Date: June 4 th a t 1 4:00 1 7:00

Transcription

1 Kommunika tionssyste m FK, Examina tion G Date: June 4 th a t 4:00 7:00 KTH/IMIT/LCN No help material is allowed. You may answer questions in English or Swedish. Please answer each question on a separate page. Fill in the table on page for each question you have addressed. The grading of the exam will be completed no later than June After grading, the exams will be available for inspection no later than August 003. Deadline for written complaints is August Course responsible is Olof Hagsand, phone Your name: Your social security number (personnummer): Your major (utbildningslinje):.. Total Points:. Grade:..

2 Question Answered Potential points Received points Total 60

3 . General (5p) a) Formulate the End- to- End argument (you do not have to quote it)? (p) A specific application- level function should not be implemented in the lower levels of the system. b) Name at least two consequences of the End- to- End argument. (p) The bulk of functions are placed at the edge of the network. The complexity of the core network is reduced. Generality in the network increases the chances that new applications can be added. c) Place each of the following protocols in the correct layer of the TCP/IP protocol stack: DHCP, IP, ARP, SNMP, RTP, IGMP, (p) DHCP, SNMP, RTP Application layer IP, IGMP Network layer ARP Link layer. Link Layer (5p) a) Name at least two advantages with PPP compared to SLIP. (p) Support for multiple protocols on a link CRC on every frame Dynamic negotiation of IP address of each end Link control with facilities for negotiating many data- link options b) How does proxy ARP work? (p) A node (proxy server) answers ARP requests on another node s behalf. c) What is gratuitous ARP? (p) When a host sends an ARP request of its own address. d) RARP can be used as a bootstrap protocol. Name at least one limitation with RARP used in this setting. (p) The use of a link- layer broadcast prevents routers from forwarding the RARP request, and the information returned is minimal: just the system s IP address. 3. Link Layer / Spanning Tree Protocol (5p) Briefly describe the Spanning Tree algorithm. Your answer should cover: a) Why the protocol is needed (p) To avoid forwarding loops in a bridged network. b) The process of constructing the tree (4p) ) An ID number is assigned to each bridge, and a cost to each port. ) Bridge with smallest ID is elected root bridge. 3) Each bridge determines its root port, the port that has the least root path

4 cost to the root. 4) One designated bridge is chosen for each segment. 5) Include the following ports in the spanning tree: root port plus any ports on which self has been elected designated bridge 6) Data traffic is forwarded only to and from ports selected for inclusion in the spanning tree. 4. IPv4 Addressing (5p) All other subnets on site... One subnet A B C Internet A site has been allocated the class B network The site is shown in the figure above, where one router routes between three interfaces A, B, and C. The site is partitioned into several subnetworks. One of those subnetworks includes host , and is connected to interface B. The other subnetworks are connected to interface A. a) The site needs to partition its class B network into at least 53 subnetworks by using fixed partitioning and classful addressing. How large is the (minimal) subnet mask and how many hosts can be on each network? (p) The minimal subnet mask is 6 bits, so that ^6 = 64 subnetworks can be addressed. There may be ^0- =0 hosts on each network (03 is also accepted as an answer). b) Construct the net- directed broadcast address for the sub- network to which host is connected. (p) c) Using classless addressing and CIDR notation, how is that network denoted? (p) / d) Draw a minimal routing table of the router using CIDR notation. Each router entry shall only contain the network in CIDR notation and an outgoing interface. (p) /0 C /6 A / B

