ASK changes the amplitude of the carrier. FSK changes the frequency of the carrier. PSK changes the phase of the carrier. QAM changes both the amplitude and the phase of the carrier.
We can say that the most susceptible technique is ASK because the amplitude is more affected by noise than the phase or frequency. A constellation diagram can help us define the amplitude and phase of a signal element, particularly when we are using two carriers.
In a constellation diagram, a signal element type is represented as a dot. The bit or combination of bits it can carry is often written next to it. The diagram has two axes. The horizontal X axis is related to the in-phase carrier; the vertical Y axis is related to the quadrature carrier.
The two components of a signal are called I and Q. The I component, called in- phase, is shown on the horizontal axis; the Q component, called quadrature, is shown on the vertical axis.
It is also called the modulation of an analog signal; the baseband analog signal modulates the carrier to create a broadband analog signal. AM changes the amplitude of the carrier b. FM changes the frequency of the carrier c. PM changes the phase of the carrier We can say that the most susceptible technique is AM because the amplitude is more affected by noise than the phase or frequency.
See Figure 5. We have two signal elements with peak amplitudes 1 and 3. The phase of both signal elements are the same, which we assume to be 0 degrees. We have two signal elements with the same peak amplitude of 2. However, there must be degrees difference between the two phases. We assume one phase to be 0 and the other degrees.
We have four signal elements with the same peak amplitude of 3. However, there must be 90 degrees difference between each phase. We assume the first phase to be at 45, the second at , the third at , and the fourth at degrees. Note that this is one out of many configurations. ASK b. QPSK d. As long as the differences are 90 degrees, the solution is satisfactory. We have four phases, which we select to be the same as the previous case.
For each phase, however, we have two amplitudes, 1 and 3 as shown in the figure. The phases can be at 0, 90, , and As long as the differences are 90 degrees, the solution is satisfac- tory.
This is ASK. There are two peak amplitudes both with the same phase 0 degrees. The distance between each dot and the origin is 3. However, we have two phases, 0 and degrees. This is also BPSK. The peak amplitude is 2, but this time the phases are 90 and degrees.
The number of points define the number of levels, L. The number of bits per baud is the value of r. This means that that we need a QAM technique to achieve this data rate. We calculate the number of channels, not the number of coexisting stations.
Multiplexing is the set of techniques that allows the simultaneous transmission of multiple signals across a single data link. In multiplexing, the word link refers to the physical path. The word channel refers to the portion of a link that carries a transmission between a given pair of lines. One link can have many n channels. TDM is used to combine digital signals; the time is shared.
To maximize the efficiency of their infrastructure, telephone companies have tradi- tionally multiplexed analog signals from lower-bandwidth lines onto higher-band- width lines. To maximize the efficiency of their infrastructure, telephone companies have tradi- tionally multiplexed digital signals from lower data rate lines onto higher data rate lines. WDM is common for multiplexing optical signals because it allows the multiplex- ing of signals with a very high frequency.
In multilevel TDM, some lower-rate lines are combined to make a new line with the same data rate as the other lines. Multiple slot TDM, on the other hand, uses multiple slots for higher data rate lines to make them compatible with the lower data rate line.
Pulse stuffing TDM is used when the data rates of some lines are not an integral multiple of other lines. In synchronous TDM, each input has a reserved slot in the output frame. This can be inefficient if some input lines have no data to send.
In spread spectrum, we spread the bandwidth of a signal into a larger bandwidth. Spread spectrum techniques add redundancy; they spread the original spectrum needed for each station.
The expanded bandwidth allows the source to wrap its message in a protective envelope for a more secure transmission. The frequency hopping spread spectrum FHSS technique uses M different car- rier frequencies that are modulated by the source signal. At one moment, the signal modulates one carrier frequency; at the next moment, the signal modulates another carrier frequency.
The direct sequence spread spectrum DSSS technique expands the bandwidth of the original signal. It replaces each data bit with n bits using a spreading code. To multiplex 10 voice channels, we need nine guard bands. The required band- Each output frame carries 1 bit from each source plus one extra bit for synchro- b.
Each frame carries 1 bit from each source. In each frame 20 bits out of 21 are useful. Each output frame carries 2 bits from each source plus one extra bit for syn- b. Each frame carries 2 bit from each source. The output data rate here is slightly less than the one in Exercise In each frame 40 bits out of 41 are useful.
