EP2120 Internetworking/Internetteknik IK2218 Internets Protokoll och Principer

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1 EP2120 Internetworking/Internetteknik IK2218 Internets Protokoll och Principer Homework Assignment 1 (Solutions due 20:00, Mon., 10 Sept. 2018) (Review due 20:00, Wed., 12 Sept. 2018) 1. IPv4 Addressing (30/100) a) What is the best fit netmask (i.e., resulting in as few host addresses as possible) for a network with 31 hosts in it? (5p) You need 31 IP addresses for the hosts, plus a network address plus a directed broadcast address = 33 addresses required. Additionally, you might need one more IP address for the router, but the router could be one of the hosts. Whether or not the router is a host, you need at least 33 addresses, and hence 6 bits are needed.thus the netmask is b) What is the maximum number of hosts you can have in a /26 network? (5p) /26 means that you have 32-26=6 bits for host addresses, which means 2 6 =64 addresses in the net, out of which one is the network address, and one is the directed broadcast address, so there can be up to 62 hosts. If the router needs its own IP address then 61 hosts. c) Split up the network /22 into three networks, including one /23 network and two /24 networks! Provide the results in binary and in dotted decimal notation including the prefix length. (5p) /22 means that the first 22 bits of the address are the network address. Hence the first 22 bits of the three networks will have to be To solve the problem, we can split up the /22 network into two 23/networks, and further divide one of the /23 networks into two 24/networks. Bit 23 of the first /23 network will be 0, while that of the second /23 network will be 1. We can split the second /23 network (for which bit 23 is 1) into two /24 networks, then bits 23 and 24 of the two /24 networks will be 10 and 11. The first network in binary format will be /23, the second network /24, and the third network /24. In dotted decimal format the networks are /23, /24, and /24. An equally good solution is to split up the first /23 network into two /24 networks. d) What is the directed broadcast address of the network /27? (5p) The directed broadcast address of this network is the address with all bits set to 1 between positions 28 and 32. That is, e) What is the limited broadcast address of the network /27? (5p) The limited broadcast address is the address with all bits set to 1. Consequently, it is f) Use the services of IANA and a regional registry to figure out to whom the IP network /18 belongs. Provide the name of the organization and the AS number. (5p) Based on the RIPE whois database ( the network belongs to KTH Royal Institute of Technology. The AS number is AS2839.

2 2. Address allocation (30/100) Consider the network above, a routed network in an organization s enterprise network. The organization built a core network (network F) connected to a central router (R1), and connected a router (R2) with (long-haul) switched Ethernet (network E). The access routers (R3 to R6) are connected to a set of local offices (networks A to D). The host H1, connected to network F, performs various traffic monitoring tasks. All networks use Ethernet on the link layer. The enterprise allocated prefix /22 for its internal addresses. Make an address allocation using /22 in the network by assigning a sub-block to each network A-F in the following way: a) The networks A through C require 124 hosts each, while network D requires 220 hosts. Create a minimal block for each local office A through D. Start with the lowest address for network A. b) There are no unnumbered point-to-point links: all Ethernet networks have IP sub-networks and all nodes (routers and hosts) have an IP address on all their network interfaces. All nodes need to be reachable from any other host. c) The address allocation should be such that the sub-networks can be aggregated. d) Each sub-network should not be larger than necessary in order to accommodate all hosts in the sub-network. Based on your address allocation, provide the network addresses of networks A to F, and the required entries of the forwarding table of router R1! Choose appropriate IP addresses for the router interfaces that appear as next hop in the forwarding table. Give a sketch of your reasoning to support your solution. (30p) Any solution should be accepted that is correct in the sense that packets destined to ANY of the subnets and to any of the IP addresses assigned to actual nodes (host or router) can be forwarded. The solution does not necessarily have to give a detailed explanation, but a good explanation is a merit, of course and helps to provide partial points should the solution not be entirely correct.

