Download - Chap 05 Network Layer
-
8/2/2019 Chap 05 Network Layer
1/104
TCP/IP Protocol Suite 1
Chapter
Outline
5.1 Introduction
5.2 Classful Addressing5.3 Classless Addressing
5.4 Special Addresses
5.5 NAT
-
8/2/2019 Chap 05 Network Layer
2/104
TCP/IP Protocol Suite 2
An IPv4 address is 32 bits long.
Note
The IPv4 addresses are uniqueand universal.
Note
-
8/2/2019 Chap 05 Network Layer
3/104
TCP/IP Protocol Suite 3
The address space of IPv4 is232or 4,294,967,296.
Note
Numbers in base 2, 16, and 256 arediscussed in Appendix B.
Note
-
8/2/2019 Chap 05 Network Layer
4/104
TCP/IP Protocol Suite 4
Binary to Decimal Conversion
Bit 7 6 5 4 3 2 1 0
27= 26= 25= 24= 23= 22= 21= 20=
Deci 128 64 32 16 8 4 2 1
Binary 1 0 1 0 1 1 1 0
Deci 128 0 32 0 8 4 2 0
Binary (10101110)2 = (128+32+8+4+2)10
Decimal = 174
Example: Convert the binary number (10101110)2 into decimalequivalent
-
8/2/2019 Chap 05 Network Layer
5/104
TCP/IP Protocol Suite 5
Figure 5.1 Dotted-decimal notation
IPv4Least Address= 00000000. 00000000. 00000000. 00000000 = 0.0.0.0
Max Addr = 11111111 .11111111. 11111111 .11111111 = 255.255.255.255
Number of address = 232 = 4,294,967,296
Or Address Space
-
8/2/2019 Chap 05 Network Layer
6/104
TCP/IP Protocol Suite 6
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.Solution
Last address 146 . 102 . 32 . 255 (-)First Address 146 . 102 . 29 . 0
--------------------------
0 . 0 . 3 . 255
--------------------------
Example 5.5
-
8/2/2019 Chap 05 Network Layer
7/104
TCP/IP Protocol Suite 7
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
-
8/2/2019 Chap 05 Network Layer
8/104
TCP/IP Protocol Suite 8
Figure 5.2 Bitwise NOT operation
-
8/2/2019 Chap 05 Network Layer
9/104
TCP/IP Protocol Suite 9
Example 5.7
-
8/2/2019 Chap 05 Network Layer
10/104
TCP/IP Protocol Suite 10
Figure 5.3 Bitwise AND operation
-
8/2/2019 Chap 05 Network Layer
11/104
TCP/IP Protocol Suite 11
Example 5.8
Shortcut#1 for AND operation
When at least one of the number is
0 or 255selects the smaller byte
or one of them if equal
Shortcut#2 for AND operation
When none of the two bytes iseither 0 or 255,
write each byte as the sum of
eight terms, where each term is a
power of 2.
select the smaller term in each (or
one of them if equal)
and add them to get the result.
-
8/2/2019 Chap 05 Network Layer
12/104
TCP/IP Protocol Suite 12
Figure 5.4 Bitwise OR operation
-
8/2/2019 Chap 05 Network Layer
13/104
TCP/IP Protocol Suite 13
Example 5.9
Shortcut#1 for OR operation
When at least one of the number is
0 or 255selects the Largest byte
or one of them if equal
Shortcut#2 for OR operation
When none of the two bytes is
either 0 or 255,
write each byte as the sum ofeight terms, where each term is a
power of 2.
select the larger term in each (or
one of them if equal)
and add them to get the result.
-
8/2/2019 Chap 05 Network Layer
14/104
TCP/IP Protocol Suite 14
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.
