Vaani1 Senior User. Points 43 Location Bangalore Student of Engineering. LAN technologies Ethernet. Flow and error control techniques, switching. Basics of Wi-Fi. Network security: authentication, basics of public key and private key cryptography, digital signatures and certificates, firewalls.
Part I considers data transport the data plane. Part II covers protocols used to discover and use topology and reachability information the control plane. Principles that underlie technologies such as Software Defined Networks SDNs are considered throughout, as solutions to problems faced by all networking technologies.
This can happen if the re- ceiver gets delayed in sending the acknowledgement frame, because its CPU is overloaded processing other jobs in the system. For frame sizes above bits, stop-and-wait is rea- sonably efficient.
It can happen. Suppose that the sender transmits a frame and a garbled ac- knowledgement comes back quickly. The main loop will be executed a sec- ond time and a frame will be sent while the timer is still running. The propagation time is 18 ms. At T1 speed, which is 1. Therefore, the first frame fully arrives The acknowledgement takes another 18 msec to get back, plus a small negligible time for the acknowledgement to arrive fully.
In all, this time is A frame takes 0. Seven-bit sequence numbers are needed. Let the window size be W. The protocol would be incorrect. Suppose that 3-bit sequence numbers are in use. Consider the following scenario: A just sent frame 7. B gets the frame and sends a piggybacked ACK. A gets the ACK and sends frames 0—6, all of which get lost. B times out and retransmits its current frame, with the ACK 7.
Look at the situation at A when the frame with r. The modified between would return true, causing A to think the lost frames were being acknowledged. It might lead to deadlock. Suppose that a batch of frames arrived cor- rectly and was accepted. The receiver would advance its window. Now sup- pose that all the acknowledgements were lost.
The sender would eventually time out and send the first frame again. The receiver would then send a NAK. If this packet were lost, from that point on, the sender would keep timing out and sending a frame that had already been accepted, but the receiver would just ignore it. Setting the auxiliary timer results in a correct acknowledge- ment being sent back eventually instead, which resynchronizes. It would lead to deadlock because this is the only place that incoming ac- knowledgements are processed.
Without this code, the sender would keep timing out and never make any progress. Consider the following scenario. A sends 0 to B. A times out and repeats 0, but now B expects 1, so it sends a NAK.
If A merely resent r. Suppose A sent B a frame that arrived correctly, but there was no reverse traf- fic. After a while A would time out and retransmit. B would notice that the sequence number was incorrect, since it would be below FrameExpected. Consequently, it would send a NAK, which carries an acknowledgement num- ber. Each frame would be sent exactly two times. This implementation fails. The even sequence numbers use buffer 0 and the odd ones use buffer 1.
This mapping means that frames 4 and 0 both use the same buffer. Suppose that frames 0—3 are received and acknowledged. If 4 is lost and 0 arrives, it will be put in buffer 0 and arrived [0] will be set to true.
The loop in the code for FrameArrival will be executed once, and an out-of-order message will be delivered to the host. This protocol requires MaxSeq to be odd to work properly.
However, other implementations of sliding window protocols do not all have this property. Thus, the cycle is msec. With a kbps channel and 8-bit sequence numbers, the pipe is always full. The number of retransmissions per frame is about 0. The total overhead is The data rate here is bits in msec, or about bps. With a window size of 7 frames, transmission time is msec for the full window, at which time the sender has to stop. At msec, the first ACK arrives and the cycle can start again.
The data rate is 47, In other words, if the window size is greater than msec worth of transmission, it can run at full speed. For a window size of 10 or greater this condition is met, so for any window size of 10 or greater e. This corresponds to four frames, or bits on the cable. PPP was clearly designed to be implemented in software, not in hardware as bit-stuffing protocols such as HDLC nearly always are.
With a software im- plementation, working entirely with bytes is much simpler than working with individual bits. In addition, PPP was designed to be used with modems, and modems accept and transmit data in units of 1 byte, not 1 bit. At its smallest, each frame has 2 flag bytes, 1 protocol byte, and 2 checksum bytes, for a total of 5 overhead bytes per frame. For maximum overhead, 2 flag bytes, 1 byte each for address and control, 2 bytes for protocol and 4 bytes for checksum.
