Networking: Error Recovery Method - Math Problem Example

NAME OF UNIVERSITY SIT202 COMPUTER NETWORKS ASSIGNMENT 3 NAME LECTURER: DATE: QUESTION ONE 2 points: Compare and contrast the go-back-N recovery method with the selective retransmission error recovery method. The GO-BACK-N error recovery method and the selective retransmission error recovery method are methods of data transmission that uses time-outs and acknowledgement signals to achieve reliable transmission of data over un-reliable network service, commonly the internet. An acknowledgement is a message sent by the receiver to the transmitter to indicate that it has correctly received a data frame or packet. A timeout is a reasonable point of time, usually after the sender sends the frame or packet. Both of these error recovery methods use Automatic Repeat Request(ARQ) to request for retransmission if a packet fails to arrive successfully. However, in GO-BACK-N ARQ protocol, the sender continues to send the frames without receiving the ACK signal from the receiver. The receiver on the other hand, receives the next frame following the sequence number and sends that number with every ACK signal it sends back. The receiver does not accept frames that don’t contain the SEO number it expects. After sending all the signals, the sender will detect that all the packets since the last lost packet are outstanding, and will go back and resend all the packets since the last ACK signal it received. Selective Repeat ARQ uses a different approach to correct transmission errors. Whereas in Go-Back-N method, all packets after the last unacknowledged frame is retransmitted, selective repeat retransmits only the unacknowledged frame. This makes it more efficient than Go-Back-N method, although more complicated to implement. This method also allows packets that have been received out of order to be stored by the receiver and later on re-ordered into their correct order, which is different from the Go-Back-N method where out of order packets are treated as errors and retransmission requested. QUESTION TWO Error detection is of the most crucial importance for network users. a) 1 point: Propose a generator. b) 1 point: Calculate CRC value for the following message: 101110101. c) 1 point: Show the whole message sent by the sender. a) CRC-8-ATM(HEC) polynomial x8+x2+x+1. . b) CRC value for 101110101 Step 1.Coefficient of the highest poser of x in x8+x2+x+1. And adding 1to the leftmost part results in 100000111 1 0 1 1 1 0 1 0 1 0 0 0 0 0 0 0 0 message - 1 0 0 0 0 0 1 1 1 divisor 0 0 1 1 1 0 0 1 0 0 0 0 0 0 0 0 0 result - 0 0 0 0 0 0 0 0 0 0 0 1 1 1 0 0 1 0 0 0 0 0 0 0 0 0 - 1 0 0 0 0 0 1 1 1 0 0 0 1 1 0 0 1 1 1 1 0 0 0 0 0 0 - 1 0 0 0 0 0 1 1 1 0 0 0 0 1 0 0 1 1 0 0 1 0 0 0 0 0 - 1 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 1 1 0 1 0 1 0 0 0 0 - 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 1 0 1 0 0 0 0 - 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 1 0 1 0 0 0 0 - 1 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 1 0 1 0 1 1 1 1 0 CRC Value - 1 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 1 0 1 1 0 0 1 Remainder (8 bits) is 01011001, is the CRC value c) The complete message sent by the sender is the message with the CRC value, i.e 101110101 01011001 QUESTION THREE Define, compare and contrast the sliding window protocol with the stop-and-wait protocol in error correction The sliding window protocol operates on the principle that both the sender and receiver keep a ‘window’ of acknowledgement and frames to transmit. The message to be sent is broken down into packets by the sender, where each packet is assigned a unique sequential number. The sender will keep the value of an expected ACK for every packet, whereas the receiver keeps a value of the expected frame to receive. When the sender receives an ACK from the receiver for a frame sent, it advances the window (or sends the next frame in the sequence) , whereas when the receiver receives the expected frame, it sends an ACK for that frame and advance the window. In this case, both the receiver and sender maintain a window for messages for which no ACK have been received.The term ‘window’ depicts the total number of packets and ACK yet to be send and received. The stop and wait protocol requires that the receiver send an acknowledge (ACK) message in turn for every frame received. This is practically useful where the sender and receiver operate at different transmission speeds. If the transmitter does not receive the ACK within a specified time, it retransmits the data packet. The sender will not send the next packet on the line, until it receives the ACK message from the receiver. This ensures that no information is dropped or lost during transmission. It typically uses an Automatic Repeat Request (ARQ). This protocol resembles a simplified version of the sliding window, only the difference being that it uses one window. Unlike the sliding window protocol, in the stop and wait, the sender will not advance to send the next frame if it does not receive the ACK for the previous frame sent. QUESTION FOUR 2 points: Consider sending an object of size M = 100 Kbytes from server to client. Let S = 536 bytes and RTT = 100 msec. Suppose the transport protocol uses static windows with window size W. A) For a transmission rate of 28kbps, determine the minimum possible latency. Determine the minimum window size that achieves this latency. Total number of frames/windows = total size of