Network layers protocols
1. Match between the protocol to its main function
Protocol name function
(1) used to transfer packets.
 IP
(2) used to report unexpected events occur during packet processing.
 IMCP
(3) used to find the destination’s physical address (Ethernet address) to send a frame,
 ARP
By broadcasting a packet asking who owns the destination’s IP address
 DHCP
(4) used to request new IP and dynamic configurations.
2. IP
is the protocol that is used for data transfer.
3. ICMP is a control protocol used for reporting unexpected events occur during packet processing.
4. ARP
is a control protocol used to find the destination’s physical address (Ethernet address) to send a
frame, by broadcasting a packet askig who owns the destination’s IP address
5. DHCP Is a control protocol to request new IP and dynamic configuration.
6. A network on the Internet has a subnet mask of 255.255.240.0. What is the maximum number of hosts
it can handle?
Number of bits in the host part = 12  number of hosts = 212-2 =4K -2
7. A router has just received the following new IP addresses: 57.6.96.0/21, 57.6.104.0/21, 57.6.112.0/21,
and 57.6.120.0/21 and all of them use the same outgoing line, what can they be aggregated to?
57.6.96.0 = 0011 1001 0000 0110 0110 0000 0000 0000
57.6.104.0 = 0011 1001 0000 0110 0110 1000 0000 0000
57.6.112.0 = 0011 1001 0000 0110 0111 0000 0000 0000
57.6.120.0 = 0011 1001 0000 0110 0111 1000 0000 0000
Aggregated prefix = 0011 1001 0000 0110 0110 0000 0000 0000 = 57.6.96.0/19
8. A router has the following (CIDR) entries in its routing table:
Address/mask
Next hop
Subnet mask
135.46.56.0/22
Interface 0 255.255.252.0
135.46.60.0/22
Interface 1 255.255.252.0
192.53.40.0/23
Router 1
255.255.254.0
default
Router 2
-For each of the following IP addresses, what does the router do if a packet with that address arrives?
(a) 135.46.63.10
(b) 135.46.57.14
(c) 135.46.52.2
(d) 192.53.40.7
(e) 192.53.56.7
(a) 135.46.63.10 AND 255.255.252.0 = 135.46.60.0  Interface 1
(b) 135.46.57.14 AND 255.255.252.0 = 135.46.56.0  Interface 0
(c) 135.46.52.2 AND 255.255.252.0 = 135.46.52.0  Router 2
(d) 192.53.40.7 AND 255.255.254.0 = 192.53.40.0  Router 1
(e) 192.53.56.7 AND 255.255.254.0 = 192.53.56.0  Router 2
9. A large number of consecutive IP addresses are available starting at 198.16.0.0. Suppose that four
organizations, A, B, , and D, request 4000, 2000, , and 8000 addresses, respectively, and in that order.
For each of these, give the first IP address assigned, the last IP address assigned, and the mask in the
w.x.y.z/s notation.
The total number of requested IPs = 4000+2000+8000
= 14000 <= 214  # bits in the host part = 14  # bits in the network part = 18
The original prefix : 198.16.0.0/17
Subnet A: 4000 <= 212  # bits in the host part = 12  # bits in the network part = 20
Possible subnets
1100 0110 0001 0000 0000 0000 0000 0000 --- the valid first subnet
1100 0110 0001 0000 0001 0000 0000 0000
1100 0110 0001 0000 0010 0000 0000 0000
1100 0110 0001 0000 0011 0000 0000 0000
Subnet A: 198.16.0.0/20
Subnet B: 2000 <= 211  # bits in the host part = 11  # bits in the network part = 21
Possible subnets
1100 0110 0001 0000 0000 0000 0000 0000 --- taken
1100 0110 0001 0000 0000 1000 0000 0000 --- taken
1100 0110 0001 0000 0001 0000 0000 0000 --- the first valid subnet
1100 0110 0001 0000 0001 1000 0000 0000
1100 0110 0001 0000 0010 0000 0000 0000
1100 0110 0001 0000 0010 1000 0000 0000
1100 0110 0001 0000 0011 0000 0000 0000
1100 0110 0001 0000 0011 1000 0000 0000
Subnet B: 198.16.16.0/21
Subnet C: 8000 <= 213  # bits in the host part = 13  # bits in the network part = 19
Possible subnets
1100 0110 0001 0000 0000 0000 0000 0000 --- taken
1100 0110 0001 0000 0010 0000 0000 0000 --- the first valid subnet
Subnet C: 198.16.32.0/19