Computer network

Q1 (a) A host with Ip Addresses 128.23.67.3 sends a message to a host with IP Address 193.45.23.7. Does the message travel through any router? Assume no sub netting.

Ans)

Yes Class B to C. This message must travel through a router because the netids of the two addresses

are different (128.23 versus 193.45.23).Since the address 128.23.67.3 lies in the range of class B, and the address 193.45.23.7 lies in the range of class A.

Q1 (b) Why ATM cells have a small and fixed size?Ans)

1. Allows better statistical multiplexing of information on a given medium than does the use of larger packets of variable length.

2. With longer variable-length packets, high-speed switching would be more complex.

3. The limited amount of information in the header allows ATM cells to be processed at a very high rate from 150 Mbps to several Gbps.

4. Provides speed, flexibility and compatibility with many different applications and physical media.

5. The reduced size of the internal buffers guarantees a minimal delay and delay jitter. This is why ATM can handle both constant rate traffic (audio, video) and variable rate traffic (data) easily.

6. T o limit the queuing delays in internal buffers.

Q1 (c) A channel has a bit rate of 4kbps and a propagation delay of 20 msec. For what range of frame sizes does stop-and-wait give an efficiency of at least 50 percent?

Ans) You need to use two formulas - firstly a= propagation delay/transmission time. We'll denote this by tprop and ttrans.

tprop is given in the question, it is 20 x 10-3

ttrans is Distance/Data RateSo ttrans = L / 4 x 10

3

ttrans = 80/L

(Because the powers of ten cancel each other out so you just multiply 20 x 4.we use powers of ten because we need the same format as the propogation delaya=80/L

2. Use the formula 1/1+2a >= .5 = 1/1+ 160/L>=.5First 1>= .5(1+160/L)

second 1>= .5 + 80/Lthird 80/L >= 1 x .5fourth .5 >= 80/L

L >= 80/.5L >= 160Efficiency will be 50% when the time to transmit the frame equals the round trip propagation delay. At a transmission rate of 4bits/ms, 160 bits takes 40 ms. For frame sizes above 160 bits, stop-and-wait is reasonably efficient.

Q1(d) Compare LAN standards 802.3,802.4,802.5.

Ans)802.3 Ethernet802.3 is the standard which Ethernet operates by. It is the standard for CSMA/CD (Carrier Sense Multiple Access with Collision Detection). This standard encompasses both the MAC and Physical Layer standards.CSMA/CD is what Ethernet uses to control access to the network medium (network cable). If there is no data, any node may attempt to transmit, if the nodes detect a collision, both stop transmitting and wait a random amount of time before retransmitting the data.The original 802.3 standard is 10 Mbps (Megabits per second). 802.3u defined the 100 Mbps (Fast Ethernet) standard, 802.3z/802.3ab defined 1000 Mbps Gigabit Ethernet, and 802.3ae define 10 Gigabit Ethernet.Commonly, Ethernet networks transmit data in packets, or small bits of information. A packet can be a minimum size of 72 bytes or a maximum of 1518 bytes.The most common topology for Ethernet is the star topology.

802.5 Token Ring

Token ring is designed to use the ring topology and utilizes a token to control the transmission of data on the network.

The token is a special frame which is designed to travel from node to node around the ring. When it does not have any data attached to it, a node on the network can modify the frame, attach its data and transmit. Each node on the network checks the token as it passes to see if the data is intended for that node, if it is; it accepts the data and transmits a new token. If it is not intended for that node, it retransmits the token on to the next node.The token ring network is designed in such a way that each node on the network is guaranteed access to the token at some point. This equalizes the data transfer on the network. This is different from an Ethernet network where each workstation has equal access to grab the available bandwidth, with the possible of a node using more bandwidth than other nodes.Originally, token ring operated at a speed of about 4 Mbps and 16 Mbps. 802.5t allows for 100 Mbps speeds and 802.5v provides for 1 Gbps over fibber.

Token ring can be run over a star topology as well as the ring topology.There are three major cable types for token ring: Unshielded twisted pair (UTP), Shielded twisted pair (STP), and fibber.

The IEEE 802.4 standard describes a token bus

Token bus is a network implementing the token ring protocol over a "virtual ring" on a coaxial cable A token is passed around the network nodes and only the node possessing the token may transmit. If a node doesn't have anything to send, the token is passed on to the next node on the virtual ring. Each node must know the address of its neighbor in the ring, so a special protocol is needed to notify the other nodes of connections to, and disconnections from, the ring. Token bus combines the physical configuration of Ethernet and the collision free feature of Token ring.

 This is an application of the concepts used in token ring networks. The main difference is that the endpoints of the bus do not meet to form a physical ring.

Token Bus

Q1(e) Explain ARP and RARPAns)

Address Resolution Protocol: A TCP/IP protocol used to obtain a node's physical address. A client station broadcasts an ARP request onto the network with the IP address of the target node it wishes to communicate with, and the node with that address responds by sending back its physical address so that packets can be transmitted. ARP returns the layer 2 address for a layer 3 address.

