Tcp ipChap 05

8/12/2019 Tcp ipChap 05

1/113

TCP/IP Protocol Suite 1Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Chapter 5

IPv4 Addresses

8/12/2019 Tcp ipChap 05

2/113

TCP/IP Protocol Suite 2

OBJECTIVES: To introduce the concept of an address space in general and the

address space of IPv4 in particular. To discuss the classful architecture and the blocks of addresses

available in each class.

To discuss the idea of hierarchical addressing and how it has

been implemented in classful addressing. To explain subnetting and supernetting for classful architecture.

To discuss classless addressing, that has been devised to solve the

problems in classful addressing.

To discuss some special blocks and some special addresses ineach block.

To discuss NAT technology and show how it can be used to

alleviate of address depletion.

8/12/2019 Tcp ipChap 05

3/113


Chapter

Outline

5 1 Introduction

5 2 Classful Addressing5 3 Classless Addressing

5 4 Special Addresses

5 5 NAT

8/12/2019 Tcp ipChap 05

4/113


5-1 INTRODUCTION

The identifier used in the IP layer of the TCP/IPprotocol suite to identify each device connected tothe Internet is called the Internet address or IPaddress. An IPv4 address is a 32-bit address thatuniquely and universally defines the connection of ahost or a router to the Internet; an IP address is theaddress of the interface.

8/12/2019 Tcp ipChap 05

5/113


Topics Discussed in the Section

Notation

Range of Addresses

Operations

8/12/2019 Tcp ipChap 05

6/113


An IPv4 add ress is 32 bits long.

Note

The IPv4 add resses are un iqueand un iversal.

Note

8/12/2019 Tcp ipChap 05

7/113


The address space of IPv4 is

232or 4,294,967,296.

Note

Numbers in base 2, 16, and 256 arediscussed in Appendix B .

Note

8/12/2019 Tcp ipChap 05

8/113


Figure 5.1 Dotted-decimal notation

8/12/2019 Tcp ipChap 05

9/113


Change the following IPv4 addresses from binary notation todotted-decimal notation.a.10000001 00001011 00001011 11101111b.11000001 10000011 00011011 11111111c.11100111 11011011 10001011 01101111d.11111001 10011011 11111011 00001111SolutionWe replace each group of 8 bits with its equivalent decimalnumber see Appendix B) and add dots for separation:a.129.11.11.239b.193.131.27.255c.231.219.139.111d.249.155.251.15

Example 5.1

8/12/2019 Tcp ipChap 05

10/113


Change the following IPv4 addresses from dotted-decimalnotation to binary notation.a.111.56.45.78b.221.34.7.82c.241.8.56.12d.75.45.34.78SolutionWe replace each decimal number with its binary equivalent:a.01101111 00111000 00101101 01001110b.11011101 00100010 00000111 01010010c.11110001 00001000 00111000 00001100d.01001011 00101101 00100010 01001110

Example 5.2

8/12/2019 Tcp ipChap 05

11/113

8/12/2019 Tcp ipChap 05

12/113


Change the following IPv4 addresses from binary notation tohexadecimal notation.a.10000001 00001011 00001011 11101111b.11000001 10000011 00011011 11111111SolutionWe replace each group of 4 bits with its hexadecimalequivalent. Note that 0X or 0x) is added at the beginning or thesubscript 16 at the end.a.0X810B0BEF or 810B0BEF16b.0XC1831BFF or C1831BFF16

Example 5.4

8/12/2019 Tcp ipChap 05

13/113


Find the number of addresses in a range if the first address is146.102.29.0 and the last address is 146.102.32.255.SolutionWe can subtract the first address from the last address in base256 see Appendix B). The result is 0.0.3.255 in this base. Tofind the number of addresses in the range in decimal), weconvert this number to base 10 and add 1 to the result..

