Matoshri College of Engineering & R. C. Nashik Department of Computer Engineering. Subject :- Computer Network. [310245] Semester:- 6 Subject I/c :- Mr. Ranjit Gawande.
Model Ans. Paper for Computer Network 2015 Pat. Unit1 Physical Layer Q1.A What are different logical arrangement of devices in a network ? Explain any 4 arrangement in detail with suitable diagram.
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Ans. Star Mesh ring bus Q1.B What are different network category ? For data communication system within a building or campus which network category you will suggest ? Explain it with suitable diagram?
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Ans. LAN WAN MAN PAN For above problem LAN will be used. Q2A For a data communication system which spanning state or countries which network category you will prefer?
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Ans. Q2B Why are protocols needed?
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Q2C Describe in short importance of & working of following commands 1) Ping 2) Netstat 3) Traceroute 4) IPConfig Ans. Ping:- The ping program use to find if a host is alive and responding. We use ping here to see how it uses ICMP packets. The source host sends ICMP echo-request messages (type: 8, code: 0); the destination, if alive, responds with ICMP echo-reply messages. The ping program sets the identifier field in the echo-request and echo-reply message and stans the sequence number from 0; this number is incremented by 1 each time a new message is sent. Netstat :- It is used for Print network connections, routing tables, interface statistics, masquerade connections, and multicast memberships Netstat describe as Netstat prints information about the Linux networking subsystem. The type of information printed is controlled by the first argument. Traceroute The traceroute program in UNIX or tracert in Windows can be used to trace the route of a packet from the source to the destination. Traceroute program to simulate the loose ______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
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source route and strict source route options of an IP datagram. This program in conjunction with ICMP packets.
ifconfig - configure a network interface Ifconfig is used to configure the kernel-resident network interfaces. It is used at boot time to set up interfaces as necessary. After that, it is usually only needed when debugging or when system tuning is needed. If no arguments are given, ifconfig displays the status of the currently active interfaces. If a single interface argument is given, it displays the status of the given interface only; if a single -a argument is given, it displays the status of all interfaces, even those that are down. Otherwise, it configures an interface. Q3.A What are advantages and disadvantages of each network topology
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Ans. Q3.B What is the difference between half-duplex and full duplex transmission modes ?
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Ans. Q3.C What is DSSS? What is the role of chips generator in hopping the signals in DSSS?
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Ans. Q4.A Explain all version of 802.11 standard & compare. Ans. Following are the versions :- 802.11a, 802.11b/g/n, and/or 802.11ac wireless standards collectively known as Wi-Fi technologies. (Bluetooth and various other wireless (but not Wi-Fi) technologies also exist, each designed for specific networking applications. 802.11 In 1997, (IEEE) created the first WLAN standard. 802.11 only supported a maximum network bandwidth of 2 Mbps - too slow for most applications. For this reason, ordinary 802.11 wireless products are no longer manufactured. 802.11b IEEE expanded on the original 802.11 standard in July 1999, creating the 802.11b specification. 802.11b supports bandwidth up to 11 Mbps, comparable to traditional Ethernet. 802.11b uses the same unregulated radio signaling frequency (2.4 GHz) as the original 802.11 standard. Being unregulated, 802.11b gear can incur interference from microwave
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ovens, cordless phones, and other appliances using the same 2.4 GHz range. However, by installing 802.11b gear a reasonable distance from other appliances, interference can easily be avoided. Pros of 802.11b - lowest cost; signal range is good and not easily obstructed Cons of 802.11b - slowest maximum speed; home appliances may interfere on the unregulated frequency band. 802.11a While 802.11b was in development, IEEE created a second extension to the original 802.11 standard called 802.11a. Because 802.11b gained in popularity much faster than did 802.11a, some folks believe that 802.11a was created after 802.11b. In fact, 802.11a was created at the same time. Due to its higher cost, 802.11a is usually found on business networks whereas 802.11b better serves the home market. 802.11a supports bandwidth up to 54 Mbps and signals in a regulated frequency spectrum around 5 GHz. This higher frequency compared to 802.11b shortens the range of 802.11a networks. The higher frequency also means 802.11a signals have more difficulty penetrating walls and other obstructions. Pros of 802.11a - fast maximum speed; regulated frequencies prevent signal interference from other devices Cons of 802.11a - highest cost; shorter range signal that is more easily obstructed In 2002 and 2003, WLAN products supporting a newer standard called 802.11g emerged on the market. 802.11g attempts to combine the best of both 802.11a and 802.11b. 802.11g supports bandwidth up to 54 Mbps, and it uses the 2.4 Ghz frequency for greater range. 802.11g is backwards compatible with 802.11b, meaning that 802.11g access points will work with 802.11b wireless network adapters and vice versa. Pros of 802.11g - fast maximum speed; signal range is good and not easily obstructed Cons of 802.11g - costs more than 802.11b; appliances may interfere on the unregulated signal frequency Q4.B Draw a hybrid topology with a ring backbone and three bus networks
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Ans. Q5.A How are OSI and ISO related to each other?
