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12. The wireless LANs that we studied used protocols such as MACA instead of
总共 2 n-1 个父节点,所以, using CSMA/CD. Under what conditions, if any, would it be possible to use
CSMA/CD instead?(E) 因为 2n >>1,所以 p≈2- n
在共享父节点的条件下遍历树,从第二级开始每一级访问两个节点,这样遍历树 如果所有站的发射有效范围都很大,以至于任一站都可以收到所有其他站发送的 所走过的节点总数 n1= 1+2+…+2+2=1=2n,接下来,我们考察两个发送站共享祖父 信号,那么任一站都可以与其他站以广播方式通信。在这样的条件下,CSMA/CD
可 节点的概率 p2和遍历树所走过的节点总数 n2。此时在每个父节点下面仅可能有一
以工作的很好。 个
站发送。两个发送站共享一个指定的祖父节点的概率是 1/ C22n-1。
共有 2n-2 个祖父节点
遍历树比 1 n
减少两个节点,即
级祖先
13. What properties do the WDMA and GSM channel access protocols have in common? See Chap. 2 for GSM. (E) 两种协议都使用 FDM 和 TDM 结合的方法,它们都可以提供专用的频道(波长),
并且都划分时隙,实现 TDM。 14. Six stations, A through F, communicate using the MACA protocol. Is it
possible that two transmissions take place simultaneously? Explain your answer.(E) 是的。想像一下它们都在一条直线上并且每个站都只能连到它最近的邻居,那么 A 可以发送给 B 同时 E 正发送给 F
通过类似的分析和计算,可以得到,两个发送站共享曾祖父节点(属 n-3
节点)的概率是 p3= 2-n+2
遍历树所经过的节点总数比 n2又少两个节点,
15. A seven-story office building has 15 adjacent offices per floor. Each office contains a wall socket for a terminal in the front wall, so the sockets form a rectangular grid in the vertical plane, with a separation of 4 m between sockets, both horizontally and vertically. Assuming that it is feasible to run a straight cable between any pair of sockets, horizontally, vertically, or diagonally, how many 因此,最坏的情形是 2n+1 个时隙(共享父节点),对应于 i=0;
最好的情形是 3 个时隙,对应于 i=n-1 (两个发送站分别位于左半树和右半meters of cable are needed to connect all sockets using 树),
l. (a) a star configuration with a single router in the middle? m. (b) an 802.3 LAN? (E)
(a) 从一到七层记数。在星形配置中,路由器在第四层中央。需要铜线的站个数 7
所以平均时隙数等于
*15 – 1=104 sites. 这些铜线的总长度 =1832 meters.
,总共
该表达式可以简化为
(b) 对 802.3, 7 水平铜线每层需要 56 m 长,加上竖直方向的共 24 m 416 m.
16. What is the baud rate of the standard 10-Mbps Ethernet?(E)
最小为 3 个,最大为 n+2 个,给出的答案最大不对。
最大的情况为两个站点兄弟,共父母。 两个黄色的园代表要发数据的站点
所以在这种情况下的时槽为:数的路径长度加 2 即: n+2
以太网使用曼彻斯特编码,这就意味着发送的每一位都有两个信号周期。标准以 太网的数据率为 10Mb/s,因此波特率是数据率的两倍,即 20MBaud。 17. Sketch the Manchester encoding for the bit stream: 0001110101.(E)
信号是一个二植方波高 (H) 和 低(L),形式为 LHLHLHHLHLHLLHHLLHHL.
