1. Datornätverk A – lektion 7 Forts. Kapitel 9: SDH och SONET. Kapitel 10: Felhantering.

2. Optical Hierarchies The old PCM hierarchy was non-synchronous ○Different multiplexors may have slightly different clock frequency. ○The whole hierarchy had to be unpacked in view to access or monitor a single telephone call, which was expensive. SDH and SONET use synchronous communication ○Clocked by a central master clock. SDH (Synchronous Digital Hierarchy) ○A standard for TDM in Europe SONET (Synchronious Optical NETwork) ○A standard for TDM used in United States IP-over-SDH/SONET allows several ISP:s to share the same fiber cable independently.

3. Table 9.1 SONET rates STSOCRate (Mbps)SPE (Mbps)User (Mbps) STS-1OC-151.8450.1249.536 STS-3OC-3155.52150.336148.608 STS-9OC-9466.56451.008445.824 STS-12OC-12622.08601.344594.432 STS-18OC-18933.12902.016891.648 STS-24OC-241244.161202.6881188.864 STS-36OC-361866.231804.0321783.296 STS-48OC-482488.322405.3762377.728 STS-192OC-1929953.289621.6049510.912

4. Figure 9.14 STS multiplexing

5. Figure 9.12 Data rate

6. Figure 9.11 Frame format SPE = Synchronous Payload Envelope

7. Figure 9.13 VT types

8. Data Link Layer PART III

9. Position of the data-link layer

10. LLC and MAC sublayers

11. IEEE standards for LANs

12. Chapter 10 Error Detection and Correction

13. Data can be corrupted during transmission. For reliable communication, errors must be detected and corrected. Note:

14. 10.1 Types of Error Single-Bit Error Burst Error

15. In a single-bit error, only one bit in the data unit has changed. Note:

16. 10.1 Single-bit error

17. A burst error means that 2 or more bits in the data unit have changed. Note:

18. 10.2 Burst error of length 5

19. Error detection uses the concept of redundancy, which means adding extra bits for detecting errors at the destination. Note:

20. 10.3 Redundancy

21. 10.4 Detection methods

22. In parity check, a parity bit is added to every data unit so that the total number of 1s is even (or odd for odd-parity). Note:

23. Example 1 Suppose the sender wants to send the word world. In ASCII (7-bit code) the five characters are coded as 1110111 1101111 1110010 1101100 1100100 Even parity is used. The following shows the actual bits sent: 11101110 11011110 11100100 11011000 11001001

24. Example 3 Now suppose the word world in Example 1 is received by the receiver without being corrupted in transmission. 11101110 11011110 11100100 11011000 11001001 The receiver counts the 1s in each character and comes up with even numbers (6, 6, 4, 4, 4). The data are accepted. Now suppose the word world in Example 1 is corrupted during transmission. 11111110 11011110 11101100 11011000 11001001 The receiver counts the 1s in each character and comes up with even and odd numbers (7, 6, 5, 4, 4). The receiver knows that the data are corrupted, discards them, and asks for retransmission.

25. Simple parity check can detect all single-bit errors. It can detect burst errors only if the total number of errors in each data unit is odd. Note:

26. In two-dimensional parity check, a block of bits is divided into rows and a redundant row of bits is added to the whole block. Note:

27. 10.6 Two-dimensional parity

28. Example 4 Suppose the following block is sent: 10101001 00111001 11011101 11100111 10101010 However, it is hit by a burst noise of length 8, and some bits are corrupted. 10100000 00001001 11011101 11100111 10101010 When the receiver checks the parity bits, some of the bits do not follow the even-parity rule and the whole block is discarded. 10100000 00001001 11011101 11100111 10101010

29. 10.7 CRC generator and checker CRC = Cyclic Redundency Check

30. ”Vanlig” division 17. 1 5 2 5 0 3 Haltande liknelse med CRC: Om vi subtraherar täljaren (17) med resten (2) vi ett tal som är jämnt delbart med nämnaren (5). Om mottagaren tar emot något som inte är delbart med 5 har sannolikt att bitfel uppstått. (Nämnare) Nominator (Täljare) Reminder (Rest) (Kvot)

31. Modulo 2 Arithmetic In modulo 2 arithmetic addition and substruction are identical to EXCLUSUVE OR (XOR) operation. Multiplication and division are the same as in base-2 arithmetic without carries in addition or borrows in substraction. 0 XOR 0 = 0 0 XOR 1 = 1 1 XOR 0 = 1 1 XOR 1 = 0 Examples: 1011 XOR 0101 = 1110 1001 XOR 1101 = 0100

32. 10.8 Binary division in a CRC generator The number of 0s is one less than the number of bits in G (divisor)

33. 10.8 Binary division in a CRC generator The first bit in the quotient is 1 and one times the divisor results in this Obtained by XOR-ing 1001 and 1101 The first bit in the numerator is 1. Then the first bit in the quotient is 1.

