1. Chapter 4: VLSM, Summarization And Troubleshooting TCP/IP
Prepared by: Adeel Ahmad
Lect. CNET Department.
Updated by: Syed Ameen Quadri.
Jazan University, Jazan.

2. Chapter 4 Objectives The CCENT Topics Covered in this chapter include:
IP addressing (IPv4 / IPv6)
Identify the appropriate IPv4 addressing scheme using VLSM and summarization to satisfy addressing requirements in a LAN/WAN environment.
Troubleshooting
Troubleshoot and correct common problems associated with IP addressing and host configurations. 2

3. What is VLSM? VLSM Stands for Variable Length Subnet Mask.
It means that we can vary the length of subnet mask.
We know that subnetting is splitting of a network into multiple networks, so as to avoid the wastage of IP Address.
VLSM is also a part of subnetting only, or you may say that it is an advance step of subnetting.
By using VLSM we can even reduce the wastage of IP Address.
To understand this concept, we will take an example.

4. Here is our network…
In this scenario, we have some LANs and some are connected to WAN, total 5 networks. That is 4 LAN and 1 WAN.
Requirement :
1 LAN 30 IPs Address, 1 LAN 20s IP Address, 1 LAN 10s IP Address, 1 LAN 04 IPs Address & 1 WAN only 2 IPs address.

5. Network requirement diagram
Required Host - 10
IP /24
Required Host - 04
Required Host - 02
Required Host - 30
Required Host - 20

6. First step…
The IP address given in the example is /24, it is a class C, with default mask.
So we have divided it with size greater then or equal to 30. so we will get the subnet as follows.
/27
/27
/27
/27
/27
By using this method, we have divided one IP into 5 different subnets, that is –
32, 64, 96, 128 & 160

7. Solution continue….
Then also wastage of IP is there, because in one network we need only 2 host, we have all subnets of 30 hosts, that means 28 are waste.
To reduce it again further, the procedure is VLSM, by using this method we can even reduce the wastage of IPs.
And it can be used in more efficient way.

8. Revise the old formula. Number of Subnet / Host = 2n – 2
Where n = Number of bits taken from network
Block size = 256 – Mask
And remember this table for quick reference.

9. Our equation is…. Total 5 subnet. 1 subnet - 30 Hosts

10. Start subnetting for 30 Hosts
Our IP address is /24
Default subnet mask is and its binary is
Now we need Number of Hosts required = > 30
2n – 2 = > 30, so we will take different values of n and check it for the condition of >= 30.
21 – 2 = 0 < 30, Not satisfied, 22 – 2 = 2 < 30, Not satisfied
23 – 2 = 6 < 30, Not satisfied, 24 – 2 = 14 < 30, Not satisfied
25 – 2 = 30 = 30, Satisfied, From this the value of n = 5, then Mask is
that is
New subnet mask is and in CIDR it is /27
And the block size is 256 – 224 = 32
So the subnet will be…

11. New subnet will be like this…
Subnet 0 – /27 (Reserved)
Subnet 1 Net Add. – /27
Host – /27
Host – /27
Subnet 1 Br. Add. – /27
Subnet 2 Net Add. –
This subnet we have created for 30 Hosts.

12. Our equation is…. Total 5 subnet. 1 subnet - 30 Hosts [Over]

13. Next subnet for 20 Hosts… Our IP address is 200.10.20.64 /27
Default subnet mask is and its binary is
Now we need Number of Hosts required = > 20
2n – 2 = > 20, so we will take different values of n and check it for the condition of >= 20.
21 – 2 = 0 < 20, Not satisfied, 22 – 2 = 2 < 20, Not satisfied
23 – 2 = 6 < 20, Not satisfied, 24 – 2 = 14 < 20, Not satisfied
25 – 2 = 30 > 20, Satisfied, From this the value of n = 5, then Mask is
that is
New subnet mask is and in CIDR it is /27
And the block size is 256 – 224 = 32
So the subnet will be…

14. Second subnet… Subnet 1 Br. Add – 200.10.20.63 /27
Subnet 2 Net Add. – /27
Host – /27
Host – /27
Subnet 2 Br. Add. – /27
Subnet 3 Net Add – /27
This subnet we have created for 30 Hosts, which is near to 20 Hosts.

15. Up to this point we created 2 subnets
/27
/27
/27
/27
For 30 Hosts
For 20 Hosts

16. Our equation is…. Total 5 subnet. 1 subnet - 30 Hosts [Over]

17. Next subnet for 10 Hosts… Our IP address is 200.10.20.96
Default subnet mask is and its binary is
Now we need Number of Hosts required = > 10
2n – 2 = > 10, so we will take different values of n and check it for the condition of >= 10.
21 – 2 = 0 < 10, Not satisfied, 22 – 2 = 2 < 10, Not satisfied
23 – 2 = 6 < 10, Not satisfied,
24 – 2 = 14 > 10, Satisfied, From this the value of n = 4, then Mask is
that is
New subnet mask is and in CIDR it is /28
And the block size is 256 – 240 = 16
So the subnet will be…

18. Second subnet… Subnet 2 Br. Add – 200.10.20.95 /27
Subnet 3 Net Add. – /28
Host – /28
Host – /28
Subnet 3 Br. Add. – /28
Subnet 4 Net Add. –
This subnet we have created for 16 Hosts, which is near to 10 Hosts.

