pn = n (1 ), n = 0 ,1, n +1 4&5-7 The M/M /1 Queue Average number of customers in system n =0 n =0 N = npn = (1 ) n = (1 ) n n N = (1 ) n n =0 = = (1 ) Littles ... empty =1/ 3, T =1 N=T/ =1/ 3 p1 =1/ 3 Example 2: LAA violated Poisson arrivals Service time of customer i: Si= Ti +1, < Upon arrival: system is always empty Average time the system has customer: p1= ... (k) (t) = MSi (t)MSi 1 (t) : : : MSiĂk +1 (t)MRiĂk (t) = i Then: MTi (t) = X ảk +1 k=0 = ạĂt (1 Ă ẵ)ẵk = (1 Ă ẵ) ạĂá ạĂá = (ạ Ă á) Ă t Ă t Ă ạĂt ạĂt ạĂt ảk +1 X ảk k=0 ạĂt 4&5- 41 Moment Generating...