Data Link Control

Definition of Terms

• Data link control deals with the design and procedures for communication between two adjacent nodes: node-to-node communication.
• The two main functions of the data link layer are data link control and media access control.
• Data link control functions include framing, flow and error control, and software-implemented protocols that provide smooth and reliable transmission of frames between nodes.
• Flow control is the regulation of the sender’s data rate so that the receiver buffer does not become overwhelmed.
• Error control is both error detection and error correction.  
• Frames can be of fixed or variable size. In fixed-size framing, there is no need for defining the boundaries of frames; in variable-size framing, we need a delimiter (flag) to define the boundary of two frames.
• Variable-size framing uses two categories of protocols: byte-oriented (or character-oriented) and bit-oriented. In a byte-oriented protocol, the data section of a frame is a sequence of bytes; in a bit-oriented protocol, the data section of a frame is a sequence of bits.
• In byte-oriented (or character-oriented) protocols, we use byte stuffing; a special byte added to the data section of the frame when there is a character with the same pattern as the flag.
• Byte stuffing is the process of adding 1 extra byte whenever there is a flag or escape character in the text.
• In bit-oriented protocols, we use bit stuffing; an extra 0 is added to the data section of the frame when there is a sequence of bits with the same pattern as the flag.
• Bit stuffing is the process of adding one extra 0 whenever five consecutive 1s follow a 0 in the data, so that the receiver does not mistake the pattern 0111110 for a flag.
• Flow control refers to a set of procedures used to restrict the amount of data that the sender can send before waiting for acknowledgment. Error control refers to methods of error detection and correction.
• For the noiseless channel, we discussed two protocols: the Simplest Protocol and the Stop-and-Wait Protocol. The first protocol has neither flow nor error control; the second has no error control. 
• In the Simplest Protocol, the sender sends its frames one after another with no regards to the receiver. 
• In the Stop-and-Wait Protocol, the sender sends one frame, stops until it receives confirmation from the receiver, and then sends the next frame.
• For the noisy channel, we discussed three protocols: Stop-and-Wait ARQ, Go-Back- N, and Selective Repeat ARQ. 
• The Stop-and-Wait ARQ Protocol, adds a simple error control mechanism to the Stop-and-Wait Protocol.
• Error correction in Stop-and-Wait ARQ is done by keeping a copy of the sent frame and retransmitting of the frame when the timer expires.
• In Stop-and-Wait ARQ, we use sequence numbers to number the frames. The sequence numbers are based on modulo-2 arithmetic.
• In Stop-and-Wait ARQ, the acknowledgment number always announces in modulo-2 arithmetic the sequence number of the next frame expected.
• Stop-and-Wait ARQ is a special case of Go-Back-N ARQ in which the size of the send window is 1.
• In the Go-Back-N ARQ Protocol, we can send several frames before receiving acknowledgments, improving the efficiency of transmission.
• In Go-Back-N ARQ, multiple frames can be in transit at the same time. If there is an error, retransmission begins with the last unacknowledged frame even if subsequent frames have arrived correctly. Duplicate frames are discarded.
• In the Go-Back-N Protocol, the sequence numbers are modulo 2^m, where m is the size of the sequence number field in bits. 
• The send window is an abstract concept defining an imaginary box of size 2^m − 1 with three variables: S_f, S_n, and S_size. The send window can slide one or more slots when a valid acknowledgment arrives.
• In Go-Back-N ARQ, the size of the send window must be less than 2^m; the size of the receiver window is always 1.
• In the Selective Repeat ARQ protocol we avoid unnecessary transmission by sending only frames that are corrupted.
• In Selective Repeat ARQ, multiple frames can be in transit at the same time. If there is an error, only the unacknowledged frame is retransmitted.
• In Selective Repeat ARQ, the size of the sender and receiver window must be at most one-half of 2^m.
• Both Go-Back-N and Selective-Repeat Protocols use a sliding window. In Go-Back- N ARQ, if m is the number of bits for the sequence number, then the size of the send window must be less than 2m; the size of the receiver window is always 1. In Selective Repeat ARQ, the size of the sender and receiver window must be at most one-half of 2m.
• A technique called piggybacking is used to improve the efficiency of the bidirectional protocols. When a frame is carrying data from A to B, it can also carry control information about frames from B; when a frame is carrying data from B to A, it can also carry control information about frames from A.
• The bandwidth-delay product is a measure of the number of bits a system can have in transit. 
• High-level Data Link Control (HDLC) is a bit-oriented protocol for communication over point-to-point and multipoint links. However, the most common protocols for point-to-point access is the Point-to-Point Protocol (PPP), which is a byte-oriented protocol. 
• HDLC is a protocol that implements ARQ mechanisms. It supports communication over point-to-point or multipoint links.
• PPP is a byte-oriented protocol using byte stuffing with the escape byte 01111101.
• HDLC stations communicate in normal response mode (NRM) or asynchronous balanced mode (ABM).
• HDLC protocol defines three types of frames: the information frame (I-frame), the supervisory frame (S-frame), and the unnumbered frame (U-frame).
• HDLC handle data transparency by adding a 0 whenever there are five consecutive 1s following a 0. This is called bit stuffing. 