ARP'ing

The IP protocol broadcasts the IP address of the destination station onto the network, and the node with that address responds.ARP -Meaning of ARP "Address Resolution Protocol", is used to map ip Network addresses to the hardware (Media Access Control sub layer) addresses used by the data link protocol. The ARP protocol operates between the network layer and the data link layer in the Open System Interconnection (osi) model.

RARP- RARP (Reverse Address Resolution Protocol) is a protocol by which a physical machine in a local area network can request to learn its IP address from a gateway server's Address Resolution Protocol (ARP) table or cache. A network administrator creates a table in a local area network's gateway router that maps the physical machine (or Media Access Control - MAC address) addresses to corresponding Internet Protocol addresses. When a new machine is set up, its RARP client program requests from the RARP server on the router to be sent its IP address. Assuming that an entry has been set up in the router table, the RARP server will return the IP address to the machine which can store it for future use. RARP is available for Ethernet, Fiber Distributed-Data Interface, and token ring LANs.

Q2(a) Explain the following with their advantages and disadvantages

(1) Star Topology

Advantages:

1. Good performance2. Reliable (if one connection fails, it doesn't affect others)3. Easy to replace or remove hosts or other devices4. Easy to install and wire.5. No disruptions to the network when connecting or removing devices6. Easy to detect faults and to remove parts.

Disadvantages of Star Network Topology

1. In star network topology, data communication depends on HUB. If central hub fails, then whole network fails.

2. Since each computer will be connected with HUB by means of a separate wire, star network topology needs more cable to connect computers. 

3. It is more expensive due to more wires.4. More expensive than linear bus topologies because of the cost of the hubs5. Expensive to install6. Extra hardware required

(2) Bus Topology

Advantages of Bus Network Topology

1. It is very simple topology.2. It is easy to use.3. It needs small amount of wire for connecting computers.4. It is less expensive due to small wire needed.5. If one computer fails, it does not disturb the other computers in network. Other computers will continue to share information and other resources with other connected computers.

Disadvantages of Bus Network Topology

1. Only small number of computers can be connected in a bus network.2. Network speed slows down as the number of computer increases in bus

topology.3. Finding a fault is difficult in bus topology.4. Entire network shuts of there is a break in main cable.

(3) Ring Topology

Advantages of Ring Network Topology

1. It is relatively less expensive than a star topology network.2. In a Ring topology, every computer has an equal access to the network.3. Performs better than a bus topology under heavy network load

Disadvantages of Ring Network Topology

1. Failure of one computer in the ring can affect the whole network.2. It is difficult to find faults in a ring network topology.3. Adding or removing computers will also affect the whole network since every computer is connected with previous and next computer.4. Sending a message from one computer to another takes time according to the number of nodes between the two computers. Communication delay is directly proportional to number of nodes in the network.

(4) Mesh Topology

Advantages of Mesh Network Topology

1. Since, there are many links to transfer data, Mesh topology gets rid of the traffic problem. Data may be transferred through different links.2. If one link becomes unusable, it does not disturb the whole system. Other links can be used for communication.

3. Since each node has physical connection with other nodes, therefore, one node can transfer data to many nodes at the same time.

Disadvantages of Mesh Network Topology

It is very expensive due to implementation of multiple links for each node.2. It is difficult to install and reconfigure.3. Adding or removing a computer is difficult

Q 2(b) Why cell switching is used in ATM? Explain ATM reference model.

Ans) ATM transfers information in fixed-size units called cells. Each cell consists of 53 octets, or bytes as shown in Fig. 4.6.4. The first 5 bytes contain cell-header information, and the remaining 48 contain the payload (user information). Small, fixed-length cells are well suited to transfer voice and video traffic because such traffic is intolerant to delays that result from having to wait for a large data packet to download, among other things.

An ATM cell header can be one of two formats: UNI or NNI. The UNI header is used for communication between ATM endpoints and ATM switches in private ATM networks. The NNI header is used for communication between ATM switches. Unlike the UNI, the NNI header does not include the Generic Flow Control (GFC) field. Additionally, the NNI header has a Virtual Path Identifier (VPI) field that occupies the first 12 bits, allowing for larger trunks between public ATM switches.

ATM Cell Header Fields

The following descriptions summarize the ATM cell header fields:

• Generic Flow Control (GFC)—Provides local functions, such as identifying

multiple stations that share a single ATM interface. This field is typically not used

and is set to its default value of 0 (binary 0000).

• Virtual Path Identifier (VPI)—In conjunction with the VCI, identifies the next

destination of a cell as it passes through a series of ATM switches on the way to

its destination.

• Virtual Channel Identifier (VCI)—In conjunction with the VPI, identifies the

next destination of a cell as it passes through a series of ATM switches on the way

to its destination.

• Payload Type (PT)—Indicates in the first bit whether the cell contains user data

or control data. If the cell contains user data, the bit is set to 0. If it contains

control data, it is set to 1. The second bit indicates congestion (0 = no congestion,

1 = congestion), and the third bit indicates whether the cell is the last in a series of

cells that represent a single AAL5 frame (1 = last cell for the frame).

