Second Midterm Exam Solution CMPSCI 591 and 453: Computer Networks Spring 2005 Prof. Jim Kurose

Transcription

1 Second Midterm Exam Solution CMPSCI 591 and 53: Computer Networks Spring 2005 Prof. Jim Kurose Question 1. (5 points) In network assisted congestion control, a router actively informs an end system about network congestion, either by marking a packet, or by actively sending a message about congestion to the sender or receiver end system. ATM networks take this approach, as does SNA. TCP does not take a network-assisted approach, as it infers the network s congestion state based on packet losses that are detected via timeout. (5 points) With longest prefix matching, a router will forward a packet towards a destination using the routing table entry that matches the largest number of bits in packet s destination address. Longest prefix matching allows a router to route to a destination whose address appears in an aggregated address range that is being advertised by another network. (points) BGP does not require that a network choose one path over another to a destination. Instead, a BGP router receives path information from its neighbors and chooses the path to use based on whatever policies the network operator chooses. For example, a network operator could choose the path with the least number of AS hops or the most number of AS hops. ( points) The ARP protocol is used to match an IP address of a host on a subnet to that hosts 8 bit MAC address. ( points) With CSMA there in no bound on the amount of time it might take for a node to successfully send a packet. In token passing, the delay is bounded by the amount of time it takes for everyone to get their turn. Thus, token passing is preferred. (5 points) Bob signs the document by (1) computing a hash (message digest) of the document and (2) encrypting the hash using his private key. Bob then sends the document and the encrypted hash to Alice. Alice (1) gets the message and computes the hash, (2) gets Bob s public key and applies it to the encrypted hash sent by Bob. If the resulting decrypted hash computed in (2) matches the hast Alice computed in (1), then Alice know that Bob signed the document. [Grading note: 1 point off if no hash. Make sure public/private key use is correct] 1

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3 Problem 2 (seven points each) During time units 1-6 TCP is in slowstart, when its window increases by 1 for every segment acknowledged. During time units 6-16, TCP is in congestion avoidance, when it s window increases by 1 every RTT (or equivalently by 1 each time it receives ACKs for window s worth of segments) The sender has received three duplicate ACKs for a previously transmitted packet. Under TCP s fast retransmit, the window is cut in half, rather than starting at 1 after a loss is detected. Here, loss of a segment is detected by a timeout (as opposed to a triple duplicate ACK), and hence the window is set to 1. Problem 3. (3a, 10 points) step N D(A),p(a) D(C),p(C) D(D),p(D) D(E),p(E) D(F),p(F) 0 B 2,B 8,B INF 10,B,B 1 BA,A 6,A 10,B,B 2 BAC 5,C 10,B,B 3 BACF 5,C 6,F BACFD 6,F 5 BACFDE (3b, 10 points) The format for distance tables changed from the 2 nd to 3 rd edition of our book. So either interpretation of a distance table is fine. First, if we take a distance table to be the cost from B to each destination (column) when going through the specified neighbor (row) then the answer is: destinations NEIGHBORS A C D E F A * C E F 10* In the above, * indicates a loop (note that this definition of distance tables allows for loops in the table, although a path with a loop will never be the shortest path. 3

4 If we take the distance table to be the set of distance vectors, then the table is: destinations NEIGHBORS A B C D E F A B C E F (3c, 6 points) In reverse path forwarding, when a node receives a packet on the shortest path link from itself to B, it will flood the packet out every link. The reverse path tree is shown below. Thus When A gets a B-flood-packet directly from B it will flood to all of its neighbors, C D. When C gets a B-flood-packet from A it will flood to all of its neighbors, B,D (maybe OK to leave B out here) When D gets a B-flood-packet from C it will flood to all of its neighbors, A,E. When E gets a B-flood-packet from F it will flood to all of its neighbors, B,D (maybe OK to leave B out here) When F gets a B-flood-packet directly from B it will flood to all of its neighbors, E A 2 2 B 10 8 C E F 1 2 D Problem. The /23 signifies that the network part of the host address is the leftmost 23 bits Subnet A requires at least bits of addressing, subnet B requires at least bits of addressing, and subnet C requires at least 6 bits of addressing. Let the first 3 bytes of the address for all of the host be X.Y.Z. The address for hosts in subnet C are in the range X.Y.Z.00CCCCCC, where the last byte of the address begins with two zero s and the rest of the 6 bits are used to address hosts in B. Note that the second bit in the last byte is a 0. For subsets A and B, this bit will be a 1.

5 The address for hosts in subnet B are in the range X.Y.Z.010BBBBB. The last byte of the address begins with 010 and the final 5 bits can be used to address the hosts in B (note that even though only are needed there is not way to allocate only since the second bit in the last byte of the address must be a 1. The address for hosts in subnet A are in the range X.Y.Z.011AAAAA. The last byte of the address begins with 011 (which differs from the leading 010 for subnet B, and the leading 00 for subnet C) and the final 5 bits can be used to address the hosts in A (note that even though only are needed there is not way to allocate only since the second bit in the last byte of the address must be a 1. The size of the single aggregated network that is advertised is thus X.Y.Z.0/25 the last seven bits are used to address hosts in subnets A, B and C. Problem c: (Let the host in A have IP address (where we have abused notation and shown the last byte in binary format) and MAC address aa:aa:aa:aa:aa:aa Let the router interface into subnet A have IP address , and MAC address bb:bb:bb:bb:bb:bb. Let the host in C have IP address (where we have abused notation and shown the last byte in binary format) and MAC address cc:cc:cc:cc:cc:cc:cc Let the router interface into subnet C have IP address , and MAC address dd:dd:dd:dd:dd:dd:dd The IP datagram from the host in subset A to the router interface in subnet A has IP source and IP destination The source MAC address is aa:aa:aa:aa:aa:aa, and the destination MAC address is bb:bb:bb:bb:bb:bb Problem d: The IP datagram from the router interface in subnet C to the destination host in subnet C has IP source and IP destination the same answer as in c. The source MAC address is cc:cc:cc:cc:cc:cc:cc, and the destination MAC address is dd:dd:dd:dd:dd:dd:dd 5