Example 5.5

8/12/2019 Tcp ipChap 05

14/113


The first address in a range of addresses is 14.11.45.96. If thenumber of addresses in the range is 32, what is the lastaddress?SolutionWe convert the number of addresses minus 1 to base 256,which is 0.0.0.31. We then add it to the first address to get thelast address. Addition is in base 256.

Example 5.6

8/12/2019 Tcp ipChap 05

15/113


Figure 5.2 Bi twise NOT operation

8/12/2019 Tcp ipChap 05

16/113


Example 5.7

8/12/2019 Tcp ipChap 05

17/113


Figure 5.3 Bi twise AND operation

8/12/2019 Tcp ipChap 05

18/113


Example 5.8

8/12/2019 Tcp ipChap 05

19/113


Figure 5.4 Bi twise OR operation

8/12/2019 Tcp ipChap 05

20/113


Example 5.9

8/12/2019 Tcp ipChap 05

21/113


5-2 CLASSFUL ADDRESSING

IP addresses, when started a few decades ago,used the concept of classes. This architecture iscalled classful addressing. In the mid-1990s, a newarchitecture, called classless addressing, wasintroduced that supersedes the original architecture.In this section, we introduce classful addressingbecause it paves the way for understandingclassless addressing and justifies the rationale formoving to the new architecture. Classlessaddressing is discussed in the next section.

8/12/2019 Tcp ipChap 05

22/113



Classes

Classes and Blocks

Two-Level Addressing

Three-Level Addressing: Subnetting

Supernetting

8/12/2019 Tcp ipChap 05

23/113


Figure 5.5 Occupati on of address space

8/12/2019 Tcp ipChap 05

24/113


Figure 5.6 F inding the class of address

8/12/2019 Tcp ipChap 05

25/113


Figure 5.7 F inding the class of an address using continuous checking

1

Class: A

0Start

1

0

Class: B

1

0

Class: C

1

0

Class: D Class: E

E l 5 10

8/12/2019 Tcp ipChap 05

26/113


Find the class of each address:a. 00000001 00001011 00001011 11101111b. 11000001 10000011 00011011 11111111c. 10100111 11011011 10001011 01101111d. 11110011 10011011 11111011 00001111SolutionSee the procedure in Figure 5.7.a.The first bit is 0. This is a class A address.b.The first 2 bits are 1; the third bit is 0. This is a class Caddress.c. The first bit is 1; the second bit is 0. This is a class Baddress.d. The first 4 bits are 1s. This is a class E address.

Example 5.10

E l 5 11

8/12/2019 Tcp ipChap 05

27/113


Find the class of each address:a.227.12.14.87b.193.14.56.22c.14.23.120.8d.252.5.15.111Solutiona. The first byte is 227 between 224 and 239); the class is D.b. The first byte is 193 between 192 and 223); the class is C.c. The first byte is 14 between 0 and 127); the class is A.d. The first byte is 252 between 240 and 255); the class is E.

Example 5.11

Fi 5 8 N tid d h tid

8/12/2019 Tcp ipChap 05

28/113


Figure 5.8 Netid and hostid

Fi 5 9 Bl k i Cl A

8/12/2019 Tcp ipChap 05

29/113


Figure 5.9 Blocks in Class A

8/12/2019 Tcp ipChap 05

30/113


Mill ions of c lass A addresses

are wasted.

Note

Fi 5 10 Bl k i Cl B

8/12/2019 Tcp ipChap 05

31/113


Figure 5.10 Blocks in Class B

8/12/2019 Tcp ipChap 05

32/113


Many c lass B addresses are wasted.

Note

Figure 5 11 Blocks in Class C

8/12/2019 Tcp ipChap 05

33/113


Figure 5.11 Blocks in Class C

8/12/2019 Tcp ipChap 05

34/113


Not so many organizat ions are so smal l

to have a class C block.

Note

Figure 5 12 The single block in Class D

8/12/2019 Tcp ipChap 05

35/113


Figure 5.12 The single block in Class D

8/12/2019 Tcp ipChap 05

36/113


Class D addresses are made of one

block, used for m ul t icast ing .