• Cell Loss Priority (CLP)—Indicates whether the cell should be discarded if it

encounters extreme congestion as it moves through the network. If the CLP bit

equals 1, the cell should be discarded in preference to cells with the CLP bit equal

to 0.

• Header Error Control (HEC)—Calculates checksum only on the first 4 bytes of

the header. HEC can correct a single bit error in these bytes, thereby preserving

the cell rather than discard

The ATM reference model is composed of the following ATM layers:

Physical layer - Analogous to the physical layer of the OSI reference model, the ATM physical layer manages the medium-dependent transmission.

ATM layer - Combined with the ATM adaptation layer, the ATM layer is roughly analogous to the data link layer of the OSI reference model. The ATM layer is responsible for the simultaneous sharing of virtual circuits over a physical link (cell multiplexing) and passing cells through the ATM network (cell relay). To do this, it uses the VPI and VCI information in the header of each ATM cell.

ATM adaptation layer (AAL) - Combined with the ATM layer, the AAL is roughly analogous to the data link layer of the OSI model. The AAL is responsible for isolating higher-layer protocols from the details of the ATM processes. The adaptation layer prepares user data for conversion into cells and segments the data into 48-byte cell payloads.

Finally, the higher layers residing above the AAL accept user data, arrange it into packets, and hand it to the AAL.

The ATM Physical Layer

The ATM physical layer has four functions: Cells are converted into a bitstream, the transmission and receipt of bits on the physical medium are controlled, ATM cell boundaries are tracked, and cells are packaged into the appropriate types of frames for the physical medium. For example, cells are packaged differently for SONET than for DS-3/E-3 media types.

The ATM physical layer is divided into two parts: the physical medium-dependent (PMD) sublayer and the transmission convergence (TC) sublayer.

The PMD sublayer provides two key functions. First, it synchronizes transmission and reception by sending and receiving a continuous flow of bits with associated timing information. Second, it specifies the physical media for the physical medium used, including connector types and cable.

The TC sub layer has four functions: cell delineation, header error control (HEC) sequence generation and verification, cell-rate decoupling, and transmission frame adaptation. The cell delineation function maintains ATM cell boundaries, allowing devices to locate cells within a stream of bits. HEC sequence generation and verification generates and checks the header error control code to ensure valid data. Cell-rate decoupling maintains synchronization and inserts or suppresses idle (unassigned) ATM cells to adapt the rate of valid ATM cells to the payload capacity of the transmission system. Transmission frame adaptation packages ATM cells into frames acceptable to the particular physical layer implementation.

ATM Adaptation Layers: AAL1

AAL1, a connection-oriented service, is suitable for handling constant bit rate sources (CBR), such as voice and videoconferencing. ATM transports CBR traffic using circuit-emulation services. Circuit-emulation service also accommodates the attachment of equipment currently using leased lines to an ATM backbone network. AAL1 requires

timing synchronization between the source and the destination. For this reason, AAL1 depends on a medium, such as SONET, that supports clocking.

ATM Adaptation Layers: AAL2

Another traffic type has timing requirements like CBR but tends to be bursty in nature. This is called variable bit rate (VBR) traffic. This typically includes services characterized as packetized voice or video that do not have a constant data transmission speed but that do have requirements similar to constant bit rate services. AAL2 is suitable for VBR traffic. The AAL2 process uses 44 bytes of the cell payload for user data and reserves 4 bytes of the payload to support the AAL2 processes.

VBR traffic is characterized as either real-time (VBR-RT) or as non-real-time (VBR-NRT). AAL2 supports both types of VBR traffic.

ATM Adaptation Layers: AAL3/4

AAL3/4 supports both connection-oriented and connectionless data. It was designed for network service providers and is closely aligned with Switched Multimegabit Data Service (SMDS). AAL3/4 is used to transmit SMDS packets over an ATM network.

An AAL 3/4 SAR PDU header consists of Type, Sequence Number, and Multiplexing Identifier fields. Type fields identify whether a cell is the beginning, continuation, or end of a message. Sequence number fields identify the order in which cells should be reassembled. The Multiplexing Identifier field determines which cells from different traffic sources are interleaved on the same virtual circuit connection (VCC) so that the correct cells are reassembled at the destination.

ATM Adaptation Layers: AAL5

AAL5 is the primary AAL for data and supports both connection-oriented and connectionless data. It is used to transfer most non-SMDS data, such as classical IP over ATM and LAN Emulation (LANE). AAL5 also is known as the simple and efficient adaptation layer (SEAL) because the SAR sublayer simply accepts the CS-PDU and segments it into 48-octet SAR-PDUs without reserving any bytes in each cell.

Q 2(c) What are advantages of using Fiber optic cable compared to copper cable.

Thinner - Optical fibers can be drawn to smaller diameters than copper wire. 

Higher carrying capacity - Because optical fibers are thinner than copper wires, more fibers can be bundled into a given-diameter cable than copper wires. This allows more phone lines to go over the same cable or more channels to come through the cable into your cable TV box.