Two Main Functions of the Data Link layer

• Data link control - deals with the design and procedures for communication between two adjacent nodes: node-to-node communication.
• Media access - control deals with procedures for sharing the link.

Two categories of protocols in Variable-size framing

• Byte-oriented protocol - data to be carried are 8-bit characters from a coding system. Character-oriented protocols were popular when only text was exchanged by the data link layers.
• Bit-oriented protocol -  the data section of a frame is a sequence of bits. Bit-oriented protocols are more popular today because we need to send text, graphic, audio, and video which can be better represented by a bit pattern than a sequence of characters.

Taxonomy of protocols discussed in this chapter

Note: You can proceed to take the multiple choice exam regarding this topic. Data Link Control - Set 1 MCQs

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Error Detection and Correction - MCQs Answers

Below are the answers key for the Multiple Choice Questions (Quiz) of Error Detection and Correction.

1. one's complement arithmetic
2. 0000
3. 0 to 10
4. one bit more than
5. Hamming distance
6. degree
7. addition and subtraction
8. The remainder
9. Simple parity check
10. block; datawords
11. burst
12. 0
13. forward
14. 3
15. an odd-number of
16. 0
17. XORing
18. retransmission
19. block; convolution
20. 0 and 1
21. 11
22. correction; detection
23. 4
24. 1111
25. 1000
26. 5
27. Checksum
28. generator
29. XOR
30. codewords
31. 6
32. x + 1
33. Cyclic
34. 2

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Error Detection and Correction - MCQs

In this particular topic you have learned that Data can be corrupted during transmission. To address this issue some applications require that errors be detected and must be corrected. If you skip the summary visit Error Detection and Correction.

Begin and Good luck!

