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1 경북대학교컴퓨터공학과 1 제 5 장서브넷팅과슈퍼넷팅 서브넷팅 (subnetting) 슈퍼넷팅 (Supernetting)

2 경북대학교컴퓨터공학과 2 서브넷팅과슈퍼넷팅 서브넷팅 (subnetting) 하나의네트워크를여러개의서브넷 (subnet) 으로분할 슈퍼넷팅 (supernetting) 여러개의서브넷주소를결합 The idea of subnetting and supernetting of classful addresses is almost obsolete.

3 경북대학교컴퓨터공학과 3 2-단계계층구조 Classful addressing : netid + hostid 먼저 netid 를사용하여네트워크에도달한후에 hostid 를사용하여호스트에도달 A, B, C 클래스는 2 단계계층구조

4 경북대학교컴퓨터공학과 4 3-단계계층구조 3 단계이상의계층구조를위하여서브넷팅활용

5 3-단계계층구조 경북대학교컴퓨터공학과 5

6 서브넷마스크 경북대학교컴퓨터공학과 6

7 경북대학교컴퓨터공학과 7 Example 15 What is the subnetwork address if the destination address is and the subnet mask is ? Solution Apply the AND operation on the address and the subnet mask. Address Subnet Mask Subnetwork Address

8 서브넷의특수주소 경북대학교컴퓨터공학과 8

9 Example B 클래스주소를가진기관이 12 개의서브넷이필요할때 12 개지만 14 개필요 (subnetid 필드가모두 1 인것과 0 인것 ) 서브넷할당을위한최소비트수 : 4 개 나머지 12 개비트는 hostid (2 12 = 4,096) 지정, 실제는 4,094 개 Mask = (240 = ) 서브넷 X.Y.0000hhhh.hhhhhhhh (X.Y.0.0) ~ X.Y.1111hhhh.hhhhhhhh (X.Y.240.0) 경북대학교컴퓨터공학과 9

10 Example 경북대학교컴퓨터공학과 10

11 경북대학교컴퓨터공학과 11 Example C 클래스주소를가진기관이 5 개의서브네트워크를필요할때 5 개지만 7 개필요 (subnetid 가모두 1 인것과 0 인것포함 ) 서브넷할당을위한최소비트수 : 3 개 (2 2 < 7 <2 3 ) 나머지 5 개의비트 hostid(2 5 = 32 개 ), 실제는 30 개호스트지정가 능 Mask = (224 = ) 서브넷 X.Y.Z.000hhhhh (X.Y.Z.0) ~ X.Y.Z.111hhhhh (X.Y.Z.224)

12 Example 경북대학교컴퓨터공학과 12

13 경북대학교컴퓨터공학과 13 가변길이서브넷팅 예 : C 클래스주소를허가받고각각 60,60,60,30,30 개 의호스트를갖는 5 개의서브넷이필요한사이트 subnetid : 2 비트할당 = 62 개호스트허용 subnetid : 3 비트할당 = 30 개호스트허용 불가능 해결책 - 가변길이서브넷팅 하나의마스크를적용한후에다른마스크를적용 마스크를 3 개의서브넷에적용 ( ) 마스크를다시 2 개의서브넷에적용 ( )

14 경북대학교컴퓨터공학과 14 가변길이서브넷팅 ( 계속 ) 서브넷필드 =00 서브넷필드 =01 서브넷필드 =10 서브넷필드 =11 2 차서브넷필드 =0 2 차서브넷필드 =1

15 슈퍼넷팅 경북대학교컴퓨터공학과 15

16 Supernet mask 경북대학교컴퓨터공학과 16

17 경북대학교컴퓨터공학과 17 Example Q: We need to make a supernetwork out of 16 class C blocks. What is the supernet mask? A: 16 개의블록필요. 마스크 = ( )

18 경북대학교컴퓨터공학과 18 Example Q: A supernet has a first address of and a supernet mask of A router receives three packets with the following destination addresses: Which packet belongs to the supernet? A: We apply the supernet mask to see if we can find the beginning address. DA = AND DA = AND DA = AND (o) (x) (x) IP packet Mask= Router Supernet

19 경북대학교컴퓨터공학과 19 Example Q: A supernet has a first address of and a supernet mask of How many blocks are in this supernet and what is the range of addresses? A: Supernet mask : = Beginning address : = So, 2 3 or 8 blocks in this supernet. The blocks are to The range is ~

20 경북대학교컴퓨터공학과 20 Example 예 : 1,000 개의주소가필요한기관 ( 시작주소 =X.Y.32.0) 슈퍼넷마스크 (252= ) 을이용하여 4 개의 C 클래스주소결합 X.Y , X.Y , X.Y , X.Y => X.Y.32.0, X.Y.33.0, X.Y.34.0, X.Y.35.0 패킷목적지주소에슈퍼넷마스크적용 최하위주소가같다면슈퍼넷에속한다

