文章目錄
- 21T2-sample 複習
- 域名
- http
- 電路交換 circuit switch
- 封包交換packet switching
- BitTorrent Tit/Tat
- UDP
- checksum
- 應用
- TCP
- RDT
- GBN and SR
- sequence number
- 四大延遲-資料傳輸大題
- 算傳播和傳輸延遲
- 不同線路算延遲
- socket
- 21T2真題
文章目錄
- 21T2-sample 複習
- 域名
- http
- 電路交換 circuit switch
- 封包交換packet switching
- BitTorrent Tit/Tat
- UDP
- checksum
- 應用
- TCP
- RDT
- GBN and SR
- sequence number
- 四大延遲-資料傳輸大題
- 算傳播和傳輸延遲
- 不同線路算延遲
- socket
- 21T2真題
21T2-sample 複習
域名
resource records (RR) 儲存在域名伺服器上（DNS），其他用戶端或者伺服器可以去dns上找到某個網址記錄，比如找到域名的ip位址，然後去通路。
視訊教程：https://www.youtube.com/watch?v=6uEwzkfViSM
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type A就是主機名和ip，value是ip位址
cname就是容易記的别名指到某個規範名字,name是很好記的名字，value就比較長的名字
ns就是權威域名伺服器的位址，上面儲存了想查詢的這個記錄具體的内容,一般value有字首ns , ns1.google.com主要域名伺服器,次級域名伺服器 ns2.gogle.com
mx就是某個郵件伺服器想給這個網址(test.com)發郵件，發到這個網站接受郵件的網址。一般value字首有mail,比如mail1.test.com，再比如smtp.gmail.com
某個主機名(google.com)的權威名稱伺服器的資訊存儲在.com TLD(top level domain)伺服器中，這樣當本地DNS伺服器查詢主機名的映射時，它們可以被定向到包含實際答案(即所請求的映射)的權威名稱伺服器。是以，谷歌域名伺服器(主要和次要)的NS記錄将存儲在.com TLD伺服器中。此外，這些名稱伺服器(即A記錄)的對應IP位址也會存儲在.com TLD伺服器。
也就是說，value帶ns的，和值帶ns的，都可以在頂級域名伺服器中找到。
© (google.com, ns1.google.com, XXX, 2 days)
(d) (google.com, ns2.google.com, XXX, 2 days)
(e) (ns1.google.com, 216.239.32.10, XXX, 2 days)
(f) (ns2.google.com, 216.239.34.1, XXX, 2 days)
© (google.com, ns1.google.com, XXX, 2 days)
(d) (google.com, ns2.google.com, XXX, 2 days)
(e) (ns1.google.com, 216.239.32.10, XXX, 2 days)
(f) (ns2.google.com, 216.239.34.1, XXX, 2 days)
有個網址(alias name)，用戶端通路，先去本地dns找對應的ip或者郵件伺服器的名稱，本地dns緩存的内容和權威dns一樣。
權威dns的資訊，儲存在tld域名伺服器上。
http
non-persistent
persistent
persistent with pipeline:先接收index，之後的objects同一時刻送出請求，所有的objects都同時過來。
電路交換 circuit switch
線路獨占，因為無法确定哪個時間段在用，是以就一直維護這段線路。
封包交換packet switching
要發資訊了，就建立一個封包，然後去發，發完馬上結束占用。是以，如果某一時刻用的人多了，超過線路帶寬，就需要排隊。
BitTorrent Tit/Tat
tit for tat 以牙還牙
最後一節
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alice 和 bob都需要下載下傳某檔案，也都有某檔案的一些塊，alice先随機選擇bob給他塊，成為4個給bob最快檔案塊的之一，那bob也會很快地優先給alice塊，是以互相受益。
但是，如果alice已經下載下傳完了，那其實以牙還牙，alice沒受到好處，找不到4個給alice最快檔案塊的人。
UDP
checksum
偶校驗：資料和校驗位加起來是偶數個1.
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應用
udp速度快，可靠性不高。dns和fps遊戲，看視訊用udp
email和bittorrent用tcp
TCP
RDT
5個特征：
TCP對最老的未确認段使用一個定時器。
根據快速重傳機制，當它收到三次重複的ACK時，發送方重新發送一個段。
tcp使用累積的ACK。累積确認
計時器過期将導緻傳輸最老的沒收到ack的段
GBN and SR
GBN: 使用累積确認的方式，不按照順序接受到的包會被丢棄。
SR: 每一個包都會被單獨确認，是以不需要按照順序。
sequence number
兩個知識點：
- 一個bytes消耗一個sequence number
- ack number 是下一個期望到來的squence number
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四大延遲-資料傳輸大題
算傳播和傳輸延遲
link type 1線路上傳播dp1: = 2X10^6 m/2X 10^8 = 0.01s
hostA往線路上打的傳輸延遲dt1(一個包): 1000 X 8 bits /(100 000 bits/s) = 0.08s
注意：！！！ 打包算傳輸延遲，一定要轉成bits的機關，而且要一個包，一個包分析
仔細地一個一個包算，一定可以。
不同線路算延遲
注意畫圖仔細
socket
ip address是點分十進制的表示，port是數字
21T2真題
- TIME ALLOWED: 1 hours and 10 minutes.