1. Checksums use _________ arithmetic.
• A)   one's complement arithmetic
• B)   two's complement arithmetic
• C)   either (a) or (b)
• D)   none of the above
2. The checksum of 1111 and 1111 is _________.
• A)   0000
• B)   1111
• C)   1110
• D)   0111
3. In modulo-11 arithmetic, we use only the integers in the range ______, inclusive.
• A)   1 to 10
• B)   0 to 10
• C)   1 to 11
• D)   none of the above
4. In cyclic redundancy checking, the divisor is _______ the CRC.
• A)   one bit less than
• B)   one bit more than
• C)   The same size as
• D)   none of the above
5. The ________ between two words is the number of differences between corresponding bits.
• A)   Hamming rule
• B)   Hamming code
• C)   Hamming distance
• D)   none of the above
6. The _______ of a polynomial is the highest power in the polynomial.
• A)   range
• B)   power
• C)   degree
• D)   none of the above
7. In modulo-2 arithmetic, __________ give the same results.
• A)   addition and subtraction
• B)   addition and multiplication
• C)   addition and division
• D)   none of the above
8. In cyclic redundancy checking, what is the CRC?
• A)   The quotient
• B)   The dividend
• C)   The divisor
• D)  
9. Which error detection method consists of just one redundant bit per data unit?
• A)   CRC
• B)   Checksum
• C)   Simple parity check
• D)   Two-dimensional parity check
10. In _____ coding, we divide our message into blocks, each of k bits, called ___.
• A)   block; blockwords
• B)   block; datawords
• C)   linear; datawords
• D)   none of the above
11. A _____ error means that two or more bits in the data unit have changed.
• A)   burst
• B)   double-bit
• C)   single-bit
• D)   none of the above
12. Adding 1 and 1 in modulo-2 arithmetic results in _________.
• A)   0
• B)   1
• C)   2
• D)   none of the above
13. In ________ error correction, the receiver corrects errors without requesting retransmission.
• A)   onward
• B)   forward
• C)   backward
• D)   none of the above
14. If the Hamming distance between a dataword and the corresponding codeword is three, there are _____ bits in error.
• A)   5
• B)   4
• C)   3
• D)   none of the above
15. A simple parity-check code can detect __________ errors.
• A)   an odd-number of
• B)   an even-number of
• C)   two
• D)   no errors
16. The Hamming distance between equal codewords is _________.
• A)   0
• B)   1
• C)   n
• D)   none of the above
17. In a linear block code, the _______ of any two valid codewords creates another valid codeword.
• A)   ANDing
• B)   XORing
• C)   ORing
• D)   none of the above
18. In ________ error correction, the receiver asks the sender to send the data again.
• A)   forward
• B)   backward
• C)   retransmission
• D)   none of the above
19. We can divide coding schemes into two broad categories: ________ and ______coding.
• A)   linear; nonlinear
• B)   block; convolution
• C)   block; linear
• D)   none of the above
20. In modulo-2 arithmetic, we use only ______.
• A)   1 and 2
• B)   0 and 1
• C)   0 and 2
• D)   none of the above
21. To guarantee correction of up to 5 errors in all cases, the minimum Hamming distance in a block code must be ________.
• A)   11
• B)   6
• C)   5
• D)   none of the above
22. The _____of errors is more difficult than the ______.
• A)   detection; correction
• B)   correction; detection
• C)   creation; correction
• D)   creation; detection
23. In block coding, if k =2 and n =3, we have _______ invalid codewords.
• A)   4
• B)   8
• C)   2
• D)   none of the above
24. The checksum of 0000 and 0000 is __________.
• A)   0000
• B)   1111
• C)   0111
• D)   1110
25. In one's complement arithmetic, if positive 7 is 0111, then negative 7 is ________.
• A)   1101
• B)   1000
• C)   1111
• D)   none of the above
26. In block coding, if n = 5, the maximum Hamming distance between two codewords is ________.
• A)   5
• B)   3
• C)   2
• D)   none of the above
27. Which error detection method uses one's complement arithmetic?
• A)   Checksum
• B)   CRC
• C)   Simple parity check
• D)   Two-dimensional parity check
28. The divisor in a cyclic code is normally called the _________.
• A)   redundancy
• B)   degree
• C)   generator
• D)   none of the above
29. In modulo-2 arithmetic, we use the ______ operation for both addition and subtraction.
• A)   OR
• B)   XOR
• C)   AND
• D)   none of the above
30. We add r redundant bits to each block to make the length n = k + r. The resulting n-bit blocks are called _________.
• A)   codewords
• B)   datawords
• C)   blockwords
• D)   none of the above
31. To guarantee the detection of up to 5 errors in all cases, the minimum Hamming distance in a block code must be _______.
• A)   11
• B)   5
• C)   6
• D)   none of the above
32. A generator that contains a factor of ____ can detect all odd-numbered errors.
• A)   x
• B)   1
• C)   x + 1
• D)   none of the above
33. _______codes are special linear block codes with one extra property. If a codeword is rotated, the result is another codeword.
• A)   Convolution
• B)   Cyclic
• C)   Non-linear
• D)   none of the above
34. The Hamming distance between 100 and 001 is ________.
• A)   0
• B)   1
• C)   2
• D)   none of the above

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Error Detection and Correction

