Question

6. In the 192.168.0.0/16 address block there are a total of __________ IP version 4 addresses.

65,536

16,777,216

4,096

1,048,576

7. Given the IPv4 address and network mask 134.124.133.139/22, identify the broadcast address in that network (the last address). [you may use the ip_addressing_basics spreadsheet, see "IPv4 Address, Network ID, etc." sheet]

134.124.132.0/22

134.124.133.255/22

134.124.132.255/24

134.124.135.255/22

8. Express the CIDR mask /18 using the dotted decimal notation.

255.255.192.0

255.192.0.0

255.255.255.0

255.255.240.0

9.

Host A has IP address/network mask: 192.168.3.222/27

Host B has IP address/network mask: 192.168.3.225/27

Given the information above, it can be inferred that:

Both host A and host B are on the same network

Host A and Host B are on different networks

Host A and Host B will have the same routing tables

Host A and Host B will be able to communicate directly without an intervening router.

10. A Host (the sending host) is trying to send a packet to another host (the destination host). The sending host knows the IP address of destination host. In deciding how to forward the packet toward destination host, the sending host will:

take the Destination IP address and perform a Logical AND operation with its own (sending hosts') Network Mask to determine if the destination host is on the same network as the sending host.

take the Destination IP address and perform a Logical AND operation with destination host's Network Mask to determine if the destination host is on the same network as the sending host.

always forward all packets to the default gateway and not be concerned with determining if the destination host is on the same network as the sending host.

never forward any packets to the default gateway and not be concerned with determining if the destination host is on the same network as the sending host.

Answer 1

6) 192.168.0.0/16 contains 16 bits for host. Hence the number of addresses is 2^16 = 65536

7) IP address 134.124.133.139 in bitwise format (the network bits are bolded):

10000110 . 01111100 . 10000101 . 10001011

The broadcast address will contain all host bits set to 1 - 10000110 . 01111100 . 10000111 . 11111111 -

134.124.135.255

8) /18 in bitwise format is - 11111111 . 11111111 . 11000000 . 00000000 - 255.255.192.0

9) 192.168.3.222/27 in bitwise form - 11000000 . 10101000 . 00000011 . 11011110

192.168.3.225/27 in bitwise form - 11000000 . 10101000 . 00000011 . 11100001

The bolded bits form the network id. Since the network ID is not the same, they are on different networks.

10)

The correct answer is -

take the Destination IP address and perform a Logical AND operation with its own (sending hosts') Network Mask to determine if the destination host is on the same network as the sending host.

If its on the same network, it will send on that interface, otherwise it will send it to the interface connected to the default gateway if a default gateway is configured.