Solutions for Chapter similar to 1 and 3

Transcription

1 Solutions for Chapter similar to 1 and In a 5-bit sequence with 32 codes, there are 8 codes that start with 00 and there are 8 codes that ends with 00. Between them and are common. So the restricted number of codes is 8+8-2=14. In the other words, the number of allowed codes is 32-14=18. Therefore, we can map all 4-bit sequences to the 5-bit sequences with the restrictions in the question. 5. The stuffed bits (zeros) are in bold (page 88): 6. 7.

2 8. According BISYN protocol in pags 85: DLE, DLE, DLE, ETX, ETX 9, 10 : NOT REQUIRED 11. Three bit errors are spread between one, two or three bytes. In all cases, there is at least one byte with an odd number of bit errors which can be detected by the parity bit of that byte. 12. Consider two bytes A =a7.. a1 a0 and B=b7.b1 b0. Any 4-bit error in the following form is not detected by 2-D parity. If Bits..ai..aj of A and bits..bi...bj.. of B are corrupted. This pattern of errors is not detected by both row and column parity bits. 13. If we know that only one bit is corrupted, then 2-D tells us its corresponding row and column. 14, 15, 17. NOT REQUIRED 16. with 16-bit words: If low order byte is decreased we can decrement the sum by 1. If high order byte is decreased we can decrement the sum by Similar to , 21: NOT REQUIRED

3 22. assume a NAK (Not Acknowledge) frame is sent by the receiver when an out of order frame is received. The receiver should maintain a timeout timer for this frame.

4 Consider that sender transmits a frame and then waits for a while, and also the frame is lost. In this case, the receiver has no way to know that the frame has been lost. At the end of transmission, the sender is unsure that all frames have been sent. 23. a) propagation delay = 20*10^3 / 2*10^8 = 100 microsecond b) RTT = 2*100 = 200 microsecond c) We did not consider the processing delay by the receiver. In other words, the receiver might answer with delay so that the sender times out and retransmits. 24. similar to C = 1Mbps, propagation delay = 3*10^4*10^3 m /3*10^8 m/s = 0.1 s RTT = 0.2 s SWS = C*RTT / (frame length) = 1*10^6 bit/s * 0.2 s / 1000*8 bit = 25 frames a) RWS = 1 MaxSeqNum SWS+1 = 26 5 bits are needed. b) RWS=SWS MaxSeqNum 2*SWS = 50 6 bits are needed. 26. If the receiver delays sending ACK until buffer space is available, the sender times out and retransmits unnecessarily. This could result in more congestion. 27.

5 28, 29, 30. NOT REQUIRED 31.

6 The worst case happens when ACK[0] is lost. In this case, at the end of the first sending window size, the sender retransmits DATA[0] instead of DTAT[5]. At this moment, the receiver window is DATA[5], DATA[6], DATA[7]. To avoid the mistake that the receiver takes DATA[0] as a new frame, 7 and 0 must have different codes. In other words, we need at least 8 sequence numbers. b) As in (a).

7 36. c) Based on the results from (a), MaxSeqNum SWS+RWS

8 37, 38.

9 39. Ethernet has a minimum frame size of 64 bytes. Packets smaller than this size should be padded out with extra data. Protocols above Ethernet should be able to distinguish such padding from actual data. 40. If the two hosts are in the same collision domain, all messages addressed to the common hardware address will be received by both hosts. If the higher-layer protocol has no way to determine which messages were actually intended for a host, this will cause problems at the higher layer. If the higher layer can tell which messages were intended for a host, then the only problem is additional load on each host at the link and network layers. 41.

10 42. NOT REQUIRED 43. a) C*RTT = minimum sent data. 10Mbps * 46.4 microsecond = 464 bits If C=100Mbps minimum sent data = 4640 bits minimum packet size = bit jamming = 586 bytes. b) This packet size is larger than many higher level packet sizes, resulting in considerable wasting of bandwidth due to padding with extra data in data link layer. c) The minimum packet size could be smaller if the collision domain could be smaller (the worst distance between transmitter and receiver). This results in smaller propagation delay. 44. c) NOT REQUIRED d) B1 will be dropped and B will try the next frame B : NOT REQUIRED

11 52) Bandwidth = transmit data / total transmit time = 1 packet / (N/2 + 5) time slots 0.25 packet / time slot N NOT REQUIRED