Definition of Terms

• Data can be corrupted during transmission. Some applications require that errors be detected and corrected.
• In a single-bit error, only one bit in the data unit has changed. A burst error means that two or more bits in the data unit have changed.
• To detect or correct errors, we need to send extra (redundant) bits with data.
• Redundancy is the concept of sending extra bits for use in error detection. 
• There are two main methods of error correction: forward error correction and correction by retransmission.
• We can divide coding schemes into two broad categories: block coding and convolution coding.
• In coding, we need to use modulo-2 arithmetic. Operations in this arithmetic are very simple; addition and subtraction give the same results. we use the XOR (exclusive OR) operation for both addition and subtraction.
• In modulo-N arithmetic, we use only the integers in the range 0 to N −1, inclusive.
• In block coding, we divide our message into blocks, each of k bits, called datawords. We add r redundant bits to each block to make the length n =  k + r. The resulting n-bit blocks are called codewords.
• An error-detecting code can detect only the types of errors for which it is designed; other types of errors may remain undetected.
• The Hamming code is an error correction method using redundant bits. The number of bits is a function of the length of the data bits.
• In the Hamming code, for a data unit of m bits, use the formula 2 r >= m + r + 1 to determine r, the number of redundant bits needed.
• By rearranging the order of bit transmission of the data units, the Hamming code can correct burst errors. 
• The Hamming distance between two words is the number of differences between corresponding bits. The minimum Hamming distance is the smallest Hamming distance between all possible pairs in a set of words.
• To guarantee the detection of up to s errors in all cases, the minimum Hamming distance in a block code must be d_min = s + 1. To guarantee correction of up to t errors in all cases, the minimum Hamming distance in a block code must be d_min = 2t + 1.
• In a linear block code, the exclusive OR (XOR) of any two valid codewords creates another valid codeword.
• A simple parity-check code is a single-bit error-detecting code in which n = k + 1 with d_min = 2.  A simple parity-check code can detect an odd number of errors.
• All Hamming codes discussed in this book have d_min = 3. The relationship between m and n in these codes is n = 2m - 1. 
• Cyclic codes are special linear block codes with one extra property. In a cyclic code, if a codeword is cyclically shifted (rotated), the result is another codeword. 
• The divisor in a cyclic code is normally called the generator polynomial or simply the generator.
• In a cyclic code, those e(x) errors that are divisible by g(x) are not caught.
• If the generator has more than one term and the coefficient of x0 is 1, all single errors can be caught.
• If a generator cannot divide x^t + 1 (t between 0 and n – 1), then all isolated double errors can be detected.
• A generator that contains a factor of x + 1 can detect all odd-numbered errors.
• A category of cyclic codes called the cyclic redundancy check (CRC) is used in networks such as LANs and WANs.
• A pattern of Os and Is can be represented as a polynomial with coefficients of 0 and 1.
• Traditionally, the Internet has been using a 16-bit checksum, which uses one's complement arithmetic. In this arithmetic, we can represent unsigned numbers between o and 2ⁿ -1 using only n bits.

Error Categories:

• Single-bit error - has one bit error per data unit.
• Burst error - has two or more bit errors per data unit.

Three common redundancy methods

• Parity check- An extra bit (parity bit) is added to the data unit. The parity check can detect only an odd number of errors; it cannot detect an even number of errors. In the two-dimensional parity check, a redundant data unit follows n data units. 
• Cyclic redundancy check (CRC) - a powerful redundancy checking technique, appends a sequence of redundant bits derived from binary division to the data unit. The divisor in the CRC generator is often represented as an algebraic polynomial. 
• Checksum -  used in the Internet by several protocols although not at the data link layer.

At least three types of error cannot be detected by the current checksum

• First, if two data items are swapped during transmission, the sum and the checksum values will not change. 
• Second, if the value of one data item is increased (intentionally or maliciously) and the value of another one is decreased (intentionally or maliciously) the same amount, the sum and the checksum cannot detect these changes.
• Third, if one or more data items is changed in such a way that the change is a multiple of 2¹⁶ − 1, the sum or the checksum cannot detect the changes.

Two Main Methods of Error Correction

• Forward error correction- the receiver tries to correct the corrupted codeword.
• Error correction by retransmission - the corrupted message is discarded (the sender needs to retransmit the message).

In block coding, errors be detected by using the following two conditions:

• a. The receiver has (or can find) a list of valid codewords.
• b. The original codeword has changed to an invalid one.

Note: You can proceed to take the multiple choice exam regarding this topic. Error Detection and Correction - Set 1 MCQs

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Telephone and Cable Networks - MCQs Answers

Below are the answers key for the Multiple Choice Questions (Quiz) of Using Telephone and Cable Networks for Data Transmission.

