HANNAM UNIVERSITY 1 Chapter 12 다중 접속 (Multiple Access)

HANNAM UNIVERSITY 12 장 다중접속 (Multiple Access) 12.1 임의접속 12.2 제어접속 12.3 채널화 12.4 요약 2

HANNAM UNIVERSITY 두 개의 계층으로 나뉘어진 데이터 링크층 두 개의 계층으로 나뉘어진 데이터 링크층  위 층 : 데이터 링크 제어  아래층 : 공유하는 매체에 대한 접근 제어 책임 3

HANNAM UNIVERSITY 다중 접속 프로토콜 다중 접속 프로토콜  다중접속 (Multiple Access) ■ 노드나 지국이 멀티포인트나 공통 링크를 사용할 때 링크에 대한 접속을 조절할 수 있는 다중접속 프로토콜이 필요 4

HANNAM UNIVERSITY RANDOM ACCESS In random access or contention methods, no station is superior to another station and none is assigned the control over another. No station permits, or does not permit, another station to send. At each instance, a station that has data to send uses a procedure defined by the protocol to make a decision on whether or not to send. ALOHA Carrier Sense Multiple Access Carrier Sense Multiple Access with Collision Detection Carrier Sense Multiple Access with Collision Avoidance Topics discussed in this section: 5

HANNAM UNIVERSITY 임의접속 (Random Access)  임의접속 ■ 각 지국은 다른 어느 지국에 의해 제어받지 않은 매체 접근 권리 를 가지고 있음  충돌을 피하기 위한 절차 ■ 언제 지국이 매체에 접속할 수 있는가 ? ■ 만약 매체가 사용된다면 지국은 무엇을 할 수 있는가 ? ■ 어떤 방법으로 지국은 전송의 실패와 성공을 파악할 수 있는가 ? ■ 만약 매체 충돌이 발생한다면 지국은 무엇을 할 수 있는가 ? 6

HANNAM UNIVERSITY 임의접속 (Random Access)( 계속 )  임의접속 방법의 진화 7

HANNAM UNIVERSITY 임의접속 (Random Access)( 계속 )  다중접속 (MA) ■ 최초의 다중접속 방법 ALOHA ■ 9,600bps 를 가진 무선 LAN 에 사용되도록 설계 8

HANNAM UNIVERSITY 임의접속 (Random Access)( 계속 )  ALOHA 규칙 ■ 다중접속 : 송신할 프레임을 가진 모든 지국은 프레임을 송신 ■ 확인응답 : 프레임 전송 후 확인응답을 기다리고 시간 내에 확인 응답을 받지 못하면 프레임을 잃어버렸다고 간주하고 재전송을 시도 9

HANNAM UNIVERSITY Figure 12.3 순수 ALOHA 네트워크에서 프레임 10

HANNAM UNIVERSITY Figure 12.4 순수 ALOHA Protocol 의 절차 11

HANNAM UNIVERSITY The stations on a wireless ALOHA network are a maximum of 600 km apart. If we assume that signals propagate at 3 × 10 8 m/s, we find T p = (600 × 10 5 ) / (3 × 10 8 ) = 2 ms. Now we can find the value of T B for different values of K. a. For K = 1, the range is {0, 1}. The station needs to| generate a random number with a value of 0 or 1. This means that T B is either 0 ms (0 × 2) or 2 ms (1 × 2), based on the outcome of the random variable. Example

HANNAM UNIVERSITY b. For K = 2, the range is {0, 1, 2, 3}. This means that T B can be 0, 2, 4, or 6 ms, based on the outcome of the random variable. c. For K = 3, the range is {0, 1, 2, 3, 4, 5, 6, 7}. This means that T B can be 0, 2, 4,..., 14 ms, based on the outcome of the random variable. d. We need to mention that if K > 10, it is normally set to 10. Example 12.1(continued) 13

HANNAM UNIVERSITY Figure 12.5 Vulnerable time for pure ALOHA protocol 14

HANNAM UNIVERSITY A pure ALOHA network transmits 200-bit frames on a shared channel of 200 kbps. What is the requirement to make this frame collision-free? Solution Average frame transmission time T fr is 200 bits/200 kbps or 1 ms. The vulnerable time is 2 × 1 ms = 2 ms. This means no station should send later than 1 ms before this station starts transmission and no station should start sending during the one 1-ms period that this station is sending. Example