Less signal degradation - The loss of signal in optical fiber is less than in copper wire

Light signals – Unlike electrical signals in copper wires, light signals from one fiber do not interfere with those of other fibers in the same cable. This means clearer phone conversations or TV reception

Low power - Because signals in optical fibers degrade less, lower-power transmitters can be used instead of the high-voltage electrical transmitters needed for copper wires. Again, this saves your provider and you money. 

Digital signals - Optical fibers are ideally suited for carrying digital information, which is especially useful in computer networks. 

Non-flammable - Because no electricity is passed through optical fibers, there is no fire hazard

.Lightweight - An optical cable weighs less than a comparable copper wire cable. Fiber-optic cables take up less space in the ground. 

Flexible - Because fiber optics are so flexible and can transmit and receive light, they are used in many flexible digital cameras for the following purposes:

Speed: Fiber optic networks operate at speeds up to 10 gigabits per second or higher, as opposed to 1.54 megabits per second for copper. A fiber optic system is now capable of transmitting the equivalent of an entire encyclopedia (24 volumes) of information in one second. Fiber can carry information so fast that you could transmit three television episodes in one second.

Bandwidth: Taken in bulk, it would take 33 tons of copper to transmit the same amount of information handled by 1/4 pound of optical fiber.

Resistance: Fiber optic cables have a greater resistance to electromagnetic noise such as radios, motors or other nearby cables. Because optical fibers carry beams of light, they are free of electrical noise and interference.

Q 3(a) In class A, the first address is 20.0.0.0.What is the 220000th address.

For address 220,000 in x.y.z.t notation is 0.3.91.96.In order to solve the problem we add this to 20.0.0.0, and the 220,001th address is 20.3.91.96.

So, the 220,000th address is 20.3.91.95.

Q 3(b) Give an advantage and disadvantage of NT12 in an ISDN network.

NT12

Q 3(c) Compare and ISO-OSI with TCP/IP reference model.

SIMILARITIES

The main similarities between the two models include the following:

They share similar architecture. -    Both of the models share a similar architecture.  This can be illustrated by the fact that both of them are constructed with layers.

They share a common application layer.- Both of the models share a common "application layer".  However in practice this layer includes different services depending upon each model.

Both models have comparable transport and network layers.- This can be illustrated by the fact that whatever functions are performed between the presentation and network layer of the OSI model similar functions are performed at the Transport layer of the TCP/IP model.

Knowledge of both models is required by networking professionals.- According to article obtained from the internet networking professionals "need to know both models". (Source:

Both models assume that packets are switched.- Basically this means that individual packets may take differing paths in order to reach the same destination.

 DIFFERENCES 

The main differences between the two models are as follows:

TCP/IP Protocols are considered to be standards around which the internet has developed.  The OSI model however is a "generic, protocol- independent standard." (www.netfact.com/crs)

TCP/IP combines the presentation and session layer issues into its application layer.

TCP/IP combines the OSI data link and physical layers into the network access layer.

TCP/IP appears to be a more simpler model and this is mainly due to the fact that it has fewer layers.

TCP/IP is considered to be a more credible model- This is mainly due to the fact because TCP/IP protocols are the standards around which the internet was developed therefore it mainly gains creditability due to this reason.  Where as in contrast networks are not usually built around the OSI model as it is merely used as a guidance tool.

The OSI model consists of 7 architectural layers whereas the TCP/IP only has 4 layers.

Q 4(a) Explain the working of 3-bit sliding window protocol with suitable example.

The simplex stop and wait ARQ protocol using sequence numbers and sequence number acknowledgments will ensure an error free communications channel for higher levels. However, waiting for an acknowledgment for each frame in turn is very wasteful and a more efficient alternative is to use a sliding window protocol which enables a number of frames to be transmitted and separately acknowledged.

In a sliding window protocol each outbound frame contains a sequence number in the range 0 to some maximum (MaxSeq). If n bits are allocated in the header to store a sequence number the number range would be from 0 to 2" - 1, eg if a 3 bit number is used the sequence numbers would range from 0 to 7. The sender and receiver maintain a window:  

Sending window

is a list of consecutive frame sequence numbers that can be sent by the sender or that have been sent and acknowledgments are waited for. When an ack arrives and all previous frames have already been acknowledged the window can be advanced and a new message obtained from the host to be transmitted with the next highest available sequence number. If a ack arrives for a frame that is not within the ‘window’ it is discarded, eg an ‘extra’ ack for a frame that has already been acknowledged.

Receiving window

is a list of sequence numbers for frames that can be accepted by the receiver. When a valid frame arrives and all previous frames have already arrived the window is advanced. If a frame arrives that is not within the ‘window’ it is discarded.

Although the IMPs (interface message processor) now have more freedom in the order in which frames are sent or received the higher layers of the destination host must get the messages in the same order that the source host supplied them. In addition, the physical layer must still appear to be a ‘piece of wire’ to the data link layer.