19. Up to this point we created 3 subnets
/27
/27
/27
/27
/28
/28
For 30 Hosts
For 20 Hosts
For 10 Hosts

20. Our equation is…. Total 5 subnet. 1 subnet - 30 Hosts [Over]

21. Next subnet for 04 Hosts… Our IP address is 200.10.20.112
Default subnet mask is and its binary is
Now we need Number of Hosts required = > 04
2n – 2 = > 04, so we will take different values of n and check it for the condition of >= 04.
21 – 2 = 0 < 04, Not satisfied, 22 – 2 = 2 < 04, Not satisfied
23 – 2 = 6 > 04, Satisfied, From this the value of n = 3, then Mask is
that is
New subnet mask is and in CIDR it is /29
And the block size is 256 – 240 = 8
So the subnet will be…

22. Second subnet… Subnet 3 Br. Add – 200.10.20.111 /29
Subnet 4 Net Add. – /29
Host – /29
Host – /29
Subnet 4 Br. Add. – /29
Subnet 5 Net Add. –
This subnet we have created for 08 Hosts, which is near to 04 Hosts.

23. Up to this point we created 4 subnets
/27
/27
/27
/27
/28
/28
/29
/29
For 30 Hosts
For 20 Hosts
For 10 Hosts
For 04 Hosts

24. Our equation is…. Total 5 subnet. 1 subnet - 30 Hosts [Over]

25. Next subnet for 02 Hosts… Our IP address is 200.10.20.120
Default subnet mask is and its binary is
Now we need Number of Hosts required = > 02
2n – 2 = > 02, so we will take different values of n and check it for the condition of >= 02.
21 – 2 = 0 < 02, Not satisfied,
22 – 2 = 2 , Satisfied, From this the value of n = 2, then Mask is
that is
New subnet mask is and in CIDR it is /30
And the block size is 256 – 252 = 4
So the subnet will be…

26. Second subnet… Subnet 4 Br. Add – 200.10.20.119 /29
Subnet 5 Net Add. – /30
Host – /30
Host – /30
Subnet 5 Br. Add. – /30
Subnet 6 Net Add. –
This subnet we have created for 02 Hosts.
In this way by using VLSM we have done subnetting to save the wastage of IPs……
2 Hosts

27. Up to this point we created 4 subnets
/27
/27
/27
/27
/28
/28
/29
/29
For 30 Hosts
For 20 Hosts
For 10 Hosts
For 04 Hosts
For 02 Hosts

28. We have completed subnetting by VLSM.
Total 5 subnet.
1 subnet - 30 Hosts [Over]
1 subnet - 20 Hosts [Over]
1 subnet - 10 Hosts [Over]
1 subnet - 04 Hosts [Over]
1 subnet - 02 Hosts [Over]

29. Figure 4.2: Classless network design
Now remember that we can use different size masks on each router interface. If we use a /30 on our WAN links and a /27, /28, and /29 on our LANs, we’ll get 2 hosts per WAN interface and 30, 14, and 6 hosts per LAN interface

30. Figure 4.3: The VLSM table

31. Figure 4.4: VLSM network Ex – 1.
In Figure 5.4, we have four WAN links and four LANs connected together, so we need to create a VLSM network that will save address space. Looks like we have two block sizes of 32, a block size of 16, and a block size of 8, and our WANs each have a block size of 4.

32. Figure 4.5: VLSM table Ex – 1.

33. Figure 4.6: VLSM network Ex – 2.
Figure 5.6 shows a network with 11 networks, two block sizes of 64, one of 32, five of 16, and three of 4.

34. Figure 4.7: VLSM table Ex – 2.

35. Figure 4.8: Summary address used in an internetwork
Figure 4.8 shows how a summary address would be used in an internetwork.

36. Figure 4.9: Summarization Ex – 4.
The Ethernet networks connected to router R1 are being summarized to R2 as /20. Which IP addresses will R2 forward to R1 according to this summary?

37. Figure 4.10: Summarization Ex – 5.
Okay, last one. In Figure 4.10, there are five networks connected to router R1. What’s the best summary address to R2?

38. Figure 4.10: Basic IP troubleshooting

39. Here are the four troubleshooting steps Cisco recommends:
1. Open a Command window and ping This is the diagnostic, or loopback, address, and if you get a successful ping, your IP stack is considered initialized. If it fails, then you have an IP stack failure and need to reinstall TCP/IP on the host.
2. From the Command window, ping the IP address of the local host If that’s successful, your network interface card (NIC) is functioning. If it fails, there is a problem with the NIC. Success here doesn’t just mean that a cable is plugged into the NIC, only that the IP protocol stack on the host can communicate to the NIC via the LAN driver.
3. From the CMD window, ping the default gateway (router). If the ping works, it means that the NIC is plugged into the network and can communicate on the local network. If it fails, you have a local physical network problem that could be anywhere from the NIC to the router.
4. If steps 1 through 3 were successful, try to ping the remote server. If that works, then you know that you have IP communication between the local host and the remote server. You also know that the remote physical network is working.

40. Written Labs and Review Questions
Read through the Exam Essentials section together in class
Open your books and go through all the written labs and the review questions.
Review the answers in class. 40

41. Thank U!!!