Computer Networks: Homework 2 Chapter 19: –Problem 18 a. netid: 114 hostid: b. netid: hostid: 8.6 c. netid: hostid: 12 –Problem 22 a. log = 10 Extra 1s = 10 Possible subnets: 1024 Mask: /26 b − 26 = 64 Addresses per subnet

Computer Networks: Homework 2 Chapter 19: –Problem 22 (Cont’d) c. Subnet 1: The first address is the beginning address of the block or To find the last address, we need to write 63 (one less than the number of addresses in each subnet) in base 256 ( ) and add it to the first address (in base 256). First address in subnet 1: Number of addresses: Last address in subnet 1: d. Subnet 1024: To find the first address in subnet 1024, we need to add 65,472 (1023 × 64) in base 256 ( ) to the first address in subnet 1. We have = Now we can calculate the last address in subnet 500 as we did for the first address. First address in subnet 1024: Number of addresses: Last address in subnet 1024:

Computer Networks: Homework 2 Chapter 19: –Problem 26 The total number of addresses in this block is = The ISP can divide this large block in several ways depending on the predicted needs of its customers in the future. We assume that the future needs follow the present pattern. In other words, we assume that the ISP will have customers that belong to one of the present groups. We design four ranges: group 1, group 2, group 3, and one reserved range of addresses as shown in Figure.

Computer Networks: Homework 2 Chapter 19: –Problem 26 (Cont’d) Group 1 In the first group, we have 200 businesses. We augment this number to 256 (the next number after 200 that is a power of 2) to let 56 more customers of this kind in the future. The total number of addresses is 256 × 128 = For this group, each customer needs 128 addresses. This means the suffix length is log = 7. The prefix length is then 32 − 7 = 25. The addresses are: 1st customer: /25 to /25 2nd customer: /25 to / th customer: /25 to /25 Unused addresses to Total Addresses in group 1 = 256 × 128 = Used = 200 × 128 = Reserved: 7168, which can be assigned to 56 businesses of this size. Group 2 In the second group, we have 400 business. We augment this number to 512 (the next number after 400 that is a power of 2) to let 112 more customer of this kind in the future. The total number of addresses is = 512 × 16 = For this group, each customer needs 16 addresses. This means the suffix length is 4 log 2 16 = 4. The prefix length is then 32 − 4 = 28. The addresses are: 1st customer: /28 to /28 2nd customer: /28 to / th customer: /28 to /28 Unused addresses to Total Addresses in group 2 = 512 × 16 = 8192 Used = 400 × 16 = 6400 Reserved: 1792, which can be assigned to 112 businesses of this size.

Computer Networks: Homework 2 Chapter 19: –Problem 26 (Cont’d) Group 3 In the third group, we have 2000 households. We augment this number to 2048 (the next number after 2000 that is a power of 2) to let 48 more customer of this kind in the future. The total number of addresses is = 2048 × 4 = For this group, each customer needs 4 addresses. This means the suffix length is 2 log 2 4 = 2. The prefix length is then 32 − 2 = 30. The addresses are: 1st customer: /30 to /30 2nd customer: /30 to / th customer: /30 to /30 Unused addresses to Total Addresses in group 3 = 2048 × 4 = 8192 Used = 2000 × 4 = 8000 Reserved: 192, which can be assigned to 48 households. Reserved Range In the reserved range, we have address that are totally unused. Note that we have unused addresses in each group and a large range of unused addresses in the reserved range. Summary: The following shows the summary of used and unused addresses:

Computer Networks: Homework 2 Chapter 20: –Problem 14 The first byte number can be calculated from the offset itself. If the offset is 120, that means that 120  8 or 960 bytes (bytes 0 through 959) were sent before this fragment. The first byte number is therefore 960. The last byte number can be calculated by adding the total length field and subtracting one. –Problem 18 The datagram must contain 16 bytes of data: 36 byte total length – (5 HLEN field  4) = 36 – 20 = 16 bytes –Problem 20 The identification field is incremented for each non-fragmented datagram. If the first is 1024, then the last is = 1123

Computer Networks: Homework 2 Chapter 21: –Problem 18 The checksum is 0xD399 or as calculated below:

Computer Networks: Homework 2 Chapter 21: –Problem 24