Effi- ciency is better than the one in Exercise We can assume that we have only 6 input lines. Each frame needs to carry one character from each of these lines. We combine six kbps sources into three kbps. Now we have seven kbps channel. Each output frame carries 1 bit from each of the seven kbps line. Frame b.
Each frame carries 1 bit from each kbps source. We can also synchronizing bits. The frame carries 4 bits from each of the first two sources and 3 bits from each b. Each frame carries 4 bit from each kbps source or 3 bits from each kbps. We can also synchronization bits. Now we have two sources, each of Kbps. The frame carries 1 bit from each source. Here the output bit rate is greater than the sum of the input rates kbps because of extra bits added to the second source. Each frame carries one extra bit.
See Figure 6. Figure 6. This means that the The Barker chip is 11 bits, which means that it increases the bit rate 11 times. The transmission media is located beneath the physical layer and controlled by the physical layer. The two major categories are guided and unguided media. Guided media have physical boundaries, while unguided media are unbounded.
The three major categories of guided media are twisted-pair, coaxial, and fiber- optic cables. Twisting ensures that both wires are equally, but inversely, affected by external influences such as noise. Refraction and reflection are two phenomena that occur when a beam of light travels into a less dense medium.
When the angle of incidence is less than the crit- ical angle, refraction occurs. The beam crosses the interface into the less dense medium. When the angle of incidence is greater than the critical angle, reflection occurs. The beam changes direction at the interface and goes back into the more dense medium. The inner core of an optical fiber is surrounded by cladding.
The core is denser than the cladding, so a light beam traveling through the core is reflected at the boundary between the core and the cladding if the incident angle is more than the critical angle. We can mention three advantages of optical fiber cable over twisted-pair and coax- ial cables: noise resistance, less signal attenuation, and higher bandwidth. In sky propagation radio waves radiate upward into the ionosphere and are then reflected back to earth.
In line-of-sight propagation signals are transmitted in a straight line from antenna to antenna. Omnidirectional waves are propagated in all directions; unidirectional waves are propagated in one direction. See Table 7. Table 7. As the Table 7. If we consider the bandwidth to start from zero, we can say that the bandwidth decreases with distance.
For example, if we can tol- erate a maximum attenuation of 50 dB loss , then we can give the following list- ing of distance versus bandwidth. We can use Table 7. As Table 7. This means all three figures represent the a.
The wave length is the inverse of the frequency if the propagation speed is same thing. We can change the wave length to frequency. For example, the value nm can be written as THz. The curve must be flipped horizontally. Therefore, we have: a. See Figure 7. Figure 7. The incident angle 40 degrees is smaller than the critical angle 60 degrees. We have refraction. The light ray enters into the less dense medium.
The incident angle 60 degrees is the same as the critical angle 60 degrees. The light ray travels along the interface. The incident angle 80 degrees is greater than the critical angle 60 degrees. We have reflection. The light ray returns back to the more dense medium. Switching provides a practical solution to the problem of connecting multiple devices in a network. It is more practical than using a bus topology; it is more effi- cient than using a star topology and a central hub.
Switches are devices capable of creating temporary connections between two or more devices linked to the switch. The three traditional switching methods are circuit switching, packet switching, and message switching.
The most common today are circuit switching and packet switching. There are two approaches to packet switching: datagram approach and virtual- circuit approach.
In a circuit-switched network, data are not packetized; data flow is somehow a continuation of bits that travel the same channel during the data transfer phase. In a packet-switched network data are packetized; each packet is somehow an indepen- dent entity with its local or global addressing information. The address field defines the end-to-end source to destination addressing. The address field defines the virtual circuit number local addressing.
In a space-division switch, the path from one device to another is spatially separate from other paths. The inputs and the outputs are connected using a grid of elec- tronic microswitches. In a time-division switch, the inputs are divided in time using TDM. A control unit sends the input to the correct output device. TSI time-slot interchange is the most popular technology in a time-division switch. It used random access memory RAM with several memory locations.
The RAM fills up with incoming data from time slots in the order received. Slots are then sent out in an order based on the decisions of a control unit. In multistage switching, blocking refers to times when one input cannot be con- nected to an output because there is no path available between them—all the possi- ble intermediate switches are occupied.
One solution to blocking is to increase the number of intermediate switches based on the Clos criteria. A packet switch has four components: input ports, output ports, the routing pro- cessor, and the switching fabric. An input port performs the physical and data link functions of the packet switch.