Note

Figure 5 13 The single block in Class E

8/12/2019 Tcp ipChap 05

37/113


Figure 5.13 The single block in Class E

8/12/2019 Tcp ipChap 05

38/113


The only block of c lass E add resses was

reserved fo r futu re pu rposes.

Note

8/12/2019 Tcp ipChap 05

39/113


The range of add resses allocated to an

organizat ion in classfu l add ress ing

was a block of addresses in

Class A , B , or C.

Note

Figure 5 14 Two-level addressing in classful addressing

8/12/2019 Tcp ipChap 05

40/113


Figure 5.14 Two level addressing in classful addressing

Example 5 12

8/12/2019 Tcp ipChap 05

41/113


Example 5.12

Figure 5.15 I nformation extraction in classful addressing

8/12/2019 Tcp ipChap 05

42/113


Figure 5.15 I nformation extraction in classful addressing

netid

First address

000 ... 0

Example 5 13

8/12/2019 Tcp ipChap 05

43/113


An address in a block is given as 73.22.17.25. Find the numberof addresses in the block, the first address, and the lastaddress.SolutionFigure 5.16 shows a possible configuration of the network thatuses this block.1. The number of addresses in this block is N = 232n =16,777,216.2. To find the first address, we keep the leftmost 8 bits and setthe rightmost 24 bits all to 0s. The first address is

73.0.0.0/8, in which 8 is the value of n.3.To find the last address, we keep the leftmost 8 bits and setthe rightmost 24 bits all to 1s. The last address is73.255.255.255.

Example 5.13

Figure 5.16 Solution to Example 5.13

8/12/2019 Tcp ipChap 05

44/113



Example 5 14

8/12/2019 Tcp ipChap 05

45/113


An address in a block is given as 180.8.17.9. Find the numberof addresses in the block, the first address, and the lastaddress.SolutionFigure 5.17 shows a possible configuration of the network thatuses this block.1. The number of addresses in this block is N = 232n =65,536.2.To find the first address, we keep the leftmost 16 bits and setthe rightmost 16 bits all to 0s. The first address is

18.8.0.0/16, in which 16 is the value of n.3.To find the last address, we keep the leftmost 16 bits and setthe rightmost 16 bits all to 1s. The last address is18.8.255.255.

Example 5.14


8/12/2019 Tcp ipChap 05

46/113


g p

Example 5.15

8/12/2019 Tcp ipChap 05

47/113


An address in a block is given as 200.11.8.45. Find the numberof addresses in the block, the first address, and the lastaddress.SolutionFigure 5.17 shows a possible configuration of the network thatuses this block.1.The number of addresses in this block is N = 232n= 256.2.To find the first address, we keep the leftmost 24 bits and setthe rightmost 8 bits all to 0s. The first address is200.11.8.0/16, in which 24 is the value of n.3.To find the last address, we keep the leftmost 24 bits and setthe rightmost 8 bits all to 1s. The last address is200.11.8.255/16.

Example 5.15


8/12/2019 Tcp ipChap 05

48/113


g p

Figure 5.19 Sample Internet

8/12/2019 Tcp ipChap 05

49/113


g

8/12/2019 Tcp ipChap 05

50/113


The network add ress is the ident i f ier o f a

network.

Note

Figure 5.20 Network addresses

8/12/2019 Tcp ipChap 05

51/113


Figure 5.21 Network mask

8/12/2019 Tcp ipChap 05

52/113


Figure 5.22 F inding a network address using the default mask

8/12/2019 Tcp ipChap 05

53/113


8/12/2019 Tcp ipChap 05

54/113

Example 5.17

8/12/2019 Tcp ipChap 05

55/113


Three-level addressing can be found in the telephone system ifwe think about the local part of a telephone number as anexchange and a subscriber connection:

p

in which 626 is the area code, 358 is the exchange, and 1301 isthe subscriber connection.