This totals to 10 overhead bytes. To make this frame size a multiple of 48, the number of padding bytes will be This will result in an AAL5 frame of size bytes.
This can fit in three ATM cells. The formula is the standard formula for Markov queueing given in Sec. For the three arrival rates, we get a 0.
At low load, no collis- ions are expected so the transmission is likely to be successful. This introduces half a slot time of delay.
Thus, we have two parametric equations, one for delay and one for throughput, both in terms of G. For each G value, it is possible to find the corresponding delay and throughput, yielding one point on the curve. Signal propa- gation time for 2 km is 8. So, the length of contention slot is Sig- nal propagation time for 40 km is The worst case is where all stations want to send and s is the lowest-num- bered station.
If a higher-numbered station and a lower-numbered station have packets to send at the same time, the higher-numbered station will always win the bid. Thus, a lower-numbered station will be starved from sending its packets if there is a continuous stream of higher-numbered stations ready to send their packets. Stations 2, 3, 5, 7, 11, and 13 want to send. Eleven slots are needed, with the contents of each slot being as follows: Slot 1: 2, 3, 5, 7, 11, 13 Slot 2: 2, 3, 5, 7 Slot 3: 2, 3 Slot 4: 2 Slot 5: 3 Slot 6: 5, 7 Slot 7: 5 Slot 8: 7 Slot 9: 11, 13 Slot 11 Slot 13 So, no other communication is possible in this case.
Imagine that they are in a straight line and that each station can reach only its nearest neighbors. Then A can send to B while E is sending to F. In the star configuration, the router is in the mid- dle of floor 4. Classic Ethernet uses Manchester encoding, which means it has two signal periods per bit sent.
The data rate is 10 Mbps, so the baud rate is twice that, or 20 megabaud. The signal is a square wave with two values, high H and low L. A complete trans- mission has six phases: 1. Transmit data Delay for last bit to get to the end 5. Acknowledgement sent 3. In this period, data bits are sent, for a rate of about 3. Number the acquisition attempts starting at 1. Since the header fields occupy 18 bytes and the packet is 60 bytes, the total frame size is 78 bytes, which exceeds the byte minimum.
Therefore, no padding is used. Prior to VLANs, the total was However, this only works when frame bursting is operating. Without frame bursting, short frames are padded to bits, in which case the maximum number is , Gigabit Ethernet has it and so does It is useful for bandwidth effi- ciency one preamble, etc. So, the 6 Mbps stations will get 0. A frame contains bits. Multiplying these two numbers together, we get about 1 damaged frame per second.
It depends how far away the subscriber is. If the subscriber is close, QAM is used for Mbps. For medium distances, QAM is used for 80 Mbps.
One reason is the need for real-time quality of service. If an error is discover- ed, there is no time for a retransmission. The show must go on. Forward error correction can be used here. Another reason is that on very low-quality lines e. To avoid this, forward error correction is used to in- crease the fraction of frames that arrive correctly. Like Also, like However, unlike It is impossible for a device to be master in two piconets at the same time.
Allowing this would create two problems. First, only 3 address bits are avail- able in the header, while as many as seven slaves could be in each piconet. Thus, there would be no way to uniquely address each slave. This is how slaves tell which message belongs to which piconet. If two overlap- ping piconets used the same access code, there would be no way to tell which frame belonged to which piconet.
In effect, the two piconets would be merged into one big piconet instead of two separate ones. So, a maximum of bits can be transmitted in a 3-slot frame.
Out of this, to bits are over- head bits, leaving a maximum of to bits for the data field. Bluetooth uses FHSS, just as Out of this, a maximum of bits are for data. In case of repetition encoding, data is replicated thrice, so the actual data transmitted is about bits. They do not. The dwell time in There is no need to announce this.
All Bluetooth devices have this hardwired into the chip. Bluetooth was designed to be cheap, and fixing the hop rate and dwell time leads to a simpler chip. We want to maximize the probability that one and only one tag responds in a given slot. Consulting Sec. This occurs when the reader sets Q equal to 10 slots. Consulting Fig. One key security concern is unauthorized tracking of RFID tags. This becomes quite serious if the item is sensitive in nature, for example, a passport, and the tag can be used to retrieve further information, for example, the nationality and other personal information of the person holding the passport.