1. CM; CMTS
2. circuit-switched
3. downstream
4. coaxial
5. 3
6. digital; analog
7. analog
8. SS7
9. DSL
10. twisted-pair
11. HFC
12. downstream data
13. SDSL
14. LEC
15. FDM; QAM
16. IXC
17. digital as well as analog
18. QPSK
19. HDSL
20. 2B1Q
21. QAM
22. head end
23. modulator; demodulator
24. switched/56; DDS
25. LATAs
26. DOCSIS
27. out-of-band
28. switched; leased
29. coaxial and fiber-optic
30. in-band
31. V-series

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• Data Communications and Networking (Chapter Quiz)

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Telephone and Cable Networks - MCQs

In this particular topic you have learned how the two public networks, telephone and cable TV, can be used for data transfer. If you skip the summary visit Using Telephone and Cable Networks for Data Transmission.

Begin and Good luck!

1. To use a cable network for data transmission, we need two key devices: a ______ and a _________
• A)   CT; CMTS
• B)   CM; CMTS
• C)   CM; CMS
• D)   none of the above
2. A local telephone network is an example of a _______ network.
• A)   packet-switched
• B)   message-switched
• C)   circuit-switched
• D)   none of the above
3. A traditional cable TV network transmits signals ______.
• A)   downstream
• B)   upstream
• C)   upstream and downstream
• D)   none of the above
4. The traditional cable TV system used ________cable end to end.
• A)   twisted-pair
• B)   fiber-optic
• C)   coaxial
• D)   none of the above
5. The telephone network is made of ______ major components.
• A)   4
• B)   3
• C)   2
• D)   none of the above
6. Data from a computer are _______; the local loop handles _______ signals.
• A)   analog; analog
• B)   digital; digital
• C)   digital; analog
• D)   analog; digital
7. The original telephone network, which is referred to as the plain old telephone system (POTS), was an ________ system.
• A)   analog
• B)   digital
• C)   digital as well as analog
• D)   none of the above
8. The protocol that is used for signaling in the telephone network is called ______.
• A)   SSS
• B)   SS7
• C)   POP
• D)   none of the above
9. _______technology is a set of technologies developed by the telephone companies to provide high data rate transmission.
• A)   LDS
• B)   ASL
• C)   DSL
• D)   none of the above
10. The local loop has _______ cable that connects the subscriber telephone to the nearest end office.
• A)   fiber-optic
• B)   coaxial
• C)   twisted-pair
• D)   none of the above
11. The second generation of cable networks is called a(n) _________ network.
• A)   HCF
• B)   HFC
• C)   CFH
• D)   none of the above
12. The largest portion of the bandwidth for ADSL carries _______.
• A)   upstream data
• B)   downstream data
• C)   control data
• D)   voice communication
13. _______ is suitable for businesses that require comparable upstream and downstream data rates.
• A)   SDSL
• B)   ADSL
• C)   VDSL
• D)   (b) and (c)
14. The carrier that handles intra-LATA services is called a(n) _____ .
• A)   IXC
• B)   LEC
• C)   POP
• D)   none of the above
15. DMT is a modulation technique that combines elements of _______ and _______.
• A)   FDM; QAM
• B)   FDM; TDM
• C)   PSK; FSK
• D)   QDM; QAM
16. The carrier that handles inter-LATA services is called a(n) _______.
• A)   IXC
• B)   LEC
• C)   POP
• D)   none of the above
17. The modern telephone network is now ________.
• A)   digital
• B)   digital as well as analog
• C)   analog
• D)   none of the above
18. In an HFC network, the upstream data are modulated using the _______ modulation technique.
• A)   ASK
• B)   PCM
• C)   QAM
• D)   QPSK
19. _______ was designed as an alternative to the T-1 line.
• A)   ADSL
• B)   HDSL
• C)   VDSL
• D)   SDSL
20. HDSL encodes data using _______.
• A)   2B1Q
• B)   1B2Q
• C)   4B/5B
• D)   6B/8T
21. In an HFC network, the downstream data are modulated using the _______ modulation technique.
• A)   PCM
• B)   QAM
• C)   PSK
• D)   ASK
22. Another name for the cable TV office is the _______.
• A)   head end
• B)   combiner
• C)   fiber node
• D)   splitter
23. The term modem is a composite word that refers to the two functional entities that make up the device: a signal _______ and a signal _______.
• A)   demodulator; modulator
• B)   modulator; demodulator
• C)   modern; demo
• D)   none of the above
24. The two most common digital services are ________ service and ______.
• A)   switched/56; DDS
• B)   switched/56; switched/64
• C)   DDS; swiched 64
• D)   leased; out-of-band
25. The United States is divided into many _______.
• A)   IXCs
• B)   LECs
• C)   LATAs
• D)   none of the above
26. The standard for data transmission over an HFC network is called _______.
• A)   ADSL
• B)   CMTS
• C)   DOCSIS
• D)   MCNS
27. In ________signaling, a portion of the bandwidth is used for signaling and another portion for data.
• A)   mixed
• B)   in-band
• C)   out-of-band
• D)   none of the above
28. Telephone companies provide two types of analog services: analog _______ services and analog _____services.
• A)   leased; out-of-band
• B)   out-of-band; in-band
• C)   switched; in-band
• D)   switched; leased
29. The HFC network uses _______ cable.
• A)   coaxial
• B)   twisted-pair
• C)   fiber-optic
• D)   a combination of (a) and (c)
30. In ______signaling, the same circuit is used for both signaling and data.
• A)   mixed
• B)   out-of-band
• C)   in-band
• D)   none of the above
31. Most popular modems available are based on the ________standards.
• A)   X-series
• B)   V-series
• C)   VX-series
• D)   none of the above