Example 5.18

8/12/2019 Tcp ipChap 05

56/113


Figure 5.23 shows a network using class B addresses beforesubnetting. We have just one network with almost 216 hosts.The whole network is connected, through one singleconnection, to one of the routers in the Internet. Note that wehave shown /16 to show the length of the netid class B).

p

Figure 5.23 Example 5.18

8/12/2019 Tcp ipChap 05

57/113


Example 5.19

8/12/2019 Tcp ipChap 05

58/113


Figure 5.24 shows the same network in Figure 5.23 aftersubnetting. The whole network is still connected to the Internetthrough the same router. However, the network has used aprivate router to divide the network into four subnetworks. Therest of the Internet still sees only one network; internally thenetwork is made of four subnetworks. Each subnetwork cannow have almost 214 hosts. The network can belong to auniversity campus with four different schools buildings). Aftersubnetting, each school has its own subnetworks, but still thewhole campus is one network for the rest of the Internet. Notethat /16 and /18 show the length of the netid and subnetids.

p


8/12/2019 Tcp ipChap 05

59/113


Figure 5.25 Network mask and subnetwork mask

8/12/2019 Tcp ipChap 05

60/113


Example 5.20

8/12/2019 Tcp ipChap 05

61/113


In Example 5.19, we divided a class B network into foursubnetworks. The value of n = 16 and the value ofn1= n2= n3= n4= 16 + log24 = 18.This means that the subnet mask has eighteen 1sand fourteen 0s. In other words, the subnet mask is255.255.192.0 which is different from the network mask forclass B 255.255.0.0).

p

Example 5.21

8/12/2019 Tcp ipChap 05

62/113


In Example 5.19, we show that a network is divided into foursubnets. Since one of the addresses in subnet 2 is141.14.120.77, we can find the subnet address as:

p

The values of the first, second, and fourth bytes are calculatedusing the first short cut for AND operation. The value of the thirdbyte is calculated using the second short cut for the ANDoperation.

Figure 5.26 Compari son of subnet, defaul t, and supernet mask

8/12/2019 Tcp ipChap 05

63/113


8/12/2019 Tcp ipChap 05

64/113


5-3 CLASSLESS ADDRESSING

Subnetting and supernetting in classful addressing didnot really solve the address depletion problem. Withthe growth of the Internet, it was clear that a largeraddress space was needed as a long-term solution.Although the long-range solution has already beendevised and is called IPv6, a short-term solution wasalso devised to use the same address space but tochange the distribution of addresses to provide a fairshare to each organization. The short-term solutionstill uses IPv4 addresses, but it is called classlessaddressing.


8/12/2019 Tcp ipChap 05

65/113



VariableLength Blocks

Two-Level Addressing

Block Allocation

Subnetting

Figure 5.27 Variable-length blocks in classless addressing

8/12/2019 Tcp ipChap 05

66/113


8/12/2019 Tcp ipChap 05

67/113


In c lass less addressing , the pref ix

def ines the network and the suff ix

def ines the host.

Note

Figure 5.28 Prefi x and suff ix

8/12/2019 Tcp ipChap 05

68/113


8/12/2019 Tcp ipChap 05

69/113


The pref ix leng th in class less

address ing can be 1 to 32.

Note

Example 5.22

8/12/2019 Tcp ipChap 05

70/113


What is the prefix length and suffix length if the whole Internet isconsidered as one single block with 4,294,967,296 addresses?SolutionIn this case, the prefix length is 0 and the suffix length is 32. All32 bits vary to define 232 = 4,294,967,296 hosts in this singleblock.

Example 5.23

8/12/2019 Tcp ipChap 05

71/113


What is the prefix length and suffix length if the Internet isdivided into 4,294,967,296 blocks and each block has onesingle address?SolutionIn this case, the prefix length for each block is 32 and the suffixlength is 0. All 32 bits are needed to define 232 = 4,294,967,296blocks. The only address in each block is defined by the blockitself.