5 5. ICMP and IP options (5p) a) ICMP messages are grouped into query and error messages. Error messages are sent when errors in IP datagrams are detected, except for some special cases. Name at least two such special cases. (p) A datagram carrying another ICMP Error A datagram destined to IP broadcast or multicast A datagram sent as link- layer broadcast (eg Ethernet) An IP fragment other than the first A datagram whose source address does not define a single host b) Traceroute is a tool to explore the path to a given destination. Traceroute uses two methods where ICMP messages are involved to detect each hop on the way to the destination. Describe these two methods and name the ICMP messages involved. (p) Traceroute starts with TTL= and sends IP datagrams with increasing TTL levels to the destination. The intermediate routers send ICMP time exceeded back to the source, if the TTL is decremented to zero. When the destination is reached, traceroute sends a UDP datagram with an unlikely port. The destination returns an ICMP Port unreachable. c) Using IP options, an alternative method to traceroute can be used to find the path to a given destination. Describe this method, and name at least one reason why it is of limited use. (p) The IPv4 record route option can be used as an alternative. Limitations of this approach include - The limited (0-byte) option field that can only be used for a limited (9) set of hops - Not all routers implement this functionality. 6. TCP (5p) a) What protocol is used for flow control in TCP? (p) Sliding windows. b) What is the difference between offered window and usable window in TCP? (p) The offered window is advertised by the receiver and defines how much data the receiver is ready to accept. The usable window is maintained by the sender and defines the amount of data the sender can transmit immediately. c) Someone complained about a throughput of 0,000 bits/sec on a 56,000bits/sec link with a 8-ms RTT (Round Trip Time) between the United States and Japan (47% utilization), and a throughput of 33,000 bits/sec when the link was routed over a satellite (3% utilization). Assume a 500 ms RTT for the satellite link. What does the window size appear to be for both cases? (p) Terrestial link: capacity = throughput x RTT = 0000 bits/s x 8 ms = 90 bytes Satellite link: capcity = throughput x RTT = bits/s x 500 ms = 06

6 bytes It appears that the receiving TCP advertises a K window size. d) How large should the window in the previous example be for optimal throughput over the satellite link? (p) For optimal throughput on the satellite link: capacity = bandwidth x RTT = bits/s x 500 ms = 6000 bytes. 7. Routing (5p) N 3 N R N R R 3 3 B N 4 A N 5 N 6 3 R 4 R 5 N 7 R 6 N 8 R 7 R 8 N 9 N 0 Internet Regard the network in the figure above. It consists of eight routers (R -R 8 ), with ten networks (N -N 0 ). Ethernets are denoted with horizontal bold lines. Hosts are not shown, except for hosts A and B. There are two point- to- point links connecting R - R 4 and R 3 -R 5. R 5 is connected to the rest of the Internet. Each router runs OSPF. The whole network belongs to the backbone area: Area 0. The costs are shown in the figure. a) Perform a shortest path calculation (such as by the Dijkstra s algorithm) with router R as root, and draw the resulting directed tree of the network shown in the figure. The tree should include routers and designated routers and costs on each link. (p) (see figure below)

7 b) The OSPF protocol really consists of three separate sub- protocols. Which are these protocols? Describe the purposes of each sub- protocol. (3p). The OSPF HELLO protocol: Neighbour detection. Authentication. Designated Router selection.. The OSPF Exchange protocol: Exchange Database description (DD) between neighbours. Request and receive LSAs initially from neighbours. 3. The Flooding protocol: When links change or age, send link updates recursively (flooding). Answer to 5a: R 3 N N 3 N 4 R 4 N R R 3 R 5 N 7 R 6 N 8 R 7 R 8 N 9 N 0 8. Autoconfigura tion and DNS (Domai n Name System) (5p)