21 Example 경북대학교컴퓨터공학과 21

22 경북대학교컴퓨터공학과 22 Example 12 An organization is granted the block /26. The organization needs 4 subnets. What is the subnet prefix length? Solution /26= 비트의 suffix => 블록내전체주소수 = 2 6 = 64 각서브넷은 16개의호스트주소지정 => subnet prefix = /28 Subnet 1: /28~ /28. ( hhhh) Subnet 2 : /28~ /28. ( hhhh) Subnet 3: /28~ /28. ( hhhh) Subnet 4: /28~ /28. ( hhhh)

23 경북대학교컴퓨터공학과 23

24 경북대학교컴퓨터공학과 24 Example 14 An organization is granted a block of addresses with the beginning address /24. There are = 256 addresses in this block. The organization needs to have 11 subnets as shown below: a. two subnets, each with 64 addresses. b. two subnets, each with 32 addresses. c. three subnets, each with 16 addresses. d. four subnets, each with 4 addresses. Design the subnets.

25 Example 14 (Continuted) 1. We use the first 128 addresses for the first two subnets, each with 64 addresses. Mask for each network is / sshhhhhh (ss=00,01) /26~ / /26~ /26 2. We use the next 64 addresses for the next two subnets, each with 32 addresses. Mask for each network is / ssshhhhh (sss=100,101) /27~ / /27~ /27 3. We use the next 48 addresses for the next three subnets, each with 16 addresses. Mask for each network is / sssshhhh (ssss=1100,1101,1110) /28~ / /28~ / /28~ /28 4. We use the last 16 addresses for the last four subnets, each with 4 addresses. Mask for each network is / sssssshh (ssssss=111100,111101,111110,111111) /30~ / /30~ / /30~ / /30~ / 경북대학교컴퓨터공학과 25

26 경북대학교컴퓨터공학과 26

27 경북대학교컴퓨터공학과 27 Example 15 A company has three offices: Central, East, and West. The Central office is connected to the East and West offices via private, point-to-point WAN lines. The company is granted a block of 64 addresses with the beginning address /26. The management has decided to allocate 32 addresses for the Central office and divides the rest of addresses between the two offices. R West R R Central R R East 64 addresses /26

28 경북대학교컴퓨터공학과 28 Example 15 (Continued) The company will have three subnets, one at Central, one at East, and one at West. a. The Central office uses the network address /27 ( /27). The addresses in this subnet are /27 to /27. (32 addresses) Note that three of these addresses are used for the routers and the company has reserved the last address in the sub-block. Note that the interface of the router that connects the Central subnet to the WAN needs no address because it is a point-to-point connection.

29 경북대학교컴퓨터공학과 29 Example 15 (Continued) b. The West office uses the network address /28 ( /28). The addresses in this subnet are /28 to /28. (16 addresses) Note that one of these addresses is used for the router and the company has reserved the last address in the sub-block. c. The East office uses the network address /28. ( /28) The addresses in this subnet are /28 to /28. (16 addresses) Note that one of these addresses is used for the router and the company has reserved the last address in the sub-block.

30 경북대학교컴퓨터공학과 30

31 경북대학교컴퓨터공학과 31 Example 16 An ISP is granted a block of addresses starting with /16 (65,536 addresses). The ISP needs to distribute these addresses to three groups of customers as follows: a. The first group has 64 customers; each needs 256 addresses. b. The second group has 128 customers; each needs 128 addresses c. The third group has 128 customers; each needs 64 addresses. Design the subblocks and find out how many addresses are still available after these allocations.

32 경북대학교컴퓨터공학과 32 Group 1: 64 개의서브넷, 서브넷당 256 개의주소필요 (8 비트의 suffix 와 24 비트의 prefix) 01 : /24 ~ /24 02 : /24 ~ / : /24 ~ /24 Total = = 16,384 Group 2: 128 개의서브넷, 서브넷당 128 개의주소필요 (7 비트의 suffix 와 25 비트의 prefix) 001 : /25 ~ / : /25 ~ / : /25 ~ /25 Total = = 16,384

33 경북대학교컴퓨터공학과 33 Group 3: 128 개의서브넷, 서브넷당 64 개의주소필요 (6 비트의 suffix, 26 비트의 prefix) 001 : /26 ~ / : /26 ~ / : /26 ~ /26 Total = = 8,192 Number of granted addresses: 65,536 Number of allocated addresses: 40,960 Number of available addresses: 24,576

34 경북대학교컴퓨터공학과 34

Microsoft PowerPoint - 06-IPAddress [호환 모드]

Microsoft PowerPoint - 06-IPAddress [호환 모드] Chapter 06 IP Address IP Address Internet address IP 계층에서사용되는식별자 32 bit 2 진주소 The address space of IPv4 is 2 32 or 4,294,967,296 netid 와 hostid 로구분 인터넷에서호스트와라우터를유일하게구분 IP Address Structure 2-Layer Hierarchical