Example 5.24

8/12/2019 Tcp ipChap 05

72/113


The number of addresses in a block is inversely related to thevalue of the prefix length, n. A small nmeans a larger block; alarge nmeans a small block.

Figure 5.29 Slash notation

8/12/2019 Tcp ipChap 05

73/113


8/12/2019 Tcp ipChap 05

74/113


In class less add ress ing , we need to

know one of the add resses in the block

and the pref ix length to def ine the block .

Note

Example 5.25

8/12/2019 Tcp ipChap 05

75/113


In classless addressing, an address cannot per se define theblock the address belongs to. For example, the address230.8.24.56 can belong to many blocks some of them areshown below with the value of the prefix associated with thatblock:

Example 5.26

8/12/2019 Tcp ipChap 05

76/113


The following addresses are defined using slash notations.a.In the address 12.23.24.78/8, the network mask is 255.0.0.0.The mask has eight 1s and twenty-four 0s. The prefix lengthis 8; the suffix length is 24.b. In the address 130.11.232.156/16, the network mask is255.255.0.0. The mask has sixteen 1s and sixteen 0s.Theprefix length is 16; the suffix length is 16.c. In the address 167.199.170.82/27, the network mask is

255.255.255.224. The mask has twenty-seven 1s and five0s. The prefix length is 27; the suffix length is 5.

Example 5.27

8/12/2019 Tcp ipChap 05

77/113


One of the addresses in a block is 167.199.170.82/27. Find thenumber of addresses in the network, the first address, and thelast address.SolutionThe value of n is 27. The network mask has twenty-seven 1sand five 0s. It is 255.255.255.240.a.The number of addresses in the network is 232 n= 32.b.We use the AND operation to find the first address networkaddress). The first address is 167.199.170.64/27.

Example 5.27 Continued

8/12/2019 Tcp ipChap 05

78/113


c.To find the last address, we first find the complement of thenetwork mask and then OR it with the given address: The lastaddress is 167.199.170.95/27.

Example 5.28

8/12/2019 Tcp ipChap 05

79/113


One of the addresses in a block is 17.63.110.114/24. Find thenumber of addresses, the first address, and the last address inthe block.SolutionThe network mask is 255.255.255.0.a.The number of addresses in the network is 232 24= 256.b. To find the first address, we use the short cut methodsdiscussed early in the chapter. The first address is17.63.110.0/24.


8/12/2019 Tcp ipChap 05

80/113


c.To find the last address, we use the complement of thenetwork mask and the first short cut method wediscussed before. The last address is 17.63.110.255/24.

Example 5.29

8/12/2019 Tcp ipChap 05

81/113


One of the addresses in a block is 110.23.120.14/20. Find thenumber of addresses, the first address, and the last address inthe block.SolutionThe network mask is 255.255.240.0.a.The number of addresses in the network is 232 20= 4096.b. To find the first address, we apply the first short cut tobytes 1, 2, and 4 and the second short cut to byte 3. Thefirst address is 110.23.112.0/20.


8/12/2019 Tcp ipChap 05

82/113


c.To find the last address, we apply the first short cut tobytes 1, 2, and 4 and the second short cut to byte 3. TheOR operation is applied to the complement of the mask.The last address is 110.23.127.255/20.

Example 5.30

8/12/2019 Tcp ipChap 05

83/113


An ISP has requested a block of 1000 addresses. The followingblock is granted.a. Since 1000 is not a power of 2, 1024 addresses aregranted 1024 = 210).b. The prefix length for the block is calculated as n = 32

log21024 = 22.c. The beginning address is chosen as 18.14.12.0 which isdivisible by 1024).The granted block is 18.14.12.0/22. The first address is18.14.12.0/22 and the last address is 18.14.15.255/22.

