Training Cisco 200-125 CCNA v3.0. Day 7. FAQ
Today we will look at answers to frequently asked questions that relate to previous video tutorials. About a year has passed since the first publication, and a lot of people left their comments under my video tutorials on the YouTube channel. I carefully read all your comments, but today I will show screenshots of comments received only in the last month, since it is impossible to consider all the questions that appeared during this year.
(translator's note: this video was published on October 18, 2014)
This screenshot shows one of the last comments, in which user Scott Rosales asks to post more videos.
')
He writes that he just watched all 8 videos and understood a lot more than what his teacher taught at school over the past few months, and that my lessons helped him a lot.
Francisco wrote that these are excellent videos and he shared them with his friends, and Somia very much asks to post all 30 video lessons as soon as possible. She writes that she really likes how I explain different things, and that she has been working at an Internet company for 6 months now, but she still learns something new from my video tutorial. “I can’t wait for the publication of the full course to start the exam. Thanks for the video and please quickly upload new lessons. ”
Another user writes that he is studying for a master at QUT and he really liked how I explained the concept of supernetworks. In the process of learning this was a difficult question for him, but everything became completely clear to him after he listened to me for 2 minutes.
Leo writes that he really liked how I explained the concept of OSI. Sean Lynch wrote that he was interested in the history of DEC / IMB, because he himself learned to program on the DEC 10 computer back in 1977. “This is a great video for those who want to brush up on OSI. The approach to examining the “top down” levels was a novelty for me, because instead of starting the review in order from level 1 to level 7, Imran immediately began from the application level, with which most people know best. I think this is a great way to learn, and I'm going to see all the series. ”
Raoul writes that if he watched my videos 4 months ago, he would have saved 18,000 rupees, which the company paid for networking. “These guys don't even know 40% of what you explained in your eight video tutorials.”
Rico wrote that my video helped him understand the basics of gateways and IP addresses, because they have a bad teacher on their course who could not explain this topic lucidly. It was only a part of the comments that I received in the last month, but my team reads all your comments, selects the most important ones and sends them to me for answers. If I have time, then I answer the questions myself. For me, this is a very important experience, as your comments help me to improve the quality of video lessons.
Let's get to the questions. The most popular question is “determine the network ID and broadcast address for the following IP addresses”:
Let's answer the first question: find the identifier and the broadcast address of the network where the IP address is 20.120.47.225/13. Let's turn to our "magic table."
Slash 13 means that we borrowed 5 bits in the second octet - if you calculate, 1 octet has 8 bits, then 1 borrowed bit is the 9th bit, the second is the tenth bit, the third is the eleventh bit, 4 is the twelfth bit and fifth is the thirteenth bit. Thus, / 13 means 5 bits of the second octet.
Since we are working with the second octet, in order to find the network identifier, we must equate 2,3 and 4 octets to zero. Then the identifier of the first network will have the address 20.0.0.0. To find the identifier of the second network, we need to insert the 8 - the number that is under the fifth borrowed bit into the second octet of the IP address. Thus, the identifier of the second network will be 20.8.0.0.
Now we can determine the broadcast address of the first network, which will be equal to 20.8.0.0 minus one, that is, 20.7.255.255.
If the identifier of the second network is 20.8.0.0, then its broadcast address will be 20.15.255.255, the third network - 20.16.0.0, and the broadcast address - 20.23.255.255. This is simply because the second octet of the broadcast address of the previous network is equal to the second octet of the identifier of the subsequent network minus 1:
16 -1 = 15.255.255, 24-1 = 23.255.255 and so on.
This will continue until the identifier of our network reaches the value of 20.120.0.0, because the ID of the next network will be 20.128.0.0. Thus, the network identifier for the IP address specified in the example condition will be 20.120.0.0, and its broadcast address will be 20.127.255.255, where 127.255.255 means 2.128.0.0. minus, that is, the ID of the next network minus one.
Our IP address 20.120.47.225/13 is located in the address range from 20.120.0.0 to 20.127.255.255, so these addresses are the answer to the question, what is the network identifier and the broadcast address of the network containing our IP address.
I want to tell you about the method of the shortest path. No matter what the position of the borrowed bit is, in our example it was 8, all these numbers are 1,2,4,8, 16, etc. - are factors of the number 128 and can not exceed it. Therefore, I look at the second octet of our IP address, equal to 120, and find out whether it is located after 128 or up to 128. In our case, it is located before 128. What am I doing? I subtract 128 from the number 8, that is, the location of the 5th borrowed bit, and get 120.0.0. Then the next network will have the value of the second to fourth octets 128.0.0. Due to this, it can be said that the first network ID will be 20.120.0.0, and its broadcast address will be 20.127.255.255.