-
8/2/2019 Chap 05 Network Layer
15/104
TCP/IP Protocol Suite 15
Figure 5.5 Occupation of address space
-
8/2/2019 Chap 05 Network Layer
16/104
TCP/IP Protocol Suite 16
Figure 5.6 Finding the class of address
-
8/2/2019 Chap 05 Network Layer
17/104
TCP/IP Protocol Suite 17
Figure 5.7 Finding 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
-
8/2/2019 Chap 05 Network Layer
18/104
TCP/IP Protocol Suite 18
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
-
8/2/2019 Chap 05 Network Layer
19/104
TCP/IP Protocol Suite 19
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
-
8/2/2019 Chap 05 Network Layer
20/104
TCP/IP Protocol Suite 20
Figure 5.8 Netid and hostid
-
8/2/2019 Chap 05 Network Layer
21/104
TCP/IP Protocol Suite 21
Figure 5.9 Blocks in Class A
27 =
0
-
8/2/2019 Chap 05 Network Layer
22/104
-
8/2/2019 Chap 05 Network Layer
23/104
TCP/IP Protocol Suite 23
Figure 5.10 Blocks in Class B
214=
10
-
8/2/2019 Chap 05 Network Layer
24/104
TCP/IP Protocol Suite 24
Many class B addresses are wasted.
Note
-
8/2/2019 Chap 05 Network Layer
25/104
TCP/IP Protocol Suite 25
Figure 5.11 Blocks in Class C
110
221=
-
8/2/2019 Chap 05 Network Layer
26/104
TCP/IP Protocol Suite 26
Not so many organizations are so smallto have a class C block.
Note
Fi 5 12 Th i l bl k i Cl D
-
8/2/2019 Chap 05 Network Layer
27/104
TCP/IP Protocol Suite 27
Figure 5.12 The single block in Class D
-
8/2/2019 Chap 05 Network Layer
28/104
TCP/IP Protocol Suite 28
Class D addresses are made of oneblock, used for multicasting.
Note
Fi 5 13 Th i l bl k i Cl E
-
8/2/2019 Chap 05 Network Layer
29/104
TCP/IP Protocol Suite 29
Figure 5.13 The single block in Class E
-
8/2/2019 Chap 05 Network Layer
30/104
-
8/2/2019 Chap 05 Network Layer
31/104
TCP/IP Protocol Suite 31
The range of addresses allocated to anorganization in classful addressing
was a block of addresses in
Class A, B, or C.
Note
Fig re 5 14 T l l add ssi g i lassf l add ssi g
-
8/2/2019 Chap 05 Network Layer
32/104
TCP/IP Protocol Suite 32
Figure 5.14 Two-level addressing in classful addressing
Figure 5 15 Information extraction in classful addressing
-
8/2/2019 Chap 05 Network Layer
33/104
TCP/IP Protocol Suite 33
Figure 5.15 Information extraction in classful addressing
netid
First address
000 ... 0
Example 5 13
-
8/2/2019 Chap 05 Network Layer
34/104
TCP/IP Protocol Suite 34
An address in a block is given as 73.22.17.25. Find the number of
addresses in the block, the first address, and the last address.
Solution
73.22.17.15 belongs class B with n=8
1. The number of addresses in this block is
N = 232n = 232
8 =224 = 16,777,216.2. To find the First address,
we keep the leftmost 8 bits and
set the rightmost 24 bits all to 0s.
The first address is 3.0.0.0/8, in which 8 is the value ofn.
3. To find the last address, we keep the leftmost 8 bits and
set the rightmost 24 bits all to 1s.
The last address is 73.255.255.255.
Example 5.13
-
8/2/2019 Chap 05 Network Layer
35/104
-
8/2/2019 Chap 05 Network Layer
36/104
Figure 5 17 Solution to Example 5 14
-
8/2/2019 Chap 05 Network Layer
37/104
TCP/IP Protocol Suite 37
Figure 5.17 Solution to Example 5.14
Example 5 15
-
8/2/2019 Chap 05 Network Layer
38/104
TCP/IP Protocol Suite 38
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 is200.11.8.0/24, 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 is200.11.8.255/24.
