CS 344/444 Computer Network Fundamentals Midterm Exam Spring /07/2007

Transcription

1 CS 344/444 Computer Network Fundamentals Midterm Exam Spring /07/2007 Question 344 Points 444 Points Score (444 only) 20 otal: Name: Course: CS344 CS444 Instructions: 1. You have 60 minutes to finish 2. Question 5 is only for 449 students 3. Closed books, closed notes. Write all your answers using the pages on this exam (use back pages if needed) Calculators allowed his exam has 8 pages, including this cover page Page 1

2 Question 1 (10 points) Answer the following using rue/false. You do not need to explain your answers. 1. One of the functionalities that transport protocols provide is reliable transfer over unreliable channels. 2. he Selective Repeat ARQ protocol is more efficient than the Go Back N ARQ protocol. 3. Efficiency of the oken Ring MAC protocol is higher than Ethernet when network utilization is high. 4. ARP is used to bind MAC addresses to IP addresses. 5. A router works at the Networking layer of the protocol stack. 6. Routing on the Internet is divided in two tiers. 7. CSMA/CD is an appropriate MAC protocol for ad hoc wireless networks. F 8. Link State routing protocols suffer from the "count to infinity" problem. F 9. Reassembly of IP fragments happens only at the destination host. 10. Each router in OSPF broadcasts its routing table to all the other routers in the network. F Page 2

3 Question 2: Suppose node A is connected to B via an intermediate router R. he A R link is instantaneous, but the R B link transmits only one packet each second, one at a time (so two packets take 2 seconds). Assume A sends to B using the sliding window protocol with SWS=4. For ime=0,1,2, state what packets arrive at and are sent from A and B. How large does the queue at R grow? his is question 2.37 in homework 2. Page 3

4 Question 3: For the network below, give global distance vector tables for all nodes in the network when: Note: Each row in the distance vector table should contain destination and distance to that destination. A 15 B 1 E D 2 C 3 F (a) (10 points) Each node knows only the distance to its immediate neighbors. Page 4

5 A B 15 B C +OO D 5 D E +OO F +OO B A 15 A D +OO E 1 E F 8 F C A +OO B 2 B D 2 D E +OO D A 5 A B +OO E +OO F +OO E A +OO B 1 B C +OO D +OO F A +OO B 8 B C 3 C D +OO E 3 E Page 5

6 (b) (10 points) Each node has reported the information it had in the preceding step to its immediate neighbors A B 15 B C 7 D D 5 D E 16 B F 23 B B A 15 A D 4 C E 1 E F 4 E C A 7 D B 2 B D 2 D E 3 B D A 5 A B 4 C E +OO F 5 C E A 16 B B 1 B C 3 B D +OO F A 19 B B 4 E C 3 C D 5 C E 3 E Page 6

7 (c) (10 points) Step (b) happens a second time A B 9 D C 7 D D 5 D E 10 D F 10 D B A 9 C D 4 C E 1 E F 4 E C A 7 D B 2 B D 2 D E 3 B D A 5 A B 4 C E 5 C F 5 C E A 16 B B 1 B C 3 B D 5 B F A 10 C B 4 E C 3 C D 5 C E 3 E Page 7

8 Question 4: (30 points) Consider the situation involving the creation of a routing loop in the network below, when the A E link goes down. List a sequence of table updates among A, B, and C, pertaining the destination E, that leads to a loop. Assume that table updates are done one at a time, that the split horizon technique is used by all participants, and that A sends an initial report of E's unreachability to B before C. You man ignore updates that don't result in changes. B A E C D F G (1) A sends E s unreachability to B and C (packets A1 and A2) (2) C sends the periodic update containing E s reachability to B (packet C1) (3) B sends the periodic update containing E s reachability to C (packet B1) (4) Packets A1 and A2 arrive at B and C respectively (5) Packet C1 reaches B (6) Packet B1 reaches C (7) B sends E s reachability to A (packet B2) (8) C sends E s reachability to A (packet C2) (9) Packet B2 arrives at A (10)Packet C2 arrives at A Now, A s table contains: A E 4 C B s table contains: B E 3 C C s table contains: Page 8

9 C E 3 B Page 9

10 Question 5: (CS444 Only) Consider the shortest path problem from every node to node 1 and Dijkstra's algorithm. Suppose that we have already calculated the shortest path from every node to node 1, and assume that a single arc length increases. Modify Dijkstra's algorithm to recalculate as efficiently as you can the shortest paths. he last run of the Dijkstra s algorithm gives the OldDistance(1, j ) where j 1,2,..., n. And also for each node j, there is a OldPath( j ) that contains the node sequence of the optimal path from 1 to j. For example, OldPath(5) might be [1,4,9,3] and it means the optimal path from 1 to 5 is 1 >4 >9 >3 >5. Suppose the length of the arc between nodes p and q increases. Node j need to be checked only when q Path( j ) or p ᅫ Path( j). he algorithm below will find out those nodes and put them into set entative and recheck them. Clear set Confirmed and entative. minpq = min(olddistance(1,q), OldDistance(1,p)); for i = 1 to n if (OldDistance(1, i) < minpq) Add i to set Confirmed ; NewDistance(1, i) = OldDistance(1, i); NewPath(i) = OldPath(i); else Add i to set entative; for each i in set entative tmppath = reverse OldPath(i); needmove = 1; for each j in tmppath if (j == q j == p) needmove = 0; break; if (OldDistance(1, j)<= minpq) Page 10

11 break; if (needmove) Add i to set Confirmed ; remove i from set entative; NewDistance(1, i) = OldDistance(1, i); NewPath(i) = OldPath(i); Start Dijkstra's algorithm with Confirmed and entative. Page 11