If you do not know how to do these calculations in your mind, just do what I said earlier - just add the value of the location of the borrowed bit to the octet each time, getting the ID of the next network until you reach the value given in the example. Now let's go to the second example.
We need to determine the network ID and the broadcast address of the network containing the IP address 220.20.17.5/27.
Slash 27 means that we are dealing with the fourth octet, since 3x8 = 24, and the number 27 exceeds this value by 3, that is, it is in the fourth octet.
Thus, we borrowed 3 bits from the 4th octet, and their area of ​​location is 32. We can say that / 25 means 1 borrowed bit, / 26 means 2 bits, and / 27 - three bits, that is, we have 3 bit of the fourth octet. This means that to find the sequence of network identifiers we must add 32 to the fourth octet.
Let's start with the IP address 220.20.17.0 - this will be the ID of the first network, then the broadcast address will be 220.20.17.31, that is, the value of the 4th octet of the broadcast address of the first subnet will be (32-1).
The identifier of the second network is formed by adding 32 to the fourth octet (0 + 32) = 32 and will look like 220.20.17.32. Let's look at the problem condition: our address ends with 5, and the number 5 is in the range between 0 and 31, where 0 is the last octet of the network identifier, and 31 is the last octet of the broadcast address.
Thus, we do not need to calculate anything else - the answers to this problem are the addresses 220.20.17.0 and 220.20.17.3.
Let's move on to the next example. Here we need to determine the parameters of the network containing the IP address 10.10.7.17/19. First of all, we need to determine which octet belongs to / 19 - this is the 3rd octet, because two octets end in (2x8) = 16, which means that the number 19 is located in the 3rd octet. We see that here, as in the previous example, 3 bits were borrowed, since 19 = 16 +3. As in the previous example, in this case / 17 means 1 borrowed bit, / 18 - two bits and / 19 - 3 bits.
Thus, to determine the identifier of the first subnet, we must substitute 0 into the value of the 3rd octet and obtain an address of the form 10.10.0.0. Then the identifier of the second network will be 10.10.32.0, hence the broadcast address of the first subnet, which is less by 1, will be equal to 10.10.31.255.
Now we compare these parameters with our IP-address to check if it is in their range. We see that 10.10.7.17 is indeed located between 10.10.0.0 and 10.10.31.255, which are respectively the identifier and the broadcast address of the network containing it.
Consider the 4th example of the problem with the address 192.8.3.1/18. Many people think this is the wrong address. Consider in which case this address is correct, because it is a trick question. Here we have a case of a classless address, that is, we do not have a rigid framework for class addressing.
Although this address starts at 192, it cannot be assigned to class C, which has a 24-bit mask. I just want to remember that there are classless addresses, in this case, we may have, for example, IP address 192 with a slash 9, and this is completely normal. You should treat these addresses the same way as class addresses, that is, start looking at it from the end where / 18 is located, not paying attention to the first octet. As we know, / 18 denotes bits borrowed from the 3rd octet.
Thus, our subnet ID will be 192.8.0.0. The table shows that 2 bits were borrowed here (2x8 = 16 +2 = 18). This means that in the third octet of each subsequent network you need to add 64. So, if the first subnet identifier is 192.8.0.0, then the second network identifier will take the value 192.8.64.0, then the broadcast address of the first subnet will be less than one address and will take the value 192.8.63.255. Looking at the IP address from the example, we will see that it is located in the range between these two addresses, which means they are the answer to the question of the problem.
Now consider the last example with the address 172.1.4.5/20. A forward of 20 means that 4 bits from the 3rd octet are borrowed from us. Therefore, the first subnet identifier must be 172.1.0.0., Since in this case the 3 and the following octets are 0.
Looking at the table, we will see that the identifier of the second subnet is obtained by adding 16 to the third octet and will be equal to 172.1.16.0, which means that the broadcast address of the first subnet will be one less, that is, 172.1.15.255. Comparing our address with these values, we will determine that it is between them, that is, the addresses we have defined are the answer for this task.