The output port performs the same functions as the input port, but in the reverse order. The routing processor performs the function of table lookup in the network layer. The switching fabric is responsible for moving the packet from the input queue to the output queue. We assume that the setup phase is a two-way communication and the teardown phase is a one-way communication. These two phases are common for all three cases. In case a, we have ms.
The ratio for case c is the smallest because we use one setup and teardown phase to send more data. We assume that the transmission time is negligible in this case.
This means that we suppose all datagrams start at time 0. In a circuit-switched network, end-to-end addressing is needed during the setup and teardown phase to create a connection for the whole data transfer phase.
After the connection is made, the data flow travels through the already-reserved resources. The switches remain connected for the entire duration of the data transfer; there is no need for further addressing. In a datagram network, each packet is independent. The routing of a packet is done for each individual packet. Each packet, therefore, needs to carry an end- to-end address.
There is no setup and teardown phases in a datagram network connectionless transmission. The entries in the routing table are somehow permanent and made by other processes such as routing protocols. In a virtual-circuit network, there is a need for end-to-end addressing during the setup and teardown phases to make the corresponding entry in the switching table.
The entry is made for each request for connection. During the data trans- fer phase, each packet needs to carry a virtual-circuit identifier to show which virtual-circuit that particular packet follows. A datagram or virtual-circuit network handles packetized data. For each packet, the switch needs to consult its table to find the output port in the case of a datagram network, and to find the combination of the output port and the virtual circuit iden- tifier in the case of a virtual-circuit network.
In a circuit-switched network, data are not packetized; no routing information is carried with the data. The whole path is established during the setup phase.
In circuit-switched and virtual-circuit networks, we are dealing with connections. A connection needs to be made before the data transfer can take place. In the case of a circuit-switched network, a physical connection is established during the setup phase and the is broken during the teardown phase. In the case of a virtual-circuit network, a virtual connection is made during setup and is broken during the tear- down phase; the connection is virtual, because it is an entry in the table.
These two types of networks are considered connection-oriented. In the case of a datagram network no connection is made. Any time a switch in this type of network receives a packet, it consults its table for routing information. This type of network is con- sidered a connectionless network. The switching or routing in a datagram network is based on the final destination address, which is global.
The minimum number of entries is two; one for the final destination and one for the output port. Here the input port, from which the packet has arrived is irrelevant. The switching or routing in a virtual-circuit network is based on the virtual circuit identifier, which has a local jurisdiction.
This means that two different input or output ports may use the same virtual circuit number. Therefore, four pieces of information are required: input port, input virtual circuit number, output port, and output virtual circuit number. Packet 1: 2 Packet 2: 3 Packet 3: 3 Packet 4: 2 Packet 1: 2, 70 Packet 2: 1, 45 Packet 3: 3, 11 Packet 4: 4, 41 In a datagram network, the destination addresses are unique. They cannot be duplicated in the routing table.
In a virtual-circuit network, the VCIs are local. A VCI is unique only in rela- tionship to a port. In other words, the port, VCI combination is unique. This means that we can have two entries with the same input or output ports. However, we cannot have two entries with the same port, VCI pair. When a packet arrives at a router in a datagram network, the only information in the packet that can help the router in its routing is the destination address of the packet.
The table then is sorted to make the searching faster. When a packet arrives at a switch in a virtual-circuit network, the pair input port, input VCI can uniquely determined how the packet is to be routed; the pair is the only two pieces of information in the packet that is used for routing.
The table in the virtual-circuit switch is sorted based on the this pair. However, since the number of port numbers is normally much smaller than the number of virtual circuits assigned to each port, sorting is done in two steps: first according to the input port number and second according to the input VCI. However, we need to know that a regular multiplexer discussed in Chapter b. However, we need to know that a regular demultiplexer discussed in See Figure 8.
Figure 8. Only four simultaneous connections are possible for each crossbar at the first stage. This means that the total number of simultaneous connections is Only six simultaneous connections are possible for each crossbar at the first stage.
The number of cross- can be left unused. Some of the input lines tion. We can see that there is no blocking involved because each 8 input line has 15 intermediate crossbars. With less than , cross- points we can design a three-stage switch. The total number of crosspoints is , We give two solutions. We first solve the problem using only crossbars and then we replace the cross- bars at the first and the last stage with TSIs.
