计算机网络第三章作业
1. An upper-layer packet is split into 10 frames, each of which has an 80% chance of arriving undamaged. If no error control is done by the data link protocol, how many times must the message be sent on average to get the entire thing through?
6. To provide more reliability than a single parity bit can give, an error-detecting coding scheme uses one parity bit for checking all the odd-numbered bits and a second parity bit for all the even-numbered bits. What is the Hamming distance of this code?
17. In the discussion of ARQ protocol in Section 3.3.3, a scenario was outlined that resulted in the receiver accepting two copies of the same frame due to a loss of acknowledgement frame. Is it possible that a receiver may accept multiple copies of the same frame when none of the frames (message or acknowledgement) are lost?
18. A channel has a bit rate of 4 kbps and a propagation delay of 20 msec. For what range of frame sizes does stop-and-wait give an efficiency of at least 50%?
30. Consider an error-free 64-kbps satellite channel used to send 512-byte data frames in one direction, with very short acknowledgements coming back the other way. What is
the maximum throughput for window sizes of 1, 7, 15, and
127? The earth-satellite propagation time is 270 msec.
3-1
10
P=0.8=0.107
E??ip(1?p) p?1ni-1
=1?p1?9.3 0.107 3-6
奇数位的任何一个位错误都可以改变奇偶校验位，与偶数位相同，两个错误发生在奇数位或偶数位，不会改变奇偶校验位,也意味着Hamming的距离为2。
3-17
比特流是10011101生成多项式1001生成多项式阶位是3，所以后面补3个0得到10011101000用10011101000整除1001得到余数100，所以最后位串为10011101100.如果第三位变反后，得出的数字不能整除1001.但如果出错后的位串仍能被1001整除，则错误无法被检测出来，最简单的就是所有位都为0.
3-18
要使效率达到50%，必须使传输数据包的时间等于来回的传输延迟。在4kbps的速率下，160bit需要40ms，也就是2倍的传输延迟。所以帧大于160bit时，才能获得50%的效率。
3-30
假设开始时间为t=0，第一个帧发送时间为512/64=8ms。t=8+270=278ms时到达。T=278+270=548ms时，ACK返回。此时的数据吞吐量为512/548=934bps。 若窗口大小为7时，在548ms内，可以传输
7*512=3548bit，吞吐量为3548/548=6.54kbps。 若窗口大小为15时，在548ms内，可以传输15*512=7680bit，吞吐量为7680/548=14kbps。
若窗口大小为127时，已经超出卫星信道全速，所以吞吐量为64kbps。