计算机网络第四版习题答案(中文版)(21)

计算机网络第四版习题答案(中文版)

第 4 章 介质访问子层

4-1 For this problem, use a formula from this chapter, but first state the formula. Frames arrive randomly at a 100-Mbps channel for transmission. If the channel is busy when a frame arrives, it waits its turn in a queue. Frame length is exponentially distributed with a mean of 10,000 bits/frame. For each of the following frame arrival rates, give the delay experienced by the average frame, including both queueing time and transmission time.

(a) 90 frames/sec. (b) 900 frames/sec. (c) 9000 frames/sec.

(time delay, T // a channel of capacity C bps // with an arrival rate of

frames/sec)

这个公式是4.1.1段落给出的Markov排队问题的标准公式。也就是，。这里C =108 、

，

sec。对于这三种到达速率，我们得出的是(a) 0.1 msec,(b) 0.11 msec, (c) 1 msec. 对于c

的情况，我们操作一个带来10倍延迟的排队系统

4-2 A group of N stations share a 56-kbps pure ALOHA channel. Each station outputs a 1000-bit frame on an average of once every 100 sec, even if the previous one has not yet been sent (e.g., the stations can buffer outgoing frames). What is the maximum value of N?

N个站共享一个56kbps的纯ALOHA信道。每个站平均每100秒输出一个1000位的帧，及时前面的帧还没有被送出，它也这样进行（比如这些站可以将送出的帧缓存起来）。请问N的最大值是多少？

答：对于纯的ALOHA，可用的带宽是0.184×56 Kb/s =10.304 Kb/ s。每个站需要的带宽为1000/100=10b/s。而N=10304/10≈1030 所以，最多可以有1030 个站，即N 的最大值为1030。

4-3 Consider the delay of pure ALOHA versus slotted ALOHA at low load. Which one is less? Explain your answer.

答：对于纯的ALOHA，发送可以立即开始。对于分隙的ALOHA，它必须等待下一个时隙。这样，平均会引入半个时隙的延迟。因此，纯ALOHA 的延迟比较小。

4-4 Ten thousand airline reservation stations are competing for the use of a single slotted ALOHA channel. The average station makes 18 requests/hour. A slot is 125 µ

sec. What is the approximate total channel load?

每个终端每200（=3600/18）秒做一次请求，总共有10 000 个终端，因此，总的负载是200 秒做10000 次请求。平均每秒钟50 次请求。每秒钟8000 个时隙，所以平均每个时隙的发送次数为50/8000=1/160。

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