计算机网络试卷及答案-英文版

HUST Examination Answer Sheet

Course: Computer Networks (closed-book exam) Jan.2014 Department of Electronics and Information Engineering

Solution

[Question 1] Network architecture

(1) Draw the architecture diagrams of ISO/OSI reference model and TCP/IP network. (2) State the differences between the two architectures.

(3) Mark the names of data unit and network equipment in the bottom 4 layers of ISO/OSI diagram.

Solution:

（1）ISO/OSI reference model TCP/IP model

Differences:

a) TCP/IP model is the actual architecture in the Internet, while OSI model is a theoretical

architecture.

b) TCP/IP model does not imply strict layering, while OSI model implies strict layering. c) In TCP/IP model, IP serves as the focal point for the architecture.

d) TCP/IP model allows for arbitrarily many different network technologies , ranging from

Ethernet to wireless to single point-to-point links.

e) TCP/IP model emphasize implementations of proposed protocols. (2)

Physical : bit, repeater or hub

Data link : frame, Ethernet switch or bridge Internet : packet, router

Transport : message, gateway

[Question 2] Principles of network design

Select ONE of the following principles, tell its main ideas and provide an example. (1) Keep it simple and stupid

(2) Complex edge and simple core (3) Smart sender and dumb receiver

Solution:

(1) KISS: Keep It Simple and Stupid

It means that you should make things simple in the designing. One example following it : Ethernet

(2) Complex edge and simple core

It means that the hosts are very complex and have many functions while the nodes are very simple and have few functions.

One example following it : The design of router, or the functions of TCP and IP

(3) Smart sender and dumb receiver

It means that the function of sender is more complex than that of the receiver, which is help to improve the robustness and performance of communication protocol. One example following it: The flow control of TCP protocol

[Question 3] Error detection

(1) Tell the main idea of error detection and error correction in communication.

(2) Given the CRC polynomial x 4 + x 3 + 1, if the original message is 10110011010, what is the CRC message to send?

(3) Suppose the first bit of the message in (2) is inverted due to the noise in transmission. How can the receiver detect it via CRC verification?

Solution:

(1)1087431()M x x x x x x x =+++++, 43()1C x x x =++. So k=4.

a) Multiply M(x) by 2k

to get 14

12

11

8

7

5

()T x x x x x x x =+++++,

b) Then divide T(x) by C(x) to get the remainder 0000.

c) The message that should be transmitted is 101100110100000.

(2)The message received is 001100110100000.Divide it by C(x), then the remainder is 1100.So it is not divisible by C(x).So the receiver knows that an error has occurred.

(1) What are the essential components to realize reliable transmission?

(2) Suppose two computers are communicated via Stop-and-wait protocol. The link bandwidth is 5kbps, and the one-way propagation delay is 20ms. To reach 80% or higher link utilization, what is the minimal frame size for this communication?

(3) If it is upgraded to Sliding-Window protocol. To reach the same goal with (2), what is the minimal window size, how many bits are required to describe the frame sequence in window? (Suppose the frame size is 1 or 100Byte)

Solution:

(1)ACK, and timer

Actual_throughput = Data / Total_Delay

Total_Delay = RTT + Data / BW

Link_Utilization = 100 * Actual_throughput / BW

So: Data/throughput = RTT + Data/BW

(BW/throughput - 1) Data = BW*RTT

Data = BW*RTT/(1/Utilization -1)

For the stop-and-wait protocol, for each RTT only one frame is sent,

Thus Frame_size = Data = BW*RTT/(1/Utilization - 1) = 5kbps * 40ms / (1/0.8 - 1)

= 200 bit / 0.25 = 800 bit = 100 Byte

If the students ignore the data transmission delay, their answer is

frame_size = BW*RTT*Utilization = 5000 bit/sec * 40 / 1000 sec * 0.8 = 160 bit = 20 Byte

In this case, at least -2 score.

(3)

For the sliding-window protocol, in each RTT the data window can be transmitted at most.