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Ans. Q5.B What is the difference between a port address, a logical address, and a physical address?
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Ans. Q6.A What is the difference between network layer delivery and transport layer delivery? Ans.
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Q6.B In which layer process to process communication occur? List the layers of the Internet model
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Q7.A Explain following network devices 1) Bridge 2) Router 3) Switch
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Q7.B Write short notes on 1) Components of optical network, 2) SDN
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Ans. 2) SDN (Software Defined Networking ) Software Defined Networking (SDN) is to run a controller that can install low-level packet-handling rules in the underlying switches using a standard protocol. This controller can run various network-management applications, such as dynamic access control, seamless user mobility, traffic engineering, server load balancing, energy-efficient networking. SDN is a framework to allow network administrators to automatically and dynamically manage and control a large number of network devices, services, topology, traffic paths, and packet handling (quality of service) policies using high-level languages and APIs. Management includes provisioning, operating, monitoring, optimizing, and managing FCAPS (faults, configuration, accounting, performance, and security) in a multi-tenant environment. Key: Dynamic , Quick Legacy approaches such as CLI were not quick particularly for large networks The physical separation of the network control plane from the forwarding plane, and where a control plane controls several devices.” 1. Directly programmable 2. Agile: Abstracting control from forwarding 3. Centrally managed 4. Programmatically configured 5. Open standards-based vendor neutral The need SDN:1. Virtualization: Use network resource without worrying about where it is physically located, how much it is, how it is organized, etc. 2. Orchestration: Should be able to control and manage thousands of devices with one command. 3. Programmable: Should be able to change behavior on the fly.
Q8.A What waves ranging frequencies in Radio waves and Micro waves ? What are advantage and disadvantages of radio waves? Ans.
Radio wave:- 3kHz and 1 GHz Microwaves 1 and 300 GHz Adv:- AM radio can receive signals inside a building. Radio wave can propagate in the sky mode , can travel long distances, can penetrate walls.
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Disadvantages;- One cannot isolate a communication to just inside or outside a building , Radio wave band is narrow just under 1GHz , thus low data rate transfer.
Q8.B What is FHSS? Explain the role of pseudorandom code generator and frequency synthesizer in FHSS with suitable example ?
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Q8.C What are different client layers of optical fiber. Ans. Q9.A What are different Line coding schemes ? Explain any one scheme in detail. Ans. Q9.B With suitable example , Explain the difference between Manchester and differential Manchester encoding. Ans. Q9.C What are different classes of transmission media? Mention the data rate (in Mbps) for CAT5 & CAT6 twisted pair cable ? Ans. Note:- This literature is only for private circulation.
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Department of Computer Engineering. Subject :- Computer Network. [310245] Semester:- 6 Subject I/c :- Mr. Ranjit Gawande.
Model Ans. Paper for Computer Network 2015 Pat. Unit2 Logical Link Control Q1.A What are major functions of Data Link Layer ? Explain each function in detail
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Ans. Q1.B What are the services provided to Network Layer?
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Ans. Q2A How framing is done in data link layer / What are methods used to carry out framing?
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Ans. Q2B Explain Asynchronous and Synchronous in serial transmission with suitable diagram
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Ans. Q2C What are types of errors in data bits transmission?
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Ans. Q3A For code word no 1 :-> 1 1 1 1 0 1 0 0 code word no 2 :-> 0 1 0 1 1 1 1 0 Calculate the hamming distance between above two code words.
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Ans. Q3B What are various error detection methods? Explain any one with suitable example?
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Ans. Q3C Write short note on 1) Cyclic Redundancy Check
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Ans. Q4A Explain error detection and correction in block coding
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Ans. Q4B Discuss the concept of redundancy in error detection and correction.