Now I will add one more example. Let's look at the IP address 172.138.75.5/12. Slash 12 means the fourth bit of the second octet of the network identifier. The table shows that it corresponds to the number 16, that is, we must consistently add this number to the second octet of each next subnet, starting with the identifier of the first subnet 172.0.0.0, until we approach the value of the second octet 138.
So, we use a product with a multiplier of 16 and reach 128. The number in the second octet is 138, that is, one is greater than 128, but less than 128 +16 = 144. Thus, we have the subnet identifier 172.128.0.0, and the next subnet will be have an id of 172.144.0.0. By subtracting one, we get the broadcast address of the previous network.
Comparing our address with these addresses, we will see that it is located between them, which means we have found the right solution.
We obtained the identifier of the desired subnet by adding 16. However, you can save time on calculations if you use the shortest path method. If this is difficult for you, you can always go back to the usual method of adding the value of the location of the borrowed bit.
I want to draw your attention to the fact that I had a group on Facebook, where you can also ask your questions, and also want to advise, if you have not done so already, subscribe to my channel on the YouTube portal. Do not forget to share these video tutorials with your friends and leave your questions and comments below. You can also write me an e-mail address, and although I don’t use it much, I’ll still try to answer all your emails. I promise that in each video lesson I will consider your most interesting questions and answer them.
Thank you for staying with us. Do you like our articles? Want to see more interesting materials? Support us by placing an order or recommending to friends, 30% discount for Habr's users on a unique analogue of the entry-level servers that we invented for you: The whole truth about VPS (KVM) E5-2650 v4 (6 Cores) 10GB DDR4 240GB SSD 1Gbps from $ 20 or how to share the server? (Options are available with RAID1 and RAID10, up to 24 cores and up to 40GB DDR4).
VPS (KVM) E5-2650 v4 (6 Cores) 10GB DDR4 240GB SSD 1Gbps before summer for free if you pay for a period of six months, you can order here .
Dell R730xd 2 times cheaper? Only we have 2 x Intel TetraDeca-Core Xeon 2x E5-2697v3 2.6GHz 14C 64GB DDR4 4x960GB SSD 1Gbps 100 TV from $ 199 in the Netherlands! Dell R420 - 2x E5-2430 2.2Ghz 6C 128GB DDR3 2x960GB SSD 1Gbps 100TB - from $ 99! Read about How to build an infrastructure building. class c using servers Dell R730xd E5-2650 v4 worth 9000 euros for a penny?
(translator's note: this video was published on October 18, 2014)
This screenshot shows one of the last comments, in which user Scott Rosales asks to post more videos.
')
He writes that he just watched all 8 videos and understood a lot more than what his teacher taught at school over the past few months, and that my lessons helped him a lot.
Francisco wrote that these are excellent videos and he shared them with his friends, and Somia very much asks to post all 30 video lessons as soon as possible. She writes that she really likes how I explain different things, and that she has been working at an Internet company for 6 months now, but she still learns something new from my video tutorial. “I can’t wait for the publication of the full course to start the exam. Thanks for the video and please quickly upload new lessons. ”
Another user writes that he is studying for a master at QUT and he really liked how I explained the concept of supernetworks. In the process of learning this was a difficult question for him, but everything became completely clear to him after he listened to me for 2 minutes.
Leo writes that he really liked how I explained the concept of OSI. Sean Lynch wrote that he was interested in the history of DEC / IMB, because he himself learned to program on the DEC 10 computer back in 1977. “This is a great video for those who want to brush up on OSI. The approach to examining the “top down” levels was a novelty for me, because instead of starting the review in order from level 1 to level 7, Imran immediately began from the application level, with which most people know best. I think this is a great way to learn, and I'm going to see all the series. ”
Raoul writes that if he watched my videos 4 months ago, he would have saved 18,000 rupees, which the company paid for networking. “These guys don't even know 40% of what you explained in your eight video tutorials.”
Rico wrote that my video helped him understand the basics of gateways and IP addresses, because they have a bad teacher on their course who could not explain this topic lucidly. It was only a part of the comments that I received in the last month, but my team reads all your comments, selects the most important ones and sends them to me for answers. If I have time, then I answer the questions myself. For me, this is a very important experience, as your comments help me to improve the quality of video lessons.
Let's get to the questions. The most popular question is “determine the network ID and broadcast address for the following IP addresses”:
Let's answer the first question: find the identifier and the broadcast address of the network where the IP address is 20.120.47.225/13. Let's turn to our "magic table."