Thus Window_size = Data = BW*RTT/(1/Utilization - 1) = 5kbps * 40ms / (1/0.8 - 1)

= 200 bit / 0.25 = 800 bit = 100 Byte

If the frame size is 100Byte, 1 frames are allowed.

To indicate the frames in both sides of sender and receiver, the sequence number should describe 2 frames, thus the [log2(2) ]=1 bit

If the students ignore the data transmission delay, their answer is

window_size = BW*RTT*Utilization = 5000 bit/sec * 40 / 1000 sec * 0.8 = 160 bit = 20 Byte

If the frame size is 100Byte, 1 frames are allowed.

To indicate the frames in both sides of sender and receiver, the sequence number should describe 2 frames, thus the [log2(2) ]=1 bit

In this case, at least -2 score.

(1) What is the main idea of CSMA/CD (Carrier Sense Multiple Access/Collision Detection) in traditional Ethernet? Can it be deployed for wireless local network, why?

(2) Suppose one traditional Ethernet has 1km cable, the signal propagation speed is 2*105km/s, and the transmission rate is designed to be 100Mbps. What is the minimal frame size to support carrier sensing?

Solution:

(1) the main idea of CSMA/CD include two parts. One is carrier sensing, which means the node should detect the channel before sending any data. If the node finds the channel is idle, it begins to transmit. Otherwise, the node stop for next round. The second issues is collision detection, which means the node should detect the channel in the duration of its data transmission. If the node finds the channel become busy, or in other word, there is a collision, it should stop transmitting immediately.

CSMA/CD cannot be deployed in wireless LAN, because the wireless radio transmitter and receiver can not work in dual mode. The wireless node cannot detect collision when it is transmitting data.

(2)the minimal frame should be transmitted throughout the whole traditional Ethernet.

Thus t = frame_size / transmit_rate = 2 * cable_length / prop_speed.

Frame_size = 100 * 106 bits/sec * 2 * 103 m / (2 * 108)m/sec = 1000 bits = 125 Byte

[Question 6] Switched network

(1) What are the differences between circuit switching and packet switching?

(2) For the following linear topology network, each link has 2ms propagation delay and 4 Mbps bandwidth.

A B C D

If we use circuit switching, circuit setup requires a 1KB message to make one round-trip on the path, which incurs a 1ms delay at each switch after the message has been completely received. Then we can send the file as one contiguous bit stream. What is the delay for circuit switching to transmit n-byte from A to D?

(3) If we use packet switching in the network of (2), we can break the file into 1KB packets, which has 24byte header and 1000byte payload. The switch takes 1ms process delay after receiving the packet, and then sent it continuously. What is the condition for packet switching to have less delay performance than circuit switching?

Solution:

(1)

(2)

T pkt = Packet_Size / Bandwidth = 1 KB / 4Mbps = 1024 *8 / (4 * 106) = 2.048 ms

T p = 2 ms,

T s = 1 ms,

T t = n B / 4Mbps

In circuit switching, Total time duration:

D = Singling_Delay + Transmission_Delay = 2 * Packet_Duration + Transmission_Delay

= 2 * (T pkt * 2 + T p * 3 + T s * 2 + T pkt) + (T p * 3 + T t )

= T pkt * 6 + T p * 9 + T s * 4 + T t

Thus, D = 2.048 * 6 + 2 * 9 + 1 * 4 + n * 8 bits / 4Mbps

= 34.288 (ms) + 2n (us)

(3)

Main idea The reliability provided by end

host

The reliability provided by the

network

Information in packet Every packet has its dest-addr

in header

Every packet has its temp VCI

locally

Forwarding action in

switch Every packet was treated

independently

The packets are processed in

the manner of VC

Packets received in

destination

Not in sequence In sequence

T pkt = Packet_Size / Bandwidth = 1 KB / 4Mbps = 1024 *8 / (4 * 106) = 2.048 ms

T p = 2 ms,

T s = 1 ms,

T t = 1.024 * n B / 4Mbps

In packet switching, Total time duration:

D = Delay_at_Switch * Switch_Num + Delay_at_last_hop

= (T p + T pkt +T s)*2 + (T p + T t )

= T pkt * 2 + T p * 3 + T s * 2 + T t

Thus, D = 2.048 * 2 + 2 * 3 + 1 * 2 + 1.024 * n B / 4Mbps

= 12.096 (ms) + 2.048 n (us)

In order to make

34.288 (ms) + 2n (us) > 12.096 (ms) + 2.048 n (us)

Thus 22.192(ms) > 0.048 n(us)

n < 22.192 * 1000 / 0.048 = 462333 Byte = 451.5 KB

[Question 7] Ethernet Switch

(1) What are the differences between hub and switch in Ethernet?

