Simple FEC is a forward error correction scheme for recovering from packet los in a VolP application. This model creates a redundant chunk by exclusive OR-ing n origina chunks and sends it along with the original ones so that the receiver can reconstruct th original n chunks if at-most one chunk is lost throughout the transferring process. Assumin that n = 4 and receiver receives four out of the five chunks sent by the sender in the followin way, reconstruct the missing (lost) chunk. • First Chunk: 11101101 • Second Chunk: 01000101 • Third Chunk: lost • Fourth Chunk: 11100010 Redundant Chunk: 11010000

Simple FEC is a forward error correction scheme for recovering from packet los in a VolP application. This model creates a redundant chunk by exclusive OR-ing n origina chunks and sends it along with the original ones so that the receiver can reconstruct th original n chunks if at-most one chunk is lost throughout the transferring process. Assumin that n = 4 and receiver receives four out of the five chunks sent by the sender in the followin way, reconstruct the missing (lost) chunk. • First Chunk: 11101101 • Second Chunk: 01000101 • Third Chunk: lost • Fourth Chunk: 11100010 Redundant Chunk: 11010000

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

See similar textbooks

Similar questions

The figure depicts K sources which are connected to the Internet via linksof capacity RS, and within the network fairly share a common link ofcapacity R to K destinations. Each destination is connected to the networkby a link of capacity RD. You can assume that there are no other links orsource-destination pairs in the network.Now suppose that source Si has an infinitely large file it wants to send todestination Di (i.e., each source sends to a different destination).a. If K=10, RS = 100 Mbps, Rd=10 Mbps, and R = 10 Gbps. What is thethroughput between each source-destination pair? Where are thebottleneck links?If we now assume that K=10, RS = 100 Mbps, Rd= 100 Mbps, and R = 0.5Gbps. What are the new throughputs between each source-destinationpair? Where are the bottle-neck links?
Computer Science Before the introduction of IEEE 802.11i, the security scheme for IEEE 802.11 was Wired Equivalent Privacy (WEP). Assume that the initialization vector (IV) is not used in WEP. Instead, RC4 is applied using K as input and producing s as output. Assume RC4(K, n) is applied for each packet and that all packets have the same length. If an attacker intercepts a single packet with ciphertext C1 and also has a known plaintext/ciphertext pair (P2, C2). Show how the attacker can find the plaintext P1. You can make use of an example by assuming some random data is being encrypted.
Consider the network scenario in the figure below. K sources are connected to the Internet via links of capacity RS, and within the network fairly share a common link of capacity R, to K destinations. Each destination is connected to the network by a link of capacity RD. You can assume that there are no other links or source-destination pairs in the network. Suppose that source Si has an infinitely large file it wants to send to destination Di (i.e., each sources sends to a different destination). Suppose that K=10, RS = 100 Mbps, RD= 54 Mbps, and R = 50 Gbps. What is the throughput between each source-destination pair?
This problem concerns the code based on the 31 - hat game. Here the message should be decomposed into blocks of length 31–5 = 26. Suppose that one of the 26 blocks in the starting message has eight 0’s followed by ten 1’s, followed by eight 0’s. That is, it is 00000000111111111100000000 (a) Find its associated L-node, which is the code word which will be transmitted. (b) Suppose that in the transmission, there is an error in the 20th character. Give a detailed account of the calculations made by the decoder to recover the original message
When Bob wants to send Alice a message M on the Internet, he breaks M into n data packets, numbers the packets consecutively, and injects them into the network. When the packets arrive at Alice’s computer, they may be out of order, so Alice must assemble the sequence of n packets in order before she can be sure she has the entire message. Describe an efficient scheme for Alice to do this. What is the running time of this algorithm? From Data Structures and Algorithms C-7.56
Calculate the total time required to transfer a 1000-KB file in the following cases, assuming an RTT of 50 ms, a packet size of 1 KBdata, and an initial 2 RTT handshake before data is sent: The bandwidth is "infinite," meaning that we take transmit time to be zero, and up to 20 packets can be sent per RTT. The bandwidth is infinite, and during the first RTT we cansend one packet, during the second RTT we can send two packets, during the third we can send four and so on.
Calculate the total time required to transfer a 1.5-MB file in the following cases, assuming an RTT of 80 ms, a packet size of 1 KBdata, and an initial 2 RTT handshake before data is sent: The link allows infinitely fast transmit, but limits bandwidth such that only 20 packets can be sent per RTT. The bandwidth is infinite, but during the first RTT we can send one packet, during the second RTT we can send two packets, during the third we can send four, etc.
(ESP) Consider that we protect a TCP SYN packet using the ESP in the transport mode. A TCP SYN packet just comprises a TCP header which is 20 bytes (i.e., without TCP option). The IP packet header is also 20 bytes long (i.e., without IP option). Moreover, the block size for encryption is 256 bits (e.g., using AES-256), and the hash output’s size for message authentication is 512 bits (e.g., using SHA-1). What is the minimum amount of padding required?
Consider the network scenario in the figure below. K sources are connected to the Internet via links of capacity RS, and within the network fairly share a common link of capacity R, to K destinations. Each destination is connected to the network by a link of capacity RD. You can assume that there are no other links or source-destination pairs in the network. Suppose that source Si has an infinitely large file it wants to send to destination Di (i.e., each sources sends to a different destination). 2. Suppose now that K=10, RS = 100 Mbps, RD= 1 Mbps, and R = 0.75 Gbps. What are the throughputs between each source-destination pair?
We demand that a commit bit be evaluated to verify whether a read request needs wait under an updated version of the timestamp protocol. Explain how the commit bit prevents aborts from cascading. Why isn't this check required when submitting requests?
BRANCHES OF COMPUTERS Data collision occurs when many senders attempt to access the same medium at the same time; what protocol(s) and at what Layer(s) are available to prevent this? Can we say for sure that these methods provide safe, collision-free transport? If not, what other ways may this issue be fixed?
Consider the following scenario in which host A is sending a file to host B over a TCP connection. Assuming that the sequence number of the first data byte sent by A is 0 and every segment always includes 1000 bytes of data, excluding the TCP header. At some point of time, bytes up to 6400 have been written into the sender’s buffer. Bytes up to 4999 have been sent out but the segment which contains bytes 2000~2999 has not arrived at host B yet. At the receiver’s side, all bytes up to 3999 have been received except for bytes 2000~2999. Bytes up to 499 have been read from the buffer by the application. Assume that the maximum size of the sender’s buffer is large enough. Consider the sliding window algorithm in TCP and answer the following questions. 1) What are the values for LastByteAcked, LastByteSent, and LastByteWritten? 2) What are the values for LastByteRead, NextByteExpected, and LastByteRcvd? 3) Assuming that the maximum size of the receiver’s buffer is 4000 byte, what would…