Q1. What is a classless IP address?
Answer: Classless is also called IP subnetting. Divide classful network addresses into smaller networks (subnet). The rules are simple: – In the classful addressing scheme, a 32-bit address is divided into network IDs and host IDs on an 8-bit boundary:
Class Network_ID Host_ID Examples……………..Leftmost_byte_range(abcd)
A……….8bit…….24bit…..1.1.1.1, 100.4.5.6………..0~127
B……….16bit……16bit……128.1.1.1, 150.1.2.3…..128~191
C…24bit…8bit…192.1.1.1, 222.1.2.3…192~223
– Classless IP allows more left bits for network identification (and fewer bits for host identification). The number of bits used by
Network ID is called network mask. For example, abcd/m says “the network ID is the leftmost m bit”. – Examples of classful and classless addresses:
IP address mask class? 100.4.5.6/8 /8 class A 100.0.0.0/8 100 4.5.6 16 777 216 100.4.5.6/16 /16 no class 100.4.0.0/16 /24 /24 no class 100.4.5.0/24 100.4.5 6 256 100.4.5.6 /25 /25 no class 100.4.5.0/25 100.4.5.0 6 128
Q2. Why no classes?
Answer: Using a hard limit to allocate bits for network identification and host identification results has a side effect: IP address starvation. Consider this example: a network has 100 computers to connect to the Internet. They are using a class C network 200.1.2.0/24. So the network grows to 500 computers. You need a class B network, say 150.1.0.0/16. However, 150.1.0.0/16 covers 65,536 class B IP addresses. This network only uses 500 of them. 65,036 IP addresses are wasted. This causes a shortage of IP addresses due to the fixed network/host limit of the classful addressing scheme. This problem can be solved by classless addressing. For example, 150.1.0.0/23 allows 512 host IDs. It covers addresses 150.1.0.0~150.1.1.255).
Q3. How to subnet and create a network ID that crosses the byte boundary?
Answer: Subnetting uses a netmask to assign the leftmost bits for the network ID. The rules are simple: – Start with a classy address (class A, B or C). – Add more bits to the classful network ID bits. Class B uses the leftmost 16 bits for network identification. His mask is /16. With classless addressing, the netmask for the class B derived network can be /16~.31. – For example,
150.1.0.0 is a class B network address. Its default mask is /16. It covers 65,536 addresses.
150.1.2.0/24 is a classless subnet derived from the class B network. Its mask is /24 and it contains 256 addresses.
150.1.2.128/25 is a classless subnet derived from a class B network. Its mask is /25 and it has 128 addresses. The following table presents classful and classless addresses in binary form:
Network ID/Netmask Binary form (32 bits) Network ID/Class Host ID number and range
– 150.1.0.0/16 is a class B address, network ID 150.1 (16 bits), contains 65 356 hosts (150.1.0.0~150.1.255.255), its binary presentation is 10010110. 00000001.00000000.00000000
– 150.1.2.0/24 has no class derived from class B, network ID 150.1.2 (24 bits), contains 256 host addresses (150.1.2.0~150.1.2.255). Its binary form is 10010110.00000001.00000010.00000000.
– 150.1.2.128/25 is classless and is derived from class B, network ID is 150.1.2.128 (25 bits), contains 128 host addresses (150.1.2.128~150.1.2.255). its binary form is 10010110. 00000001.00000010.10000000
Q4. Can we subnet to create 150.1.2.3/14? Answer: No. Subnetting begins with a classful addressing scheme. 150.1.2.3 is a class B address (150 is within the class B range: 128~191). The class B netmask is /16. Subnetting a class B network address can only use masks to reduce the size of the network or increase the mask. /16~/21 are allowed. /1~/15 are not allowed.
P5. How does a router look up a classless IP address?
Answer: Let’s use examples to explain the router operations: 1) When R1 receives the ping sent by Jack, it compares the ping destination IP address (150.1.2.3/25) with the destination networks listed in the routing table.
– 100.0.0.0/8. The leftmost 8 bits of 150.1.2.3/25 are 150, not 100. Not a coincidence.
– 150.1.0.0/16. 150.1.2.3/25 The leftmost 16 bits are 150.1. is a match R1 forwards the ping to 150.1.0.0/16. 2) When R2 receives pings, it compares the ping destination IP address (150.1.2.3/25) with the destination networks in the routing table. -150.1.0.0/24: 150.1.2.3/25 the leftmost bit of 24 is 150.1. 2, no. 150.1.0. It is not a party.
-150.1.1.0/24: 150.1.2.3/25 the leftmost bit of 24 is 150.1. 2, note 150.1.1. It is not a party.
-150.1.2.0/24: 150.1.2.3/25 the leftmost bit of 24 is 150.1. 2. It’s a match. R2 forwards the ping to the 150.1.2.0/24 network 3) When R3 receives the ping, it checks the ping destination IP address (150.1.2.3/25) against the routing table. -150.1.2.128/25: 150.1.2.3/25 The leftmost bit of 25 is 150.1.2.0 2, not 150.1.2.128, not a coincidence.
– 150.1.2.0/25: 150.1.2.3/25 the leftmost bit of 25 is 150.1.2.0. in game R3 forwards the ping to Rose.
Note: This example is clearer if read in conjunction with the complementary simulation topology.
This article is the FAQ for an interactive simulation. You can get clearer concepts by playing the simulation with visualized topology, packets, table. It is located under External Links.