Another security concern is the ability of a reader to change tag information. This can be used by an adversary to, for ex- ample, change the price of a tagged item he plans to buy. The worst case is an endless stream of byte bit frames. A store-and-forward switch stores each incoming frame in its entirety, then examines it and forwards it.
A cut-through switch starts to forward incoming frames before they have arrived completely. As soon as the destination ad- dress is in, the forwarding can begin. B2 will forward it on 1, 2 and 3. B1 will forward it on 1, 2 and 3. B1 will not see it. B2 will forward it on port 2. Store-and-forward switches store entire frames before forwarding them. After a frame comes in, the checksum can be verified. If the frame is dam- aged, it is discarded immediately.
With cut-through, damaged frames cannot be discarded by the switch because by the time the error is detected, the frame is already gone. Trying to deal with the problem is like locking the barn door after the horse has escaped.
A bridge that does not have any station directly connected to any of its ports and is part of a loop is a candidate for not being a part of the spanning tree bridges. This can happen if the shortest paths to the root for all bridges con- nected to this bridge does not include this bridge.
Hubs just connect all the incoming lines together electrically. There is nothing to configure. No routing is done in a hub. Every frame coming into the hub goes out on all the other lines. It would work. Frames entering the core domain would all be legacy frames, so it would be up to the first core switch to tag them. Similarly, on the way out, that switch would have to untag outgoing frames. File transfer, remote login, and video on demand need connection-oriented service. Virtual circuit networks most certainly need this capability in order to route connection setup packets from an arbitrary source to an arbitrary destination.
The negotiation could set the window size, maximum packet size, data rate, and timer values. A large noise burst could garble a packet badly. If the destination field or, equivalently, virtual-circuit number, is changed, the packet will be delivered to the wrong destination and accepted as genuine.
Put in other words, an occasional noise burst could change a perfectly legal packet for one destination into a perfectly legal packet for another destination. Pick a route using the shortest path. Now remove all the arcs used in the path just found, and run the shortest path algorithm again. The second path will be able to survive the failure of any line in the first path, and vice versa. It is conceivable, though, that this heuristic may fail even though two line-disjoint paths exist.
To solve it correctly, a max-flow algorithm should be used. Going via B gives 11, 6, 14, 18, 12, 8. Going via D gives 19, 15, 9, 3, 9, Going via E gives 12, 11, 8, 14, 5, 9. Taking the minimum for each destination except C gives 11, 6, 0, 3, 5, 8. The routing table is bits. Twice a second this table is written onto each line, so bps are needed on each line in each direction.
It always holds. If a packet has arrived on a line, it must be acknowledged. If no packet has arrived on a line, it must be sent there. The cases 00 has not arrived and will not be sent and 11 has arrived and will be sent back are logically incorrect and thus do not exist.
The minimum occurs at 15 clusters, each with 16 regions, each region having 20 routers, or one of the equivalent forms, e. Conceivably it might go into promiscuous mode, reading all frames dropped onto the LAN, but this is very inefficient. Instead, what is normally done is that the home agent tricks the router into thinking it is the mobile host by re- sponding to ARP requests. When the router gets an IP packet destined for the mobile host, it broadcasts an ARP query asking for the A total of 21 packets are generated.
Node F currently has two descendants, A and D. It now acquires a third one, G, not circled because the packet that follows IFG is not on the sink tree. Node G acquires a second descendant, in addition to D, labeled F. This, too, is not circled as it does not come in on the sink tree. Multiple spanning trees are possible. Node H is three hops from B, so it takes three rounds to find the route. The protocol is terrible. Let time be slotted in units of T sec. In slot 1 the source router sends the first packet.
At the start of slot 2, the second router has received the packet but cannot acknowledge it yet. At the start of slot 3, the third router has received the packet, but it cannot acknowledge it either, so all the routers behind it are still hanging. The first acknowledgement can only be sent when the destination host takes the packet from the destination router. Now the acknowledgement begins propagating back. Each packet emitted by the source host makes either 1, 2, or 3 hops.
The probability that it makes one hop is p. First, the ECN method explicitly sends a congestion notification to the source by setting a bit, whereas RED implicitly notifies the source by simply drop- ping one of its packets. Second, the ECN method drops a packet only when there is no buffer space left, whereas RED drops packets before all the buffer are exhausted. Each packet holds 48 data bytes or bits.
