宁波大学计算机网络期末试卷(英文).doc

(word完整版)宁波大学计算机网络期末试卷(英文)
I. Choose the single correct answer from following choose。 （1.5*20=30)
[Correct checked：1。5； Other wise:0]
1.Which physical media has high—speed operation and low error rate:A
[］ A。Fiber optic cable
［］ B。Coaxial cable
［］ C。Twisted pair
［］ D.Radio

2。If no free buffers in router， the arriving packets will be： B
[] A.dropped
［] B。queued
［] C。returned
[] D.marked

3. Which can provides delay measurement from source to router along end-end Internet path towards destination: A
［] A。Ping
［] B。Traceroute
[] C。Ipconfig
［］ D。Nslookup

4.In TCP/IP， which layer can make routing of datagrams from source to destination: C
［] A。Applicaion
[］ B.Transport
[］ C.Network
[］ D。Data Link

5.Web page consists of （ ） which includes several referenced objects：A
[］ A.referenced HTML—file
[］ B.host HTML-file
[］ C。path HTML-file
［] D。base HTML—file

6.What is the default persistent model in HTTP/1.1:C
［］ A。Nonpersistent HTTP
［］ B。Persistent without pipelining
［] C。Persistent with pipelining
［] D。Nonpersistent with pipelining

7.Web server maintains no information about past client requests， so HTTP is:B
［] A.stateful
［］ B.stateless
［］ C.satisfied
[］ D。unsatisfied

8。FTP client browses remote directory by sending commands over: D
[］ A。connection-less
[] B.free connection
［] C.data connection
[] D。control connection

9.Which can satisfy client request without involving origin server A
[］ A。Web caches
［] B。Write caches
［] C.TCP buffer
［] D。Router buffer

10。UDP socket identified by:B
［] A。two-tuple （source IP address, source port number)
[] B.two-tuple （dest IP address， dest port number)
[］ C.two-tuple (source IP address， dest port number）
［］ D.two—tuple (dest IP address, source port number）

11.In GBN,when receiver receive a out—of-order packet，then discard and re—ACK the packet with：
［］ A. highest in—order sequence # A
［] B. lowest in-order sequence ＃
[] C。 highest in-order port #
[] D。 lowest in—order port ＃

12。In RDT Approachs, which is designed for performance： B
[］ A. checksum
[］ B. pipeline
［] C. sequence ＃
［］ D. ACK or NAK

13。Queued datagram at front of router's queue prevents others in queue from moving forward is:A
[] A.Head-of-the—Line （HOL） blocking
[］ B。Hops—of-the—Line （HOL) blocking
[] C。Head—of-the-List (HOL） blocking
[］ D.Hops—of-the—List (HOL） blocking

14.What’s a network ？ From IP address perspective they can physically reach each other without intervening router and the device interfaces with： C
[] A。 same IP address
[］ B。 same TCP port #
[］ C。 same network part of IP address
[］ D. same host part of IP address

15。 Large IP datagram divided (“fragmented”） within network,it will be reassembled：B
[］ A. only at last router
[] B。 only at final destination
［] C. only at next router
[］ D. maybe at next router
16。Which is not a common Intra-AS routing protocols:D
［] A。RIP： Routing Information Protocol
［] B。OSPF： Open Shortest Path First
［] C.IGRP: Interior Gateway Routing Protocol
[] D。ICMP: Interior Control Message Protocol
17.Which is not a MAC Random Access protocol in Ethernet： D
[］ A.SCMA
［］ B。SCMA/CA
[］ C.SCMA/CD
[］ D。Slotted SCMA
18.In DHCP client-server scenario， which message has DHCP-options field:D
[］ A. host broadcasts “DHCP discover”
［］ B。 DHCP server responds with “DHCP offer”
［］ C。 host requests IP address: “DHCP request”
[］ D。 DHCP server sends address： “DHCP ack”
19。How to determine MAC address of host B, If knowing host B’s IP address?A
[］ A.ARP
[］ B。RARP
［] C.RAP
[］ D。RIP
20. Which device can break subnet into LAN segments: C
[］ A.IP mask
[] B.NAT
[] C.Router
［] D.Switch

II. Choose the multiple correct answer from following choose。 （2*10=20)
［All correct checked：2; Part correct checked:1； No checked:0; Full checked:0］
1.Which is the part of network structure： ABD
［] A.network edge
［] B。network core
［] C.network user
[］ D。access networks
2.How to connect end systems to edge router? BCD
［］ A。Microsoft access networks
［］ B。Residential access networks
[] C。Institutional access networks
［] D。Mobile access networks
3.What kind of transport service does an application need？ ABC
[］ A.Data loss
［] B。Timing
[] C.Bandwidth
[] D。Security
4.Electronic Mail three phases of transfer is: ABD
[] A.handshaking （greeting)
[] B.transfer of messages
[］ C。opens the 2nd TCP connection
[］ D.close
5.In TCP Connection Management， initialize TCP variables include： AB
［] A.sequence ＃
［］ B.buffers
［］ C.Sender MTU
［］ D.RcvWindow
6.How does sender perceive congestion? AB
［］ A。timeout
[] B。3 duplicate ACKs
［] C.3 duplicate data
[] D.slow start
7.TCP Congestion Control use three mechanisms： ABC
［］ A。additive increase and multiplicative decrease
［］ B。slow start
[］ C.Conservative after timeout events
[] D。additive decrease and multiplicative increase
8.What are the Key Network—Layer Functions： ABC
[］ A.forwarding
［] B.routing
［］ C.connection setup
[］ D.flow control
9。Link Layer Services include: ABD
［] A.Reliable delivery between adjacent nodes
［］ B。Flow Control between adjacent nodes
［] C.Connection Manage
[] D.Error Detection and Correction
10.MAC Protocol’s taxonomy， three broad classes is： ABC
［］ A。Channel Partitioning
［] B.Random Access
[] C.Taking turns
[］ D。Peer—to—peer
III. Fill the blank from options. (1。5＊16=24）
1）。The network protocols define format ， order of message sent and received among network entities， and action taken on message transmission， receipt。
(options： delay / format / policy / order / request / replay / actions taken / price / interface ）

