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5 Written questions

5 Matching questions

  1. Variable Length Subnet Masks (VLSMs)
  2. Able to ping but still unable to communicate
  3. 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?
  4. Why subnet?
  5. What subnet and broadcast address is the IP address 172.16.50.10 255.255.224.0 (/19) a member of?
  1. a Possible DNS problem
  2. b The subnet is 172.16.32.0, and the broadcast must be 172.16.63.25
  3. c 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.
  4. d Simplified management
  5. e each network segment can use a different subnet mask

5 Multiple choice questions

  1. RIPv1 and IGRP
  2. To test the local stack on your host, ping the loopback interface of 127.0.0.1
  3. 11111000 = 248
  4. A /25 mask is 255.255.255.128. Used with a Class B network, the third and fourth octets are used for subnetting with a total of 9 subnet bits, 8 bits in the third octet and 1 bit in the fourth octet. Since there is only 1 bit in the fourth octet, the bit is either off or on—which is a value of 0 or 128. The host in the question is in the 0 subnet, which has a broadcast address of 127 since 128 is the next subnet.
  5. A Class B network ID with a /22 mask is 255.255.252.0, with a block size of 4 in the third octet. The network address in the question is in subnet 172.16.16.0 with a broadcast address of 172.16.19.255. Only option E even has the correct subnet mask listed, and 172.16.18.255 is a valid host.

5 True/False questions

  1. What's the broadcast address of each subnet?63.255, 127.255, 191.255, 255.255

          

  2. Unable to Ping Default GatewayIP stack failure; Reinstall TCP/IP

          

  3. 192.168.100.37/28/29 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 host 17-22

          

  4. Given 172.16.0.0/18, What are the valid hosts?.1 - .126 & .129 - .254

          

  5. Classless Routing Protocolsall nodes in the network use the same subnet mask

          

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