计算机网络题目集锦

计算机网络题目集锦

第一章

↓在下列选项中，不属于网络体系结构所描述的内容是

A. 网络的层次

B. 每一层使用的协议

C. 协议的内部实现细节

D. 每一层必须完成的功能

↓把一个长度为100MB的数据块，通过带宽为1Mb/s的光纤信道，传送到1000km远的计算机上，求数据传送的总时延。假定信号在光纤中的传播速率为2.0*105km/s,忽略处理时延。若传送距离减小为1km，数据传送的总时延是多少？

↓把一个长度为1Byte的数据块，通过带宽为1Mb/s的光纤信道，传送到1000km远的计算机上，求数据传送的总时延。假定信号在光纤中的传播速率为2.0*105km/s,忽略处理时延。若带宽提高到1Gb/s，数据传送的总时延是多少？

↓在如图所示的采用“存储转发”方式的分组交换网中，所有链路的数据传输速率为100Mbps，分组大小为1000B，其中分组头大小为20B。若主机H1向主机H2发送一个大小为980000B 文件，则在不考虑分组拆装时间和传播延迟的情况下，从H1发送开始到H2接收完为止，需要的时间至少是多少?

18、

补充题：

试在下列条件下比较电路交换和分组交换。要传送的报文共X(bit),从源站到目的站共经过K 段链路，每段链路的传播时延为d 秒，数据率为b （bits/s ）.在电路交换时电路的建立时间为s 秒。在分组交换时分组的长度为p(bit),且各结点的排毒的等待时间可忽略不计，问：怎样的条件下，分组交换的时延比电路交换的时延要小？

第三章

停止等待协议

↓Problem: B sends a NAK frame back to A, after having received a data frame with errors. What happens if A always gets NAK frames ？

Solution: set a max. number for retransmission times, e.g. 8. If not successful, gives an error report to the above layer.

↓Problem: Due to poor link conditions, the frame sent by A doesn't get B at all. It gets lost ! In this case, A will never get any response from the peer.

Solution: schedule a timeout timer 超时计时器 to expire at some time after the ACK should have been returned. If the timer goes off, retransmit the frame.

↓

Problem: Retransmissions may introduce duplicate 双倍的 frames received by B, Why?

Solution: assign sequence numbers 序号 for every frame, so that B can distinguish between new frames and old copies. However, an ACK for the duplicated frame is still necessary! Why?

↓

A channel has a bit rate of 4 Kbps and a propagation delay of 20 ms.

Question: For what range of frame sizes does the stopand wait protocol give an efficiency of 50%? （见右图）

↓

Quiz: What is the size of the sending window for a simple Stop-and-Wait Protocol? Answer:

one.

magine a link that uses a geo-stationary satellite同步卫星:The data rate is 50-kbps.The round-trip delay is 500ms.What is the link utilization, if you use stopandwait protocol to send 1000-bit frames？

If we use stopandwait protocol to send a 1000-bit frame, the receiver will get

the whole frame 270msec later，The acknowledgement will take a further 250msec to get back. Out of 520msec, data is only being sent for 20msec. Only 20/520 ≈4% of the link’s capacity is being utilised.

第四章

Cable length d= 400m, Data rate R = 10 Mbit/sec, Propagation speed V= 2x108 m/sec ，The End-to-end propagation delay time t prop =？ The round-trip propagation delay is, of course, twice the end-to-end delay.

Thus the round trip delay is

With a data rate of each bit has

duration The number of bits we can fit into a round-trip propagation delay is The minimum frame length is thus 40 bits (5 bytes).

A margin of error 误差幅度 is usually added to this (often to make it a power of 2) so we might

use 64 bits (8 bytes).

The 802.3 standard frame length is at least 512 bits (64 bytes) long, which is much

longer than our minimum requirement of 64 bits (8 bytes).

sec 210210

240068μ=?=?==-V d t prop sec 42μ=?pr op t Mbps

R 10=sec

1.0000,000,1011μ===R t b bits

t t n b p b 401.042==?

=

↓Quiz: A Ethernet consists of 5 segments connected by 4 repeaters. Each segment is 500 meter long. The signal propagation speed is 2x108 m/sec. The delay caused by each repeater is 2.5 μsec.

There are two hosts, A and B, which are located at far ends of the LAN.

Please calculate the signal propagation delay from host A to B ？

↓Two nodes are communicating using CSMA/CD protocol.

Transmission rate is 100 Mbits/sec and frame size is 1500 bytes. The propagation speed is 3*108 m/sec.

Calculate the distance between the nodes such that the

time to transmit the frame = time to recognize that the collision have occurred.

Solution ：

↓

The data link layer consist of two sublayers, i.e. LLC sublayer and MAC sublayer.

↓In order to work out the minimum frame length for LANs, we have to take network scale, type of media, number of repeaters into account.