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Ans. Q4C Generate the hamming codeword for ASCII character “U”= 1010101 . Assume even parity for the hamming code.
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Ans. Q5A Explain two dimensional parity check
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Ans. Q5B What is hamming code? Generate code words using hamming code for following data words 1011 , 0101
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Ans. Q5C Explain liner code block
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Ans. Q6A What is hamming distance? Explain it with an example , Explain simple parity check code?
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Ans. Q6B Explain CRC generator and CRC checker with suitable example.
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Ans. Q6C Generate the CRC code for message 1101010101. Given generator polynomial 4 2 g(x) = x +x +1 Ans. Q7A Describe in detail Internet checksum method with suitable example. Ans. Q7B Consider the use of 1000-bit frames on a 1 Mbps satellite channel with a 270 ms delay. What is the maximum link utilization for 1) Stop and wait flow control? 2) Sliding window flow control with a window size of 7 ? 3) Sliding window flow control with a window size of 127? 4) Sliding window flow control with a window size of 255? Ans. Q7C Explain character stuffing & bit stuffing Ans. Q8A What is piggybacking? Ans. Q8B Explain the stop and wait protocol? Mentioned its major drawbacks
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Ans. Q8C Define throughput efficiency and explain how it can be increased? Ans. Q9A Draw and explain various frame types in HDLC Ans. Q9B Explain transparency and bit stuffing in HDLC. Ans. Q9C Explain the flow diagram for PPP connection. Ans. Q10A What is LCP and NCP. Ans. Q10B Write the difference between SLIP and PPP. Ans. Q10C A channel has a bit rate of 4.8 kbits/sec and a propagation delay of 20 msec. For what range of a frame size does stop and wait protocols given an efficiency of 50%
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Matoshri College of Engineering & R. C. Nashik Department of Computer Engineering. Subject :- Computer Network. [310245] Semester:- 6 Subject I/c :- Mr. Ranjit Gawande.
Model Ans. Paper for Computer Network 2015 Pat. Unit3 Medium Access Control Q1.A
Define random access and list three protocols in this category.
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Define controlled access and list three protocols in this category
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Q1.B
Ans. Ref Above Q 1 Ans Fig. Q2A
What is backoff time in pure ALHOA ? With suitable fugue and explanation show collision of frames in pure ALHOA
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Ans. Pure ALOHA dictates that when the time-out period passes, each station waits a random amount of time before resending its frame. The randomness will help avoid more collisions. We call this time the back-off time TB. Q2B
Explain steps involved in, to prevent congesting the channel in pure ALHOHA
Ans.
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Q2C
The stations on a wireless ALOHA network are a maximum of 600 km apart. If we assume that signals propagate at 3 x 108 mis, then find Tp also fine find the value of TB for different values of K. Assume maximum attempts are 5
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Ans. Number of attempts = K, Maximum propagation time = Tp , we find Tp = (600 x 105) / (3 x 108) = 2ms Now we can find the value of TB for different values of K. a. For K = 1, the range is {O, I}. The station needs to generate a random number with a value of 0 or 1. This means that TB is either 0ms (0 x 2) or 2 ms (l x 2), based on the outcome of the random variable. b. For K = 2, the range is {O, 1, 2, 3}. This means that TB can be 0, 2, 4, or 6 ms, based on the outcome of the random variable. c. For K =3, the range is to, 1,2,3,4,5,6, 7}. This means that TB can be 0,2,4, ... , 14 ms, based on the outcome of the random variable. d. We need to mention that if K > 10, it is normally set to 10. Q3A
A pure ALOHA network transmits 200-bit frames on a shared channel of 200 kbps. What is the requirement to make this frame collision-free?