We can replace the crossbar at the first and third stages with TSIs as shown in Figure 8. In other words, the input frame has 10 slots and the output frame has only 4 slots. The data in the first slot of all input TSIs are directed to the first switch, the output in the second slot are directed to the sec- ond switch, and so on. We can see the inefficiency in the first solution. Since the slots are separated in time, only one of the switches at the middle stage is active at each moment. This means that, instead of 4 crossbars, we could have used only one with the same result.
In this case we still need memory locations but only crosspoints. The telephone network is made of three major components: local loops, trunks, and switching offices. The telephone network has several levels of switching offices such as end offices, tandem offices, and regional offices. A LATA is a small or large metropolitan area that according to the divestiture of was under the control of a single telephone-service provider.
These car- riers, sometimes called long-distance companies, provide communication services between two customers in different LATAs. Signaling System Seven SS7 is the protocol used to provide signaling services in the telephone network. It is very similar to the five-layer Internet model.
Telephone companies provide two types of services: analog and digital. Dial-up modems use part of the bandwidth of the local loop to transfer data.
The latest dial-up modems use the V-series standards such as V. Telephone companies developed digital subscriber line DSL technology to pro- vide higher-speed access to the Internet. It uses a device called a digital sub- scriber line access multiplexer DSLAM at the telephone company site.
The traditional cable networks use only coaxial cables to distribute video infor- mation to the customers. The hybrid fiber-coaxial HFC networks use a combi- nation of fiber-optic and coaxial cable to do so. To provide Internet access, the cable company has divided the available bandwidth of the coaxial cable into three bands: video, downstream data, and upstream data.
The downstream-only video band occupies frequencies from 54 to MHz. The downstream data occupies the upper band, from to MHz. The upstream data occupies the lower band, from 5 to 42 MHz. The cable modem CM is installed on the subscriber premises. The cable modem transmission system CMTS is installed inside the distribution hub by the cable company.
It receives data from the Internet and passes them to the combiner, which sends them to the subscriber. Packet-switched networks are well suited for carrying data in packets. The end-to- end addressing or local addressing VCI occupies a field in each packet. Tele- phone networks were designed to carry voice, which was not packetized. A cir- cuit-switched network, which dedicates resources for the whole duration of the conversation, is more suitable for this type of communication.
The setup phase can be matched to the dialing process. After the callee responds, the data transfer phase here voice transfer phase starts. When any of the parties hangs up, the data transfer is terminated and the teardown phase starts. It takes a while before all resources are released.
In a telephone network, the telephone numbers of the caller and callee are serving as source and destination addresses. These are used only during the setup dialing and teardown hanging up phases. The delay can be attributed to the fact that some telephone companies use satellite networks for overseas communication.
In these case, the signals need to travel sev- eral thousands miles earth station to satellite and satellite to earth station. See Figure 9. Figure 9. SDSL e. VDSL The DSL technology is based on star topology with the hub at the telephone office. The local loop connects each customer to the end office. This means that there is no sharing; the allocated bandwidth for each customer is not shared with neigh- bors. The data rate does not depend on how many people in the area are transfer- ring data at the same time.
The cable modem technology is based on the bus or rather tree topology. The cable is distributed in the area and customers have to share the available band- width. This means if all neighbors try to transfer data, the effective data rate will be decreased. In a single bit error only one bit of a data unit is corrupted; in a burst error more than one bit is corrupted not necessarily contiguous. Redundancy is a technique of adding extra bits to each data unit to determine the accuracy of transmission.
In forward error correction, the receiver tries to correct the corrupted codeword; in error detection by retransmission, the corrupted message is discarded the sender needs to retransmit the message.
A linear block code is a block code in which the exclusive-or of any two code- words results in another codeword. A cyclic code is a linear block code in which the rotation of any codeword results in another codeword.
The Hamming distance between two words of the same size is the number of differences between the corresponding bits. The Hamming distance can easily be found if we apply the XOR operation on the two words and count the number of 1s in the result. The minimum Hamming distance is the smallest Hamming distance between all possible pairs in a set of words.
The single parity check uses one redundant bit for the whole data unit. In a two- dimensional parity check, original data bits are organized in a table of rows and columns. The parity bit is then calculated for each column and each row. The remainder is always one bit smaller than the divisor. The degree of the generator polynomial is one less than the size of the divisor. For example, the CRC generator with the polynomial of degree 32 uses a bit divisor.
