Pages 558 - 560 of the textbook: P18-2 (10 pts) Rewrite the following IP address

Question

Pages 558 - 560 of the textbook:

P18-2 (10 pts) Rewrite the following IP addresses using binary notation:

110.11.5.88

12.74.16.18

201.24.44.32

P18-6 (10 pts) In classless addressing, show the whole entirely available address space (IPv4) as a single block using the CIDR notation.

P18-10 (10 pts) 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

P18-12 (10 pts) Each of the following addresses belongs to a block. Find the first and the last address in each block.

14.12.72.8/24

b. 200.107.16.17/18

c. 70.110.19.17/16

P18-14 (20 pts) 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.

P18-20 (20 pts) 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.

P18-22 (20 pts)

            An ISP is granted the block 80.70.56.0/21. The ISP needs to allocate for two organizations each with 500 addresses, two organizations each with 250 addresses and three organizations each with 50 addresses.

Find the number and range of addresses in the ISP block.

Find the range of addresses for each organization and the range of unallocated addresses.

Show the outline of the address distribution and the forwarding table.

Explanation / Answer

P18-2:

110.11.5.88: 01101110.00001011.00000101.01011000

12.74.16.18: 00001100.01001010.00010000.00010010

201.24.44.32: 11001001.00011000.00101100.00100000

P18-6

CIDR IP addresses comprises of two groups of numbers. First is network address and second is host identifier (routing prefix).

It provides ability to group addresses into a block of single routing network

Number of addresses in a network = 2address length prefix length

e.g, for a IPv4 block, 192.168.100.14/24

Number of addresses = 232-24= 28= 256 addresses

192.168.100.0 to 192.168.100.255

Hence, available space 192.168.100.0 to 192.168.100.255 is represented as 192.168.100.14/24 in CIDR notation

P18-10 (10 pts):

Prefix expected here is the host identifier

Binary notation for 255.224.0.0 is 11111111 11100000 00000000 00000000

Number of 1s =11 represents the prefix, so the prefix length =11

Similary,

255.240.0.0

Binary: 11111111 11110000 00000000 00000000

Number of 1s =12

Prefix = 12

For, 255.255.255.128

Binary: 11111111 11111111 11111111 10000000

Number of 1s=25

Prefix = 25

P18-12 (10 pts):

Block: 14.12.72.8/24

This means, first 24 bits will remain same in the block and mask is 255.255.255.0

Thus, first IP address = 14.12.72.0

Number of addresses = 232-24 = 28=256

IP address range: 14.12.72.0 to 14.12.72.255

200.107.16.17/18 => mask 11111111 11111111 11000000 00000000 = 255.255.192.0

Binary of 200.107.16.17 = 11001000.01101011.00010000.00010001

First address: 11001000.01101011.00000000 00000000 = 200.107.0.0

Number of available addresses = 232-18= 214=16384

Last address: 200.107.255.255

70.110.19.17/16 => mask 11111111 11111111 00000000 00000000 = 255.255.0.0

First address: 70.110.0.0

Last address: 70.110.255.255

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