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Ans. Tfr= Average transmission time for a frame, Each station send fixed length frame. (Pure ALOHA vulnerable time = 2 x Tfr) Average frame transmission time Tfr is 200 bits/200 kbps or 1 ms. The vulnerable time is 2 x 1 ms =2 ms. This means no station should send later than 1 ms before this station starts transmission and no station should start sending during the one 1-ms period that this station is sending Q3B
A pure ALOHA network transmits 200-bit frames on a shared channel of 200 kbps. What is the throughput if the system (all stations together) produces a. 1000 frames per second b. 500 frames per second c. 250 frames per second
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Ans. G the average number of frames generated by the system during one frame transmission time. The average number of successful transmissions for pure ALOHA is S = G x e-2G. The maximum throughput Smax is 0.184, for G = 1 / 2 Solution:The frame transmission time is 200 / 200 kbps or 1 ms a. If the system creates 1000 frames per second, this is 1 frame per millisecond. The load is 1. In this case S = G x e-2G or S =0.135 (13.5 percent). This means that the throughput is 1000 X 0.135 = 135 frames. Only 135 frames out of 1000 will probably survive. b. If the system creates 500 frames per second, this is (1/2) frame per millisecond. The load is (112). In this case S = G x e-2G or S = 0.184 (18.4 percent). This means that the throughput is 500 x 0.184 =92 and that only 92 frames out of 500 will probably survive. Note that this is the maximum throughput case, percentage wise. c. If the system creates 250 frames per second, this is (1/4) frame per millisecond. The load is (1/4). In this case S = G x e-2G or S = 0.152 (15.2 percent). This means that the throughput is 250 x 0.152 = 38. Only 38 frames out of 250 will probably survive. Q3C
What are the disadvantages of pure ALOHA? How efficiency improved in Slotted ALOHA?Explain slotted ALOHA with suitable diagram.
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Ans. Pure ALOHA has a vulnerable time of 2 x Tfr . This is so because there is no rule that defines when the station can send. A station may send soon after another station has started or soon before another station has finished In slotted ALOHA we divide the time into slots of Tfr s and force the station to send only at the beginning of the time slot.
The throughput for slotted ALOHA is S =: G x e-G. The maximum throughput Smax = 0.368 when G =1. Slotted ALOHA vulnerable time = Tfr Q4A
A slotted ALOHA network transmits 200-bit frames using a shared channel with a 200-kbps bandwidth. Find the throughput if the system (all stations together) produces a. 1000 frames per second b. 500 frames per second c. 250 frames per second
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Ans. The frame transmission time is 200/200 kbps or 1 ms a. In this case G is 1. So S = G x e-G or S =0.368 (36.8 percent). This means that the throughput is 1000 x 0.0368 = 368 frames. Only 368 out of 1000 frames will probably survive. Note that this is the maximum throughput case, percentage wise. b. Here G is 1/2. In this case S =G x e-G or S = 0.303 (30.3 percent). This means that the throughput is 500 x 0.0303 =151. Only 151 frames out of 500 will probably survive. c. Now Gis 1. In this case S = G x e-G or S = 0.195 (19.5 percent). This means that the throughput4is 250 x 0.195 = 49. Only 49 frames out of 250 will probably survive. Q4B
Explain how CSMA minimize the chance of collision and, increase the performance ?
Ans. The CSMA method was developed. The chance of collision can be reduced if a station senses the medium before trying to use it. Carrier sense multiple access (CSMA) requires that each station first listen to the medium (or check the state of the medium) before sending. In other words, CSMA is based on the principle "sense before transmit" or "listen before talk." CSMA can reduce the possibility of collision, but it cannot eliminate it.
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Q4C
With suitable diagram explain how vulnerable time = propagation time Tp in CSMA
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Ans. The vulnerable time for CSMA is the propagation time Tp . This is the time needed for a signal to propagate from one end of the medium to the other. When a station sends a frame, and any other station tries to send a frame during this time, a collision will result. But if the first bit of the frame reaches the end of the medium, every station will already have heard the bit and will stop from sending frame.
Q5A
If the station finds line is idle and wish to transmit frame then in P- persistent method what are the steps involved in transmission.
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Ans. Following steps are involved in P- persistent method. 1. With probability p, the station sends its frame. 2. With probability q = 1 - p, the station waits for the beginning of the next time slot and checks the line again. a. If the line is idle, it goes to step 1. b. If the line is busy, it acts as though a collision has occurred and uses the backoff procedure Q5B
Explain difference between CSMA and CSMA/CD?