The degree of the generator polynomial is the same as the size of the remainder length of checkbits. For example, CRC with the polynomial of degree 32 creates a remainder of 32 bits. In this arithmetic, when a number needs more than n bits, the extra bits are wrapped and added to the number.
In this arithmetic, the complement of a number is made by inverting all bits. At least three types of error cannot be detected by the current checksum calcula- tion. First, if two data items are swapped during transmission, the sum and the checksum values will not change. Third, if one or more data items is changed in such a way that the The value of a checksum can be all 0s in binary. This happens when the value of the sum after wrapping becomes all 1s in binary. It is almost impossible for the value of a checksum to be all 1s.
For this to happen, the value of the sum after wrapping must be all 0s which means all data units must be 0s. First, the result of XORing two equal patterns is an all-zero pattern part b. Second, the result of XORing of any pattern with an all-zero pattern is the original non-zero pattern part c.
Third, the result of XORing of any pattern with an all-one pattern is the complement of the original non-one pattern. The codeword for dataword 10 is This codeword will be changed to if a 3-bit burst error occurs. This pattern is not one of the valid codewords, so the receiver detects the error and discards the received pattern. This pattern is not one of the valid codewords, so the receiver discards the received pattern. The code is not linear. We check five random cases.
All are in the code. We show the dataword, the codeword, the corrupted codeword, and the interpreta- tion of the receiver for each case: a. Comment: The above result does not mean that the code can never detect three errors. The last two cases show that it may happen that three errors remain unde- tected.
We show the dataword, codeword, the corrupted codeword, the syndrome, and the interpretation of each case: a. C 7,4 cannot correct two errors. C 7,4 cannot correct three errors. If we rotate one bit, the result is , which is in the code. If we rotate two bits, the result is , which is in the code. And so on. We use trial and error to find the right answer: a. To detect single bit errors, a CRC generator must have at least two terms and the coefficient of x0 must be nonzero.
It has more than one term and the coefficient of x0 is 1. It can detect a single-bit error. It will detect all burst errors of size 8 or less. This means 8 out of burst errors of size 9 c. Burst errors of size 9 are detected most of the time, but they slip by with proba- are left undetected. This means 4 out of burst errors of size 15 d. Burst errors of size 15 are detected most of the time, but they slip by with prob- are left undetected. It detects all single-bit error. It will detect all burst errors of size 32 or less.
This means out of burst c. Burst errors of size 33 are detected most of the time, but they are slip by with errors of size 33 are left undetected.
This means out of burst d. Burst errors of size 55 are detected most of the time, but they are slipped with errors of size 55 are left undetected. We need to add all bits modulo-2 XORing. However, it is simpler to count the number of 1s and make them even by adding a 0 or a 1. We have shown the parity bit in the codeword in color and separate for emphasis. Figure To separate the addresses used inside the home or business and the ones used for the Internet, the Internet authorities have reserved three sets of addresses as private addresses, shown in Table Everyone knows that these reserved addresses are for private net- works.
They are unique inside the organization, but they are not unique globally. No router will forward a packet that has one of these addresses as the destination address. The site must have only one single connection to the global Internet through a router that runs the NAT software. As Figure The router that connects the network to the global address uses one private address and one global address. The private network is transparent to the rest of the Internet; the rest of the Internet sees only the NAT router with the address All incoming packets also pass through the NAT router, which replaces the destination address in the packet the NAT router global address with the appropriate private address.
But how does the NAT router know the destination address for a packet coming from the Internet? The problem is solved if the NAT router has a trans- lation table. Using One IP Address In its simplest form, a translation table has only two columns: the private address and the external address destination address of the packet. When the router translates the source address of the outgoing packet, it also makes note of the destination address—where the packet is going.
Note that the addresses that are changed translated are shown in color. The NAT mechanism described requires that the private network start the communica- tion. The customer, however, may be a member of a private network that has many private addresses. For example, when e-mail that originates from a non- customer site is received by the ISP e-mail server, the e-mail is stored in the mailbox of the customer until retrieved.
A private network cannot run a server program for clients outside of its network if it is using NAT technology. To remove this restriction, the NAT router uses a pool of global addresses. For example, instead of using only one global address However, there are still some drawbacks. In this example, no more than four connections can be made to the same destination. Also, no private-network host can access two external server programs e.
Using Both IP Addresses and Port Numbers To allow a many-to-many relationship between private-network hosts and external server programs, we need more information in the translation table.
