Virtual Circuit Switching-Data Communication Systems-Assignment Solution, Exercises of Data Communication Systems and Computer Networks

Download Virtual Circuit Switching-Data Communication Systems-Assignment Solution and more Exercises Data Communication Systems and Computer Networks in PDF only on Docsity! CHAPTER 18 Virtual Circuit Switching: Frame Relay and ATM Solutions to Odd-Numbered Review Questions and ExercisesReview Questions 1. Frame Relay does not use flow or error control, which means it does not use the sliding window protocol. Therefore, there is no need for sequence numbers. 3. T-lines provide point-to-point connections, not many-to-many. In order to connect several LANs together using T-lines, we need a mesh with many lines. Using Frame Relay we need only one line for each LAN to get connected to the Frame Relay network. 5. Frame Relay does not define a specific protocol for the physical layer. Any proto- col recognized by ANSI is acceptable. 7. A UNI (user network interface) connects a user access device to a switch inside the ATM network, while an NNI (network to network interface) connects two switches or two ATM networks. 9. An ATM virtual connection is defined by two numbers: a virtual path identifier (VPI) and a virtual circuit identifier (VCI). 11. In an UNI, the total length of VPI+VCI is 24 bits. This means that we can define 224 virtual circuits in an UNI. In an NNI, the total length of VPI+VCI is 28 bits. This means that we can define 228 virtual circuits in an NNI. Exercises 13. We first need to look at the EA bits. In each byte, the EA bit is the last bit (the eight bit from the left). If EA bit is 0, the address ends at the current byte; if it 1, the address continues to the next byte. Address → 10110000 00010111 The first EA bit is 0 and the second is 1. This means that the address is only two bytes (no address extension). DLCI is only 10 bits, bits 1 to 6 and 9 to 12 (from left). 1 docsity.com 2 Address → 10110000 00010111 DLCI → 1011000001 → 705 15. We first need to look at the EA bits. In each byte, the EA bit is the last bit (the eight bit from the left). If the EA bit is 0, the address ends at the current byte; if it 1, the address continues to the next byte. Address → 0x7C74E1 → 01111100 01110100 11100001 The first two EA bit are 0s and the last is 1. This means that the address is three bytes (address extension). DLCI is 16 bits, bits 1 to 6, 9 to 12, and 17 to 22. Address → 01111100 01110100 111000 DLCI → 0111110111111000 → 32248 17. See Figure 18.1. 19. In AAL1, each cell carries only 47 bytes of user data. This means the number of cells sent per second can be calculated as [(2,000,000/8)/47] ≈ 5319.15. 21. a. In AAL3/4, the CS layer needs to pass 44-byte data units to SAR layer. This means that the total length of the packet in the CS layer should be a multiple of 44. We can find the smallest value for padding as follows: H + Data + Padding + T = 0 mod 44 4 + 47,787 + Padding + 4 = 0 mod 44 Padding = 33 bytes b. The number of data unit in the SAR layer is (4 + 47787 + 33 +4) / 44 = 1087 c. In AAL3/4, the number of cells in the ATM layer is the same as the number of data unit in the SAR layer. This means we have 1087 cells. 23. a. The minimum number of cells is 1. This happens when the data size ≤ 36 bytes. Padding is added to make it exactly 36 bytes. Then 8 bytes of header cre- ates a data unit of 44 bytes at the SAR layer. Figure 18.1 Solution to Exercise 17 A B DLCI: 45 DLCI: 56 DLCI: 78 DLCI: 80 DLCI: 66 DLCI: 72docsity.com