Example 5.15
-
8/2/2019 Chap 05 Network Layer
39/104
Figure 5 19 Sample Internet
-
8/2/2019 Chap 05 Network Layer
40/104
TCP/IP Protocol Suite 40
Figure 5.19 Sample Internet
-
8/2/2019 Chap 05 Network Layer
41/104
TCP/IP Protocol Suite 41
The network address is the identifier of anetwork.
Note
Figure 5.20 Network addresses
-
8/2/2019 Chap 05 Network Layer
42/104
TCP/IP Protocol Suite 42
Figure 5.20 Network addresses
Figure 5.21 Network mask
-
8/2/2019 Chap 05 Network Layer
43/104
TCP/IP Protocol Suite 43
Figure 5.21 Network mask
-
8/2/2019 Chap 05 Network Layer
44/104
TCP/IP Protocol Suite 44
Given To find Method
First
Address
and last
address
Number of addresses (Range) Subtract first and last address
Convert the result into
decimal equivalent plus 1.
IP address
(classful
addressing)
Number of addresses = 232-n
First Address = AND operation with
network address
Last Address = we keep the leftmost
mask bits and set the rightmost
remaining all bits to 1s
Identify the class of IP address
Identify the netid of the IP
address
Identify the network
address/mask by placing 1s for
netid and 0s for hostid.
Figure 5.22 Finding a network address using the default mask
-
8/2/2019 Chap 05 Network Layer
45/104
TCP/IP Protocol Suite 45
g g g f
Example 5 16
-
8/2/2019 Chap 05 Network Layer
46/104
TCP/IP Protocol Suite 46
A router receives a packet with the destination address201.24.67.32. Show how the router finds the network address ofthe packet.SolutionSince the class of the address is B, we assume that the routerapplies the default mask for class B, 255.255.0.0 to find thenetwork address.
Example 5.16
Example 5.18
-
8/2/2019 Chap 05 Network Layer
47/104
TCP/IP Protocol Suite 47
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).
Example 5.18
-
8/2/2019 Chap 05 Network Layer
48/104
TCP/IP Protocol Suite 48
NetworkSubnetwork
Computer lab # 1 Computer lab # 2
Computer lab # 4Computer lab # 3
Example 5.19
-
8/2/2019 Chap 05 Network Layer
49/104
TCP/IP Protocol Suite 49
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.
Example 5.19
Figure 5.24 Example 5.19
-
8/2/2019 Chap 05 Network Layer
50/104
TCP/IP Protocol Suite 50
Computer lab # 1 Computer lab # 2
Computer lab # 3Computer lab # 4
Figure 5.25 Network mask and subnetwork mask
-
8/2/2019 Chap 05 Network Layer
51/104
TCP/IP Protocol Suite 51
Formula to find the subnet id : nsub = n + log2s
Where, nsub: length of each subnetidn : length of netid
s : number of subnet (must power of 2)
Example 5.20
-
8/2/2019 Chap 05 Network Layer
52/104
TCP/IP Protocol Suite 52
Divided a class B network into four subnetworks.SolutionFormula to find the subnet id : nsub = n + log2s
Where, nsub: length of each subnetid
n : length of netid
s : number of subnet (must power of 2)
The value of n = 16
The value of n1 = n2 = n3 = n4 = 16 + log24 = 18.
This means that the subnet mask has eighteen 1s and fourteen 0s.
In other words, the subnet mask is 255.255.192.0
Different from the network mask for class B (255.255.0.0).
Example 5.20
Example 5.21
-
8/2/2019 Chap 05 Network Layer
53/104
TCP/IP Protocol Suite 53
Since one of the addresses in subnet 2 is 141.14.120.77, wecan find the subnet address as:
Example 5.21
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 Comparison of subnet, default, and supernet mask
-
8/2/2019 Chap 05 Network Layer
54/104
TCP/IP Protocol Suite 54
nsub=n+log2s
nsuper=n-log2c
-
8/2/2019 Chap 05 Network Layer
55/104
TCP/IP Protocol Suite 55
Type To find Method
IP address
(classful
addressing)
Number of addresses = 232-n
First Address = AND operation with
network address
Last Address = we keep the
leftmost mask bits and set the
rightmost remaining all bits to 1s
Identify the class of IP address
Identify the netid of the IP
address
Identify the network
address/mask by placing 1s for
netid and 0s for hostid.