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Using Telephone and Cable Networks for Data Transmission

Definition of Terms

• The telephone, which is referred to as the plain old telephone system (POTS), was originally an analog system. During the last decade, the telephone network has undergone many technical changes. The network is now digital as well as analog.
• A home computer can access the Internet through the existing telephone system or through a cable TV system. 
• The telephone network is made of three major components: local loops, trunks, and switching offices. It has several levels of switching offices such as end offices, tandem offices, and regional offices. 
• Telephone companies provide two types of services: analog and digital.
• The United States is divided into many local access transport areas (LATAs). The services offered inside a LATA are called intra-LATA services. The carrier that handles these services is called a local exchange carrier (LEC). The services between LATAs are handled by interexchange carriers (lXCs).
• A LATA is a small or large metropolitan area that according to the divestiture of 1984 was under the control of a single telephone-service provider.
• In in-band signaling, the same circuit is used for both signaling and data. In out-of band signaling, a portion of the bandwidth is used for signaling and another portion for data. The protocol that is used for signaling in the telephone network is called Signaling System Seven (SS7).
• Telephone companies provide two types of services: analog and digital. We can categorize analog services as either analog switched services or analog leased services. The two most common digital services are switched/56 service and digital data service (DDS).
• Data transfer using the telephone local loop was traditionally done using a dial-up modem. The term modem is a composite word that refers to the two functional entities that make up the device: a signal modulator and a signal demodulator.
• Most popular modems available are based on the V-series standards. The V.32 modem has a data rate of 9600 bps. The V32bis modem supports 14,400-bps transmission. V90 modems, called 56K modems, with a downloading rate of 56 kbps and uploading rate of 33.6 kbps are very common. The standard above V90 is called V92. These modems can adjust their speed, and if the noise allows, they can upload data at the rate of 48 kbps.
• Telephone companies developed another technology, digital subscriber line (DSL), to provide higher-speed access to the Internet. DSL technology is a set of technologies, each differing in the first letter (ADSL, VDSL, HDSL, and SDSL. ADSL provides higher speed in the downstream direction than in the upstream direction. The high-bitrate digital subscriber line (HDSL) was designed as an alternative to the T-l line (1.544 Mbps). The symmetric digital subscriber line (SDSL) is a one twisted-pair version of HDSL. The very high-bit-rate digital subscriber line (VDSL) is an alternative approach that is similar to ADSL.
• DSL supports high-speed digital communications over the existing telephone local loops. 
• ADSL technology allows customers a bit rate of up to 1 Mbps in the upstream direction and up to 8 Mbps in the downstream direction. 
• ADSL uses a modulation technique called DMT which combines QAM and FDM.  
• ADSL is an asymmetric communication technology designed for residential users; it is not suitable for businesses.
• ADSL is an adaptive technology. The system uses a data rate based on the condition of the local loop line.
• SDSL, HDSL, and VDSL are other DSL technologies. 
• Theoretically, the coaxial cable used for cable TV allows Internet access with a bit rate of up to 12 Mbps in the upstream direction and up to 30 Mbps in the downstream direction. 
• An HFC network allows Internet access through a combination of fiber-optic and coaxial cables. 
• The coaxial cable bandwidth is divided into a video band, a downstream data band, and an upstream data band. Both upstream and downstream bands are shared among subscribers. 
• DOCSIS defines all protocols needed for data transmission on an HFC network. 
• Synchronous Optical Network (SONET) is a synchronous high-data-rate TDM network for fiber-optic networks. 
• SONET has defined a hierarchy of signals (similar to the DS hierarchy) called synchronous transport signals (STSs). 
• Optical carrier (OC) levels are the implementation of STSs. 
• A SONET frame can be viewed as a matrix of nine rows of 90 octets each. 
• SONET is backward compatible with the current DS hierarchy through the virtual tributary (VT) concept. VT's are a partial payload consisting of an m-by-n block of octets. An STS payload can be a combination of several VT's.
• STSs can be multiplexed to get a new STS with a higher data range.
• Community antenna TV (CATV) was originally designed to provide video services for the community. The traditional cable TV system used coaxial cable end to end. The second generation of cable networks is called a hybrid fiber-coaxial (HFC) network. The network uses a combination of fiber-optic and coaxial cable.
• Communication in the traditional cable TV network is unidirectional.
• Communication in an HFC cable TV network can be bidirectional.
• 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 550 MHz. The downstream data occupies the upper band, from 550 to 750 MHz. The upstream data occupies the lower band, from 5 to 42 MHz.
• 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.