2).InTCP Congestion Control， after 3 duplicate ACKs CongWin is cut in half and window then grows____linearly___. But after timeout event， CongWin instead set to 1 MSS , window then grows___exponentially____, when it up to a __threshold____ again, then grows linearly.
(options: half / double / 1 MSS / 0 MSS / linearly / exponentially / threshold / top / bottom )

3）.Please fill the general format of Http request message:
HTTP request message general format
9。[Method ]
sp
10.［URL ]
sp
11.［version ］
12。［CR LP ]
…
…
13.[header field name ]
:
14。［field value ］
15. ［Cr Lf ］
16。 ［ Cr Lf ]
Entity Body

(options: header field name / URL / field value / version / method / Cr Lf/ 200 OK )
IV. Question (26）
1.As follow， LAN1 connect to LAN2 via a router：
In session 1， Host A send a HTTP connection to WEB server D，if Host A initial TCP port 1025，Host D use TCP port 80；
In session 2, Host A send a HTTP connection to WEB server B,if Host A initial TCP port 1026,Host B use TCP port 80；

Fill it: (9)
Session
Step
Source MAC
Destination
MAC
Source
IP
Destination
IP
Source
Port#
Destination
Port#
Host A ：1025
–〉 Host D:80
Host A –〉
ROUTER

ROUTER
–> Host D

Host A :1026
–> Host B：80
Host A –>
Host B

2。Read and answer:（17)
Two of the most important fields in the TCP segment header are the sequence number field and the acknowledgment number field。 These fields are a critical part of TCP’s reliable data transfer service。 But before discussing how these fields are used to provide reliable data transfer, let us first explain what exactly TCP puts in these fields.
TCP views data as an unstructured， but ordered, stream of bytes. TCP’s use of sequence numbers reflects this view in that sequence numbers are over the stream of transmitted bytes and not over the series of transmitted segments。 The sequence

number for a segment is the byte-stream number of the first byte in the segment. Let’s look at an example。 Suppose that a process in host A wants to send a stream of data to a process in host B over a TCP connection。 The TCP in host A will implicitly number each byte in the data stream.  Suppose that the data stream consists of a file consisting of 500,000  bytes， that the MSS is 1,000 bytes， and that the first byte of the data stream is numbered zero。 As shown in Figure 3.5-3, TCP constructs 500 segments out of the data stream. The first segment gets assigned sequence number 0， the second segment gets assigned sequence number 1000， the third segment gets assigned sequence number 2000， and so on。。 Each sequence number is inserted in the sequence number field in the header of the appropriate TCP segment.

Figure 3.5-3： Dividing  file data into TCP segments。
Now let us consider acknowledgment numbers。 These are a little trickier than sequence numbers。 Recall that TCP is full duplex, so that host A may be receiving data from host B while it sends data to host B （as part of the same TCP connection）. Each of the segments that arrive from host B have a sequence number for the data flowing from B to A. The acknowledgment number that host A puts in its segment is sequence number of the next byte host A is expecting from host B. It is good to look at a few examples to understand what is going on here. Suppose that host A has received all bytes numbered 0 through 535 from B and suppose that it is about to send a segment to host B. In other words， host A is waiting for byte 536 and all the subsequent bytes in host B’s data stream。 So host A puts 536 in the acknowledgment number field of the segment it sends to B。
As another example， suppose that host A has received one segment from host B containing bytes 0 through 535 and another segment containing bytes 900 through 1，000。  For some reason host A has not yet received bytes 536 through 899。 In this example, host A is still waiting for byte 536 (and beyond) in order to recreate B’s data stream. Thus, A’s next segment to B will contain 536 in the acknowledgment number field。 Because TCP only acknowledges bytes up to the first missing byte in the stream, TCP is said to provide cumulative acknowledgements.
This last example also brings up an important but subtle issue. Host A received the third segment (bytes 900 through 1，000) before receiving the second segment （bytes 536 through 899). Thus， the third segment arrived out of order. The subtle issue is: What does a host do when it receives out of order segments in a TCP connection？ Interestingly, the TCP RFCs do not impose any rules here, and leave the decision up to the people programming a TCP implementation. There are basically two choices： either （i) the receiver immediately discards out—of-order bytes; or (ii) the receiver keeps the out-of—order bytes and waits for the missing bytes to fill in the gaps. Clearly, the latter choice is more efficient in terms of network bandwidth， whereas the former choice significantly simplifies the TCP code. Throughout the remainder of this introductory discussion of TCP， we focus on the former implementation， that is， we assume that the TCP receiver discards out-of-order segments。
In Figure 3.5。3 we assumed that the initial sequence number was zero。 In truth, both sides of a TCP connection randomly choose an initial sequence number. This is done to minimize the possibility a segment that is still present in the network from an earlier， already-terminated connection  between two hosts is mistaken for a valid segment  in a later connection between these same two hosts （who also happen to be using the same port numbers as the old connection) 。

Question 1: Does TCP's use of sequence numbers over the series of transmitted segments？ （3）
Question 2: What does the sequence number for a segment means? For example. (4)
Question 3： What does the acknowledgment number means, that host A puts in its segment to host B?（3）
Question 4: How does TCPs to choose an initial sequence number? （3）
Question 5: What does a host do when it receives out of order segments in a TCP connection？ （4）

（所有答案请填在答题卡上，答在试卷上的答案一律无效)