Disadvantage - Computer Networks - Past Exam Paper, Exams of Computer Science

Download Disadvantage - Computer Networks - Past Exam Paper and more Exams Computer Science in PDF only on Docsity! Exam #2 for Computer Networks (CIS 6930) Fall 2003 >>> SOLUTIONS <<< Welcome to Exam #2 in Computer Networks (CIS 6930). You have 75 minutes. Read each problem carefully. There are eight required problems (each worth 12.5 points) and one extra credit problem worth 5 points. You may have with you a calculator, pencils, eraser, blank paper, and one 8.5 x 11 inch “formula sheet”. On this formula sheet you may have anything you want (definitions, formulas, etc.) handwritten by you. You may use both sides. Computer generated text, photocopies, and scans are not allowed on this sheet. Please submit your formula sheet with your exam. No sharing of calculators. Start each numbered problem on a new sheet of paper and do not write on the back of the sheets. Submit everything in problem order. Good luck and be sure to show your work. Problem #1 a) Compare checksums and CRCs as a means of detecting errors. Discuss the tradeoffs between checksums and CRCs. Checksums have a greater probability of undetected errors than do CRCs. That is, CRCs are better at detecting errors and will result in less undetected errors than checksums. CRCs are easily be computed in hardware, but not very easily in software. Checksums can be computed in software much faster than can CRCs. b) Compute the CRC for a given message (M) and a generator polynomial (P). M is 0111101 and P is 1011. We divide M by P and the remainder is the CRC. We use modulo-2 math with no borrows or carries. The quotient of the division is 110110 and the remainder is 10. Thus, the CRC is 010. c) Describe an error that cannot be detected with a checksum. Consider a message of at least two N-bit words. For a N-bit check sum if an odd numbered work has a value M added to it that an even numbered word has subtracted from it, then the checksum will be the same as the for the original message. d) An Ethernet frame has 8 bytes of 10101010b in the preamble. What is the purpose of this preamble? This preamble serves to delimit the frame for the receiving adapter. Specifically, the purpose of the preamble is to synchronize the receiver’s clock with the sender’s clock for the receipt of the frame. Problem #2 a) What is a Local Area Network (LAN)? Be precise. Very precisely, a LAN is a data network optimized for a medium-sized area with 10’s to 100’s of stations and 100’s to 1000’s of meter span. A LAN is owned, operated, and used by a single organization. b) What are the types of media used in LANs? The types of media are coax, unshielded twisted pair (two categories – voice grade (3) and data grade (5)), fiber, and wireless. There was also shielded twisted pair (from IBM). Coax comes in two flavors – thick and thin. c) What is the key disadvantage of a physically direct-wired LAN? Explain how a star-wired LAN remedies this disadvantage. Fault detection and isolation are very difficult in a large (large span and large number of stations) direct-wired LAN. To find and isolate a fault one must literally “walk the wire”. In a star wired LAN faults can be detected and isolated in a central location – the wiring closet. To detect and isolate a fault in a wiring closet, individual lobes of the LAN can be unplugged until the LAN returns to operation. This can be automated. d) Compute the minimum possible frame size for a CSMA/CD protocol given the following parameters. Maximum medium span is 5000 meters (signal propagation is 5 nanoseconds per meter) and the data rate is 100- Mbps. The minimum frame size is prfr tt ⋅= 2 where frt is the frame size in bits divided by the data rate in bits per second and prt is the medium span in meters multiplied by the signal propagation in seconds per meter. We solve for frame size, x, in ( ) 96 105500010100 −⋅⋅=⋅x and the result is 312.5 bytes. e) Compute the utilization (U) for a token ring LAN where all stations have frames queued to send given the following parameters. The ring length is 5000 meters (signal propagation is 5 nanoseconds per meter), number of stations is 25, mean frame length is 100 bytes, and the data rate is 100-Mbps. Would early token release increase the utilization? If yes, by how much? We first compute ( ) ( )6101008100 ⋅⋅=frt which is 8 microseconds and 91055000 −⋅⋅=prt which is 25 microseconds. Here we have prfr tt < so ( )( )NtttU prprfr += which is 30.8%. for this case, early token release would improve performance to ( )( )NtttU prfrfr += which is 88.9%. this is an almost 3 times improvement. Problem #3 a) Give the forwarding and learning algorithm for transparent bridging (TB). How are entries removed from tables? The TB forwarding and learning algorithm is executed for each received frame at each port: 1) Receive a frame 2) If (DA is in table) and (DA is local) then not forward the frame, else forward the frame 3) If (SA is in table) then check and update the direction in the table 4) If (SA is not in the table) then add SA to the table A time-out policy is used to remove unused addresses in a bridge table (i.e., addresses not checked and updated within a set period of time – typically 5 minutes). b) Compare and contrast transparent bridging and source routing. Transparent bridging (TB) does not require changes to the end nodes (e.g., to applications), source routing (SR) does. In SR end nodes must initiate route discovery and store routes. TB does not allow for parallel forwarding bridges (i.e., cycles in a network of LANs), SR does allow this. TB may be less speed scalable since it involves table look-up and updating in a bridge whereas SR requires much less processing (essentially only a matching of ring and bridge number in a RIF) in a bridge. For frequently changing topologies, SR is a mechanism of quickly determining (likely) new paths. The TB table learning and (especially) ST algorithm require considerable time to adjust to a new topology Problem #4 a) What are the attributes of a perfect transport? The perfect transport is 1) suitable for LAN and WAN, 2) suitable for carrying real-time traffic (low delay), 3) speed scalable, and 4) easy to switch.