object / frame size =(100*1024)bytes /536 =191.0045frames Transmitting 536b over 28kbps network will take 536/(28*1024)=0.018694seconds The RTT produces 100msec delay for each transmission, therefore to transmit one frame will take 0.018694seconds +100msec, producing 0.118694seconds per frame Total time taken for complete transmission = 191.0045frames * 0.118694sec/frame = 22.671seconds B) 100kbps Transmitting 536b over 100kbps network will take 536/(100*1024)=0.005234seconds The RTT produces 100msec delay for each transmission, therefore to transmit one frame will take 0.005234seconds +100msec, producing 0.105234seconds per frame Total time taken for complete transmission = 191.0045frames * 0.105234sec/ frame = 20.100 sec C) 1mbps Transmitting 536b over 1048576bitspersec network will take 536/(1048576)=0.000511169 seconds The RTT produces 100msec delay for each transmission; therefore to transmit one frame will take 0.000511169 seconds +100msec, producing 0.100511169 seconds per frame Total time taken for complete transmission = 191.0045frames * 0.100511169 sec/ frame = 19.1980 sec D) 10mbps Transmitting 536b over 10485760bitspersec network will take 536/(10485760)=0.0000511169 seconds The RTT produces 100msec delay for each transmission; therefore to transmit one frame will take 0.0000511169 seconds +100msec, producing 0.1000511169 seconds per frame Total time taken for complete transmission = 191.0045frames * 0.1000511169 sec/ frame = 19.1102sec QUESTION FIVE 3 points: Consider a broadcast channel with N nodes and a transmission rate of R bps. Suppose the broadcast channel uses polling (with an additional polling node) for multiple accesses. Suppose the amount of time from when a node completes transmission until the subsequent node is permitted to transmit (that is, the polling delay) is tpoll. Suppose that within a polling round, a given node is allowed to transmit at most Q bits. What is the maximum throughput of the broadcast channel? The length of a polling round is N(Q/R +tpoll) The number of bits transmittedin a polling round is NQ. The maximum throughput therefore is QUESTION SIX 2 points: Describe the operation of the IEEE 802.3 medium access control protocol. Identify and discuss the features of this protocol that allow it to perform well during traffic bursts. IEEE 802.3,is the most popular wired network architecture to date.This protocol uses CSMA/CD technique (Carrier Sense Multiple Access with Collision Detect). Stations with CSMA/CD can access the network at any time, and listen to the network to see if the network is busy or already in use. If the network is busy, the station that wish to transmit data waits, but if the station is not in use, it transmits. Collisions occur when two or more stations listen to the network for data traffic and hear none, and then transmit simultaneously. In such occasions, all the transmissions are corrupted, and retransmission must be made later. Backoff algorithm is used to determine when the colliding station retransmits. This protocol implements the use of collision detection. However, IEEE 802.3 does not provide any support for real life data traffic. There are no provisions made to reserve bandwidth for a certain connection, and nodes access the network using CSMA/CD. QUESTION SEVEN Define frequency-division-multiplexing (FDM) and time-division-multiplexing (TDM). a) 1 point: Compare and contrast FDM and TDM. b) 1 point: What advantages does TDM have over FDM in a circuit-switched network? A) TDM (Time Division Multiplexing) and FDM (Frequency Division Multiplexing) are methods of multiplexing multiple signals into a single carrier for transmission. Multiplexing is a process of transmitting various signals to different receivers using one transmission carrier. The basic difference between FDM and TDM is that, FDM divides the channel into two or more frequency ranges that do not overlap, while TDM divides and allocates specific time periods to data to be transmitted by various transmitters in an alternating way. In TDM, each signal that is transmitted uses all of the available bandwidth for a given time, whereas multiplexing in FDM, each signal uses a small portion of the bandwidth all the time. B) TDM offers more flexibility in circuit switched networks by allocating more time periods to the signals that need more of the bandwidth, while limiting the time periods allocated to those signals that do not require more of the available bandwidth. FDM does not have this flexibility, because it cannot dynamically change the width of the allocated frequency. QUESTION EIGHT a) B)List of equipment’s 5 file servers 5 printer servers 12 printers 132 workstations 7 switches List of equipment’s C) 5 file servers D) 5 printer servers E) 12 printers F) 132 workstations G) 7 switches QUESTION NINE 2 points: What does it mean for a wireless network to be operating in “infrastructure mode”? If the network is not in infrastructure mode, what mode of operation is it in, and what is the difference between that mode of operation and infrastructure mode? Infrastructure mode of operation in wireless networks is whereby each wireless host is connected to the larger network through a base station or access point. Wireless networks can also operate in ad-hoc mode. In this mode, the wireless hosts have no infrastructure with which to connect. In the absence of such infrastructure, the hosts themselves must provide for services such as routing, address assignment, DNS-like name