Since frames currently within the senders window could be lost the sender has to have sufficient buffer space to store n frames. When all the buffers are full (waiting for acks) the IMP must stop the host from passing more messages until a buffer is free. Must make sure that no more than 2n - 1 frames are un-ack-ed at any point,not 2n frames (even though there are 2n distinct frame numbers).

Consider 3 bit frame number, where frames numbered 0..7 (8 distinct numbers)

1. 8 un-ack-ed frames bad:transmit 8 frames numbered 0..7 If received ok, will get ack=7 If error, will get ack=7 too! (Ack up to last ok frame) Next transmission 8 frames 0..7 ack=7 comes back is that ack-ing the new batch, or just re-ack-ing the old?

2. 7 un-ack-ed frames ok:If transmit 7 frames 0..6 If error, ack=7 If received, ack=6 Then transmit 7 frames 7,0..5 If error, ack=6 If received, ack=5 

 That is, after transmitting 7 packets, there are 8 possible results: Anywhere from 0 to 7 packets could have been received successfully. This is 8 possibilities, and the transmitter needs enough information in the acknowledgment to distinguish them all.If the transmitter sent 8 packets without waiting for acknowledgment, it could find itself in a quandary similar to the stop-and-wait case: does the acknowledgment mean that all 8 packets were received successfully, or none of them?

Q 4(b) Consider an error-free 64-kbps satellite channel used to send 512-byte data frames in one direction with very short acknowledgements coming back the other way. What is the maximum throughput for window sizes of 1, 7, 15 and 127?

A 512 byte (4096 bits) data frame has a duration of 4096/64000 seconds – that is 64 msec. Assume that the satellite is at 36000 km distance. This leads to roundtrip propagation time of 240 msec. We should also add 64 msec to this to account for transmission time. Hence with a window size of 1, 4096 bits can be sent every 240+64=304 msec. This equates to the throughput of 4096 bits/304 msec or 13.5 kbps. For a window size greater than 5, the full 64 kbps is used. 3- Two adjacent nodes (A and B) use a sliding window protocol and a 3-bit sequence

OR

Sliding window – send specified amount of frames, wait for one ack for that set of frames512 bytes x 8 bits/B = 4096 bits per frame4096/64000 bps = 64 msec to send one frameRound trip delay = 540 msecWindow size 1: send 4096 bits per 540msec4096 bits / 540 msec = 7.585 x 103 bps throughputWindow size 7: 7585 x 7 = 53096 bpsWindow size 9 and greater: 7585 x 9 = 68265 bps but the maximum capacity is 64 kbps so for window sizes greater than 9 the maximum throughput is 64 kbps.

Q 5(a) Why framing is necessary? Explain framing techniques.

Framing determines start and end of packets (all we see is a string of bits) and their order by inserting packet headers The main problem is to decide where successive packets start and end. Therefore, the DLC encapsulates the packet into a frame by adding its own header and trailer.

Moreover, if there is a period of idle fills between successive ,it becomes also necessary to separate idle fills from frames. Even when idle fills are replaced by dead periods, the problem is not simplified, because often there are no dead periods.

Types of Framing:

1 Character based framing: Character based codes, such as ASCII (7 bits and 1 parity bit), provide binary representation for keyboard characters and terminal control characters. The idea in character based framing is that such codes can also provide representation for various communication characters:

One disadvantage of character based framing is that the frame must contain aninteger number of characters; otherwise, the characters cannot be read correctly.

2 Fixed length packets/framesIn networks with fixed length packets, the length field is implicit (not needed) because all packets (and hence frames) have the same size. An example of such networks is ATM, where all packets are 53 bytes long. This framing technique requires synchronization upon initialization. Another drawback is that message length often not a multiple of packet size.

3 Bit oriented framing

In bit oriented framing, a special binary flag indicates the end of the frame, and it is avoided within the frame itself by using a technique called bit stuffing. Therefore, the concept is thesame, but the main difference is that a flag.

Q 5(b) Explain Sliding window protocol. Also explain error control at DLL.

A sliding window protocol is a feature of packet-based data transmission protocols. Sliding

window protocols are used where reliable in-order delivery of packets is required, such as in

the Data Link Layer (OSI model) as well as in the Transmission Control Protocol (TCP).

Conceptually, each portion of the transmission (packets in most data link layers, but bytes in

TCP) is assigned a unique consecutive sequence number, and the receiver uses the numbers

to place received packets in the correct order, discarding duplicate packets and identifying

missing ones. The problem with this is that there is no limit of the size of the sequence

numbers that can be required.