(2) Suppose one server and nine clients are connected via hub in 10Mbps Ethernet, what is the maximal bandwidth for the client-server connection?

(3) If the hub is upgraded to switch, can the client-server connection obtain more bandwidth? if can, how much is it?

Solution:

(1)hub works in Layer2, it works as the shared media and relays the frames to all the nodes connecting to hub. Switch works in Layer2, it forwards the frames to the specific node according to the destination address embedded in the frame header.

(2)the max bandwidth in client-server connection is 10Mbps/(1+9)= 1Mbps

(3)the max bandwidth in client-server connection is 10Mbps/9 = 1.1111Mbps

Additional 0.1111 Mbps is obtained for each connection.

[Question 8] Router

(1) Somebody says that, ``the only difference between switch and router is that they do switch function based on the address in different layers.’’ Is it correct? Why?

(2) If we obtain the following information from one router. What kind of routing protocol does it

Solution:

(1)it is not fully correct. The part talking about the forwarding function is correct, while it is not the only difference. Another but not the last difference is that, router has more functions on control plane, which do routing and find the paths for packets.

(2)The routing protocol is RIP.

The routing table is:

Destination Next hop Interface

192.168.1.0/24 * Fa 0/4

192.168.2.0/24 * Fa 0/6

192.168.10.0/24 192.168.1.1 Fa 0/4

192.168.30.0/24 192.168.2.2 Fa 0/6

[Question 9] Routing algorithm

(1) State the main differences between distance vector routing and link state routing.

(2) For the network given in the following figure, provide the steps of forward search in Dijkstra algorithm for node A finding the shortest path to node E

Solution:

(1) The idea of distance vector routing is to tell the neighbors about the learned topology to all the nodes in the network. The idea of link state routing is to tell all the nodes in network about the neighborhood topology.

Step 1 2 3 4 5 6 7 8 9 Confirmed (A,0,-) (A,0,-)

(A,0,-)

(D,2,D)

(A,0,-)

(D,2,D)

(A,0,-)

(D,2,D)

(B,4,D)

(A,0,-)

(D,2,D)

(B,4,D)

(A,0,-)

(D,2,D)

(B,4,D)

(E,6,D)

(A,0,-)

(D,2,D)

(B,4,D)

(E,6,D)

(A,0,-)

(D,2,D)

(B,4,D)

(E,6,D)

(C,7,D) Tentative

(B,5,B)

(D,2,D)

(B,5,B) (B,4,D)

(E,7,D)

(E,7,D)

(E,6,D)

(C,8,D)

(C,8,D) (C,7,D)

A→E

A→D

→E

A→D

→B→E

A→D

→B→E

[Question 10] IP address allocation

A company has one C class IP address of 200.1.1.*. It has four departments.

(1) If each department has less than 25 computers. Provide a kind of IP address allocation. Give the network address, subnet mask, and the available IP address range for each department. (2) If the four departments have 72, 35, 34, 20 computers respectively. Provide the IP address allocation scheme again.

Solution:

(1)Since each department has less than 25 computers, even considering the additional two more IP address for gateway and broadcast, the 64-computer subnet is enough for them.

One IP address allocation scheme is to even divide the 256 IP addresses into 4 subnets, each subnet allows 64 hosts.

Another IP address allocation scheme is to even divide the 256 IP addresses into 4 subnets, each subnet allows 32 hosts.