translations among other tasks. QUESTION 10 3 points: Encryption is very important. a) The task is to encrypt a plaintext message using polyalphabetic encryption. For this purpose you should use two Caesar ciphers, with two keys k = 4, k = 12, and the repeating pattern C1, C2, C1, C1, C2. The message is as follows: plain text a b c d e f g h i j k l m n o p q r s t u v w x y z C1(k=4) d e f g h i j k l m n o p q r s t u v w x y z a b c C2(k=12) l m n o p q r s t u v w x y z a b c d e f g h i j k Using the repeating pattern of c1,c2,c1,c1,c2 Message computer networks are important to business Cipher text fzpsfwpuqpwzzuvxdchlxsrcwlqwwzexvlyhvv b) Which elements of this encryption must be treated securely? The keys and the repeat pattern. QUESTION ELEVEN Define, compare and contrast key management in systems that exploit symmetric encryption with systems that exploit public key encryption. Symmetric key cryptography involves two people using the same private key to encrypt and decrypt data. Public key cryptography is whereby two different keys are used for data encryption, a public key for encryption and a private key for decryption. Symmetric key cryptography uses the same key for both encryption and decryption, therefore is much faster as compared to public key cryptography. It is also much easier to implement and use, and in overall required less processing power. The only limitation of symmetric key is that the two parties communicating must agree to use the same private key prior to transmitting secure information, and at times can be impossible due to security reasons. Public key cryptography offers an advantage in such scenario because it is public or disclosed. People use public key cryptography to create a shared session key and then communicate through symmetric key cryptography using the created session key. This generally improves the performance of the shared key cryptography and offers the convenience of public key cryptography. QUESTION TWELVE Digital signatures a) 1 point: Explain the reasons for using message digests in signing messages. A message digest, also called a cryptographic message are special numbers that are effectively hashcode produced by a function that is difficult to reverse. This, when combined with a person’s private key, ensures the integrity(ensures that the message has not been tampered with) and authenticatiN of the message b) 1 point: Compare and contrast signing long messages with and without message digests. The main function of signing a message with message digest is to ensure that the message is kept unaltered, i.eensure the integrity of the message. A message can be encrypted using public key cryptography without using message digest, or conventional cryptography to ensure that only the sender and receiver of a message understands it. However, there is concerns that the message can be modified while in transit. Message digests are used to sign long messages using message digests that create short, fixed length representation off longer variable length messages. This eliminate the possibility of substituting one message for another in transit while maintaining the same digest. QUESTION THIRTEEN You have been commissioned by the CEO of a business to build a new network to support its two branches, one in Victoria and one in Queensland. The CEO has asked that all traffic between the two branches and within each branch be monitored. The CEO claims it is so that employees do not use the network for purposes other than work. Based on this description, discuss: (i) 3 points: What legal issues exist for such a network in respect to the CEO’s requirements? The Surveillance Devices Act 1998 (WA)(SD Act) regulates the use of listening, devices, optical surveillance devices andtracking devices. The SD Act may applyto workplaces where surveillance devicessuch as security cameras, closed circuitTV (CCTV), telephone monitoring and GPS systems are being used.It is generally an offence to use, install ormaintain: • listening devices to record or listen to aprivate conversation; • optical surveillance devices to recordvisually or observe a private activity;or • tracking devices to determine thegeographical location of a person. Legal issues may arise when a company chooses to monitor the activities that their employees do while on a network. This is usually justified by various countries in their laws as upholding the community standards and ensuring national security. Private companies may scrutinize and filter the content of their employee’s communication on the network. This can arise various legal issues such as Network monitoring and filtering often may result in the employees feeling that their citizen’s state-recognized civil rights have been violated. This usually includes rights that involve speech, press, privacy, religion and association. (ii) What must be specified to meet the CEO’s requirements The CEO must specify the code of ethics and work. The CEO must make it clear that the code is not a legal code, but a computer code. Software code avoids the jurisdiction pitfalls of its legal counterpart,, enforcement is automatic and in theory, absolute. The CEO should interplay the two codes as well as social norms code (iii) 1 point: If an existing employee claims this is a violation of privacy, does the employee have a case? No. (iv) If the branches were in different states of Australia, would the implications change? No. the implications would still be the same Read More