For example, suppose two hosts with addresses We discuss port numbers in Chapter Note that when the response from HTTP comes back, the combination of source address Note also that for this translation to work, the temporary port numbers and must be unique.
For example, suppose an ISP is granted addresses, but has , customers. Each of the customers is assigned a private network address. The ISP translates each of the , source addresses in outgoing packets to one of the global addresses; it translates the global destination address in incoming packets to the corresponding private address. In this section, we compare the address structure of IPv6 to IPv4. In Chapter 20, we discuss both protocols. Structure An IPv6 address consists of 16 bytes octets ; it is bits long.
An IPv6 address is bits long. In this nota- tion, bits is divided into eight sections, each 2 bytes in length. Two bytes in hexadecimal notation requires four hexadecimal digits. Therefore, the address consists of 32 hexadecimal digits, with every four digits separated by a colon, as shown in Figure In this case, we can abbreviate the address.
The leading zeros of a section four digits between two colons can be omitted. Only the leading zeros can be dropped, not the trailing zeros see Figure Note that cannot be abbreviated. Further abbreviations are possible if there are consecutive sections consisting of zeros only. We can remove the zeros altogether and replace them with a double semicolon.
Note that this type of abbreviation is allowed only once per address. If there are two runs of zero sections, only one of them can be abbreviated. Reexpansion of the abbreviated address is very simple: Align the unabbre- viated portions and insert zeros to get the original expanded address.
The designers of IPv6 divided the address into several categories. The third column shows the fraction of each type of address relative to the whole address space.
The provider-based address is generally used by a normal host as a unicast address. The address format is shown in Figure In the future, this link address will probably be the same as the node physical address. A packet sent to a multicast address must be delivered to each member of the group.
A transient group address, on the other hand, is used only tempo- rarily. Systems engaged in a teleconference, for example, can use a transient group address. However, a packet destined for an anycast address is delivered to only one of the members of the anycast group, the nearest one the one with the shortest route.
No block is assigned for anycast addresses. Unspecified 8 bits bits Loopback 8 bits 88 bits 32 bits All 0s IPv4 address c. Compatible 8 bits 72 bits 16 bits 32 bits All 0s All 1s IPv4 address d. A loopback address is used by a host to test itself without going into the network. It is used when a computer using IPv6 wants to send a message to another computer using IPv6, but the message needs to pass through a part of the network that still operates in IPv4. A mapped address is also used during transition.
However, it is used when a computer that has migrated to IPv6 wants to send a packet to a computer still using IPv4. Local Addresses These addresses are used when an organization wants to use IPv6 protocol without being connected to the global Internet.
In other words, they provide addressing for private net- works. Nobody outside the organization can send a message to the nodes using these addresses. Link local 10 bits 38 bits 32 bits 48 bits Subnet All 0s Node address b. The items in brackets [. IPv6 addresses are discussed in Section A good discussion of NAT can be found in [Dut01].
Classes A, B, and C differ in the number of hosts allowed per network. Class D is for multicasting and Class E is reserved. The number of addresses needs to be a power of 2. The starting address must be divisible by the number of addresses in the block. What is the number of bits in an IPv4 address?
What is the number of bits in an IPv6 address? What is dotted decimal notation in IPv4 addressing? What is the number of bytes in an IPv4 address represented in dotted decimal notation?
What is hexadecimal notation in IPv6 addressing? What is the number of digits in an IPv6 address repre- sented in hexadecimal notation? What are the differences between classful addressing and classless addressing in IPv4?
Explain why most of the addresses in class A are wasted. Explain why a medium-size or large-size corporation does not want a block of class C addresses. What is a mask in IPv4 addressing? What is a default mask in IPv4 addressing? What is the network address in a block of addresses? How do the subnet mask and supernet mask differ from a default mask in classful addressing?
How can we distinguish a multicast address in IPv4 addressing? How can we do so in IPv6 addressing? What is NAT? How can NAT help in address depletion?
Exercises What is the address space in each of the following systems? A system with 8-bit addresses b. A system with bit addresses c. An address space has a total of addresses. How many bits are needed to rep- resent an address? An address space uses the three symbols 0, 1, and 2 to represent addresses. If each address is made of 10 symbols, how many addresses are available in this system?
Change the following IP addresses from dotted-decimal notation to binary notation. Change the following IP addresses from binary notation to dotted-decimal notation. Find the class of the following IP addresses. Find the netid and the hostid of the following IP addresses. E-mail is not an interactive application.
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