The networks traversed on a path to a network destination

The networks traversed on a path to a network destination

Describe the benefits of variable length subnet masks (VLSMs).

 

·         VLSMs enable the creation of subnets of specific sizes and allow the division of a classless network into smaller networks that do not need to be equal in size. This makes use of the address space more efficient because many times IP addresses are wasted with classful subnetting.

 

Understand the relationship between the subnet mask value and the resulting block size and the allowable IP addresses in each resulting subnet.

 

·         The relationship between the classful network being subdivided and the subnet mask used determines the number of possible hosts or the block size. It also determines where each subnet begins and ends and which IP addresses cannot be assigned to a host within each subnet.

 

Describe the process of summarization or route aggregation and its relationship to subnetting.

 

·         Summarization is the combining of subnets derived from a classful network for the purpose of advertising a single route to neighboring routers instead of multiple routes, reducing the size of routing tables and speeding the route process.

 

Calculate the summary mask that will advertise a single network representing all subnets.

 

·         The network address used to advertise the summary address is always the first network address in the block of subnets. The mask is the subnet mask value that yields the same block size.

 

Remember the four diagnostic steps.

 

·         The four simple steps that Cisco recommends for troubleshooting are ping the loopback address, ping the NIC, ping the d

 

Identify and mitigate an IP addressing problem.

 

·         Once you go through the four troubleshooting steps that Cisco recommends, you must be able to determine the IP addressing problem by drawing out the network and finding the valid and invalid hosts addressed in your network.

 

Understand the troubleshooting tools that Understand the troubleshooting tools that

 

·         The ping 127.0.0.1 command tests your local IP stack, and tracert is a Windows command to track the path a packet takes through an internetwork to a destination. Cisco routers use the command traceroute , or just trace for short. Don't confuse the Windows and Cisco commands. Although they produce the same output, they don't work from the same prompts. The command ipconfig /all will display your PC network configuration from a DOS prompt, and arp -a (again from a DOS prompt) will display IP-to-MAC-address mapping on a Windows PC.

 

For each of the following sets of networks, determine the summary address and the mask to be used that will summarize the subnets.

1. 192.168.1.0/24 through 192.168.12.0/24
2. 172.144.0.0 through 172.159.0.0
3. 192.168.32.0 through 192.168.63.0
4. 192.168.96.0 through 192.168.111.0
5. 66.66.0.0 through 66.66.15.0
6. 192.168.1.0 through 192.168.120.0
7. 172.16.1.0 through 172.16.7.0
8. 192.168.128.0 through 192.168.190.0
9. 53.60.96.0 through 53.60.127.0
10. 172.16.10.0 through 172.16.63.0

 

·         192.168.100.25/30. A /30 is 255.255.255.252. The valid subnet is 192.168.100.24, broadcast is 192.168.100.27, and valid hosts are 192.168.100.25 and 26.
2. 192.168.100.37/28. A /28 is 255.255.255.240. The fourth octet is a block size of 16. Just count by 16s until you pass 37. 0, 16, 32, 48. The host is in the 32 subnet, with a broadcast address of 47. Valid hosts 33-46.
3. A /27 is 255.255.255.224. The fourth octet is a block size of 32. Count by 32s until you pass the host address of 66. 0, 32, 64, 96. The host is in the 64 subnet, and the broadcast range is 65-94.
4. 192.168.100.17/29. A /29 is 255.255.255.248. The fourth octet is a block size of 8. 0, 8, 16, 24. The host is in the 16 subnet, broadcast of 23. Valid hosts 17-22.
5. 192.168.100.99/26. A /26 is 255.255.255.192. The fourth octet has a block size of 64. 0, 64, 128. The host is in the 64 subnet, broadcast of 127. Valid hosts 65-126.
6. 192.168.100.99/25. A /25 is 255.255.255.128. The fourth octet is a block size of 128. 0, 128. The host is in the 0 subnet, broadcast of 127. Valid hosts 1-126.
7. A default Class B is 255.255.0.0. A Class B 255.255.255.0 mask is 256 subnets, each with 254 hosts. We need fewer subnets. If we used 255.255.240.0, this provides 16 subnets. Let's add one more subnet bit. 255.255.248.0. This is 5 bits of subnetting, which provides 32 subnets. This is our best answer, a /21.
8. A /29 is 255.255.255.248. This is a block size of 8 in the fourth octet. 0, 8, 16. The host is in the 8 subnet, broadcast is 15.
9. A /29 is 255.255.255.248, which is 5 subnet bits and 3 host bits. This is only 6 hosts per subnet.
10. A /23 is 255.255.254.0. The third octet is a block size of 2. 0, 2, 4. The subnet is in the 16.2.0 subnet; the broadcast address is 16.3.255.

 

 

On a VLSM network, which mask should you use on point-to-point WAN links in order to reduce the waste of IP addresses?

A. /27
B. /28
C. /29
D. /30
E. /31

 

·         D. A point-to-point link uses only two hosts. A /30, or 255.255.255.252, mask provides two hosts per subnet.

 

Question 2 has a Diagram. (can't be posted) In the network shown in the diagram, how many computers could be in subnet B?

 

·         C. Using a /28 mask, there are 4 bits available for hosts. Two to the fourth power minus 2 = 14.

 

In the diagram below, in order to have as efficient IP addressing as possible, which network should use a /29 mask?

A. A
B. B
C. C
D. D

 

·         D. For 6 hosts we need to leave 3 bits in the host portion since 2 to the third power = 8 and 8 less 2 is 6.With 3 bits for the host portion, that leaves 29 bits for the mask or /29.

 

To use VLSM, what capability must the routing protocols in use possess?

A. Support for multicast
B. Multiprotocol support
C. Transmission of subnet mask information
D. Support for unequal load balancing

 

·         C. To use VLSM, the routing protocols in use possess the capability to transmit subnet mask information.

