Subnet Masks - Subnets, Ranges, and Increments - how to understand this?
Hello, It seems to me that my text book does not fully explain Subnets, their Ranges, and Increments. This is the relevant extract from the notes I have made from the text book: - - - Copy notes start: - - - - Subnet Mask specifies the Network Ranges Understanding Network Ranges Subnet Mask also breaks down the Ranges (aka Increments) of the network. E.g. 255.255.255.224 allows for 6 Subnets . There should be 6 TCP/IP Ranges that go with the 6 Subnets. Following Table shows the different Ranges (aka Increments) for the different Subnet Masks. [Note: Range aka Increment, is difference between its Subnet Mask Number and next Subnet Mask Number.] Table 7.4, P.414 to P.415. Subnet Mask Ranges: Subnet Mask Number: Ranges aka Increment: 255 1 254 2 252 4 248 8 240 16 224 32 192 64 128 128 P.415: Text book explains Table 7.4 using the same example of: 255.255.255.224. Because 224 allows for 6 Subnets, the 6 Ranges are in increments of 32 (2nd Ed. P.414). [Notes: 1) 32: To get 32, subtract the next Subnet Mask Number of 192 from our above example of 224 (indicated in orange above). Thus: 224 (192) 32 Thus 224 has a Range of 32. 2) 224 allows for 6 Subnets: Text book does not explain this. 3) the 6 Ranges are in increments of 32: Text book does not explain this. ] - - - - End of copy notes - - - - Please could you explain how we know how many Subnets there are in a Subnet Mask value? Please could you also explain how we would know that the 6 Ranges to which reference is made, would be in increments of 32? To illustrate my problem about these points, please could you consider the following chapter review question and explain how we would work out the answer for this question in an examination? I do not completely understand the explaination it gives. - - - - Copy Review Question: - - - - Question: You are the network administrator for a small organization that has decided to switch the companys TCP/IP subnet mask. The organization has chosen to use 255.255.255.240. What is the increment for subnet ranges in a 240 subnet mask? Answer: 16. Explanation: On a 240 subnet mask, the ranges go by 16. Because this is a Class C, the ranges would be 16-31, 32-47, 48-63, 64-79, 80-95, 96-111, 112-127, 128-143, 144-159, 160-175, 176-191, 192-207, 208-223, and 224-239. Remember, that in a Class C address, these are not all usable. You have to subtract the first number (network ID) and the last number (broadcast). (Review Questions: 2nd Ed. Q.10, P.462, P.466.) - - - - End of Review Question - - - - I look forward to help with this query. Regards, Robert339
May 23rd, 2012 10:06pm

1. You should write the binary equivalent of mask and find which combination of ones and zeros are allowed. 2. From possible combinations you should exclude IP addresses that has special purpose, namely the subnet and multicast address. 3. Whenever you find all possible combinations of addresses, convert binary addresses to decadic ones. In first example: Rewrire 224 as binary 11100000 Which combinations of zeros and ones can hide under first three ones? 000 001 010 011 100 101 110 111 First and last have special purpose and do not count to possible subnets (subnet address and multicast address) If you write possible combinations of zeros and ones that belong to host address part - that part of last eight bits that MASK does not resolve (ones or zeros multiplied by zero gives always zero), 00000 00001 00010 00011 00100 00101 00110 00111 01000 01001 01010 01011 01100 01101 01110 01111 10000 10001 10010 10011 10100 10101 10110 10111 11000 11001 11010 11011 11100 11101 11110 11111 In summary you are searching combinations of zeros and ones that are allowed by masking (multiplication) of subnet mask and IP address and excluding adresses that have special purpose, namely the subnet and multicast addresses. Regards Milos
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May 24th, 2012 1:42am

Like Milos said you need to look at it from binary http://searchnetworking.techtarget.com/tip/How-to-subnet-Subnetting-calculations-and-shortcuts If you are asked for valid hosts in the 192.168.30.7/28 range you start out by counting off the chart to see the numerical value for the 28th bit. It is 16. Then take a look at the last octet (since 28 puts us in the last octet range) and divide it by 16. In this case we don't have to because we can see it is less than 16 but if you are given .189 rather than .7 you would divide 189 by 16 and see that it was over 11. Then multiply 11 times 16 to see the first address in that subnet -- 176. Then add 16 to 176 and you will get the first address of the next subnet (192). That means that .189 falls in the subnet 176 188. We know that the first number is the network address (192.168.30.176) and the last number is the broadcast address (255.255.255.188 -- remember we are breaking down that 256 block so we can't use the full 256 subnet, only the portion blocking our network). That means that the hosts must fall in the range of 177 187 (one after 176 and one before 188). Going back to our question of valid hosts in the 192.168.30.7/28 range they would be: 192.168.30.1 192.168.30.2 192.168.30.3 192.168.30.4 192.168.30.5 192.168.30.6 192.168.30.7 192.168.30.8 192.168.30.9 192.168.30.10 192.168.30.11 192.168.30.12 192.168.30.13 192.168.30.14 And if you counted them up you would see there are 14 of them. Remember the 2<super></super>-2 formula -- it is the bit value minus 2 (one for network and one for broadcast address or subnet mask) equals the number you can use for hosts. If you want to know how many of these 16-bit networks we can get out of that third octet, divide 256 by 16. It is 16. Don't slip and think that the number of networks you can get is the bit value-- it just works out that way with /28. For a /26 you can get four networks (0 63; 64 127; 128 191 and 192 255 (and I didn't use the number 256 because I started at zero -- the modulus (value of 256) comes from zero to 255).
May 24th, 2012 2:02am