HANNAM UNIVERSITY The throughput for pure ALOHA is S = G × e −2G. The maximum throughput S max = when G= (1/2). Note 16

HANNAM UNIVERSITY A pure ALOHA network transmits 200-bit frames on a shared channel of 200 kbps. What is the throughput if the system (all stations together) produces a frames per second b. 500 frames per second c. 250 frames per second. Solution The frame transmission time is 200/200 kbps or 1 ms. a. If the system creates 1000 frames per second, this is 1 frame per millisecond. The load is 1. In this case S = G× e −2 G or S = (13.5 percent). This means that the throughput is 1000 × = 135 frames. Only 135 frames out of 1000 will probably survive. Example

HANNAM UNIVERSITY b. If the system creates 500 frames per second, this is (1/2) frame per millisecond. The load is (1/2). In this case S = G × e −2G or S = (18.4 percent). This means that the throughput is 500 × = 92 and that only 92 frames out of 500 will probably survive. Note that this is the maximum throughput case, percentagewise. c. If the system creates 250 frames per second, this is (1/4) frame per millisecond. The load is (1/4). In this case S = G × e − 2G or S = (15.2 percent). This means that the throughput is 250 × = 38. Only 38 frames out of 250 will probably survive. Example 12.3(continued) 18

HANNAM UNIVERSITY Figure 12.6 Frames in a slotted ALOHA network 19

HANNAM UNIVERSITY The throughput for slotted ALOHA is S = G × e −G. The maximum throughput S max = when G = 1. Note 20

HANNAM UNIVERSITY Figure 12.7 Vulnerable time for slotted ALOHA protocol 21

HANNAM UNIVERSITY A slotted ALOHA network transmits 200-bit frames on a shared channel of 200 kbps. What is the throughput if the system (all stations together) produces a frames per second b. 500 frames per second c. 250 frames per second. Solution The frame transmission time is 200/200 kbps or 1 ms. a. If the system creates 1000 frames per second, this is 1 frame per millisecond. The load is 1. In this case S = G× e −G or S = (36.8 percent). This means that the throughput is 1000 × = 368 frames. Only 386 frames out of 1000 will probably survive. Example

HANNAM UNIVERSITY b. If the system creates 500 frames per second, this is (1/2) frame per millisecond. The load is (1/2). In this case S = G × e −G or S = (30.3 percent). This means that the throughput is 500 × = 151. Only 151 frames out of 500 will probably survive. c. If the system creates 250 frames per second, this is (1/4) frame per millisecond. The load is (1/4). In this case S = G × e −G or S = (19.5 percent). This means that the throughput is 250 × = 49. Only 49 frames out of 250 will probably survive. Example 12.4(continued) 23

HANNAM UNIVERSITY 임의접속 (Random Access)( 계속 )  반송파 감지 다중접속 (CSMA; Carrier Sense Multiple Access) ■ 충돌 가능성을 줄이기 위해 개발 ■ 각 지국은 전송 전 매체의 상태를 점검 ■ 충돌 가능성을 줄일수는 있지만 제거는 할 수 없음 ■ 전파지연 때문에 출동 가능성은 존재 24

HANNAM UNIVERSITY Figure 12.8 CSMA 에서 충돌의 시간 / 공간적인 방법 25

HANNAM UNIVERSITY Figure 12.9 CSMA 에서 취약 시간 26

HANNAM UNIVERSITY 임의접속 (Random Access)( 계속 )  CSMA 에서의 충돌 27

HANNAM UNIVERSITY 임의접속 (Random Access)( 계속 )  지속성 전략 (persistence strategy) ■ 매체가 사용될 때 지국의 진행 절차를 정의 ■ 비지속성 ■ 지속성 28

HANNAM UNIVERSITY 임의접속 (Random Access)( 계속 )  비지속성 전략 (nonpersistent strategy) ■ 회선을 감지해서 사용중이 아니면 즉시 전송 ■ 사용 중이면 임의의 시간 동안 대기하다 다시 회선을 감지  지속성 전략 (persistent strategy) ■ 회선을 감지해서 사용중이 아니면 즉시 전송 ■ 1- 지속성 : 회선이 사용중이 아니면 1 의 확률로 프레임을 전송 ■ P- 지속성 : 회선이 사용중이 아니면 p 의 확률로 전송하거나 1-p 의 확률로 전송하지 않음  P 가 0.2 일때 회선이 사용중이 아니면 0.2( 시간의 20%) 확률로 전 송하고 0.8( 시간의 80%) 의 확률로 전송을 중단 29