 

 

balancing 5. What summary address would cover all the networks shown and advertise a single, efficient route to Router B that won't advertise more networks than needed?

A. 172.16.0.0/24
B. 172.16.1.0/24
C. 172.16.0.0/24
D. 172.16.0.0/20
E. 172.16.16.0/28
F. 172.16.0.0/27

 

·         D. In a question like this, you need to look for an interesting octet where you can combine networks. In this example, the third octet has all our subnets so we just need to find our block size now. If we used a block of 8 starting at 172.16.0.0/19, then we cover 172.16.0.0 through 172.16.7.255. However, if we used 172.16.0.0/20, then we'd cover a block of 16 which would be from 172.16.0.0 through 172.16.15.255, which is the best answer.

 

In the diagram below what is the most likely reason the station cannot ping outside of its network?

 

·         C. The IP address of the station and the gateway are not in the same network. Since the address of the gateway is correct on the station, it is most likely the IP address of the station is incorrect.

 

 

If Host A is configured with an incorrect default gateway and all other computers and the router are known to be configured correctly, which of the following statements is TRUE?
A. Host A cannot communicate
B. Host A can communicate with other hosts in the same subnet.
C. Host A can communicate with hosts in other subnets. D. Host A can communicate with no other systems.

 

·         B. With an incorrect gateway, Host A will not be able to communicate with the router or beyond the router but will be able to communicate within the subnet.

 

Which of the following troubleshooting steps, if completed successfully, also confirms the other steps will succeed as well?

A. ping a remote computer
B. ping the loopback address
C. ping the NIC
D. ping the default gateway

 

·         Pinging the remote computer would fail if any of the other tests fail.

 

When a ping to the local host IP address fails, what can you assume?

A. The IP address of the local host is incorrect.
B. The IP address of the remote functional.
C. When a ping to the local host IP address fails, you can assume the NIC is not functional.
D. The IP stack has failed to

 

·         C. When a ping to the local host IP address fails, you can assume the NIC is not functional.

 

When a ping to the local host IP address succeeds but a ping to the default gateway IP address fails, what can you rule out? (Choose all that apply.)
A. The IP address of the local host is incorrect.
B. The IP address of the gateway is incorrect.
C. The NIC is not functional.
D. The IP stack has failed to initialize.

 

·         C, D. If a ping to the local host succeeds, you can rule out IP stack or NIC failure.

 

Which of the networks in the diagram could use a /29 mask?

A. Corporate
B. LA
C. SF
D. NY
E. None

 

·         E. A /29 mask yields only 6 addresses, so none of the networks could use it.

 

What network service is the most likely problem if you can ping a computer by IP address but not by name?

A. DNS
B. DHCP
C. ARP
D. ICMP

 

·         The most likely problem if you can ping a computer by IP address but not by name is a failure of DNS.

 

When you issue the ping command, what protocol are you using?
A. DNS
B. DHCP
C. ARP
D. ICMP

 

·         D. When you issue the ping command, you are using the ICMP protocol.

 

Which of the following commands displays the networks traversed on a path to a network destination?

A. Ping
B. Traceroute
C. Pingroute
D. Pathroute

 

·         B. The traceroute command displays the networks traversed on a path to a network destination.

 

What command generated the output shown below? Reply from 172.16.10.2: bytes=32 time<1ms TTL=128 Reply from 172.16.10.2: bytes=32 time<1ms TTL=128 Reply from 172.16.10.2: bytes=32 time<1ms TTL=128 Reply from 172.16.10.2: bytes=32 time<1ms TTL=128

A. traceroute
B. show ip route
C. ping
D. pathping

C. The ping command tests connectivity to another station. The full command is shown below.

·        
C:\> ping 172.16.10.2 Pinging 172.16.10.2 with 32 bytes of data: Reply from 172.16.10.2: bytes=32 time<1ms TTL=128 Reply from 172.16.10.2: bytes=32 time<1ms TTL=128 Reply from 172.16.10.2: bytes=32 time<1ms TTL=128 Reply from 172.16.10.2: bytes=32 time<1ms TTL=128 Ping statistics for 172.16.10.2: Packets: Sent = 4, Received = 4, Lost = 0 (0% loss), Approximate round trip times in milli-seconds: Minimum = 0ms, Maximum = 0ms, Average = 0ms

 

Which of the following network addresses correctly summarizes the three networks shown below efficiently? 10.0.0.0/16 10.1.0.0/16 10.2.0.0/16

A. 10.0.0.0/15
B. 10.1.0.0/8
C. 10.0.0.0/14
D. 10.0.0.8/16

 

·         C. The interesting octet in this 255.252.0.0 mask, we are telling the summary to use a block size of four in the 2nd octet. This will cover 10.0.0.0 through 10.3.255.255. This is the best answer.

 

What command displays the ARP table on a Cisco router?

A. show ip arp
B. traceroute
C. arp -a
D. Tracert

 

·         The command that displays the ARP table on a Cisco router is show ip arp .

 

What switch must be added to the ipconfig command on a PC to verify DNS configuration?

A. /dns
B. -dns
C. /all
D. Showall

 

·         C. The /all switch must be added to the ipconfig command on a PC to verify DNS configuration.

 

Which of the following is the 192.168.128.0 through 192.168.159.0

A. 192.168.0.0/24
B. 192.168.128.0/16
C. 192.168.128.0/19
D. 192.168.128.0/20

 

·         C. If you start at 192.168.128.0 and go through 192.168.159.0, you can see this is a block of 32 in the third octet. Since the network address is always the first one in the range, the summary address is 192.168.128.0. What mask provides a block of 32 in the third octet? The answer is 255.255.224.0, or /19.