The net data rate is then However, this answer is wrong, because during that interval, more tokens arrive. There is no guarantee. If too many packets are expedited, their channel may have even worse performance than the regular channel. No other fragmentation will occur.
Then, b is about 53,, bps. Since the information is needed to route every fragment, the option must appear in every fragment.
With a 2-bit prefix, there would have been 18 bits left over to indicate the net- work. Consequently, the number of networks would have been or , However, all 0s and all 1s are special, so only , are avail- able.
The address is The mask is 20 bits long, so the network part is 20 bits. The remaining 12 bits are for the host, so host addresses exist. When burning the address into the card, the manufac- turer has no idea where in the world the card will be used, making the address useless for routing.
In contrast, IP addresses are either assigned either stati- cally or dynamically by an ISP or company, which knows exactly how to get to the host getting the IP address. To start with, all the requests are rounded up to a power of two.
The starting address, ending address, and mask are as follows: A: They can be aggregated to It is sufficient to add one new table entry: If an incoming packet matches both This rule makes it possible to assign a large block to one outgoing line but make an exception for one or more small blocks within its range. After NAT is installed, it is crucial that all the packets pertaining to a single connection pass in and out of the company via the same router, since that is where the mapping is kept.
You say that ARP does not provide a service to the network layer, it is part of the network layer and helps provide a service to the transport layer. The issue of IP addressing does not occur in the data link layer. Data link layer proto- cols are like protocols 1 through 6 in Chap.
They move bits from one end of a line to the other. In the general case, the problem is nontrivial. Fragments may arrive out of order and some may be missing.
On a retransmission, the datagram may be fragmented in different-sized chunks. Furthermore, the total size is not known until the last fragment arrives.
Probably the only way to handle reas- sembly is to buffer all the pieces until the last fragment arrives and the size is known. Then build a buffer of the right size, and put the fragments into the buffer, maintaining a bit map with 1 bit per 8 bytes to keep track of which bytes are present in the buffer. When all the bits in the bit map are 1, the datagram is complete. As far as the receiver is concerned, this is a part of new datagram, since no other parts of it are known.
It will therefore be queued until the rest show up. If they do not, this one will time out too. An error in the header is much more serious than an error in the data. A bad address, for example, could result in a packet being delivered to the wrong host. Many hosts do not check to see if a packet delivered to them is in fact really for them. They assume the network will never give them packets in- tended for another host. Data is sometimes not checksummed because doing so is expensive, and upper layers often do it anyway, making it redundant here.
The fact that the Minneapolis LAN is wireless does not cause the pack- ets that arrive for her in Boston to suddenly jump to Minneapolis.
The best way to think of this situation is that the user has plugged into the Minneapolis LAN, the same way all the other Minneapolis users have. That the connection uses radio instead of a wire is irrelevant. With 16 bytes there are or 3. If we allocate them at a rate of per second, they will last for years. This number is times the age of the universe. The Protocol field tells the destination host which protocol handler to give the IP packet to.
Intermediate routers do not need this information, so it is not needed in the main header. Actually, it is there, but disguised. Conceptually, there are no changes. Technically, the IP addresses requested are now bigger, so bigger fields are needed. When an attempt to connect was made, the caller could be given a signal. In our original scheme, this flexibility is lacking. Since the two end points are peers, a separate application-level mechanism is needed that informs the end points at run time about which end will act as server and which end will act as client, as well as their addresses.
One way to do this is to have a separate coordinator process that provides this information to the end points before a connection between the end points is established. The tran- sition can happen immediately. One other criteria is how the client is affected by extra delay involved in pro- cess server technique.
The server for the requested service has to be loaded and probably has to be initialized before the client request can be serviced. At zero gen- eration rate, the sender would enter the forbidden zone at Look at the second duplicate packet in Fig.
When that packet arrives, it would be a disaster if acknowledgements to y were still floating around. Deadlocks are possible. For example, a packet arrives at A out of the blue, and A acknowledges it. The acknowledgement gets lost, but A is now open while B knows nothing at all about what has happened. Now the same thing happens to B, and both are open, but expecting different sequence numbers.
The problem is essentially the same with more than two armies.
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