# What is the Address Space in Each of the Following Systems

July 14, 2017 | Author: Bill Wells | Category: Network Layer Protocols, Telecommunications, Data Transmission, Internet, Internet Protocols

#### Description

Tuesday, May 03, 2011 1:20 PM

11. What is the address space in each of the following systems? a. System with 8-bit addresses i. 28 = 256 b. System with 16-bit addresses i. 216 = 65,536 c. System with 64-bit addresses i. 264 = 1.844674407 x 1019 12. An address space has a total of 1024 addresses. How many bits are needed to represent an address? i. Log21024 = 10

13. An address space uses the three symbols 0, 1, and 2 to represent addresses. If each address is made of 10 symbols, how many addresses are available in this system? i. 3 possible options for 10 symbols, so 3 10 = 59,049 16. Find the class of the following IP addresses (page 552) a. 208.34.54.12 i. Class C b. 238.34.2.1 i. Class D c. 114.34.2.8 i. Class A d. 129.14.6.8 i. Class B 18. Find the netid and the hostid of the following IP addresses. (page 553) a. 114.34.2.8 i. Netid = 114 Hostid = 34.2.8 b. 132.56.8.6 i. Netid = 132.56 Hostid = 8.6 c. 208.34.54.12 i. Netid = 208.34.54 Hostid = 12

19. In a block of addresses, we know the IP address of one host is 25.34.12.56/16. What are the first address (network address) and the last address (limited broadcast address) in this block? (page 556) a. We set the 32-n rightmost bits to 0 to get the first address: i. 25.34.0.0 b. We set the 32-n rightmost bits to 1 to get the last address: i. 25.34.255.255 20. In a block of addresses, we know the IP address of one host is 182.44.82.16/26. What are the first address (network address) and the last address in this block? (page 556) • 182.44.82.16 = 10110110 00101100 01010010 00010000 a. Set the 32-n rightmost bits to 0 to get first address: i. 10110110 00101100 01010010 00000000 ii. 182.44.82.0 b. Set the 32-n rightmost bits to 1 to get last address: i. 10110110 00101100 01010010 00111111 ii. 182.44.82.63

21. An organization is granted the block 16.0.0.0/8. The administrator wants to create 500 fixed -length subnets. a. Find the subnet mask i. 500 subnets means we need log2500 bits reserved in addition to the current mask, so 8+9 = 17 bits total. The Design of Networks Page 1

23. An organization is granted the block 211.17.180.0/24. The administrator wants to create 32 subnets. a. Find the subnet mask i. Log232 = 5, so 5+24 = /29 b. Find the number of addresses in each subnet i. 232-29 = 8 addresses per subnet c. Find the first and last address in subnet 1 i. First address: 211.17.180.0 ii. Last address: 211.17.180.7 d. Find the first and last address in subnet 32 i. First address: 211.17.180.248 ii. Last address: 211.17.180.255 24. Write the following masks in slash notation. a. 255.255.255.0 i. /24 b. 255.0.0.0 i. /8 c. 255.255.224.0 i. /19 d. 255.255.240.0 i. /20

25. Find the range of addresses in the following blocks. • Find the first address by setting the 32-n rightmost bits to 0. Find the last address by setting the 32-n bits to 1 a. 123.56.77.32/29 i. First address: 123.56.77.32 ii. Last address: 123.56.77.39 b. 200.17.21.128/27 i. First address: 200.17.21.128 ii. Last address: 200.17.21.159 c. 17.34.16.0/23 i. First address: 17.34.16.0 ii. Last address: 17.34.17.255 d. 180.34.64.64/30 i. First address: 180.34.64.64 ii. Last address: 180.34.64.67 Design of Networks Page 2

27. An ISP is granted a block of addresses starting with 120.60.4.0/22. The ISP wants to distribute these blocks to 100 organizations with each organization receiving just eight addresses. Design the subblocks and give the slash notation for each subblock. Find out how many addresses are still available after these allocations. i. The ISP was given 232-n or 210 = 1024 addresses. These can be broken up into 100 8-address blocks to be distributed to the organizations while the ISP can retain the remaining addresses for later distribution. Each organization will receive a block with a /29 subnet since 23 = 8 hosts and 32-3 = 29. • Subnet 1: 120.60.4.0/29 to 120.60.4.7 • Subnet 2: 120.60.4.8/29 to 120.60.4.15 • … • Subnet 100: 120.60.7.24/29 to 120.60.7.31 ii. We used 800 addresses (8 * 100 organizations) out of a possible 1024. So, 1024-800 = 224 addresses left 29. Show the shortest form of the following addresses. a. 2340:1ABC:119A:A000:0000:0000:0000:0000 i. 2340:1ABC:119A:A000::0 b. 0000:00AA:0000:0000:0000:0000:119A:A231 i. 0:AA::119A:A231 c. 2340:0000:0000:0000:0000:119A:A001:0000 i. 2340::119A:A001:0 d. 0000:0000:0000:2340:0000:0000:0000:0000 i. 0:0:0:2340::00

30. Show the original (unabbreviated) form of the following addresses. a. 0::0 i. 0000:0000:0000:0000:0000:0000:0000:0000 b. 0:AA::0 i. 0000:000A:000A:0000:0000:0000:0000:0000 c. 0:1234::3 i. 0000:1234:0000:0000:0000:0000:0000:0003 d. 123::1:2 i. 0123:0000:0000:0000:0000:0000:0001:0002 34. Show in Hexidecimal colon notation the IPv6 address • 129.6.12.34 = 10000001 00000110 00001100 00100010 = 81 06 0C 22 a. Compatible to the IPv4 address 129.6.12.34 i. 0::8106:0C22 b. Mapped to the IPv4 address 129.6.12.34 i. 0::FFFF:8106:0C22

Design of Networks Page 3