8/12/2019 Tcp ipChap 05

84/113


Example 5.31

8/12/2019 Tcp ipChap 05

85/113


Assume an organization has given a class A block as 73.0.0.0in the past. If the block is not revoked by the authority, theclassless architecture assumes that the organization has ablock 73.0.0.0/8 in classless addressing.

8/12/2019 Tcp ipChap 05

86/113


The restr ict ion s app l ied in al locat ing

add resses for a subnetwo rk are

parallel to the ones used to al locate

add resses fo r a network .

Note

Example 5.32

8/12/2019 Tcp ipChap 05

87/113


An organization is granted the block 130.34.12.64/26. Theorganization needs four subnetworks, each with an equalnumber of hosts. Design the subnetworks and find theinformation about each network.SolutionThe number of addresses for the whole network can be foundas N = 232 26 = 64. The first address in the network is130.34.12.64/26 and the last address is 130.34.12.127/26. Wenow design the subnetworks:1. We grant 16 addresses for each subnetwork to meet thefirst requirement 64/16 is a power of 2).2. The subnetwork mask for each subnetwork is:


8/12/2019 Tcp ipChap 05

88/113


3.We grant 16 addresses to each subnet starting from thefirst available address. Figure 5.30 shows the subblock foreach subnet. Note that the starting address in eachsubnetwork is divisible by the number of addresses inthat subnetwork.


8/12/2019 Tcp ipChap 05

89/113


Example 5.33

8/12/2019 Tcp ipChap 05

90/113


An organization is granted a block of addresses with thebeginning address 14.24.74.0/24. The organization needs tohave 3 subblocks of addresses to use in its three subnets asshown below:One subblock of 120 addresses.One subblock of 60 addresses.One subblock of 10 addresses.SolutionThere are 23 4= 256 addresses in this block. The firstaddress is 14.24.74.0/24; the last address is 14.24.74.255/24.a.The number of addresses in the first subblock is not a

power of 2. We allocate 128 addresses. The subnetmask is 25. The first address is 14.24.74.0/25; the lastaddress is 14.24.74.127/25.


8/12/2019 Tcp ipChap 05

91/113


b.The number of addresses in the second subblock is not apower of 2 either. We allocate 64 addresses. The subnetmask is 26. The first address in this block is14.24.74.128/26; the last address is 14.24.74.191/26.c.The number of addresses in the third subblock is not apower of 2 either. We allocate 16 addresses. The subnetmask is 28. The first address in this block is14.24.74.192/28; the last address is 14.24.74.207/28.d.If we add all addresses in the previous subblocks, theresult is 208 addresses, which means 48 addresses are leftin reserve. The first address in this range is 14.24.74.209.The last address is 14.24.74.255.e.Figure 5.31 shows the configuration of blocks. We haveshown the first address in each block.


8/12/2019 Tcp ipChap 05

92/113


Example 5.34

8/12/2019 Tcp ipChap 05

93/113


Assume a company has three offices: Central, East, and West.The Central office is connected to the East and West offices viaprivate, WAN lines. The company is granted a block of 64addresses with the beginning address 70.12.100.128/26. Themanagement has decided to allocate 32 addresses for theCentral office and divides the rest of addresses between the twoother offices.1.The number of addresses are assigned as follows:

2. We can find the prefix length for each subnetwork:


8/12/2019 Tcp ipChap 05

94/113


3. Figure 5.32 shows the configuration designed by themanagement. The Central office uses addresses70.12.100.128/27 to 70.12.100.159/27. The company has usedthree of these addresses for the routers and has reserved thelast address in the subblock. The East officeuses the addresses 70.12.100.160/28 to 70.12.100.175/28. Oneof these addresses is used for the router and the company hasreserved the last address in the subblock. The West office usesthe addresses 70.12.100.160/28 to 70.12.100.175/28. One ofthese addresses is used for the router and the company hasreserved the last address in the subblock. The company usesno address for the point-to-point connections in WANs.