By placing limits on the number of packets that can be transmitted or received at any given

time, a sliding window protocol allows an unlimited number of packets to be communicated

using fixed-size sequence numbers. The term "window" on transmitter side represents the

logical boundary of the total number of packets yet to be acknowledged by the receiver. The

receiver informs the transmitter in each acknowledgment packet the current maximum

receiver buffer size (window boundary). The TCP header uses a 16 bit field to report the

receive window size to the sender. Therefore, the largest window that can be used is

2^16=65k bytes. In slow-start mode, the transmitter starts with low packet count and

increases the number of packets in each transmission after receiving acknowledgment

packets from receiver. For every ack packet received, the window slides by one packet

(logically) to transmit one new packet. When the window threshold is reached, the

transmitter sends one packet for one ack packet received. If the window limit is 10 packets

then in slow start mode the transmitter may start transmitting one packet followed by two

packets (before transmitting two packets, one packet ack has to be received), followed by

three packets and so on until 10 packets. But after reaching 10 packets, further

transmissions are restricted to one packet transmitted for one ack packet received. In a

simulation this appears as if the window is moving by one packet distance for every ack

packet received. On the receiver side also the window moves one packet for every packet

received. The sliding window method ensures that traffic congestion on the network is

avoided. The application layer will still be offering data for transmission to TCP without

worrying about the network traffic congestion issues as the TCP on sender and receiver side

implement sliding windows of packet buffer. The window size may vary dynamically

depending on network traffic.

For the highest possible throughput, it is important that the transmitter is not forced to stop

sending by the sliding window protocol earlier than one round-trip delay time (RTT). The

limit on the amount of data that it can send before stopping to wait for

an acknowledgment should be larger than the bandwidth-delay product of the

communications link. If it is not, the protocol will limit the effective bandwidth of the link.

1. Sender window might grow as it receives more frames to send and still has ones un-ack'ed. Starts with nothing to send, then NA gives it frames to send.Later, window may shrink as frames are ack-ed and NA has no more.

2. Receiver window constant size Receiver window size 1 means will only accept them in order. Size n means will receive out of order (e.g. receive later ones after earlier frame is lost) and then must buffer them before sending to NB (must send to NB in order).

Examples of Error Detecting methods: Parity bit: Simple example of error detection technique is parity bit. The parity bit

is chosen that the number of 1 bits in the code word is either even( for even parity) or odd (for odd parity). For example when 10110101 is transmitted then for even parity an 1 will be appended to the data and for odd parity a 0 will be appended. This scheme can detect only single bits. So if two or more bits are changed then that can not be detected.

Longitudinal Redundancy Checksum: Longitudinal Redundancy Checksum is an error detecting scheme which overcomes the problem of two erroneous bits. In this concept of parity bit is used but with slightly more intelligence. With each byte we send one parity bit then send one additional byte which have the parity corresponding to the each bit position of the sent bytes. So the parity bit is set in both horizontal and vertical direction. If one bit get flipped we can tell which row

and column have error then we find the intersection of the two and determine the erroneous bit. If 2 bits are in error and they are in the different column and row then they can be detected. If the error are in the same column then the row will differentiate and vice versa. Parity can detect the only odd number of errors. If they are even and distributed in a fashion that in all direction then LRC may not be able to find the error.

Cyclic Redundancy Checksum (CRC): We have an n-bit message. The sender adds a k-bit Frame Check Sequence (FCS) to this message before sending. The resulting (n+k) bit message is divisible by some (k+1) bit number. The receiver divides the message ((n+k)-bit) by the same (k+1)-bit number and if there is no remainder, assumes that there was no error. How do we choose this number? For example, if k=12 then 1000000000000 (13-bit number) can be chosen, but this is a pretty crappy choice. Because it will result in a zero remainder for all (n+k) bit messages with the last 12 bits zero. Thus, any bits flipping beyond the last 12 go undetected. If k=12, and we take 1110001000110 as the 13-bit number (incidentally, in decimal representation this turns out to be 7238). This will be unable to detect errors only if the corrupt message and original message have a difference of a multiple of 7238. The probablilty of this is low, much lower than the probability that anything beyond the last 12-bits flips. In practice, this number is chosen after analyzing common network transmission errors and then selecting a number which is likely to detect these common errors.

Q 6(a) Explain network devices repeaters, switches, hubs, and bridges.

Repeaters and hubs

A repeater is an electronic device that receives a signal, cleans it of unnecessary noise, regenerates it, and retransmits it at a higher power level, or so that the signal can cover longer distances without degradation. In most twisted pair Ethernet configurations, repeaters are required for cable that runs longer than 100 meters. A repeater with multiple ports is known as a hub. Repeaters work on the Physical Layer of the OSI model. Repeaters require a small amount of time to regenerate the signal. This can cause a propagation delay which can affect network communication when there are several repeaters in a row. Many network architectures limit the number of repeaters that can be used in a row (e.g. Ethernet's 5-4-3 rule).

Bridges

A network bridge connects multiple network segments at the data link layer (layer 2) of the OSI model. Bridges broadcast to all ports except the port on which the broadcast was received. However, bridges do not promiscuously copy traffic to all ports, as hubs do, but learn which MAC addresses are reachable through specific ports. Once the bridge associates a port and an address, it will send traffic for that address to that port only.

Bridges learn the association of ports and addresses by examining the source address of frames that it sees on various ports. Once a frame arrives through a port, its source address is stored and the bridge assumes that MAC address is associated with that port. The first time that a previously unknown destination address is seen, the bridge will forward the frame to all ports other than the one on which the frame arrived.