- TOTAL MARKS AVAILABLE: 20 marks worth 20% of the total marks for the course.
- ALL QUESTIONS MUST BE ANSWERED.
-
MARKS AVAILABLE FOR EACH QUESTION ARE SHOWN IN THE EXAM. THERE IS NO
NEGATIVE MARKING, IN THAT THE MINIMUM MARK FOR EACH QUESTION IS ZERO.
- THE EXAM IS OPEN BOOK, OPEN NOTES. USE OF CALCULATORS IS PERMITTED.
-
STUDENTS ARE ADVISED TO READ THE EXAMINATION QUESTION BEFORE
ATTEMPTING TO ANSWER THE QUESTION.
- THIS EXAM CANNOT BE COPIED, FORWARDED, OR SHARED IN ANY WAY.
-
STUDENTS ARE REMINDED OF THE UNSW RULES REGARDING ACADEMIC INTEGRITY
AND PLAGIARISM.
-
YOUR WORK WILL BE SAVED PERIODICALLY THROUGHOUT THE EXAM AND WILL BE
AUTOMATICALLY SUBMITTED PROVIDED YOU ARE CONNECTED TO THE INTERNET.
09/07/2021
Sample Mid Term Exam
https://unsw.inspera.com/admin#author/test/873040302/33
We know smtp.gmail.com is the mail server of gmail.com and ns1.google.com and
ns2.google.com are the authoritative DNS servers for google.com. The questions
below are about the following DNS Resource Records (a - f) where the entry types
are masked (XXX).
(a) (gmail.com, smtp.gmail.com, XXX, 2 days)
(b) (smtp.gmail.com, 108.177.125.10, XXX, 2 days)
© (google.com, ns1.google.com, XXX, 2 days)
(d) (google.com, ns2.google.com, XXX, 2 days)
(e) (ns1.google.com, 216.239.32.10, XXX, 2 days)
(f) (ns2.google.com, 216.239.34.1, XXX, 2 days)
Answer the 6 multiple-choice questions. You may select multiple choices for each
answer. However, note that selecting additional choices beyond the correct
answer(s) will be considered incorrect. Partial marks may be allocated as noted in
the marking scheme. The lowest possible mark is 0.
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1 DNS Q1
-
Which of the provided Resource Records are Type A entries?
Select one or more alternatives:
(b)
(f)
None of the provided records
(e)
©
(a)
(d)
Maximum marks: 0.5
Type A records provide a mapping from hostname to IP address. Thus the correct answer is (b), (e)
and (f). 0.167 mark for each correct answer and -0.167 mark for each incorrect answer. 0 if “none of
the provided records” is chosen. Minimum possible mark is 0.
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Sample Mid Term Exam
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2 DNS Q2
Which of the provided Resource Records are Type MX entries?
Select one or more alternatives:
(b)
(a)
©
(f)
(e)
(d)
None of the provided records
Maximum marks: 0.5
MX refers to the mail server record, thus the correct answer is (a). 0.5 mark for the correct answer and
-0.1 for each incorrect answer. 0 if “none of the provided records” is chosen. Minimum possible mark is 0.
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3 DNS Q3
Which of the provided Resource Records are Type NS entries?
Select one or more alternatives:
(f)
None of the provided records
(a)
(b)
(d)
(e)
©
Maximum marks: 0.5
NS refers to a name server record which provides the name of the nameserver responsible for the
hostname/domain.
Thus the correct answer is © and (d) , which are the primary and secondary nameserver for google.com.
0.25 for each correct answer and -0.125 answer for each incorrect answer. 0 if “none of the provided
records” is chosen. Minimum possible mark is 0.
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4 DNS Q4
Which of the provided Resource Records are type CNAME entries?
Select one or more alternatives:
None of the provided records
©
(b)
(a)
(f)
(d)
(e)
Maximum marks: 0.5
CNAME refers to a canonical name record for the hostname. None of the provided records are of this
type. 0.5 mark for the correct answer. -0.5 mark if any of the other answers is chosen. Minimum possible
mark is 0.
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5 DNS Q5
Which of the provided Resource Records are stored in the .com TLD servers?