Subnetting
Classful
Addressing
Subnet id Network : n+log2S
S No. of subnets required (mustbe in power of 2)
Identify the class of IP address
Identify the netid of the IP
address (n)
-Sbunetting
-Classless
Addressing
Prefix of network: n+log2(N/Ni)
N: total number of address granted
Ni : number of address per subnet
-
8/2/2019 Chap 05 Network Layer
56/104
TCP/IP Protocol Suite 56
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.
Topics Discussed in the Section
-
8/2/2019 Chap 05 Network Layer
57/104
TCP/IP Protocol Suite 57
Topics Discussed in the Section
Variable Length Blocks
Two-Level Addressing
Block Allocation
Subnetting
Figure 5.27 Variable-length blocks in classless addressing
-
8/2/2019 Chap 05 Network Layer
58/104
TCP/IP Protocol Suite 58
-
8/2/2019 Chap 05 Network Layer
59/104
TCP/IP Protocol Suite 59
In classless addressing, the prefixdefines the network and the suffix
defines the host.
Note
Figure 5.28 Prefix and suffix
-
8/2/2019 Chap 05 Network Layer
60/104
TCP/IP Protocol Suite 60
-
8/2/2019 Chap 05 Network Layer
61/104
TCP/IP Protocol Suite 61
The prefix length in classlessaddressing can be 1 to 32.
Note
Example 5.22
-
8/2/2019 Chap 05 Network Layer
62/104
TCP/IP Protocol Suite 62
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.
p
Example 5.23
-
8/2/2019 Chap 05 Network Layer
63/104
TCP/IP Protocol Suite 63
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.
p
Example 5.24
-
8/2/2019 Chap 05 Network Layer
64/104
TCP/IP Protocol Suite 64
The number of addresses in a block is inversely related to thevalue of the prefix length, n. A small n means a larger block; alarge n means a small block.
Figure 5.29 Slash notation
-
8/2/2019 Chap 05 Network Layer
65/104
TCP/IP Protocol Suite 65
-
8/2/2019 Chap 05 Network Layer
66/104
TCP/IP Protocol Suite 66
In classless addressing, we need toknow one of the addresses in the blockand the prefix length to define the block.
Note
Example 5.25
-
8/2/2019 Chap 05 Network Layer
67/104
TCP/IP Protocol Suite 67
In classless addressing, an address cannot as such 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
-
8/2/2019 Chap 05 Network Layer
68/104
TCP/IP Protocol Suite 68
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
-
8/2/2019 Chap 05 Network Layer
69/104
TCP/IP Protocol Suite 69
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.
167.199.170.64/27
255.255.255.240
167.199.170.82
Example 5.27 Continued
-
8/2/2019 Chap 05 Network Layer
70/104
TCP/IP Protocol Suite 70
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.
-
8/2/2019 Chap 05 Network Layer
71/104
Example 5.28 Continued
-
8/2/2019 Chap 05 Network Layer
72/104
TCP/IP Protocol Suite 72
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
-
8/2/2019 Chap 05 Network Layer
73/104
TCP/IP Protocol Suite 73
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.
Example 5.29 Continued
-
8/2/2019 Chap 05 Network Layer
74/104
TCP/IP Protocol Suite 74
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
-
8/2/2019 Chap 05 Network Layer
75/104
TCP/IP Protocol Suite 75
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.
First address 18.14.12.0/22 18.14.00001100.00000000
Last Address 18.14.15.255/22 18.14.00001111.11111111
-
8/2/2019 Chap 05 Network Layer
76/104
TCP/IP Protocol Suite 76
Example 5.31
-
8/2/2019 Chap 05 Network Layer
77/104
TCP/IP Protocol Suite 77
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.