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Ans. The CSMA method does not specify the procedure following a collision. Carrier sense multiple access with collision detection (CSMA/CD) augments the algorithm to handle the collision. In this method, a station monitors the medium after it sends a frame to see if the transmission was successful. If so, the station is finished. If, however, there is a collision, the frame is sent again
The first bits transmitted by the two stations involved in the collision. Although each station continues to send bits in the frame until it detects the collision, stations A and C are involved in the collision At time t 1, station A has executed its persistence procedure and starts sending the bits of its frame. At time t2, station C has not yet sensed the first bit sent by A. Station C executes its persistence procedure and starts sending the bits in its frame, which propagate both to the left and to the right. The collision occurs sometime after time t2' Station C detects a collision at time t3 when it receives the first bit of A's frame. Q5C
A network using CSMA/CD has a bandwidth of 10 Mbps. If the maximum propagation time
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(including the delays in the devices and ignoring the time needed to send a jamming signal) is 25.6 μs, what is the minimum size of the frame? Ans. The frame transmission time is Tfr = 2 x TP = 51.2 μs. This means, in the worst case, a station needs to transmit for a period of 51.2 μs, to detect the collision. The minimum size of the frame is 10 Mbps x 51.2 μs, =512 bits or 64 bytes. This is actually the minimum size of the frame for Standard Ethernet. Q6A
What are the limitation of CSMA/CD ?
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Ans. The basic idea behind CSMA/CD is that a station needs to be able to receive while transmitting to detect a collision. When there is no collision, the station receives one signal: its own signal. When there is a collision, the station receives two signals: its own signal and the signal transmitted by a second station. To distinguish between these two cases, the received signals in these two cases must be significantly different. In other words, the signal from the second station needs to add a significant amount of energy to the one created by the first station. In a wired network, the received signal has almost the same energy as the sent signal because either the length of the cable is short or there are repeaters that amplify the energy between the sender and the receiver. This means that in a collision, the detected energy almost doubles. However, in a wireless network, much of the sent energy is lost in transmission. The received signal has very little energy. Therefore, a collision may add only 5 to 10 percent additional energy. This is not useful for effective collision detection Q6B
How to avoid collision in wireless network? Which strategy you will suggest? Mentioned the advantages of suggested strategy over other available strategy.
Ans. We need to avoid collisions on wireless networks because they cannot be detected. Carrier sense multiple access with collision avoidance (CSMA/CA) was invented for this network. Collisions are avoided through the use of CSMA/CA's three strategies: the inter-frame space, the contention window, and acknowledgments
Inter-frame Space (IFS) First, collisions are avoided by deferring transmission even if the channel is found idle. When an idle channel is found, the station does not send immediately. It waits for a period of time called the inter-frame space or IFS. Even though the channel may appear idle when it is sensed, a distant station may have already started transmitting. The distant station's signal has not yet reached this station. The IFS time allows the front of the transmitted signal by the distant station to reach this station. If after the IFS time the channel is still idle, the station can send, but it still needs to wait a time equal to the contention time. In CSMA/CA, the IFS can also be used to define the priority of a station or a frame. Contention Window:- The contention window is an amount of time divided into slots. A station that is ready to send chooses a random number of slots as its wait time. The number of slots in the window changes according to the binary exponential back-off strategy. In CSMA/CA, if the station finds the channel busy, it does not restart the timer of the contention window; it stops the timer and restarts it when the channel becomes idle. Acknowledgment:- There still may be a collision resulting in destroyed data. In addition, the data may be corrupted during the transmission. The positive acknowledgment and the time- out timer can help guarantee that the receiver has received the frame.
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Procedure:- The channel also needs to be sensed during the contention time. For each time slot of the contention window, the channel is sensed. If it is found idle, the timer continues; if the channel is found busy, the timer is stopped and continues after the timer becomes idle again. Q6C Explain the role & importance of binary exponential back-off in pure ALOHA. Ans.
The back-off time TB is a random value that normally depends on K (the number of attempted unsuccessful transmissions). The formula for TB depends on the implementation. One common formula is the binary exponential back-off. In this method, for each retransmission, a multiplier in the range 0 to 2K - 1 is randomly chosen and multiplied by Tp (maximum propagation time) or T Π , (the average time required to send out a frame) to find TB' Note that in this procedure, the range of the random numbers increases after each collision. The value of K max is usually chosen as 15.
Q7A
Draw and explain neat diagram to show the relationship of IEEE 802 Standard to the traditional OSI model.