Like Milos said you need to look at it from binary http://searchnetworking.techtarget.com/tip/How-to-subnet-Subnetting-calculations-and-shortcuts If you are asked for valid hosts in the 192.168.30.7/28 range you start out by counting off the chart to see the numerical value for the 28th bit. It is 16. Then take a look at the last octet (since 28 puts us in the last octet range) and divide it by 16. In this case we don't have to because we can see it is less than 16 but if you are given .189 rather than .7 you would divide 189 by 16 and see that it was over 11. Then multiply 11 times 16 to see the first address in that subnet -- 176. Then add 16 to 176 and you will get the first address of the next subnet (192). That means that .189 falls in the subnet 176 188. We know that the first number is the network address (192.168.30.176) and the last number is the broadcast address (255.255.255.188 -- remember we are breaking down that 256 block so we can't use the full 256 subnet, only the portion blocking our network). That means that the hosts must fall in the range of 177 187 (one after 176 and one before 188). Going back to our question of valid hosts in the 192.168.30.7/28 range they would be: 192.168.30.1 192.168.30.2 192.168.30.3 192.168.30.4 192.168.30.5 192.168.30.6 192.168.30.7 192.168.30.8 192.168.30.9 192.168.30.10 192.168.30.11 192.168.30.12 192.168.30.13 192.168.30.14 And if you counted them up you would see there are 14 of them. Remember the 2<super></super>-2 formula -- it is the bit value minus 2 (one for network and one for broadcast address or subnet mask) equals the number you can use for hosts. If you want to know how many of these 16-bit networks we can get out of that third octet, divide 256 by 16. It is 16. Don't slip and think that the number of networks you can get is the bit value-- it just works out that way with /28. For a /26 you can get four networks (0 63; 64 127; 128 191 and 192 255 (and I didn't use the number 256 because I started at zero -- the modulus (value of 256) comes from zero to 255).
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May 24th, 2012 2:09am

Hello Milos and Brano, Thank you for your replies and I appreciate the trouble and time you have both taken with the information you have provided. I would find it useful if you could explain two things which the text book does not explain. It states: 1) 224 allows for 6 Subnets. 2) 255.255.255.224 allows for 6 Subnets . There should be 6 TCP/IP Ranges that go with the 6 Subnets. For each value like 224, how do we know or discover, the number of applicable Subnets? How do we know or discover, how many Ranges go with the subnets? Finally, (and I apologise if this ought to be rather obvious from the information you have provided, but I am still going through it), could you please summarize how I would use your information to answer the example Review Question to which I referred in my original message. In particular, what information exactly, is it, that I ought to commit to memory in order to answer this sort of question in the 70:680 examination? Regards, Robert339
May 24th, 2012 6:32pm

Branco has already spelled out what the range of each of the four subnets is for a /26 subnet. I suggest that you read his post again. Surely you can do the calculations for other subnet lengths yourself. If you can't you should be able to find a website which will do it for you. Bill
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May 24th, 2012 7:32pm

Branco has already spelled out what the range of each of the four subnets is for a /26 subnet. I suggest that you read his post again. Surely you can do the calculations for other subnet lengths yourself. If you can't you should be able to find a website which will do it for you. Bill
May 24th, 2012 7:33pm

Hello, I want to thank all who contributed to this thread. Milos's response of 24 May 2012 should not be marked as an answer, since it is not a satisfactory answer to the whole of my query. Bill's response of 24 May 2012 should not be marked as an answer, since it does not provide any further satisfactory explanation for the previous responses. It also does not relate correctly to my question, which sought the information to be learnt and used during an examination. Regards, Robert339
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June 5th, 2012 3:53pm

Hello, I want to thank all who contributed to this thread. Milos's response of 24 May 2012 should not be marked as an answer, since it is not a satisfactory answer to the whole of my query. Bill's response of 24 May 2012 should not be marked as an answer, since it does not provide any further satisfactory explanation for the previous responses. It also does not relate correctly to my question, which sought the information to be learnt and used during an examination. Regards, Robert339
June 5th, 2012 4:00pm

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