Bridges come in three basic types:

Local bridges: Directly connect LANs Remote bridges: Can be used to create a wide area network (WAN) link between

LANs. Remote bridges, where the connecting link is slower than the end networks, largely have been replaced with routers.

Wireless bridges: Can be used to join LANs or connect remote stations to LANs.

Switches

A network switch is a device that forwards and filters OSI layer 2 datagrams (chunks of data communication) between ports (connected cables) based on the MAC addresses in the packets.[15] A switch is distinct from a hub in that it only forwards the frames to the ports involved in the communication rather than all ports connected. A switch breaks the collision domain but represents itself as a broadcast domain. Switches make forwarding decisions of frames on the basis of MAC addresses. A switch normally has numerous ports, facilitating a star topology for devices, and cascading additional switches.[16] Some switches are capable of routing based on Layer 3 addressing or additional logical levels; these are called multi-layer switches. The term switch is used loosely in marketing to encompass devices including routers and bridges, as well as devices that may distribute traffic on load or by application content (e.g., a Web URL identifier).

Repeater  Forwards every frame it receives it is a generator,not an amplifier(i.e it removes noise & regenerates signal )

Bi-directional in nature

Useful in increasing ethernet size/length

Maximum of 5 Repeaters in an Ethernet

Hub 

basically a multiport repeater

can be used to divide a single LAN into multiple levels of Hierarchy

Bridge connect similar/dissimilar LANS Designed to store and forward frame

Protocol independent

Transparent to End Stations

Operates in Layer-1 & Layer-2

uses a table for filtering/routing

does not change the Mac address in the frame

Q 6(b) Explain MAC sublayer of token ring.

802.5 Token Ring protocols specify that the MAC sub-layer must supply a 48-bit (6 byte) address. The MAC address is most frequently represented as 12 hexadecimal digits. The MAC address

uniquely identifies a specificnetwork device and MAC addresses must be unique on a given LAN (a network of computing devices in a single subnet of IP addresses). The first 12-bit portion of the MAC address identifies the vendor of the network device, the last 12-bit portion identifies the unique id of the device itself. When looking at a hexadecimal representation of the MAC address, the first six hexadecimal digits identify the vendor and the last six hexadecimal digits identify the specific network interface card.

Token Ring and IEEE 802.5 support two basic frame types: tokens and data/command frames. Tokens are 3 bytes in length and consist of a start delimiter, an access control byte, and an end delimiter. Data/command frames vary in size, depending on the size of the Information field. Data frames carry information for upper-layer protocols, while command frames contain control information and have no data for upper-layer protocols.

Token Frame Fields

Start Delimiter Access Control Ending delimiter

Token Frame contains three fields, each of which is 1 byte in length:

• Start delimiter (1 byte): Alerts each station of the arrival of a token (or data/command frame). This field includes signals that distinguish the byte from the rest of the frame by violating the encoding scheme used elsewhere in the frame.

• Access-control (1 byte): Contains the Priority field (the most significant 3 bits) and the Reservation field (the least significant 3 bits), as well as a token bit (used to differentiate a token from a data/command frame) and a monitor bit (used by the active monitor to determine whether a frame is circling the ring endlessly).

• End delimiter (1 byte): Signals the end of the token or data/command frame.This field also contains bits to indicate a damaged frame and identify the frame that is the last in a logical sequence.

Data/Command Frame Fields

StartDelimiter

AccessControl

FrameControl

Destination address

Source address

Data Frame check sequence

EndDelimiter

FrameStatus

Data/command frames have the same three fields as Token Frames, plus several others. The Data/command frame fields are described below:

• Frame-control byte (1 byte)—Indicates whether the frame contains data or control information. In control frames, this byte specifies the type of control information.

• Destination and source addresses (2-6 bytes)—Consists of two 6-byte address fields that identify the destination and source station addresses.

• Data (up to 4500 bytes)—Indicates that the length of field is limited by the ring token holding time, which defines the maximum time a station can hold the token.

• Frame-check sequence (FCS- 4 byte)—Is filed by the source station with a calculated value dependent on the frame contents. The destination station recalculates the value to determine whether the frame was damaged in transit. If so, the frame is discarded.

• Frame Status (1 byte)—This is the terminating field of a command/data frame. The Frame Status field includes the address-recognized indicator and frame-copied indicator.

Q 6(c) Explain FDDI

Fiber distributed data interface (FDDI), which is an optical data communication standard used for long distance networks provides communication with fiber optic lines up to 200 kilometers at a speed of 100 megabit per second (Mbps). 

FDDI uses a dual-ring architecture with traffic on each ring flowing in opposite directions called counter-rotating. The dual-rings consist of a primary and a secondary ring. The primary ring offers up to 100 Mbps capacity. During normal operation, the primary ring is used for data transmission, and the secondary ring remains idle and available for backup. The primary purpose

of the dual rings is to provide superior reliability and robustness. FDDI was later extended to FDDI-2 for long distance voice and multimedia communication.

FDDI can have two types of Network Interface Cards, A and B, that connect to it. 