34. 10.8 Binary division in a CRC generator Transmitted data: 1 0 0 1 0 0 0 0 1

35. 10.9 Binary division in CRC checker

36. Another Example The transmitted data is 11010110111110

37. 10.11 A polynomial representing a divisor 7th order polynomial 8 bit divisor

38. Polynomyal representation Based upon discrete mathetematics, where bit strings are treated as polynomials. k bit divisor is represented as (k-1) degree polynomial with coefficients 0 and 1. Example: 1010110 has 7 bits It can be represented as a polynomial of 6 th degree 1·x 6 + 0·x 5 + 1·x 4 +0·x 3 + 1·x 2 + 1·x 1 + 0·x 0 = x 6 + x 4 + x 2 + x

39. Table 10.1 Standard polynomials NamePolynomialApplication CRC-8x 8 + x 2 + x + 1ATM header CRC-10x 10 + x 9 + x 5 + x 4 + x 2 + 1ATM AAL ITU-16x 16 + x 12 + x 5 + 1HDLC ITU-32 x 32 + x 26 + x 23 + x 22 + x 16 + x 12 + x 11 + x 10 + x 8 + x 7 + x 5 + x 4 + x 2 + x + 1 LANs

40. Example 5 It is obvious that we cannot choose x (binary 10), x 2 (binary 100), etc, because then the reminder is 0. We cannot chose a polynomial that is divisible by x, for example x 2 + x (binary 110). However, we can choose x + 1 (binary 11) because it is not divisible by x, but is divisible by x + 1. We can also choose x 2 + 1 (binary 101) because it is divisible by x + 1 (binary division).

41. How CRC Operates The sender wants to send k bits message The sender and the receiver must agree in advance on n+1 bit string called generator polynomial (divisor), G. ○G can be represented as n-degree polynomial n redundant bits are added to the k bits message. They are called CRC bits. Data bits to be sent CRC bits k bitsn bits

42. How CRC Operates (Cont.) The redundant bits are chosen in such a way that the resulting k+n bit string is exactly divisible (with a reminder=0) by G using modulo 2 arithmetic. The receiver divides the received data together with the CRC bits by G using modulo 2 arithmetic. ○If the reminder is 0, then the string is considered to be without errors ○If the reminder is not 0, the data unit is with errors and it is rejected

43. Example 6 The CRC-12 x 12 + x 11 + x 3 + x + 1 which has a degree of 12, will detect all burst errors affecting an odd number of bits, will detect all burst errors with a length less than or equal to 12, and will detect, 99.97 percent of the time, burst errors with a length of 12 or more.

44. 10.12 Checksum

45. 10.13 Data unit and checksum

46. The sender follows these steps: The unit is divided into k sections, each of n bits.The unit is divided into k sections, each of n bits. All sections are added using one’s complement to get the sum.All sections are added using one’s complement to get the sum. The sum is complemented and becomes the checksum.The sum is complemented and becomes the checksum. The checksum is sent with the data.The checksum is sent with the data. Note:

47. Example 10 1 1 1 1 00110 10001 11001 10111 1000111 10 01001 10110 n=5 bit checksum. 20 bit data gives k=20/n=4 sections (rows). The addition starts in the last column. The bits are carried in the columns before. The 6 th and 7 th bit are added The sum is complemented Carry from column 1 Column: 7 6 5 4 3 2 1 Carry from column 2 Carry from column 3 Carry from column 4 Carry from column 5 Sum Checksum

48. Example 7 Suppose the following block of 16 bits is to be sent using a checksum of 8 bits. 10101001 00111001 The numbers are added using one’s complement 10101001 00111001 ------------ Sum 11100010 Checksum 00011101 The pattern sent is 10101001 00111001 00011101

49. Example 8 Now suppose the receiver receives the pattern sent in Example 7 and there is no error. 10101001 00111001 00011101 When the receiver adds the three sections, it will get all 1s, which, after complementing, is all 0s and shows that there is no error. 10101001 00111001 00011101 Sum11111111 Complement 00000000 means that the pattern is OK.

50. The receiver follows these steps: The unit is divided into k sections, each of n bits.The unit is divided into k sections, each of n bits. All sections are added using one’s complement to get the sum.All sections are added using one’s complement to get the sum. The sum is complemented.The sum is complemented. If the result is zero, the data are accepted: otherwise, rejected.If the result is zero, the data are accepted: otherwise, rejected. Note:

51. Example 9 Now suppose there is a burst error of length 5 that affects 4 bits. 10101111 11111001 00011101 When the receiver adds the three sections, it gets 10101111 11111001 00011101 Partial Sum 1 11000101 Carry 1 Sum 11000110 Complement 00111001 the pattern is corrupted.

52. Felrättande koder FEC = Forward Error Correction = Felrättande koder. Två typer: 1.Faltningskoder (convolutional codes). Ex:Vid Faltningskod med kodtakt (code rate) 1/3 infogas en redundant bit mellan varje bit i nyttomeddelandet. 2.Blockkoder Ex: Vid Read Salomon-kod med beteckningen RS(128,16), har man 128 bytes nyttodata och 16 bytes redundant felupptäckande kod i varje block.