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The IEEE has subdivided the data link layer into two sublayer: logical link control (LLC) and media access control (MAC). IEEE has also created several physical layer standards for different LAN protocols. Data Link Layer As we mentioned before, the data link layer in the IEEE standard is divided into two sublayer: LLC and MAC. Logical Link Control (LLC) In IEEE Project 802, flow control, error control, and part of the framing duties are collected into one sublayer called the logical link control. Framing is handled in both the LLC sublayer and the MAC sublayer. Q7B
Draw and explain labeled classification of Ethernet evolution through with four generations?
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The original Ethernet was created in 1976 at Xerox's Palo Alto Research Center (PARC). Since then, it has gone through four generations: Standard Ethernet (lot Mbps), Fast Ethernet (100 Mbps), Gigabyte Ethernet (l Gbps), and Ten-Gigabyte Ethernet (l0 Gbps), as shown in Figure above Q7C
Explain IEEE 802.3 MAC frame format in detail.
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The Ethernet frame contains seven fields: preamble, SFD, DA, SA, length or type of protocol data unit (PDU), upper-layer data, and the CRe. Ethernet does not provide any mechanism for acknowledging received frames, making it what is known as an unreliable medium. Acknowledgments must be implemented at the higher layers. The format of the MAC frame is shown in Figure above. Preamble. The first field of the 802.3 frame contains 7 bytes (56 bits) of alternating Os and Is that alerts the receiving system to the coming frame and enables it to synchronize its input timing. Start frame delimiter (SFD). The second field (l byte: 10101011) signals the beginning of the frame. The SFD warns the station or stations that this is the last chance for synchronization. The last 2 bits is 11 and alerts the receiver that the next field is the destination address. Destination address (DA). The DA field is 6 bytes and contains the physical address of the destination station or stations to receive the packet. Source address (SA). The SA field is also 6 bytes and contains the physical address of the sender of the packet. We will discuss addressing shortly. Length or type. The IEEE standard used it as the length field to define the number of bytes in the data field. Both uses are common today. Data. This field carries data encapsulated from the upper-layer protocols. It is a minimum of 46 and a maximum of 1500 bytes CRC. The last field contains error detection information, in this case a CRC-32 Q8A
What are the various categories of standard Ethernet.
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The Standard Ethernet defines several physical layer implementations; four of the most common, are shown in Figure above. Encoding and Decoding All standard implementations use digital signaling (baseband) at 10 Mbps. At the sender, data are converted to a digital signal using the Manchester scheme; at the receiver, the received signal is interpreted as Manchester and decoded into data. Manchester encoding is self-synchronous, providing a transition at each bit interval
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Q8B
What is the maximum length of the coaxial cable in 10 base 5 Ethernet.
Ans. 10Base5, thick Ethernet, or Thicknet. The nickname derives from the size of the cable, which is roughly the size of a garden hose and too stiff to bend with your hands. 10Base5 was the first Ethernet specification to use a bus topology with an external transceiver (transmitter/receiver) connected via a tap to a thick coaxial cable The maximum length of the coaxial cable must not exceed 500 m, otherwise, there is excessive degradation of the signal. If a length of more than 500 m is needed, up to five segments, each a maximum of 500-meter, can be connected using repeaters. Q8C
What is the max length of each segment in 10 Base 2 Ethernet.
Ans. 10 Base 2, thin Ethernet, or Cheapernet. 10 Base 2 also uses a bus topology, but the cable is much thinner and more flexible. The cable can be bent to pass very close to the stations. In this case, the transceiver is normally part of the network interface card (NIC), which is installed inside the station. Installation is simpler because the thin coaxial cable is very flexible. However, the length of each segment cannot exceed 185 m (close to 200 m) due to the high level of attenuation in thin coaxial cable. Q9A
In which Ethernet scheme physical star topology used and the stations are connected to a hub via two pairs of twisted cable? Justify your answer.
Ans. 10 Base T Ethernet scheme will be used for above scenario, 10Base-T uses a physical star topology. The stations are connected to a hub via two pairs of twisted cable,two pairs of twisted cable create two paths (one for sending and one for receiving) between the station and the hub. Any collision here happens in the hub. Compared to l0Base5 or l0Base2, the hub actually replaces the coaxial cable as far as a collision is concerned. The maximum length of the twisted cable here is defined as 100 m, to minimize the effect of attenuation in the twisted cable. Q9B
Explain with suitable diagram, what are the advantage of a bridge in the collision domain.
