Chapter 18 Problems

In classless addressing, show the whole address space as a single block using the CIDR notation.

2^32 addresses, if prefix = 0 N = 2^32, the entire address space can be identified by 0.0.0.0

Show the n leftmost bits of the following network-addresses/masks that can be used in a forwarding table. a.) 170.40.11.0/24 b.) 110.40.240.0/22 c.) 70.14.0.0./18

Convert address to binary and count out the bits given as the prefix length: a.) 10101010 00101000 00001011 00000000 n = 24 => 10101010 00101000 00001011 b.) 01101110 00101000 11110000 00000000 n = 22 => 01101110 00101000 111100 c.) 01001000 00001110 00000000 00000000 n = 18 => 01001000 00001110 00

Change each of the following masks to a prefix length: a.) 255.224.0.0 b.) 255.240.0.0 c.) 255.255.255.128

Convert to binary and count the number of bits before the first 0: a.) 11111111 11100000 0... => n = 11 b.) 11111111 11110000 0... => n = 12 c.) 11111111 11111111 11111111 10000000 => n = 25

Which of the following cannot be a mask in CIDR? a.) 255.225.0.0 b.) 255.192.0.0 c.) 255.255.255.6

Convert to binary and find any where a one comes after a zero: 'a' and 'c' cannot be CIDR addresses

Explain how DHCP can be used when the size of the block assigned to an organization is less than the number of hosts in the organization.

DHCP dynamically assigns IP addresses to those who need them and unassign when their operation is complete.

An ISP is granted the block 80.70.56.0/21. The ISP needs to allocate addresses for two organizations each with 500 addresses, two organizations each with 250 addresses, and three organizations each with 50 addresses. a.) Find the number and range of addresses in the ISP block. b.) Find the range of addresses for each organization and the range of unallocated addresses.

Follow same process as p18-21

Combine the following three blocks of addresses into a single block: a.) 16.27.24.0/26 b.) 16.27.24.64/26 c.) 16.27.24.128/25

Follow the same process as p18-12 to find first and last address, then find the range. 64 + 64 + 128 = 256 total addresses n = 32-logBASE(2)(256) = 24 => 16.27.24.0/24 to 14.27.24.255/24

Can router R1 in Figure 18.35 receive a packet with destination address 140.24.7.194? What will happen to the packet if this occurs?

If it comes from org1-3 it gets forwarded to R2. Cannot come from R2

A large organization with a large block address (12.44.184.0/21) is split into one medium-size company using the block address (12.44.184.0/22) and two small organizations. If the first small company uses the block (12.44.188.0/23), what is the remaining block that can be used by the second small company? Explain how the datagrams destined for the two small companies can be correctly routed to these companies if their address blocks still are part of the original company.

In notebook

Assume we have an internet with a 12-bit address space. The addresses are equally divided between eight networks (N0 to N7). The internetwork communication is done through a router with eight interfaces (m0 to m7). Show the internet outline and the forwarding table (with two columns: prefix in binary and the interface number) for the only router that connects the networks. Assign a network address to each network.

In notebook

Assume we have an internet with a 9-bit address space. The addresses are divided between three networks (N0 to N2), with 64, 192, and 256 addresses respectively. The internetwork communication is done through a router with three interfaces (m0 to m2). Show the internet outline and the forwarding table (with two columns: prefix in binary and the interface number) for the only router that connects the networks. Assign a network address to each network.

In notebook

Assume we have an internet with an 8-bit address space. The addresses are equally divided between four networks (N0 to N3). The internetwork communication is done through a router with four interfaces (m0 to m3). Show the internet outline and the forwarding table (with two columns: prefix in binary and the interface number) for the only router that connects the networks. Assign a network address to each network.

In notebook

Compare NAT and DHCP. Both can solve the problem of a shortage of addresses in an organization, but by using different strategies.

NAT utilizes a private network and a limited number of private addresses for local communication. DHCP dynamically assigns addresses as it sees fit.

Assume router R2 in Figure 18.35 receives a packet with destination address 140.24.7.42. How is the packet routed to its final destination?

R2 finds network address using its prefix and subnet mask and searches forwarding table. Continues until 140.24.7.0/24 matches and is then routed to R1. R1 does the same search and matches with Org. 1, which the packet is sent to.