8 a) Many configuration problems in TCP/IP are solved automatically by DHCP and BOOTP. DHCP is a follow- up of BOOTP. Which is the major advantage of DHCP over BOOTP? (p) BOOTP is not a dynamic solution and requires a predetermined mapping between MAC addresses and IP addresses. DHCP can provide dynamic configuration of IP addresses by giving the client a temporary address from a pool of available addresses. b) DNS uses a hierarchical structure for, among other things, translation between names and IP addresses in the Internet. Several name servers may have to cooperate to find out the IP address of a particular hostname. Describe the two different modes of resolution that can be used within DNS. (p) Iterative Resolution: the host has to ask the next name server in the hierarchy. Recursive Resolution: the name server asks the next name server in the hierarchy. c) What is the purpose of the in-addr.arpa zone in the DNS hierarchy? (p) To make reverse lookup (IP address to hostname) possible without having to search the complete DNS tree. 9. IPv4 Multicast (5p) a) Briefly outline how IPv4 multicast addresses are mapped to Ethernet multicast addresses. Describe any limitations with this mapping that may have consequences to host interfaces. (You do not need to state the exact bit patterns used) (p) The 3 low- order bits of the IP multicast IP address are placed in the 3 loworder bits of the 48-bit Ethernet multicast address. Since IP multicast addresses uses 8 bits to denote addresses, the mapping is not unique but gives a 3: overlap. This means that the Ethernet layer cannot filter frames uniquely, the IP layer needs to do additional filtering, typically in software. b) IGMP handles signalling between hosts and multicast routers. Briefly outline the IGMP (version ) messages and their operation: In which situations are they used, who sends them, and to whom are they sent? (p) Membership query sent by multicast routers to query membership on a network. Two situations: general query sent to , or specific group queries to specified address. Membership report sent by hosts to multicast routers to report group membership. Leave group Sent by hosts to multicast routers when leaving a multicast group. c) DVMRP (Distance- Vector Multicast Routing Protocol) is used to propagate multicast routes. Briefly outline the operation of DVMRP. (p) DVMRP uses Truncated Reverse Path forwarding with pruning and grafting. That is, it broadcasts to all networks and then prunes branches with no members. Prune messages are propagated upwards toward the source. When new members join, graft messages extend the delivery tree.

9 0. Differen ti a t e d Services and Inte gra t e d Services (5p) a) Within the Differentiated Services architecture, the AF PHB (Assured Forwarding Per- Hop Behavior) has been defined. Typically, the ingress node performs traffic conditioning on incoming packets according to the SLA (Service Level Agreement) with the upstream domain. The traffic conditioning includes the following four functions: Marking, Metering, Classification, and Shaping. In what order are these functions performed? (p) ) Classification ) Metering 3) Marking 4) Shaping b) Suppose that the following flows, specified with token bucket traffic specifications, have been accepted by an IntServ (Integrated Services) capable router: R (rate in packets/second) B (bucket depth in no of packets) All flows are in the same direction and the router forwards 0 packets per second. Note that the example is unrealistic in its use of packets, instead of bytes. What is the maximum delay a packet may face? (p) Max delay is given by max queue length, which is the sum of all buckets. B tot = +6+3 = 0. Max delay = B tot /link capacity = 0/0 = second. c) What is the maximum number of packets from the third flow (r = 8, B = 3) that the router would send over.0 seconds, assuming the router sends packets at its maximum rate uniformly? (p) Max no of packets over seconds for the third flow = rt + B = 8x + 3 = 9 pkts.. IPv6 (5p) a) Show the shortest form of the following IPv6 address: 340:0000:0000:000F:7000:9A:A00:0000 (p) 340::F:7000:9A:A00:0 b) What is the difference between fragmentation in IPv6 versus IPv4? (p) In IPv4, packets can be fragmented by the host and by routers along the path between sender and receiver (hop- by-hop fragmentation). In IPv6, only the sending host is allowed to perform fragmentation. The sending host should learn the path MTU through path MTU discovery, or transmit packets that are small enough to fit any MTU limit. c) Name two main issues in IPv4 that were addressed by IPv6? (p) ) The address space in IPv4 was not considered large enough ) The routing tables in IPv4 were getting too large

10 3) Security improvements needed 4) Better support for autoconfiguration (plug- and- play). IPsec (5p) Outline the IPsec architecture. You should cover: a) In what layer IPsec is implemented (p) IPsec is implemented in the network layer as a part of the IP implementation. b) The three protocols included in IPsec, and the purpose of each protocol. (p). ESP (Encapsulating Security Payload - defines the encryption of the IP payload ). AH (Authentication Header - defines the authentication method). 3. ISAKMP (Internet Security Association and Key Management Protocol - manages the exchange of cryptographic keys). c) How do two nodes agree on security schemes? (p) A sender and receiver agree on a set of security schemes and establishes a security association (SA). This includes: Encryption algorithms, Keys, lifetime, addresses, etc. d) Briefly explain the two modes used by IPsec. When are the two modes used? How do they differ? (p) The two modes are transport and tunnel mode. Transport mode has a SA between two end- hosts, while tunnel mode has a SA between two routers. Tunnel mode encapsulates the original datagram within an IPsec encapsulated header, while transport mode inserts the IPsec headers between the original header and its payload.