Ans. The advantage of a bridge is the separation of the collision domain. Figure below shows the collision domains for an unbridged and a bridged network. One can see that the collision domain becomes much smaller and the probability of collision is reduced tremendously. Without bridging, 12 stations contend for access to the medium; with bridging only 3 stations contend for access to the medium.
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Q9C
Explain how the limitations of 10Base5 and l0Base2 are overcome using full duplex Ethernet.
Ans.
One of the limitations of 10Base5 and l0Base2 is that communication is half-duplex (l0Base-T is always full-duplex); a station can either send or receive, but may not do both at the same time. The next step in the evolution was to move from switched Ethernet to fullduplex switched Ethernet. The full-duplex mode increases the capacity of each domain from 10 to 20 Mbps. Figure above shows a switched Ethernet in full-duplex mode. Note that instead of using one link between the station and the switch, the configuration uses two links: one to transmit and one to receive. Q10A What are the design goal of Fast Ethernet? Explain. Ans. Fast Ethernet was designed to compete with LAN protocols such as FDDI or Fiber Channel (or Fibre Channel). IEEE created Fast Ethernet under the name 802.3u. Fast Ethernet is backward-compatible with Standard Ethernet, but it can transmit data 10 times faster at a rate of 100 Mbps. The goals of Fast Ethernet can be summarized as follows: 1. Upgrade the data rate to 100 Mbps. 2. Make it compatible with Standard Ethernet. 3. Keep the same 48-bit address. 4. Keep the same frame format. 5. Keep the same minimum and maximum frame lengths. Q10B Show the different category of Fast Ethernet implementation at the physical layer? Which encoding method used in each category ? Explain in detail. Ans. Fast Ethernet implementation at the physical layer can be categorized as either two-wire or four-wire. The two-wire implementation can be either category 5 UTP (l00Base-TX) or fiber-optic cable (l00Base-FX). The four-wire implementation is designed only for category 3 UTP (l00Base-T4)
100Base-TX:- Encoding scheme , (Multi-Level Transmit) MLT-3 is not a self-synchronous line coding scheme, 4B/5B block coding is used to provide bit synchronization by preventing the occurrence of a long sequence of Os and Is (see Chapter 4). This creates a data rate of 125 Mbps, which is fed into MLT-3 for encoding. 100Base-FX :- uses two pairs of fiber-optic cables, NRZ-I has a bit synchronization problem for long sequences of 0s (or 1s, based on the encoding) .To overcome this problem, ______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
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the designers used 4B/5B block encoding as we described for 100Base-TX. The block encoding increases the bit rate from 100 to 125 Mbps, which can easily be handled by fiberoptic cable.
Q10C
Enlist the designe goals of Gigabit Ethernet
Ans. The need for an even higher data rate resulted in the design of the Gigabit Ethernet protocol (1000 Mbps). The IEEE committee calls the Standard 802.3z. The goals of the Gigabit Ethernet design can be summarized as follows: 1. Upgrade the data rate to 1 Gbps. 2. Make it compatible with Standard or Fast Ethernet. 3. Use the same 48-bit address. 4. Use the same frame format. 5. Keep the same minimum and maximum frame lengths. 6. To support autonegotiation as defined in Fast Ethernet. Q11 A Draw and explain Topologies of Gigabit Ethernet also give the detail of categories of Gigabit Ethernet. Ans.
Gigabit Ethernet is designed to connect two or more stations. If there are only two stations, they can be connected point-to-point. Three or more stations need to be connected in a star topology with a hub or a switch at the center. Another possible configuration is to connect several star topologies or let a star topology be part of another as shown in Figure above.
Gigabit Ethernet can be categorized as either a two-wire or a four-wire implementation. The two-wire implementations use fiber-optic cable (1000Base-SX, short-wave, or l000Base-LX, long-wave), or STP (1000Base-CX). The four-wire version uses category 5 ______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
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twisted-pair cable (l000Base- T).l000Base-T was designed in response to those users who had already installed this wiring for other purposes such as Fast Ethernet or telephone services. Q11 B Explain the architectural standard of services define under IEEE 802.11 Ans. IEEE 802.11 defines the basic service set (BSS) as the building block of a wireless LAN. A basic service set is made of stationary or mobile wireless stations and an optional central base station, known as the access point (AP) The BSS without an AP is a stand-alone network and cannot send data to other BSSs. It is called an ad hoc architecture. In this architecture, stations can form a network without the need of an AP; they can locate one another and agree to be part of a BSS. A BSS with an AP is sometimes referred to as an infrastructure network.