↓某局域网采用CSMA/CD 协议实现介质访问控制，数据传输速率为10Mbps ，主机甲和主机乙之间的距离为2km ，信号传播速度是200 000km/s 。请回答下列问题，要求说明理由或写出计算过程。

（1）若主机甲和主机乙发送数据时发生冲突，则从开始发送数据时刻起，到两台主机均检测到冲突止，最短需经过多长时间？最长需经过多长时间？（假设主机甲和主机乙发送数据的过程中，其它主机不发送数据）

（2）若网络不存在任何冲突与差错，主机甲总是以标准的最长以太网数据帧（1518字节）向主机乙发送数据，主机乙每成功收到一个数据帧后立即向主机甲发送一个64字节的确认帧，主机甲收到确认帧后方可发送下一个数据帧。此时主机甲的有效数据传输速率是多少？（不考虑以太网帧的前导码）

46102.110

1008

1500-?=??==R L t frame prop frame trip round t

t T ?=→2_5

41062102.12--?=?==frame prop t t V d

t prop =()()

km V t d prop 1810181031063

85=?=???=?=-

、当A向B发送数据时，C也要向B发送信息。

但A和C相距太远，C因检测不到A的信号而认

为自己可以发送。C是A的隐蔽发送站。

隐蔽发送站问题的解决办法：信道预约

1.A向B发送数据帧之前，先发一个控制帧RTS (Request To Send请求发送)。C收

不到此帧；

2.如果B处信道空闲，则回答一个控制帧

CTS(Clear To Send允许发送). A和C均可

收到.

3.A收到CTS后，开始发送数据帧，C回避。

第五章

↓To which class do the following IP addresses belong ?

(1) 128.36.199.3 (2) 21.12.240.17

(3) 183.194.76.253 (4) 192.12.69.248

↓

↓ IP Address: 130.97.16.132Subnet Mask: 255.255.255.192 Net-id=? Host-id=?

↓

↓A set of IP address assignments.

若路由器R因为拥塞丢弃IP分组，则此时R可向发出该IP分组的源主机发送的ICMP报文类型是c

A.路由重定向

B. 目的不可达

C.源抑制

D. 超时

38.convert the Ip address whose hexadecimal representation is C22F1582 to dotted decimal notation

（C22F1582）16=（1100 0010 0010 1111 0001 0101 1000 0010）2

地址是：194.47.21.130

39.A network on the Internet has a subnet mask of 255.255.240.0。what is the maximum number of hosts it can handle？

查表可得255.255.240.0 的前缀长为20

所以主机数=2 32-20=2 12=4096

41. A router has just received the following new IP address：57.6.96.0/21,57.6.104.0/21 ，57.6.104.0/21and57.6.120.0/21，If all of them use the same outgoing line，can they be aggregated？if so，to what ，if not ，why not？

57=（0011 1001）26=（0000 0110）296=（0110 0000）2

104=（0110 1000）2 112=（0111 0000）2 120=（0111 1000）2

共同前缀19位0011 1001 0000 0110 011

聚合的CIDR地址为57.6.96.0/19

42.The set of Ip address from 29.18.0.0 to 19.18.128.255 has been aggregated to 29.18.0.0/17。However，there is a gap of 1024 unassigned address from 29.18.60.0 to 29.18.63.255 that are now suddenly assigned to a host using a different outgoing line 。is it now necessary to split up the aggregate address into its constituent blocks，add the new block to the table，and then see if any reaggregation is possible？if not ，what can be done instead？

43.A router has the following（CIDR）entries in its routing table：

Address/mask next hop

135.46.56.0/22 interface 0

135.46.60.0/22 interface 1

192.5340.0/23 Router 1

Default Router 2

For each of the following IP address，what does the router do if packet with that address arrives？

（a）135.46.63.10 （b）135.46.57.14 （c）135.46.52.2（d）192.53.40.7（e）192.53.56.7

第六章

设TCP使用慢开始和拥塞避免实现拥塞控制，TCP 的阈值sstresh的初始值为12(单位为报文段)。当拥塞窗口上升到16时网络发生了超时，试分别求出第1轮次到第15轮次传输的各拥塞窗口的大小。

TCP的拥塞窗口cwnd大小与传输轮次的关系如下表所示：

（1）试画出拥塞窗口与传输轮次的关系曲线图

（2）指明TCP工作在慢开始阶段的时间间隔

（3）指明TCP工作在拥塞避免阶段的时间间隔

（4）在第16轮次和第22轮次之后发送方是通过收到三个重复确认还是通过超时检测到丢失了报文段？

（5）在第1轮次、第18轮次和第24轮次发送时，阈值ssthresh分别被设置为多少？（6）在第几轮次发送出第70个报文段？

（7）假定在第26轮次之后收到了三个重复的确认，因而检测出了报文段的丢失，那么拥塞窗口cwnd和阈值ssthresh应设置多大？