Each of the following addresses belongs to a block. Find the first and the last address in each block. a.) 14.12.72.8/24 b.) 200.107.16.17/18 c.) 70.110.19.17/16

Transform into binary then AND with the mask to find first address. OR with the NOT of the mask to get the last address: a.) Addr.: 00001110 00001100 01001000 00001000 Mask: 11111111 11111111 11111111 00000000 Addr. AND Mask = 000011110 00001100 01001000 00000000 = 14.12.72.0 Addr. OR NOT Mask = 00001110 00001100 01001000 11111111 = 14.12.72.255 b.) and c.) Look at the process for part a and you'll figure it out.

Change each of the following prefix lengths to a mask in dotted-decimal notation: a.) n = 0 b.) n = 14 c.) n = 30

a.) 00000000 00000000 00000000 00000000 = 0.0.0.0 b.) 11111111 11111100 00000000 00000000 = 255.252.0.0 c.) 11111111 11111111 11111111 11111100 = 255.255.255.252

Rewrite the following IP addresses using binary notation: a.) 110.11.5.88 b.) 12.74.16.18 c.) 201.24.44.32

a.) 01101110 00001011 00000101 01011000 b.) 00001100 01001010 00010000 00010010 c.) 11001001 00011000 00101100 00100000

Find the class of the following classful IP addresses: a.) 130.34.54.12 b.) 200.34.2.1 c.) 245.34.2.8

a.) 130 is between 128 and 191 => Class B b.) 200 is between 192 and 223 => Class C c.) 245 is between 240 and 254 => Class E

What is the size of the address space in each of the following systems? a.) A system in which each address is only 16 bits. b.) A system in which each address is made of six hexadecimal digits. c.) A system in which each address is made of four octal digits.

a.) 2^16 b.) 4 bits per hex digits: 6 x 4 = 24 bits => 2^24 addresses c.) 3 bits per octal digit: 4 x 3 = 12 bits => 2^12 addresses

Rewrite the following IP addresses using dotted-decimal notation: a.) 01011110 10110000 01110101 00010101 b.) 10001001 10001110 11010000 00110001 c.) 01010111 10000100 00110111 00001111

a.) 94.176.117.21 b.) 137.142.208.49 c.) 86.132.55.15

Find the class of the following classful IP addresses: a.) 01110111 11110011 10000111 11011101 b.) 11101111 11000000 11110000 00011101 c.) 11011111 10110000 00011111 01011101

a.) First byte begins with 0 => Class A b.) First byte begins with 1110 => Class D c.) First byte begins with 110 => Class C

An ISP is granted the block 16.12.64.0/20. The ISP needs to allocate addresses for 8 organizations, each with 256 addresses. a.) Find the number and range of addresses in the ISP block. b.) Find the range of addresses for each organization and the range of unallocated addresses.

a.) Follow the same process as p18-12: 4096 addresses, 16.12.64.0 to 16.12.79.255 b.) 1: 16.12.64.0 to 16.12.64.255 2: 16.12.65.0 to 16.12.65.255 ... 8: 16.12.71.0 to 16.12.17.255 Unassigned: 16.12.72.0 to 16.12.79.255

In classless addressing, what is the size of the block (N) if the value of the prefix length (n) is one of the following? a.) n = 0 b.) n = 14 c.) n = 32

a.) N = 2^(32 - 0) = 2^32 b.) N = 2^(32 - 14) = 2^18 c.) N = 2^(32 - 32) = 1

An organization is granted the block 130.56.0.0/16. The administrator wants to create 1024 subnets. a.) Find the number of addresses in each subnet. b.) Find the subnet prefix. c.) Find the first and the last address in the first subnet. d.) Find the first and the last address in the last subnet.

a.) N = 2^(32 - 16) - 65535, 65535 / 1024 = 64 addresses per subnet b.) n = 32 - logBASE(2)(64) = 26 c.) 130.56.0.0/26 to 130.56.0.63/26 d.) 130.56.0.0 + 0.0.255.192 = 130.56.255.192/26 to 130.56.255.255/26

In classless addressing, what is the value of the prefix length (n) if the size of the block (N) is one of the following? a.) N = 1 b.) N = 1024 c.) N = 2^32

a.) n = 32 - logBASE(2)(1) = 32 b.) n = 32 - logBASE(2)(1024) = 22 c.) n = 32 - logBASE(2)(2^32) = 0