How should the variable long subnet mask be divided???

Solution: Ideas: According to the topic, one is 500, the remaining three are almost 500, the first two are 500, and the rest is subdivided; 500 requires nine host bits, and the child subnet mask is.

2 The rest of the machine is carved from 500 subnets.

This is the optimal division of subnets for calculation.

IP address and subnet mask division knowledge.

Extend the network bit length, binary of course.

The number of hosts has been changed from 0 to 1.

For example, the first 25 network numbers remain the same, and the number of the last seven hosts is changed from 0 to 1.

The lower 7 bits are the host number.

The range is: Or.

The actual available ranges are to delete the host and all 1, one is the subnet address and the other is the subnet broadcast answering address.

So the actual range of usability:

The place where the host auspicious number is 0 is the subnet address, and all 1s are warnings.

How to assign the IP of the duplicate mask of the variable long subnet, first consider dividing the subnet under the condition that the mask remains unchanged, because there are three departments, so at least four subnets need to be divided, nsubnet<=2 x; That is, 3<=2 x, and x=2 is obtained, so at least 2 bits need to be borrowed from the host bit, because the subnet is debited from the host bit. Because the number of bits of the host of the Class C address is 8 bits, after lending 2 bits to the subnet, the host bits are left with 6 bits, 2 6-2 = 62, and the 62 host IPs cannot meet the requirements of 100 people in department A, so it does not work. So we have to consider changing the mask, the next life rotten allocation host because the C class IP mask is, after the borrowing to get the mask, when the borrowing can be divided into two subnets 0000 0000 and stove back bucket 1000 0000, in the following calculation, only 0000 0000 subnets are used, 1000 0000 is reserved.

When one is borrowed, borrow another one from departments B and C, and the mask is obtained, and two subnets can be obtained: 1000 0000 and 1100 0000. Note:

Here the first subnet assigned is 1000 0000 and the later assigned subnet 1000 0000 is not the same because the mask is different, although similar IPs may be generated. I hope mine can help you. If my answer is helpful to you, please give a thumbs up (in the lower left corner), I look forward to your like, your efforts are very important to me, and your support is also the motivation for my progress.

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The standard mask is:

There's something wrong with your mask, isn't it wrong?

500 masks must be 23 bits, 250 masks 24 bits, 100 25 bits, 50 26 bits, but five departments with five different masks, it will look very confusing, so it is recommended that you use 23 bits directly. Your 19-bit mask is enough anyway.

You've got a lot of addresses left.

Solution: Ideas: According to the title, one 500 units, the remaining three are almost 500, you can divide two big 500 first, and then subdivide the rest; 500 units require 9 host bits, then the subnet mask is.

This results in the optimal subnet division.

If you have more than 255 computers in one subnet, why not use a Class B IP?

network address. The first department, the second department, the third department, the fourth department, the fifth department. This is the most IP-saving.

However, your IP address resources seem to be quite abundant, so you don't need to save so much.

It is a block of Class B format address, mask 19, which means that this Class B address is divided into 8 segments, each segment contains 32 Class C addresses, and the total number of IP addresses in the segment is 256*32=8192. Obviously, one of the paragraphs is a resource for this topic.

The first sector had a requirement of 500, which meant that the available resources needed to be divided into 16 segments, each containing 512 IPs. A segment is then assigned to the department. Since 2 4 = 16, the mask needs to be added by 4 bits to become 23.

The second sector requirement was 250, which meant that the available resources needed to be divided into 32 segments, each containing 256 IPs. A segment is then assigned to the department. Since 2 5 = 32, the mask needs to be added by 5 bits to become 24.

The requirement for the third sector is 100, which means that the available resources need to be divided into 64 segments, each containing 128 IPs. A segment is then assigned to the department. Since 2 6 = 64, the mask needs to be added by 6 bits to become 25.

The requirement for Sector IV was 50, which meant that the available resources needed to be divided into 128 segments, each containing 64 IPs. A segment is then assigned to the department. Since 2 7 = 128, the mask needs to be added by 7 bits to become 26.

The requirement for the fifth sector was 28, which meant that the available resources needed to be divided into 256 segments, each containing 32 IPs. A segment is then assigned to the department. Since 2 8 = 256, it means that the mask needs to be added by 7 bits to become 27.

According to the subject requirements, each subnet is almost less than 254 hosts, and 254 hosts need 8 hosts, so the subnet mask is:

2. Because the administration department is divided into two departments, it is necessary to divide the second subnet (into two small subnets.

4. Because the design department is divided into two departments, but one of them has more than 126 hosts, only two subnets can be used.

Divided into 5 departments, each department is 55, 49, 24, 26, 13 people, the program is not unique, for example:

1. Network address, mask, available host address from, broadcast address.

2. Network address, mask, available host address from to, broadcast address.

3. Network address, mask, available host address from, broadcast address.

4. Network address, mask, available host address from to, broadcast address.

5. Network address, mask, available host address from to, broadcast address.

To plan a subnet, you must first know that the subnet is formed by the square of 2.

So for 500 addresses, greater than 500, the closest is 2 9 = 512, and then subtract 9 from 32 of the IP to get the answer 23.

It is important to note that you have to subtract 2 addresses from the answer you get, and each subnet has a network address and a broadcast address, which cannot be used by users.

So if the problem requires you to plan 255 addresses, you need to use 23. Because 24 (256 addresses) minus 2 254 usable addresses.

The following are the CIDRs that have been classified

23 for 500 (510 usable addresses) and 23 for 300

264 with 23

253 with 24 (254 usable addresses) 169 with 24

77 with 25 (126 usable addresses).

50 with 26 (62 usable addresses).

10 with 28 (14 usable addresses).

First of all, according to the number of hosts required, add 2, and see how many times the power of the integer of the smallest 2 that can accommodate this number is 2, which is the host bit.

32 subtracts the host bit to get the network bit. Setting all network bits to 1 is the required subnet mask.

In the above example, the number of hosts in each subnet is 52, 57, 47, 24, and 27, respectively.

The common denominator with 47 is greater than 2 to the 6th power, but less than 2 to the 7th power, so the host bit is 6, then the network bit is 32-6=26. That is to say, the subnet mask is 26 bits, expressed in binary, converted to decimal.

The common denominator between 24 and 77 is greater than 2 to the 5th power, but less than 2 to the 6th power, so the host bit is 5, then the network bit is 32-5=27. That is to say, the subnet mask is 27 bits, expressed in binary, converted to decimal.

Know this, and you can count the rest.

The meaning of a variable-long subnet mask is that the length of the subnet mask can be different for different subnets in a network. In order to achieve the purpose of saving IP addresses. Distinguished from fixed-length subnet masks.