1) Class A Network Interface Cards connect to both rings while class B Network Interface Cards connect to only one ring.

2) Only class A cards can be used to heal broken rings. Thus the number of class A cards define the fault tolerant characteristics of the network. 

3) When an error occurs the nearest computer routes frames from the inner ring to the outer ring.

Q 7 (a) Imagine two LAN bridges, both connecting a pair of 802.4 networks . The first bridge is faced with 1000 512-byte frames per second that must be forwarded. The second is faced with 200 4096-byte frames per second. Which bridge do you think will need the faster CPU? Discuss.

The bridge with 1000 512-byte frames will require a faster CPU since it has to do 1000 forwarding lookup and learning operations compared to 200 for the other bridge. It is the number if frame headers not the size of the frame which governs how busy the bridge microprocessor is. The 100 frames/sec bridge woul need a faster CPU. Although the other one has a higher throughput, the 1000 frames/sec bridge has more interrupts, more process switches, more frames passed and more of everything that needs the CPU.

Q 7(b) Explain the disadvantages of CDMA?

Main disadvantage of CDMA is the degradation if the number of user will be increases an RF

channel because in cdma 10 users can acess at a time.

One major problem in CDMA technology is channel pollution, where signals from too many cell sites are present in the subscriber’s phone but none of them is dominant. When this situation arises the quality of the audio degrades. Another disadvantage in this technology when compared to GSM is the lack of international roaming capabilities. The ability to upgrade or change to another handset is not easy with this technology because the network service information for the phone is put in the actual phone unlike GSM which uses SIM card for this. One another disadvantage is the limited variety of the handset, because at present the major mobile companies use GSM technology.  Some other drawbacks :

• Due to its proprietary nature, all of CDMA's flaws are not known to the engineering community. • CDMA is relatively new, and the network is not as mature as GSM. • CDMA cannot offer international roaming, a large GSM advantage.

Q 7(c) Explain ALOHA

 

Q 8(a) What is the maximum burst length on an 155.2Mbps ATM ABR connection whose PCR value is 200,000 & whose L value is 25 micro sec?

Q 8(b) Give three examples of protocol parameters that might be negotiated when a connection is set up.

The negotiation can be on setting the following :

1) Window size2) Maximum packet size (MTU or MSS)3) Timer values.4) The protocol parameters that may be negotiated during the setup of a connection

could be A) Maximum packet sizeB) Maximum transmission/reception speedC) Quality of service standards.

Q 9 (a) Give two examples for which connection oriented service is appropriate. Now give two examples for which connectionless service is best.

Connection-oriented communication is a data communication mode in telecommunications whereby the devices at the end points use a protocol to establish an end-

to-end logical or physical connection before any data may be sent. Connection-oriented protocol services are often but not always reliable network services, that provide acknowledgment after successful delivery, and  automatic repeat request functions in case of missing data or detected bit-errors.

Circuit mode communication, for example the public switched telephone network, ISDN, SONET/SDH and optical mesh networks, are examples of connection-oriented communication. Circuit mode communication provides guarantees that data will arrive with constant bandwidth and at constant delay.

Packet mode communication may also be connection-oriented, which is called virtual circuit mode communication. Due to the packet switching, the communication may suffer from variable bit rate and delay, due to varying traffic load and packet queue lengths. . Examples of connection-oriented packet mode communication, i.e. virtual circuit mode communication:

The Transmission Control Protocol (TCP) is a connection-oriented reliable protocol that is based on a datagram protocol (the IP protocol).

X.25 is a connection-oriented reliable network protocol.

Frame relay is a connection-oriented unreliable data link layer protocol.

GPRS

Asynchronous Transfer Mode

Multiprotocol Label Switching

The alternative to connection-oriented transmission is connectionless communication, also known as datagram communication, in which data is sent in form of packets from one end point to another without prior arrangement or signaling. Each data packet must contain complete address information, since packets are routed individually and independently of each other, possibly transmitted along different network paths.

Connectionless protocols are usually described as stateless because the end points have no protocol-defined

List of connectionless protocols

Hypertext Transfer Protocol

IP 

UDP

ICMP

IPX

TIPC

NetBEUI

DNS

SNMP

Q 9(b) Give an argument why the leaky bucket algorithm should allow just one packet per tick, independent of how large the packet is.

. When variable sized packet are being used, it is often better to allow a fixed number of bytes per tick rather than just one packet the leaky bucket consists of a finite queue. When a packet aarives if there ia a room on the queue it is appended to the queue, otherwise it is discarded. At every clock tick, one packet is transmitted. The host is aalowed to put one packet per clock tick onto the network. Again this can be enforced by the interface card or by OS this algorithm is mostly used for traffic shaping. If leaky bucket allow more than one packet then congestion will occurs. If we send more than one packet per tick, independent of packet size, then eventually you might see a potential increase of processing going on at the nearest router by the leaky bucket. The more packets you are putting on the line per tick for the router to handle, the more time it has to give to handle multiple packets per clock tick. The router might get behind it’s work and become congested.Therefore, we would only want to put out one packet per clock tick.