TDC 375-701 Student Name: Homework 2 Due 2009-09-23 From TCP/IP Protocol Suite, 4th Edition, answer: 5.9, Questions 1, 4, 8, 12, 21, 27, 28, 34 Q 5.9.1. a. 2^8 or 256 addresses b. 2^16 or 65,536 addresses c. 2^64 or 18,446,744,073,709,551,616 addresses note: while some calculators may give you an answer of 1.84467441 × 10^19, this is not a precise number, but I'll accept it since I don't expect you to calculate this by hand. Q. 5.9.4. a. 01110010.00100010.00000010.00001000 b. 10000001.00001110.00000110.00001000 c. 11010000.00100010.00110110.00001100 d. 11101110.00100010.00000010.00000001 Q. 5.9.8. a. 127.240.103.125 b. 175.192.248.29 c. 223.176.31.93 d. 239.247.199.29 Q. 5.9.12. NOTE: You can use the book's base 256 method, which I find a bit goofy personally. Another option is to simply convert the IP address to a 32-bit integer and do the math. Be sure to add one, because you are including the starting and ending addresses in the range, not just taking the difference. 14.7.24.0 as a 32-bit integer = 235345920 14.14.34.255 as a 32-bit integer = 235807487 235807487 - 235345920 + 1 = 461,568 Q. 5.9.21. NOTE: the book question has an error. It incorrectly refers to the last address in the netblock as the "limited" broadcast address. The limited broadcast address is always 255.255.255.255 for any subnetwork. The last address in the block is correctly referred to as the "directed" broadcast address. I doubt anyone was fooled by this, but if you counted the addresses from the first address to the "limited" broadcast address noting the error, I'd give you credit, in fact I might have given you a bonus point if you pointed this out. :-) first address: 25.34.0.0 last address: 25.34.255.255 Q. 5.9.27. a. /24 b. /8 c. /19 d. /20 Q. 5.9.28. Note: its asks for the range of addresses in the netblock, you should include the network and broadcast address. It doesn't indicate any reason why you wouldn't. You may not be turning them into routed subnets, so if they are just allocated blocks they might all be used and fall within a larger covering routed netblock. a. 123.56.77.32 - 123.56.77.39 b. 200.17.21.128 - 200.17.21.159 c. 17.34.16.0 - 17.34.17.255 d. 180.34.64.64 - 180.34.64.67 Q. 5.9.34. NOTE: I don't have the answers for the text and its not clear to to me if the author intended to use the address 120.60.4.0/20. Its either a typo or some sort of weird trick, where he actually wants you to start addressing at 120.60.4.0 instead of at 120.60.0.0/20, which is how the block would be assigned in practice. An ISP would never get assigned a subset of addresses within a larger CIDR block as worded by this question. However, based on the poorly worded question, if you do it either way and still get a reasonable answer thats convinces me you were approaching this by allocated /29 blocks to customers that is fine by me. Here is one way to answer it assuming the question had a typo: 120.60.0.0/29 120.60.0.8/29 120.60.0.16/29 ... 120.60.0.248/29 (up to 32 blocks of 8 so far) 120.60.1.0/29 120.60.1.8/29 ... 120.60.1.248/29 (up to 64 blocks of 8 so far) ... 120.60.2.248/29 (up to 96 blocks of 8 so far) 120.60.3.0/29 (97) 120.60.3.8/29 (98) 120.60.3.16/29 (99) 120.60.3.24/29 (100) Remaining addresses, 120.60.3.32 to 120.60.15.255 = 224 + (256 * 12) = 224 + 3072 = 3296 Also answer the following questions: Q. If a given network prefix is 140.192.6.0/26, what is the directed broadcast address in dotted decimal notation? A. 140.192.6.63 Q. Can the the following pairs of netblocks be aggregated into a single /23 prefix (a supernet)? Just write Yes or No for each pair. a. 140.192.0.0/24, 140.192.1.0/24 (yes) b. 140.192.1.0/24, 140.192.2.0/24 (no) c. 140.192.63.8/24, 140.192.64.1/24 (no) d. 140.192.64.49/24, 140.192.65.254/24 (yes) $Id: homework2.txt,v 1.4 2009/09/29 00:38:12 jtk Exp $