HANNAM UNIVERSITY Figure Behavior of three persistence methods 30

HANNAM UNIVERSITY Figure Flow diagram for three persistence methods 31

HANNAM UNIVERSITY 임의접속 (Random Access)( 계속 )  충돌 검출 반송파 감지 다중 접송 (CSMA/CD; Carrier sense multiple access with collision detection) ■ 충돌을 처리하는 절차를 더함 ■ 충돌 발생시 재전송을 요구 ■ 두번 째 충돌을 줄이기 위해 대기 ■ 지속적인 백오프 방법에서 대기 시간  0 과 2 N × 최대전송시간 사이만큼 대기 (N: 전송 시도 회수 ) 32

HANNAM UNIVERSITY Figure Collision of the first bit in CSMA/CD 33

HANNAM UNIVERSITY Figure Collision and abortion in CSMA/CD 34

HANNAM UNIVERSITY A network using CSMA/CD has a bandwidth of 10 Mbps. If the maximum propagation time (including the delays in the devices and ignoring the time needed to send a jamming signal, as we see later) is 25.6 μs, what is the minimum size of the frame? Solution The frame transmission time is T fr = 2 × T p = 51.2 μs. This means, in the worst case, a station needs to transmit for a period of 51.2 μs to detect the collision. The minimum size of the frame is 10 Mbps × 51.2 μs = 512 bits or 64 bytes. This is actually the minimum size of the frame for Standard Ethernet. Example

HANNAM UNIVERSITY Figure Flow diagram for the CSMA/CD 36

HANNAM UNIVERSITY Figure Energy level during transmission, idleness, or collision 37

HANNAM UNIVERSITY Figure Timing in CSMA/CA 38

HANNAM UNIVERSITY In CSMA/CA, the IFS can also be used to define the priority of a station or a frame. Note 39

HANNAM UNIVERSITY In CSMA/CA, if the station finds the channel busy, it does not restart the timer of the contention window; it stops the timer and restarts it when the channel becomes idle. Note 40

HANNAM UNIVERSITY Figure Flow diagram for CSMA/CA 41

HANNAM UNIVERSITY CONTROLLED ACCESS In controlled access, the stations consult one another to find which station has the right to send. A station cannot send unless it has been authorized by other stations. We discuss three popular controlled- access methods. Reservation Polling Token Passing Topics discussed in this section: 42

HANNAM UNIVERSITY 제어 접속 (Controlled Access)  제어접속 (Controlled Access) ■ 어느 지국이 송신 권한을 가지고 있는지 서로 협력 ■ 예약 (Reservation) ■ 폴링 (Polling) ■ 토큰전달 (Token passing) 43

HANNAM UNIVERSITY Figure Reservation access method  예약 (Reservation) ■ 지국은 데이터를 송신하기 전에 예약을 필요로 함 ■ N 개의 지국이 존재하면 N 개의 예약된 미니 슬롯 (mini slot) 들 이 예약 프레임 안에 존재 ■ 예약을 한 지국은 데이터 프레임을 예약 프레임 뒤에 전송 44

HANNAM UNIVERSITY 제어접속 (Controlled Access)( 계속 )  폴링 (Polling) ■ 주국과 종국으로 구성되어져 있는 토폴로지에서 동작 ■ 폴링 (polling) ■ 주국이 데이터 수신을 원할 때 종국 장치에게 문의 ■ 선택 (selection) ■ 주국이 데이터 송신을 원할 때 종국장치에게 알려줌  선택 (selection) ■ 주국이 언제든지 송신할 것이 있을 사용 ■ 예정된 전송을 위해 주국은 종국의 준비 상태에 대한 확인 응답을 대 기 ■ 주국은 전송 예정된 장치의 주소를 한 필드에 포함하고 선택 프레임 (SEL) 을 만들어 전송  폴 ■ 주국이 종국으로부터 전송을 요청하는데 사용 45

HANNAM UNIVERSITY Figure Select and poll functions in polling access method 46

HANNAM UNIVERSITY 제어접속 (Controlled Access)( 계속 )  토큰 전달 (Token passing) ■ 토큰을 가진 지국이 데이터 송신할 권한을 가짐  토큰 전달 네트워크 47

