A full Class C IP address range 210 31 233 0, subnet mask 255 255 255 192

A total of 248 are divided into 4 subnets, and the host IP available for the first subnet is 1 62

0 indicates the CIDR block of the subnet, and 63 indicates the broadcast IP address of the subnet

The available host IP for the second subnet is 65 126

64 indicates the CIDR block of the subnet, and 127 indicates the broadcast IP of the subnet

The third subnet has an available host IP of 129 190

128 indicates the CIDR block of the subnet, and 191 indicates the broadcast IP address of the subnet

The 4th subnet has 193 254 available host IPs

192 indicates the CIDR block of the subnet, and 255 indicates the broadcast IP address of the subnet

So 255-8=248 IPs are available for allocation.

This is a class C IP, the broadcast address is up to 254 machines can be connected to the intranet, because the intranet IP can be any can be or , the intranet mainly uses the subnet mask, and the computer relies on this to detect the IP, so it doesn't matter how many computers you connect to the intranet.

Just set the IP address of the gateway's private NIC to .

The subnet mask can be set to you.

The last share will do.

You don't need to use the entire IP segment, just one can be used by countless intranet machines!

If the network number is, then there are 4 subnets.

Each subnet has 64-2=62 valid IP addresses available.

It seems to be the way it was said on the first floor.

That's OK.

Is the broadcast address on the first floor correct?It should be and".

If there are subnet masks of four IP addresses in the same Class C network, it belongs to a 26 subnet.

The specific calculation method is as follows:

Convert the subnet mask to binary form:.

The 24 bits in front of the mask represent the network address, and the 6 bits in the back segment represent the host address.

Because there are 4 IP addresses, you need at least 4 host addresses.

In a 6-bit binary, at least 3 bits need to be used for 4 host addresses (because 3-bit binary can represent 0-7, which is exactly enough for 4 host addresses).

So the subnet mask of this network needs to be shifted at least 3 bits to the left, becoming, ie.

However, if you use this subnet mask, the network address will be subdivided into 8 subnets, each with only 4 host addresses, which is not enough to meet the demand. Therefore, a smaller subnet mask needs to be used.

Move all the 1s in the 3-bit binary 3 bits to the left to get 8, which means that the network needs at least 8 host addresses. In 6 bits, a 5-bit binary is also required to represent 8 host addresses.

Therefore, the subnet mask for this network should be, i.e., it corresponds to a prefix length of 29.

Therefore, this network belongs to a 29-subnet.

withSubnet maskConductedand operationsto get the network address.

Calculation process. Convert to binary:

Convert to binary:

The two are carried out and calculated, and the operation is carried out: 1 and 1 get 1, and 1 and 0 get 0 The final result:

Convert to decimal as:

The default subnet mask for a Class C IP address is that 254 hosts can be connected.

The subnet mask of your address is the concept of a supernet. This means merging several C-type addresses into a single subnet.

The subnet mask is, and it means that it gives you 16 Class C addresses, and these 16 IP addresses are contiguous. By shortening their subnet masks, they are merged together to form a large subnet. This supernet has a total of 16x256 = 4096 addresses, and there can be a total of 4094 host addresses.

You have two subnets, the first of which is.

Available address: The subnet mask is indeed 1 placeholder, not 01, I don't know why the landlord said it was 01, it should be "10000000", about the subnet number, the landlord's problem is not very clear, This subnet mask tells us, The back 7 bits represent the host bit.

The CIDR block can also be subdivided into 127-2=125 subnets. Remove two CIDR blocks that cannot be divided. b

The IP address is converted into a 32-bit binary system, the subnet mask is converted into a 32-bit binary system, the bit of the subnet mask is 1 corresponding to the bit on the IP address is unchanged, and the bit of the subnet mask is 0 corresponding to the bit change on the IP address to 0, and then it is converted back to the decimal system.

This question is correct, you think it is wrong! The reason for this subnet mask is the subnetting! According to the subnet mask, the number of bits in the subnet can be calculated as 5, and the number of subnets that can be provided is 2 5-2=30

The host IP mask network address.

Broadcast address. Available host addresses.

1. 128 in the mask is represented by binary system: 1000 0000, so the host bit has 7 bits.

2. 64 in IP is represented by binary system: 0100 0000, and the high position is 0, that is, the network bit is 0

3. So, the domain address is.

So this subnet mask is valid.

The two subnets are divided into two subnets: and The valid IP ranges of these two subnets are:

Ditto. Uncategorized addressing, is Class A The default mask is.

26 refers to the mask of , which divides the subnet.

The IP address is the network address of the subnet segment.

For Class C IP addresses with a subnet mask of , the number of subnets that can be provided is 32.

The first 21 bits of a Class C IP address identify the network number, and the variable 8 bits identify the host number, with the first three digits being the first three digits"110"。The value of the first segment of a Class C address is between 192 223, and the first, second, and third segments together represent the network number. The last segment identifies the host number on the network.

Therefore, the binary of the fourth segment 248 is 11111000, the subnet bit is 5 bits, and the number of subnets that can be divided is 2 5, that is, 32.

Mainly look at the last 248, the binary is 11111000, so the subnet bit is 5 bits, and the number of subnets that can be divided is 2 5, that is, 32.

The binary of b248 is 11111000, and the network bits occupy 5 bits, so the number of subnets is: 2 to the 5th power = 32

, that is, 1000, the first 5 1s represent the subnet bits, so the number of subnets provided is 2 5 = 32

The binary of 248 is 11111000, and the network bits occupy 5 bits, so the number of subnets is: 2 to the 5th power, because the network address needs to remove the largest and smallest two special addresses, so the number of subnets is: 32-2=30

Summary. This is a standard private network address, a total of 256 addresses, the starting address is to, of which the first and last two addresses cannot be assigned to computers and other devices, because they are special addresses, domain addresses, that is, network addresses, and broadcast addresses. Therefore, the actual available addresses are from to, a total of 254.

A Class C address: , the subnet mask is , what is the address range of the network segment where the IP address is located? How many IP addresses are in this CIDR block? What are the network numbers and broadcast numbers, and what are the address ranges that can be assigned to the host?

This is a standard private network address, a total of 256 addresses, the starting address is to, of which the first and last two addresses cannot be assigned to computers and other devices, because they are special addresses, domain addresses, that is, network addresses, and broadcast addresses. Therefore, the actual available addresses are from to, a total of 254.

What about its network number and radio number.

The broadcast address is (all the address bits in the IP address are 1), and the IP address is converted to binary 100101 then its broadcast address is: 111111 is converted to decimal.

The number of hosts is 2 to the power of 6 (number of bits of host address) minus 2 (network address and broadcast address) = 622 6 -2 = 62 minus 2 refers to: 111111