Q11 C Explain DCF and PCF under IEEE 802.11 MAC sublayers? Ans.
IEEE 802.11 defines two MAC sublayers: the distributed coordination function (DCF) and point coordination function (PCF). Figure above shows the relationship between the two MAC sublayers, the LLC sublayer, and the physical layer. DCF :- One of the two protocols defined by IEEE at the MAC sublayer is called the distributed coordination function (DCF). PCF :The point coordination function (PCF) is an optional access method that can be implemented in an infrastructure network (not in an ad hoc network). It is implemented on top of the DCF and is used mostly for time-sensitive transmission. PCF has a centralized, contention-free polling access method. The AP performs polling for stations that are capable of being polled. The stations are polled one after another, sending any data they have to the AP
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Q12 A Explain all the cases in IEEE 802.11 addressing mechanism Ans. The IEEE 802.11 addressing mechanism specifies four cases, defined by the value of the two flags in the FC field, To DS and From DS. Each flag can be either 0 or I, resulting in four different situations. The interpretation of the four addresses (address I to address 4) in the MAC frame depends on the value of these flags, as shown in Table below
address 1 is always the address of the next device. Address 2 is always the address of the previous device. Address 3 is the address of the final destination station if it is not defined by address I. Address 4 is the address of the original source station if it is not the same as address 2. Case 1: 00 In this case, To DS = 0 and From DS = O. This means that the frame is not going to a distribution system (To DS = 0) and is not coming from a distribution system (From DS = 0). The frame is going from one station in a BSS to another without passing through the distribution system. Case 2: 01 In this case, To DS = 0 and From DS = 1. This means that the frame is coming from a distribution system (From DS = 1). The frame is coming from an AP and going to a station. The ACK should be sent to the AP. Case 3: 10 In this case, To DS = 1 and From DS = O. This means that the frame is going to a distribution system (To DS = 1). The frame is going from a station to an AP. The ACK is sent to the original station. Case 4:11 In this case, To DS = 1 and From DS = 1. TIus is the case in which the distribution system is also wireless. The frame is going from one AP to another AP in a wireless distribution system. Q12 B What is hidden station problem ? What is the solution to hidden station problem? Ans. Figure shows Station B has a transmission range shown by the left oval (sphere in space); every station in this range can hear any signal transmitted by station B. Station C has a transmission range shown by the right oval (sphere in space); every station located in this range can hear any signal transmitted by C. Station C is outside the transmission range of B; likewise, station B is outside the transmission range of C. Station A, however, is in the area covered by both Band C; it can hear any signal transmitted by B or C.
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The solution to hidden station problem :-
The solution to the hidden station problem is the use of the handshake frames (RTS and CTS) Figure shows that the RTS message from B reaches A, but not C. However, because both Band C are within the range of A, the CTS message, which contains the duration of data transmission from B to A reaches C. Station C knows that some hidden station is using the channel and refrains from transmitting until that duration is over.
Q12 C What is the use of handshaking in exposed station problem? Explain in detail. Ans.
In this problem a station refrains from using a channel when it is, in fact, available. In Figure above station A is transmitting to station B. Station C has some data to send to station D, which can be sent without interfering with the transmission from A to B. However, station C is exposed to transmission from A; it hears what A is sending and thus refrains from sending . The handshaking messages RTS and CTS cannot help in this case, Station C hears the RTS from A, but does not hear the CTS from B. Station C, after hearing the RTS from A, can wait for a time so that the CTS from B reaches A; it then sends an RTS
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to D to show that it needs to communicate with D. Both stations B and A may hear this RTS, but station A is in the sending state, not the receiving state. Station B, however, responds with a CTS. The problem is here. If station A has started sending its data, station C cannot hear the CTS from station D because of the collision; it cannot send its data to D. It remains exposed until A finishes sending its data.
PRACTICE QUESTION (HOME WORK) 1.
What is the difference between a BSS and an ESS?
2.
Discuss the three types of mobility in a wireless LAN
3.
How is OFDM different from FDM?
4.
What is the access method used by wireless LANs?
5.
What is the purpose of the NAV?
6.
Compare and contrast CSMA/CD with CSMA/CA.
7.
compare and contrast the fields in IEEE 802.3 and 802.11
8. 9. 10.
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