NAME: ________________________

IP Subnetting, Variable Length Subnet Masks (VLSMs), & Troubleshooting IP Test

Question Types


Start With


Question Limit

of 81 available terms

Upgrade to
remove ads

5 Written Questions

5 Matching Questions

  1. Additional Cisco Troubleshooting
  2. The network address of 172.16.0.0/19 provides how many subnets and hosts?
  3. Subnet Mask
  4. Unable to Ping Loopback
  5. How many subnets?
  1. a Tracert, Show ip arp
  2. b IP stack failure; Reinstall TCP/IP
  3. c 2^x where x is the number of masked bits (or 1's)
  4. d A CIDR address of /19 is 255.255.224.0. This is a Class B address, so that is only 3 subnet bits but provides 13 host bits, or 8 subnets, each with 8,190 hosts.
  5. e 32-bit value that allows the recipient of IP packets to distinguish the network ID portion of the IP address from the host ID portion of the IP address

5 Multiple Choice Questions

  1. all nodes in the network use the same subnet mask
  2. /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
  3. Optimized network performance
  4. First, if you have two hosts directly connected, as shown in the graphic, then you need a crossover cable. A straight-through cable won't work. Second, the hosts have different masks, which puts them in different subnets. The easily solution is just to set both masks to 255.255.255.0 (/24).
  5. A point-to-point link uses only two hosts. A /30, or 255.255.255.252, mask provides two hosts per subnet.

5 True/False Questions

  1. Your router has the following IP address on Ethernet0: 172.16.2.1/23. Which of the following can be valid host ID's on the LAN interface attached to the router?The routers IP address on the E0 interface is 172.16.2.1/23, which is a 255.255.254.0. This makes the third octet a block size of 2. The routers interface is in the 2.0 subnet, the broadcast address is 3.255 because the next subnet is 4.0. The valid host range is 2.1 through 3.254. The router is using the first valid host address in the range.

          

  2. /2711100000 = 224

          

  3. What is the subnetwork number of a host with an IP address of 172.16.66.0/21?A /21 is 255.255.248.0, which means we have a block size of 8 in the third octet, so we just count by 8 until we reach 66. The subnet in this question is 64.0. The next subnet is 72.0, so the broadcast address of the 64 subnet is 71.255.

          

  4. Given 172.16.0.0/18, How many subnets?4

          

  5. Given 192.168.10.0/28, How many hosts per subnet?2

          

Create Set