Summary of IP computing problem of soft test computer network principle

Transfer from http://www.cnblogs.com/jyh317/archive/2013/04/14/3018650.html

1. IP Address

Classification:

①a Class IP Address

②b Class IP Address

③c Class IP Address

④d Class IP Address

⑤e Class IP Address

2, Knowledge points

① full 0 (0.0.0.0) represents the broadcast address of any network, full 1 (255.255.255.255) of the current subnet.

② Subnet Mask: It is a bitmask that indicates which bits of an IP address identify the subnet on which the host resides and which bits identify the host. That is, it is used to divide an IP address into two parts: network address and host address.

A subnet mask is a "full 1" bit mode that masks the network portion of an IP address. The default subnet mask for Class 5 IP addresses is:

3, Sub-network division

Subnetting is implemented by borrowing several bits of the IP address to act as the subnet address to divide the original network into several subnets.

Three elements:

① determining the number of subnets

② Finding the subnet mask

③ determining the address range for each subnet

Examples:

Each subnet of an organization with a class A address needs to allocate a maximum of 1000 IP addresses, trying to find the subnet mask.

∵ requires a maximum of 1000 IP addresses and takes into account the full 0 and all 1 host addresses

∴ a minimum of 1002 IP addresses

∴2^9<1002<2^10, the number of host bits is 10

∵ is a Class A IP address

∴ number of subnet Bits =a class IP address host bits -10=24-10=14

So the subnet mask (network address occupies 14 bits of the host address): 1111 1111. 1111 1111.111100.0000 0000→ (decimal) 255.255.252.0

Derivative soft question type:

(1) "The first half of 2009.66" The subnet mask for a Class B network is 255.255.224.0, then the network is divided into a () subnet.

∵ Subnet mask 255.255.224.0 binary →1111 1111.1111 1111. 1110 0000.0000 0000

Class B default Mask binary →1111 1111.1111 1111. 0 0000.0000 0000

∴ as shown in bold above, the network address occupies the three-bit address of the host address.

∴ is divided into 2^3=8 subnets

4, constitute the super net

Merge several Class C nets into a larger range of addresses.

Method: Change Some 1 of the network address to 0. This is typically used in routing aggregation, or CIDR. That is, the number of digits in the network number to add to the host number of digits.

Although the soft examinations will not appear the specific knowledge point, but this kind of calculation will be out.

Several rules that make up the network:

① the number of address blocks that make up the network must be 2^n (N=1, 2, 3 ...).

The address block (②) that makes up the network must be a contiguous block of addresses.

The third byte of the first address of the ③ block must be able to be divided evenly by the number of blocks (the start address must be divisible by the number of addresses N).

Soft question type:

(1) "2011 in the first half. 70" The subnet mask is () when the 4 contiguous Class C networks are converged into a single network

4 subnets into an ultra-network, 4=2^2, so to the network address of the two-bit address to the host address, that is, the network address of two bits from 1 into 0.

That is, the original Class C default mask binary is represented as 1111 1111.1111 1111.1111 1111.0000 0000→255.255.255.0 (two of the last 1 bits of the network address becomes 0)

1111 1111.1111 1111.1111 1100.0000 0000→255.255.252.0 (correct answer)

5, CIDR notation

CIDR uses the slash notation to become CIDR notation, which is to add a slash "/" after the IP address, and then write the number of bits that the network prefix occupies (this number corresponds to the number of bits 1 in the subnet mask)

such as "the first half of 2010.66" IP Address Block 222.125.80.128/26 contains () An available host address, where the minimum address is (), and the maximum address is ().

The ∵ slash followed by 26 is the number of bits of the network prefix, so the number of bits for the host number is 32-26=6.

∴ Available Host addresses total (2^6-2) = 62.

　　

Summary, this year, the soft examinations computer network accounted for a small score, the main basic knowledge, but also occasionally a few years will be a few calculation problems, such as 2005, 2006, 2007, 2009, 2010, 2011, 2012 are out of the calculation, the question is my blog's two types of expansion, Although there is only one point, but after reading my blog summary, draw a little time to review, should be no problem.

Summary of IP computing problem of soft test computer network principle