Slash 13 means that we borrowed 5 bits in the second octet - if you calculate, 1 octet has 8 bits, then 1 borrowed bit is the 9th bit, the second is the tenth bit, the third is the eleventh bit, 4 is the twelfth bit and fifth is the thirteenth bit. Thus, / 13 means 5 bits of the second octet.
Since we are working with the second octet, in order to find the network identifier, we must equate 2,3 and 4 octets to zero. Then the identifier of the first network will have the address 20.0.0.0. To find the identifier of the second network, we need to insert the 8 - the number that is under the fifth borrowed bit into the second octet of the IP address. Thus, the identifier of the second network will be 20.8.0.0.
Now we can determine the broadcast address of the first network, which will be equal to 20.8.0.0 minus one, that is, 20.7.255.255.
If the identifier of the second network is 20.8.0.0, then its broadcast address will be 20.15.255.255, the third network - 20.16.0.0, and the broadcast address - 20.23.255.255. This is simply because the second octet of the broadcast address of the previous network is equal to the second octet of the identifier of the subsequent network minus 1:
16 -1 = 15.255.255, 24-1 = 23.255.255 and so on.
This will continue until the identifier of our network reaches the value of 20.120.0.0, because the ID of the next network will be 20.128.0.0. Thus, the network identifier for the IP address specified in the example condition will be 20.120.0.0, and its broadcast address will be 20.127.255.255, where 127.255.255 means 2.128.0.0. minus, that is, the ID of the next network minus one.
Our IP address 20.120.47.225/13 is located in the address range from 20.120.0.0 to 20.127.255.255, so these addresses are the answer to the question, what is the network identifier and the broadcast address of the network containing our IP address.
I want to tell you about the method of the shortest path. No matter what the position of the borrowed bit is, in our example it was 8, all these numbers are 1,2,4,8, 16, etc. - are factors of the number 128 and can not exceed it. Therefore, I look at the second octet of our IP address, equal to 120, and find out whether it is located after 128 or up to 128. In our case, it is located before 128. What am I doing? I subtract 128 from the number 8, that is, the location of the 5th borrowed bit, and get 120.0.0. Then the next network will have the value of the second to fourth octets 128.0.0. Due to this, it can be said that the first network ID will be 20.120.0.0, and its broadcast address will be 20.127.255.255.
If you do not know how to do these calculations in your mind, just do what I said earlier - just add the value of the location of the borrowed bit to the octet each time, getting the ID of the next network until you reach the value given in the example. Now let's go to the second example.
We need to determine the network ID and the broadcast address of the network containing the IP address 220.20.17.5/27.
Slash 27 means that we are dealing with the fourth octet, since 3x8 = 24, and the number 27 exceeds this value by 3, that is, it is in the fourth octet.
Thus, we borrowed 3 bits from the 4th octet, and their area of ​​location is 32. We can say that / 25 means 1 borrowed bit, / 26 means 2 bits, and / 27 - three bits, that is, we have 3 bit of the fourth octet. This means that to find the sequence of network identifiers we must add 32 to the fourth octet.
Let's start with the IP address 220.20.17.0 - this will be the ID of the first network, then the broadcast address will be 220.20.17.31, that is, the value of the 4th octet of the broadcast address of the first subnet will be (32-1).
The identifier of the second network is formed by adding 32 to the fourth octet (0 + 32) = 32 and will look like 220.20.17.32. Let's look at the problem condition: our address ends with 5, and the number 5 is in the range between 0 and 31, where 0 is the last octet of the network identifier, and 31 is the last octet of the broadcast address.
Thus, we do not need to calculate anything else - the answers to this problem are the addresses 220.20.17.0 and 220.20.17.3.
Let's move on to the next example. Here we need to determine the parameters of the network containing the IP address 10.10.7.17/19. First of all, we need to determine which octet belongs to / 19 - this is the 3rd octet, because two octets end in (2x8) = 16, which means that the number 19 is located in the 3rd octet. We see that here, as in the previous example, 3 bits were borrowed, since 19 = 16 +3. As in the previous example, in this case / 17 means 1 borrowed bit, / 18 - two bits and / 19 - 3 bits.
Thus, to determine the identifier of the first subnet, we must substitute 0 into the value of the 3rd octet and obtain an address of the form 10.10.0.0. Then the identifier of the second network will be 10.10.32.0, hence the broadcast address of the first subnet, which is less by 1, will be equal to 10.10.31.255.