Three Major Components of Telephone System

• Local loops - a twisted-pair cable that connects the subscriber telephone to the nearest end office or local central office. The local loop, when used for voice, has a bandwidth of 4000 Hz (4 kHz). The existing local loops can handle bandwidths up to 1.1 MHz.
• Trunks - transmission media that handle the communication between offices. A trunk normally handles hundreds or thousands of connections through multiplexing. Transmission is usually through optical fibers or satellite links.
• Switching offices - A switch connects several local loops or trunks and allows a connection between different subscribers.

Telephone Line Bandwidth

Bandwidth division in ADSL

Summary of DSL technologies

Division of coaxial cable band by CATV

A SONET system can use the following equipment:

• STS multiplexer - combines several optical signals to make an STS signal. 
• Regenerator - removes noise from an optical signal. 
• Add/drop multiplexer - adds STSs from different paths and removes STSs from a path.

Note: You can proceed to take the multiple choice exam regarding this topic. Using Telephone and Cable Networks for Data Transmission - Set 1 MCQs

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Switching - MCQs Answers

Below are the answers key for the Multiple Choice Questions (Quiz) of Set 1 MCQs of Switching .

1. two
2. datagram switching
3. 19
4. datagram switching
5. space-division
6. virtual-circuit
7. less than 40,000
8. circuit-switched
9. 4
10. datagram switching
11. virtual-circuit
12. TSI
13. virtual-circuit
14. four
15. 40,000
16. 3
17. destination
18. three
19. datagram
20. TST
21. circuit switching
22. 10
23. physical
24. multistage
25. datagram switching
26. three
27. banyan
28. destination
29. 15,200
30. crossbar

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• Data Communications and Networking (Chapter Quiz)

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Switching - MCQs

In this particular topic you have learned the three fundamental switching methods: circuit switching, packet switching, and message switching. If you skip the summary visit Switching which is a method in which communication devices are connected to one another efficiently.

Begin and Good luck!