18. Sketch the differential Manchester encoding for the bit stream of the previous problem. Assume the line is initially in the low state.(E)
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对于 1km 电缆,单程传播时间为 1/200000=5×10-6 s=5 ,往返传播时间为 2t
工作,最小帧19. A 1-km-long, 10-Mbps CSMA/CD LAN (not 802.3) has a propagation speed of =10 。为了能够按照 CSMA/CD 。以 1Gb/s 200 m/μsec. Repeaters are not allowed in this system. Data frames are 256 bits long, 的发射时间不能小于 10
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including 32 bits of header, checksum, and other overhead. The first bit slot after a 速率工作,10 可以发送的比特数等于:(10*10)/(1*10)=10000
因此,最小帧是 10000 bit = 1250 字节长。 successful transmission is reserved for the receiver to capture the channel in order to send a 32-bit acknowledgement frame. What is the effective data rate, excluding 22. An IP packet to be transmitted by Ethernet is 60 bytes long, including all its overhead, assuming that there are no collisions?(M) headers. If LLC is not in use, is padding needed in the Ethernet frame, and if so, -6依题知一公里的在铜缆中单程传播时间为 1/200000=5×10 s=5 usec,往返传播时 how many bytes?(E) 间为 2t =10 usec,一次完整的传输分为 6 步: 最小的以太网帧是 64 bytes,包含了以太网地址帧头,类型/长度域,以及校验发送者侦听铜缆时间为 10usec,若线路可用 发送数据帧传输时间为 256 bits / 10Mbps = 25.6 usec 数据帧最后一位到达时的传播延迟时间为 5.0usec 接收者侦听铜缆时间为 10 usec,若线路可用 和。 由于帧头域占用 18 bytes,并且分组是 60 bytes,总帧长是 78 bytes,这已经超过了 64-byte 的最小限制。 因此,不必再填充 z 了。 形式为 HLHLHLLHHLLHLHHLHLLH.
23. Ethernet frames must be at least 64 bytes long to ensure that the transmitter is still going in the event of a collision at the far end of the cable. Fast Ethernet has the 接收者发送确认帧用时 3.2 usec same 64-byte minimum frame size but can get the bits out ten times faster. How is it 确认帧最后一位到达时的传播延迟时间为 5.0 usec possible to maintain the same minimum frame size?(E) 总共 58.8sec,在这期间发送了 224 bits 的数据,所以数据传输率为 3.8 Mbps. 将快速以太网的电缆长度至为以太网的 1/10 即可。 20. Two CSMA/CD stations are each trying to transmit long (multiframe) files. 24. Some books quote the maximum size of an Ethernet frame as 1518 bytes After each frame is sent, they contend for the channel, using the binary exponential instead of 1500 bytes. Are they wrong? Explain your answer.(E) backoff algorithm. What is the probability that the contention ends on round k, and 以太网一帧中数据占用是 1500 bytes,但是把目的地地址,源地址,类型/长度域 what is the mean number of rounds per contention period?(M) 以及校验和域也算上,帧总长就为 1518 bytes 把获得通道的尝试从 1 开始编号。第 i 次尝25. The 1000Base-SX specification states that the clock shall run at 1250 MHz, 试分布在 2i-1个时隙中。因此,i 次 even though gigabit Ethernet is only supposed to deliver 1 Gbps. Is this higher speed 尝试碰撞的概率是 2-(i-1),开头 k-1 次尝试失败,紧接着第 kto provide for an extra margin of safety? If not, what is going on here?(E) 次尝试成功的概率是: 传输数据用 10 位来表示 8 位的真实数据,编码的利用率是 80%,一秒钟可以
即:
传 送 1250 mb 的数据,相当于 125*106码字。 位有效数据,所 以实际的数据传输率是 1000 mb/sec. 每个码字代表的是 8上式可简化为:
所以每个竞争周期的平均竞争次数是∑?kpk
26. How many frames per second can gigabit Ethernet handle? Think carefully and take into account all the relevant cases. Hint: the fact that it is gigabit Ethernet matters.(E)
最小的以太网帧是 64bytes = 512 bits,所以依题 1 Gbps 的带宽可得 1,953,125
21. Consider building a CSMA/CD network running at 1 Gbps over a 1-km cable
=2
with no repeaters. The signal speed in the cable is 200,000 km/sec. What is the *106 frames/sec,然而,这只是在充满最小的帧时是这样,如果没有充满帧,填充短 minimum frame size?(E)
帧至 4096 bits,这时每秒处理的帧的最大数量为 244,140 bytes,对于最大的帧长
12,144 bits,每秒处理的帧的最大数量为 82,345 frames/sec.
27. Name two networks that allow frames to be packed back-to-back. Why is this
feature worth having?(E)
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