-
8/2/2019 Chap 05 Network Layer
78/104
TCP/IP Protocol Suite 78
The restrictions applied in allocatingaddresses for a subnetwork are
parallel to the ones used to allocateaddresses for a network.
Note
Example 5.32
-
8/2/2019 Chap 05 Network Layer
79/104
TCP/IP Protocol Suite 79
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 is 130.34.12.64/26 and the lastaddress is 130.34.12.127/26. We now 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:
N: total number of address granted
Ni : number of address per subnet
Example 5.32 Continued
-
8/2/2019 Chap 05 Network Layer
80/104
TCP/IP Protocol Suite 80
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.
Figure 5.30 Solution to Example 5.32
-
8/2/2019 Chap 05 Network Layer
81/104
TCP/IP Protocol Suite 81
Example 5.33
-
8/2/2019 Chap 05 Network Layer
82/104
TCP/IP Protocol Suite 82
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 232 24 = 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.
Example 5.33 Continued
-
8/2/2019 Chap 05 Network Layer
83/104
TCP/IP Protocol Suite 83
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.
Figure 5.31 Solution to Example 5.33
-
8/2/2019 Chap 05 Network Layer
84/104
TCP/IP Protocol Suite 84
Example 5.34
-
8/2/2019 Chap 05 Network Layer
85/104
TCP/IP Protocol Suite 85
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:
Example 5.34 Continued
-
8/2/2019 Chap 05 Network Layer
86/104
TCP/IP Protocol Suite 86
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.
Figure 5.32 Example 5.34
-
8/2/2019 Chap 05 Network Layer
87/104
TCP/IP Protocol Suite 87
-
8/2/2019 Chap 05 Network Layer
88/104
Example 5.35 Continued
-
8/2/2019 Chap 05 Network Layer
89/104
TCP/IP Protocol Suite 89
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: first step
-
8/2/2019 Chap 05 Network Layer
90/104
TCP/IP Protocol Suite 90
Figure 5.34 Solution to Example 5.35: second step
-
8/2/2019 Chap 05 Network Layer
91/104
TCP/IP Protocol Suite 91
5-4 SPECIAL ADDRESSES
-
8/2/2019 Chap 05 Network Layer
92/104
TCP/IP Protocol Suite 92
5 4 SPECIAL ADDRESSES
In classful addressing some addresses werereserved for special purposes. The classlessaddressing scheme inherits some of these specialaddresses from classful addressing.
-
8/2/2019 Chap 05 Network Layer
93/104
Figure 5.35 Example of using the all-zero address
-
8/2/2019 Chap 05 Network Layer
94/104
TCP/IP Protocol Suite 94
Source: 0.0.0.0Destination: 255.255.255.255
Packet
Figure 5.36 Example of limited broadcast address
-
8/2/2019 Chap 05 Network Layer
95/104
TCP/IP Protocol Suite 95
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
-
8/2/2019 Chap 05 Network Layer
96/104
TCP/IP Protocol Suite 96
Transport layer
Application layer
Network layer
Process 1 Process 2
Destination address:127.x.y.z
Packet
-
8/2/2019 Chap 05 Network Layer
97/104
TCP/IP Protocol Suite 97
Figure 5.38 Example of a directed broadcast address
-
8/2/2019 Chap 05 Network Layer
98/104
TCP/IP Protocol Suite 98
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
-
8/2/2019 Chap 05 Network Layer
99/104
TCP/IP Protocol Suite 99
5 5 NAT
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).
Topics Discussed in the Section
-
8/2/2019 Chap 05 Network Layer
100/104
TCP/IP Protocol Suite 100
Address Translation
Translation Table
Figure 5.39 NAT
-
8/2/2019 Chap 05 Network Layer
101/104
TCP/IP Protocol Suite 101
Figure 5.40 Address resolution
-
8/2/2019 Chap 05 Network Layer
102/104
TCP/IP Protocol Suite 102
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
-
8/2/2019 Chap 05 Network Layer
103/104
TCP/IP Protocol Suite 103
-
8/2/2019 Chap 05 Network Layer
104/104