3 One possible solution is given below. Assign the networks as follows: A: / hosts B: / hosts C: / hosts D: / hosts E: /29 up to 6 hosts/routers F: /29 up to 6 hosts/routers We assume that within the subnet allocated to network F is the IP address assigned to R1, is the IP address assigned to R2, is the IP address assigned to R6, and is the IP address assigned to H1. R1 s forwarding table (not including the forwarding rules for the rest of the network ): Destination Next hop Flags Interface / UG North / UG North /29 U North 3. IPv4 forwarding (20/100) A router has the forwarding table shown below. Determine the next-hop address and the outgoing interface for the packets arriving to the router with destination addresses as given in points (a)-(e). Indicate whether the delivery at this router is direct or indirect. Destination Next hop Flags Interface /8 - U m / UG m / UG m /23 - U m / UG m /28 - U m / UG m / UG m2 a) (4p) on m0, direct delivery b) (4p) on m2, default route c) (4p) on m0, indirect delivery d) (4p) on m1, direct delivery e) (4p) on m0, indirect delivery

4 4. IPv4 and IPv6 datagram formats (20/100) a) Which header fields in IPv4 and IPv6 are used to indicate the protocol used in the higher layer (3p) In IPv4, the 8-bit protocol type field indicates the higher layer protocol. In IPv6, the 8-bit next header field is used either for specifying the next extension header, or the higher layer protocol. b) What happens when some bits in the IP header are changed on the link between two routers in a network that supports only IPv4? What if the network supports only IPv6? (3p) In IPv4, a checksum is computed on the IPv4 header hop-by-hop to check the integrity of the header. Therefore, each router will verify the checksum before processing the datagram. If the checksum verification fails, the datagram is discarded. In IPv6, no checksum is computed. Instead, IPv6 will depend on the checksums computed in the data link layer and in the transport layer to detect errors. c) In the IPv4 header, options are included as fields with a variable length of up to 40 bytes at the end of the header, before the payload. What is the advantage of this format? In IPv6 options have been replaced by extension headers. What are the advantages of this new format? (4p) In IPv4 including the options as fields with variable length reduces the overhead when no options are needed. In IPv6 extension headers remove the 40 byte limitation on additional options. Another advantage is that extension headers need only be processed at the destination node (except for the hop-by-hop options extension header). d) Fragmentation in IPv4 and in IPv6 is supported by three fields (identification, offset and the MF bit (the DF bit disables it)). Is this solution better than assigning a sequence number to every fragment in the node (host or router) that does the fragmentation? Is there any difference between IPv4 and IPv6 in terms of the suitability of this alternative solution? In an IPv6 network, what happens if a router needs to forward a datagram that exceeds the specified MTU on the outgoing link? (5p) Yes, using the offset and MF bit is better than assigning a sequence number to every fragment. First of all, if fragmentation were performed with the help of the only sequence number, the destination host would not know whether to wait for more fragments before reassembling the datagram. Furthermore, knowing the offset of every fragment allows the destination host to store the fragment in the right position of the buffer, even if previous fragments have not yet reached the destination. A sequence number would not allow fragmenting a fragment in a router, as there would be no sequence number that can be assigned. This would affect IPv4. If an IPv6 router receives a datagram with a length exceeding the MTU, the router will not attempt to fragment it. Instead, it will discard the datagram and notify the sender. The sender will have to fragment the datagram according to the specified MTU. Hence the last issue does not affect IPv6. If you use a sequence number and a

5 number if fragments field then sequence numbers may work. The solution may not be efficient though, as you need two numbers (how many bits for each?), and upon reception of a fragment you would not know where to put the fragment in the receive buffer. Hence providing the offset is more practical even for IPv6. e) Consider an IPv4 network (called Network1), and two hosts, Host A and Host B connected to this network. Assume that a process on Host A sends a UDP payload of length 30 bytes to Host B in a datagram. If we compare the IPv4 header of the datagram received at Host B and that of the datagram sent by host A, what fields can be different? (5p) Since the length of the IPv4 datagram is less than 68 bytes (assuming options are not used), which is the minimum IPv4 MTU requirement, the datagram will not be fragmented. Every time the datagram is forwarded by a router, the TTL will be decreased and the checksum will be updated. Furthermore, the source address or destination address can be changed if the datagram is forwarded by a network address translation device (we have not yet talked about NAT, hence this is not part of the answer).