8/12/2019 Tcp ipChap 05

95/113


Example 5.35

8/12/2019 Tcp ipChap 05

96/113


An ISP is granted a block of addresses starting with190.100.0.0/16 65,536 addresses). The ISP needs to distributethese addresses to three groups of customers as follows:The first group has 64 customers; each needs approximately256 addresses.The second group has 128 customers; each needsapproximately 128 addresses.The third group has 128 customers; each needs approximately64 addresses.We design the subblocks and find out how many addresses arestill available after these allocations.


8/12/2019 Tcp ipChap 05

97/113


SolutionLet us solve the problem in two steps. In the first step, weallocate a subblock of addresses to each group. The totalnumber of addresses allocated to each group and the prefixlength for each subblock can found as

Figure 5.33 shows the design for the first hierarchical level.Figure 5.34 shows the second level of the hierarchy. Note thatwe have used the first address for each customer as the subnetaddress and have reserved the last address as a specialaddress.

Figure 5.33 Solution to Example 5.35: f irst step

8/12/2019 Tcp ipChap 05

98/113


Figure 5.34 Solution to Example 5.35: second step

8/12/2019 Tcp ipChap 05

99/113


5-4 SPECIAL ADDRESSES

8/12/2019 Tcp ipChap 05

100/113


5 4 SPECIAL ADDRESSES

In classful addressing some addresses werereserved for special purposes. The classlessaddressing scheme inherits some of these specialaddresses from classful addressing.


8/12/2019 Tcp ipChap 05

101/113


Special Blocks

Special Addresses in each Block

Figure 5.35 Example of using the all-zero address

8/12/2019 Tcp ipChap 05

102/113


Source: 0.0.0.0Destination: 255.255.255.255

Packet

Figure 5.36 Example of l imi ted broadcast address

8/12/2019 Tcp ipChap 05

103/113


221.45.71.20/24 221.45.71.178/24

221.45.71.64/24 221.45.71.126/24

Network

Figure 5.37 Example of loopback address

8/12/2019 Tcp ipChap 05

104/113


Transport layer

Application layer

Network layer

Process 1 Process 2

Destination address:127.x.y.z

Packet

8/12/2019 Tcp ipChap 05

105/113


Figure 5.38 Example of a directed broadcast address

8/12/2019 Tcp ipChap 05

106/113


221.45.71.0/24

221.45.71.20/24 221.45.71.178/24

221.45.71.64/24 221.45.71.126/24

Network:

Packet

5-5 NAT

8/12/2019 Tcp ipChap 05

107/113


The distribution of addresses through ISPs hascreated a new problem. If the business grows or thehousehold needs a larger range, the ISP may not beable to grant the demand because the addressesbefore and after the range may have already beenallocated to other networks. In most situations,however, only a portion of computers in a smallnetwork need access to the Internet simultaneously.A technology that can help in this cases is networkaddress translation NAT).


8/12/2019 Tcp ipChap 05

108/113


Address Translation

Translation Table

Figure 5.39 NAT

8/12/2019 Tcp ipChap 05

109/113


Figure 5.40 Address resolution

8/12/2019 Tcp ipChap 05

110/113


Internet

Site using private addresses

172.18.3.1

172.18.3.2

172.18.3.20

Source: 172.18.3.1 Source: 200.24.5.8

Destination: 200.24.5.8Destination: 172.18.3.1

Figure 5.41 Translation

8/12/2019 Tcp ipChap 05

111/113


NAT Table with only IP addresses

8/12/2019 Tcp ipChap 05

112/113


If using only one global address

Only one private-network host to access the sameexternal host

If using a pool of global addresses (e.g. 4 addr)

No more than 4 connections can be made to thesame destination

No private-network host can access two externalserver programs (e.g. HTTP and TELNET) at the sametime???

Two private-network hosts cannot access the sameexternal server program at the same time (by usingthe same global address)

8/12/2019 Tcp ipChap 05

113/113

Tcp ipChap 05

Documents

Transcript of Tcp ipChap 05