Select one or more alternatives:
(b)
(f)
None of the provided records
(a)
(e)
©
(d)
Maximum marks: 0.5
Information of the authoritative name servers of a hostname is stored in the TLD servers so that when a
local DNS server queries for the mapping for the hostname, they can be directed to the authoritative
name servers which would contain the actual answer (i.e. the mapping requested). Thus, the NS records
for the Google name servers (primary and secondary) would be stored in the .com TLD server. In
addition, the corresponding IP addresses of these name servers (i.e. the A records) would also be stored
in the .com TLD server.
Thus, the correct answer is ©, (d), (e) and (f). 0.125 for each correct answer and -0.25 for each incorrect
answer. 0 if “none of the provided records” is chosen. Minimum possible mark is 0.
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6 DNS Q6
Which of the provided Resource Records are stored at an authoritative name server?
Select one or more alternatives:
(a)
(d)
(f)
(e)
©
None of the provided records
(b)
Maximum marks: 0.5
The MX record for the gmail mail server and the corresponding A record that holds the IP address for this
server would be stored at the authoritative name servers (i.e. ns1.google.com and ns2.google.com). These
are the “final” answers that are provided in response to a MX query for the gmail mail server.
Thus, the correct answer is (a) and (b). 0.25 for each correct answer and -0.125 mark for each incorrect
answer. 0 if “none of the provided records” is chosen. Minimum possible mark is 0.
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Assume a webpage comprised of 10 objects which includes the index.html file, 8
embedded images and one embedded audio clip. The 10 objects are so small that:
(i) their transmission time is negligible and (ii) each object can be completely
transmitted in one TCP segment. Consider a client wishing to download the
webpage.
You are asked to make the following assumptions:
the round trip time between the client and all servers is T
the time to set up and tear down a TCP connection is S and F, respectively.
You must account for both these times in your computations. Note
that, S includes the 3-way handshake (SYN, SYN-ACK, ACK) and F includes
the time for sending FINs and ACKs from both endpoints.
there are no packet losses.
the client knows the IP address of all servers (i.e. neglect DNS resolution
delay).
neither the client nor any of the servers support parallel TCP connections.
Answer the following 5 questions. No explanations are required. Simply write the
expression for each answer which should ONLY contain the
variables T, S and F (e.g., 20T+100S+50F) in the space provided.
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7 HTTP Q1
Assume that the client uses non-persistent HTTP for downloading the web page. What is the time
required to complete the transfer of the web page (including the time for setting up and tearing
down each TCP connection involved)?
Fill in your answer here
Maximum marks: 0.6
In non-persistent HTTP, every object is downloaded over a fresh TCP connection. Since parallel
connections are not supported, this would mean the ten objects are fetched serially.
The time required to fetch one object = time to setup TCP connection + RTT for sending GET request
and receiving response + time to tear down TCP connection = S + T + F.
Thus, the total time = 10S +10T + 10F.
10S +10T + 10F
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8 HTTP Q2
Assume that the client uses persistent HTTP without pipelining for downloading the web page.
What is the time required to complete the transfer of the web page (including the time for setting
up and tearing down each TCP connection involved)?
Fill in your answer here
Maximum marks: 0.6
In this instance, all objects can be fetched over one single TCP connection but serially (one after the
other).
Thus the total time = time to setup TCP connection + 10 x (RTT for sending GET request and receiving
the object) + time to tear down TCP connection = S + 10T + F.
S + 10T + F.
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9 HTTP Q3
Assume that the client uses persistent HTTP with pipelining for downloading the web page. What
is the time required to complete the transfer of the web page (including the time for setting up and
tearing down each TCP connection involved)?
Fill in your answer here
Maximum marks: 0.6
In this instance, since pipelining is used once the index page is fetched and the client knows of the 9
embedded objects, these 9 objects can be requested back-to-back (simultaneously) and the corresponding
objects would also be received back-to-back.
Thus the total time = time to setup TCP connection + RTT for sending GET request for the index page and
receiving that page + RTT for sending 9 GET requests for embedded objects and receiving them + time to
tear down TCP connection = S + T + T + F = S + 2T + F.
S + 2T + F
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10 HTTP Q4
Now assume that all 10 objects are located on 10 different servers (one object on each server).
The client can only have one active TCP connection at any given time. Assume that the round
trip time between the client and each of the 10 servers is T. Neglect DNS queries. Assume that
the client uses persistent HTTP with pipelining for downloading the web page. What is the time
required to complete the transfer of the web page (including the time for setting up and tearing
down each TCP connection involved)?
Fill in your answer here
Maximum marks: 0.6
Now each object is fetched from a different server. Since parallel connections are not supported, this
would mean these objects have to be fetched serially (one after the one).
The time to fetch one object from one server = time to setup TCP connection + RTT for sending GET
request for the object and receiving that object + time to tear down TCP connection = S + T + F.
