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ECE 461 – Internetworking Problem Sheet 1 Instructions (read carefully): • Try to solve the problems on your own. • Solutions will be discussed in tutorials. Problem 1. Describe how the entire address space of IPv4 classes A, B, C, D and E can be expressed with a network prefix in CIDR notation. Provide one network prefix for each class. Class A addresses start with “0” à 0.0.0.0/1 Class B addresses start with “10” à 128.0.0.0/2 Class C addresses start with “110” à 192.0.0.0/3 Class D addresses start with “1110” à 224.0.0.0/4 Class E addresses start with “1111” à 240.0.0.0/4 Problem 2. Consider the 128.100.112.0/21 block of IP addresses. This block of addresses must be divided into four subnetworks that have each at least 500 IP addresses. a) Provide the subnetmask in dotted decimal notation for 128.100.112.0/21. b) Give the subnet mask of the four new subnetworks. c) Specify the network address and the network prefix for each subnetwork. d) Specify the broadcast IP address for each subnetwork. (a) The subnet mask has “1” on the first 21 bits, and “0” on the last 11 bits. In binary with dots at the byte boundary this is: 11111111.11111111.11111000.00000000 In dotted decimal notation, this is: 255.255.248.0 (b) For 500 IP addresses, we need a /23 prefix (This yields 512 addresses, of which 510 can be assigned to hosts or routers). In binary with dots at the byte boundary this is: 11111111.11111111.11111110.00000000 In dotted decimal notation, this is: 255.255.254.0 (c) Binary representation of the network prefixes Given Network Prefix Binary notation (prefix in bold). 128.100.112.0/21 1000 0000.0110 0100.0111 0 000.0000 0000 The four created networks are

Given Network Prefix Binary notation (prefix in bold). 128.100.112.0/23 1000 0000.0110 0100.0111 000 0.0000 0000 128.100.114.0/23 1000 0000.0110 0100.0111 001 0.0000 0000 128.100.116.0/23 1000 0000.0110 0100.0111 010 0.0000 0000 128.100.118.0/23 1000 0000.0110 0100.0111 011 0.0000 0000 (d) Broadcast address has “1” on all bits of the interface identifier. Dotted decimal notation Binary notation (prefix in bold). 128.100.113.255 1000 0000.0110 0100.0111 000 1.1111 1111 128.100.115.255 1000 0000.0110 0100.0111 001 1.1111 1111 128.100.117.255 1000 0000.0110 0100.0111 010 1.1111 1111 128.100.119.255 1000 0000.0110 0100.0111 011 1.1111 1111 Problem 3. Select a subnet mask for 10.0.0.0/8 so that there will be at least 16,000 subnets with at least 700 host addresses on each subnet. To find the number of subnets, we need to select a number “n” such that 2 n ≥ 16,000. There must be enough host bits h remaining so that 2 h -2 ≥ 700. A subnet mask of 255.255.252.0 provides 16,382 subnets of the class A address and 1022 host addresses on each subnet. (Note this is the only subnetmask which will work). Problem 4. Assume that you have been assigned the 200.35.1.0/24 network block. a) Define an extended-network-prefix that allows the creation of 20 hosts on each subnet. A minimum of five bits are required to define 20 hosts so the extended network prefix is a /27 (27 = 32-5). b) What is the maximum number of hosts that can be assigned to each subnet? The maximum number of hosts on each subnet is 2 5 -2, or 30. c) What is the maximum number of subnets that can be defined? The maximum number of subnets is 2 3 , or 8. d) Specify the subnets of 200.35.1.0/24 in binary format and dotted decimal notation. Subnet #0: 11001000.00100011.00000001. 000 00000 = 200.35.1.0/27 Subnet #1: 11001000.00100011.00000001. 001 00000 = 200.35.1.32/27 Subnet #2: 11001000.00100011.00000001. 010 00000 = 200.35.1.64/27 Subnet #3: 11001000.00100011.00000001. 011 00000 = 200.35.1.96/27 Subnet #4: 11001000.00100011.00000001. 100 00000 = 200.35.1.128/27 Subnet #5: 11001000.00100011.00000001. 101 00000 = 200.35.1.160/27 Subnet #6: 11001000.00100011.00000001. 110 00000 = 200.35.1.192/27 Subnet #7: 11001000.00100011.00000001. 111 00000 = 200.35.1.224/27

d) Yes, 142.150.0.0/16 is assigned to UofT (see ARIN records). So, subnetting is used. Note that this requires information not provided in the information above. I don’t need to know if subnetting is used when managing my own network. e) The information is used for correct “wiring” of Ethernet switches to the IP router, and for configuring the IP router. f) From ARIN records, the original network address is 142.150.0.0/16. So, 11 bits are used to designate the subnet. g) This is the address of an IP router on the 142.150.235.0/27 subnet. All systems on this subnet should use the address 142.150.235.1 as default gateway. h) No, I assume that the router is provided (at address 142.150.235.1). i) I can assign 4 networks with a 29-bit prefix, with network addresses: 142.150.235.0/29 142.150.235.8/29 142.150.235.16/29 142.150.235.24/29 There can be 6 IP addresses on each subnet (3 bits = 8 addresses. But cannot use `000’ (network address) and `111’ (broadcast address))