HANNAM UNIVERSITY Figure Logical ring and physical topology in token-passing access method 48

HANNAM UNIVERSITY 제어접속 (Controlled Access)( 계속 )  토큰 전달 절차 49

HANNAM UNIVERSITY 채널화 (CHANNELIZATION) Channelization is a multiple-access method in which the available bandwidth of a link is shared in time, frequency, or through code, between different stations. In this section, we discuss three channelization protocols. Frequency-Division Multiple Access (FDMA) Time-Division Multiple Access (TDMA) Code-Division Multiple Access (CDMA) Topics discussed in this section: 50

HANNAM UNIVERSITY We see the application of all these methods in Chapter 16 when we discuss cellular phone systems. Note 51

HANNAM UNIVERSITY 채널화 (channelization)( 계속 )  주파수 분할 다중접속 (FDMA) ■ 사용 사능한 대역폭은 모든 지국들에 의해 공유 ■ 각 지국들은 할당된 대역을 사용하여 데이터를 전송 ■ 각각의 대역은 특정 지국을 위해 예약되어 있음 FDMA 에서 대역폭은 채널들로 나누어진다. 52

HANNAM UNIVERSITY Figure Frequency-division multiple access (FDMA) 53

HANNAM UNIVERSITY In FDMA, the available bandwidth of the common channel is divided into bands that are separated by guard bands. Note 54

HANNAM UNIVERSITY 채널화 (channelization)( 계속 )  코드 분할 다중접속 (CDMA) ■ 링크이 전체 대역폭을 하나의 채널에서 점유 ■ 모든 지국들은 시분할 없이 동시에 데이터를 송신 가능 CDMA 는 하나의 채널로 모든 전송을 동시에 운송한다. 55

HANNAM UNIVERSITY Figure Time-division multiple access (TDMA) 56

HANNAM UNIVERSITY In TDMA, the bandwidth is just one channel that is timeshared between different stations. Note 57

HANNAM UNIVERSITY 채널화 (channelization)( 계속 )  코드 분할 다중접속 (CDMA) ■ 링크이 전체 대역폭을 하나의 채널에서 점유 ■ 모든 지국들은 시분할 없이 동시에 데이터를 송신 가능 CDMA 는 하나의 채널로 모든 전송을 동시에 운송한다. 58

HANNAM UNIVERSITY In CDMA, one channel carries all transmissions simultaneously. Note 59

HANNAM UNIVERSITY Figure Simple idea of communication with code 60

HANNAM UNIVERSITY Figure Chip sequences 61

HANNAM UNIVERSITY Figure Data representation in CDMA 62

HANNAM UNIVERSITY Figure Sharing channel in CDMA 63

HANNAM UNIVERSITY Figure Digital signal created by four stations in CDMA 64

HANNAM UNIVERSITY Figure Decoding of the composite signal for one in CDMA 65

HANNAM UNIVERSITY Figure General rule and examples of creating Walsh tables 66

HANNAM UNIVERSITY The number of sequences in a Walsh table needs to be N = 2 m. Note 67

HANNAM UNIVERSITY Find the chips for a network with a. Two stations b. Four stations Solution We can use the rows of W 2 and W 4 in Figure 12.29: a. For a two-station network, we have [+1 +1] and [+1 −1]. b. For a four-station network we have [ ], [+1 −1 +1 −1], [+1 +1 −1 −1], and [+1 −1 −1 +1]. Example

HANNAM UNIVERSITY What is the number of sequences if we have 90 stations in our network? Solution The number of sequences needs to be 2 m. We need to choose m = 7 and N = 2 7 or 128. We can then use 90 of the sequences as the chips. Example

HANNAM UNIVERSITY Prove that a receiving station can get the data sent by a specific sender if it multiplies the entire data on the channel by the sender’s chip code and then divides it by the number of stations. Solution Let us prove this for the first station, using our previous four-station example. We can say that the data on the channel D = (d 1 ⋅ c 1 + d 2 ⋅ c 2 + d 3 ⋅ c 3 + d 4 ⋅ c 4 ). The receiver which wants to get the data sent by station 1 multiplies these data by c 1. Example

HANNAM UNIVERSITY When we divide the result by N, we get d 1. Example 12.8(continued) 71

HANNAM UNIVERSITY 요약 12.4 요약 72