(2) Since one of the department has more requirements than 64, then the even distribution scheme

[Question 11] TCP protocol

(1) Somebody says that, ``because of the reliable transmission service in layer 2, there is no need to provide such service again within TCP protocol in layer 4’’. Is it correct? Why?

(2) State the main rules of TCP connection setup according to the following figure. Explain every word and number in the figure.

Solution:

(1) It is wrong. TCP is based on the un-reliable IP layer, which only provides best effort service. If TCP wants to provide reliable transmission service, it has to realize this by itself.

(2)TCP use three hand-shakes to setup the connection.

According to the figure, there are two nodes. The sender is with IP address of 192.168.1.163 and the receiver 192.168.1.165 respectively.

●At first, the sender send a request ``SYN’’to the receiver to setup the connection. This

message is with the sequence number of 424CF1DC;

●Secondly, the receiver reply an acknowledgment message ``SYN/ACK’’ to the sender. This

message has two sequence numbers. The seq in the ACK is 424CF1DD, which is to confirm the last ``SYN’’ from sender. The seq in the SYN is 30318555, which is a new message from the receiver.

●Thirdly, the sender reply an acknowledgment message ``ACK’’ to the receiver. The seq in the

ACK is 30318556, which is to confirm the last ``SYN’’ from receiver.

In the end, both the sender and receiver knows that the other side is ready for this TCP connection.

[Question 12] TCP congestion control

(1) Both flow control and congestion control in TCP are realized by window based packet control. How can TCP get the window sizes in these two mechanisms?

(2) Assume that TCP implements an extension that allows window sizes much larger than 64 KB. Suppose that you are using this extended TCP over a 1Gbps link with a latency of 150ms. TCP packet size is 1KB, and the max receive window is 1 MB. Suppose there is no real congestion and packet loss in transmission. How many RTTs does it take until slow start opens the send window to 1 MB? How long does it take to send the complete file? ( Suppose file size is 10MB )

Solution:

(1) In flow control, TCP sender knows the window size by the field of advertise-window replied from the receiver. In congestion control, TCP sender learns the window size adaptively by AIMD( Additive Increase and Multiplicative Decrease) mechanism responding to the packet loss

event.

(2)

When TCP realizes congestion control mechanism, its effective send window size will be min (CongestionWindow, AdvertizedWindow). In original design of TCP header, the field of AdvertizedWindow is 16 bit, which is 216=26*210=64 KB. So the maximum effective window of original TCP sender is 64KB. The assumption of the first sentence in this question relaxes such constraint for TCP.

In slow start, the send window starts from w0=1 packet, which is 1 KB. For each RTT after a successful transmission, the window size will be doubled. After i RTT, it will be 2i * w0. Let 2i * 1KB = 1MB, so

i = log2(1MB/1KB) = log2(210) = 10.

It will take 10 RTTs to reach 1MB send window.

Case 1: if the receiver window remains as 1MB

In the first 10 RTTs, total (1 + 2 + 4 + … + 210) * 1KB has been transmitted.

Which is (211 - 1) * 1KB = 2 MB - 1 KB, the rest file is 10MB - (2MB - 1KB) = 8MB + 1KB

In the reset transmission, each RTT can only support 1MB transmission.

Thus, additional 9 RTTs are required. Total 19 RTT = 19 * 150ms = 2.85 s

Case 2: if the receiver window can be changed.

Since there is no congestion and loss, the maximum send window will be the bandwidth * delay for this TCP connection.

w max= 1G bps * 150ms = 109 * 150 * 10-3

= 150 *106 bit = 18.75 * 106 byte = 17.88 MB.

So, this 10 MB file can be transferred before reaching w max. In another word, it can be sent in its slow start phase. Assume x RTT is required to send this file, then:

(1 + 2 + 4 + … + 2x) * 1KB ≥ 10 MB

2 * 2x–1 ≥ 10 * 1024

x ≥ log2(10241) – 1 = 12.3

Thus x = 13, it will take 13 RTTs to transfer this file. 13RTT = = s

Total delay = 13RTT + Filesize/BW = 13 * 150ms+ 10MB / 1Gbps =1.95 + 0.08 * 1.024^2=2.03 s