1. Packet-switched networks can also be divided into ______subcategories: virtual-circuit networks and datagram networks.
• A)   five
• B)   three
• C)   two
• D)   four
2. In __________, each packet is treated independently of all others.
• A)   circuit switching
• B)   datagram switching
• C)   frame switching
• D)   none of the above
3. Based on the Clos criteria, if N = 200, then k must be equal to or greater than ______.
• A)   19
• B)   21
• C)   31
• D)   41
4. In _________, resources are allocated on demand.
• A)   circuit switching
• B)   datagram switching
• C)   frame switching
• D)   none of the above
5. In _______ switching, the paths in the circuit are separated from one another spatially.
• A)   time-division
• B)   two-dimensional
• C)   space-division
• D)   three-dimensional
6. In a ________ network, two types of addressing are involved: global and local.
• A)   datagram
• B)   virtual-circuit
• C)   circuit-switched
• D)   none of the above
7. In a three-stage space division switch, if N = 200, the number of crosspoints is ______.
• A)   40,000
• B)   less than 40,000
• C)   greater than 40,000
• D)   greater than 100,000
8. A ________ network is made of a set of switches connected by physical links, in which each link is divided into n channels.
• A)   circuit-switched
• B)   line-switched
• C)   frame-switched
• D)   none of the above
9. In a banyan switch, for 8 inputs and 8 outputs, we have _____ microswitches at each stage.
• A)   2
• B)   3
• C)   4
• D)   8
10. In _________, there is no resource allocation for a packet.
• A)   circuit switching
• B)   datagram switching
• C)   frame switching
• D)   none of the above
11. A _________ network is a cross between a circuit-switched network and a datagram network. It has some characteristics of both.
• A)   packet-switched
• B)   frame-switched
• C)   virtual-circuit
• D)   none of the above
12. The most popular technology in time-division switching is called the _________.
• A)   TSI
• B)   STI
• C)   ITS
• D)   none of the above
13. A switched WAN is normally implemented as a _______ network.
• A)   virtual-circuit
• B)   datagram
• C)   circuit-switched
• D)   none of the above
14. We can say that a packet switch has _______ types of components.
• A)   four
• B)   three
• C)   two
• D)   none of the above
15. In a one-stage space division switch, if N = 200, the number of crosspoints is ______.
• A)   20,000
• B)   40,000
• C)   30,000
• D)   10,000
16. In a banyan switch, for 8 inputs and 8 outputs, we have _____ stages.
• A)   2
• B)   3
• C)   4
• D)   8
17. A switch in a datagram network uses a routing table that is based on the ______ address.
• A)   destination
• B)   source
• C)   local
• D)   none of the above
18. Traditionally, _____ methods of switching have been important.
• A)   six
• B)   five
• C)   four
• D)   three
19. The network layer in the Internet is designed as a __________ network.
• A)   circuit-switched
• B)   datagram
• C)   virtual-circuit
• D)   none of the above
20. A ________ switch combines space-division and time-division technologies to take advantage of the best of both.
• A)   SSS
• B)   TST
• C)   TTT
• D)   none of the above
21. In _______, the resources need to be reserved during the setup phase; the resources remain dedicated for the entire duration of data transfer phase until the teardown phase.
• A)   frame switching
• B)   datagram switching
• C)   circuit switching
• D)   none of the above
22. Based on the Clos criteria, if N = 200, then n must be equal to or greater than ____.
• A)   40
• B)   30
• C)   20
• D)   10
23. Circuit switching takes place at the ________ layer.
• A)   physical
• B)   data line
• C)   network
• D)   transport
24. A ________ switch combines crossbar switches in several (normally three) stages.
• A)   multiple path
• B)   multiple crossbar
• C)   multistage
• D)   none of the above
25. In _______ there are no setup or teardown phases.
• A)   circuit switching
• B)   datagram switching
• C)   frame switching
• D)   none of the above
26. We can divide today's networks into ____ broad categories.
• A)   five
• B)   four
• C)   three
• D)   two
27. A ________ switch is a multistage switch with microswitches at each stage that route the packets based on the output port represented as a binary string.
• A)   TSI
• B)   banyan
• C)   crossbar
• D)   none of the above
28. The _______ address in the header of a packet in a datagram network normally remains the same during the entire journey of the packet.
• A)   destination
• B)   source
• C)   local
• D)   none of the above
29. Based on the Clos criteria, if N = 200, then the minimum number of crosspoints is greater than or equal to _______.
• A)   42,000
• B)   20,000
• C)   18,000
• D)   15,200
30. The simplest type of switching fabric is the ______ switch.
• A)   crossbar
• B)   crosspoint
• C)   TSI
• D)   STS

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• Data Communications and Networking (Chapter Quiz)

Note: Check your works. Switching - MCQs Answers

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Integration by Parts - Set 1 Answers key

Below are the answers for Integration by Parts - Set 1 Problems.

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Review Integration by Partial Fractions

In case you want to review the the Integration by Parts lectures to fully understand how this method works. Kindly follow the link above.

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Integration by Parts - Set 1 Problems

Below are some exercises that will challenge yourself to integrate using the method of Integration by Parts. If you want review the topic visit Integration by Partial Fractions. Evaluate the integral.

Begin and good luck!

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Integration Formulas

Important: To make this exercises easy for you to work on, I compiled the integration formulas that you can find by following the link Table of Integral.

To check your works, the answers key for this problems set can be found by following this link. Integration by Parts - Set 1 Answers key

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