Now we compare these parameters with our IP-address to check if it is in their range. We see that 10.10.7.17 is indeed located between 10.10.0.0 and 10.10.31.255, which are respectively the identifier and the broadcast address of the network containing it.
Consider the 4th example of the problem with the address 192.8.3.1/18. Many people think this is the wrong address. Consider in which case this address is correct, because it is a trick question. Here we have a case of a classless address, that is, we do not have a rigid framework for class addressing.
Although this address starts at 192, it cannot be assigned to class C, which has a 24-bit mask. I just want to remember that there are classless addresses, in this case, we may have, for example, IP address 192 with a slash 9, and this is completely normal. You should treat these addresses the same way as class addresses, that is, start looking at it from the end where / 18 is located, not paying attention to the first octet. As we know, / 18 denotes bits borrowed from the 3rd octet.
Thus, our subnet ID will be 192.8.0.0. The table shows that 2 bits were borrowed here (2x8 = 16 +2 = 18). This means that in the third octet of each subsequent network you need to add 64. So, if the first subnet identifier is 192.8.0.0, then the second network identifier will take the value 192.8.64.0, then the broadcast address of the first subnet will be less than one address and will take the value 192.8.63.255. Looking at the IP address from the example, we will see that it is located in the range between these two addresses, which means they are the answer to the question of the problem.
Now consider the last example with the address 172.1.4.5/20. A forward of 20 means that 4 bits from the 3rd octet are borrowed from us. Therefore, the first subnet identifier must be 172.1.0.0., Since in this case the 3 and the following octets are 0.
Looking at the table, we will see that the identifier of the second subnet is obtained by adding 16 to the third octet and will be equal to 172.1.16.0, which means that the broadcast address of the first subnet will be one less, that is, 172.1.15.255. Comparing our address with these values, we will determine that it is between them, that is, the addresses we have defined are the answer for this task.
Now I will add one more example. Let's look at the IP address 172.138.75.5/12. Slash 12 means the fourth bit of the second octet of the network identifier. The table shows that it corresponds to the number 16, that is, we must consistently add this number to the second octet of each next subnet, starting with the identifier of the first subnet 172.0.0.0, until we approach the value of the second octet 138.
So, we use a product with a multiplier of 16 and reach 128. The number in the second octet is 138, that is, one is greater than 128, but less than 128 +16 = 144. Thus, we have the subnet identifier 172.128.0.0, and the next subnet will be have an id of 172.144.0.0. By subtracting one, we get the broadcast address of the previous network.
Comparing our address with these addresses, we will see that it is located between them, which means we have found the right solution.
We obtained the identifier of the desired subnet by adding 16. However, you can save time on calculations if you use the shortest path method. If this is difficult for you, you can always go back to the usual method of adding the value of the location of the borrowed bit.
I want to draw your attention to the fact that I had a group on Facebook, where you can also ask your questions, and also want to advise, if you have not done so already, subscribe to my channel on the YouTube portal. Do not forget to share these video tutorials with your friends and leave your questions and comments below. You can also write me an e-mail address, and although I don’t use it much, I’ll still try to answer all your emails. I promise that in each video lesson I will consider your most interesting questions and answer them.
Thank you for staying with us. Do you like our articles? Want to see more interesting materials? Support us by placing an order or recommending to friends, 30% discount for Habr's users on a unique analogue of the entry-level servers that we invented for you: The whole truth about VPS (KVM) E5-2650 v4 (6 Cores) 10GB DDR4 240GB SSD 1Gbps from $ 20 or how to share the server? (Options are available with RAID1 and RAID10, up to 24 cores and up to 40GB DDR4).
VPS (KVM) E5-2650 v4 (6 Cores) 10GB DDR4 240GB SSD 1Gbps before summer for free if you pay for a period of six months, you can order here .
Dell R730xd 2 times cheaper? Only we have 2 x Intel TetraDeca-Core Xeon 2x E5-2697v3 2.6GHz 14C 64GB DDR4 4x960GB SSD 1Gbps 100 TV from $ 199 in the Netherlands! Dell R420 - 2x E5-2430 2.2Ghz 6C 128GB DDR3 2x960GB SSD 1Gbps 100TB - from $ 99! Read about How to build an infrastructure building. class c using servers Dell R730xd E5-2650 v4 worth 9000 euros for a penny?
Source: https://habr.com/ru/post/454456/
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