Thus the total time = 10 (S + T + F)
10S + 10T + 10F
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11 HTTP Q5
Now assume that the index page and 7 embedded images are on one server, while the
remaining image and audio clip are on another server. The client can only have one active TCP
connection at any given time. Assume that the round trip time between the client and both
servers is T. Neglect DNS queries. Assume that the client uses persistent HTTP with pipelining
for downloading the web page. What is the time required to complete the transfer of the web
page (including the time for setting up and tearing down each TCP connection involved)?
Fill in your answer here
Maximum marks: 0.6
In this instance, all objects from each server can be fetched over a single TCP connection established
with that server.
The client would first fetch the index page and would become aware of the embedded objects. The 7
objects hosted on this same server (as the index page) are then fetched in one go. Next, the client
fetches the other two objects from the other server in one go.
Thus the total time = time to setup TCP connection with the first server + RTT for sending GET request
for the index page and receiving that page + RTT for sending 7 GET requests for embedded objects and
receiving them + time to tear down TCP connection + time to setup TCP connection with the second
server + RTT for sending 2 2 GET requests for embedded objects and receiving them = S + T + T + F +
S + T + F = 2S + 3T + 2F.
2S + 3T + 2F
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Suppose a number of users share a 4 Mbps link. Also, suppose that each user
transmits continuously at 2 Mbps when transmitting, but each user transmits only
20% of the time.
Answer the 3 questions.
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12 Packet/Circuit Switching Q1
When circuit switching is used, how many users can be supported? No explanation is required.
Simply enter the numeric value in the space provided:
(2)
Maximum marks: 0.25
In circuit switching, the percentage of time a user is active is irrelevant. A circuit needs to be established
for each active user. Since each user requires 2Mbps and the link capacity is 4Mbps, 2 users can be
supported.
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13 Packet/Circuit Switching Q2
Now suppose packet switching is used. Why will there be essentially no queuing delay before the
link if two or fewer users transmit at the same time? Why will there be queuing delay if three
users transmit at the same time? Be brief (2 sentences at most for each question).
Fill in your answer here
Format
Σ
Words: 0
Maximum marks: 0.75
Since each user requires 2 Mbps when transmitting, if two or fewer users transmit simultaneously, a
maximum of 4 Mbps will be required. Since the available bandwidth of the shared link is 4 Mbps, there
will be no queuing delay before the link. (0.25 marks)
Whereas, if three users transmit simultaneously, the bandwidth required will be 6 Mbps which is more
than the available bandwidth of the shared link. In this case, there will be queuing delay before the link.
(0. 5 marks)
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14 Packet/Circuit Switching Q3
Suppose with packet switching, there are three users. Find the probability that at any given time,
all three users are transmitting simultaneously. No explanation is required.Simply enter the
numeric value in the space provided:
(0.008)
Maximum marks: 0.5
The probability that all three users are transmitting simultaneously =(0.2)^3 = 0.008
09/07/2021
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15 BitTorrent Tit/Tat Short Answer
BitTorrent uses a “tit-for-tat” incentive mechanism for selecting peers to whom a particular peer
would upload chunks. Consider a peer who has finished downloading the file but wishes to
continue seeding the file to other peers (i.e. continue uploading chunks of that file) participating in
the torrent. Will “tit-for-tat” still be useful for this peer? Explain why or why not in 2-3 sentences.
Answers without explanations will not receive marks.
Fill in your answer here
Format
Σ
Words: 0
Maximum marks: 1
No. Tit-for-tat is not useful. This peer is no longer downloading any chunks of this file.
So the peer will never be able to determine the top (four) peers that are providing her
chunks at the best rate. Recall that “tit-for-tat” selects these top four peers to upload
chunks.
NOT EXPECTED IN ANSWER. THIS IS JUST FYI: In this instance, BitTorrent
selects the peers who have the best upload rate (in general). This ensures that
chunks get uploaded faster, and they get replicated faster.
09/07/2021
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16 UDP Checksum
Assume that the UDP checksum is only computed over the data (i.e. ignore all other UDP
headers and pseudo IP headers from the computation). Assume that the UDP segment format is
(checksum, data). Assume that a UDP sender sends a segment (0010, 1110) and the UDP
receiver receives (0011, 1110). Which of the following is true of the UDP receiver?
Select one alternative:
It thinks that the segment is corrupted and discards the segment.
It thinks only the checksum is corrupted and delivers the correct data to the application.
It concludes that nothing is wrong with the segment.
It explodes.
Maximum marks: 1
Recall that the checksum algorithm can detect errors but not locate the precise bits that are in
error.The receiver can thus detect an error, but it would know which bit(s) is/are in error. Thus, it
must discard the segment. The correct answer is thus (a).
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17 TCP RDT
Which of the following is true about how TCP implements reliable data transfer? (Multiple choices
may be correct. Selecting additional choices beyond those that is/are correct will be considered
as incorrect)
Select one or more alternatives:
TCP uses multiple timers
TCP receiver always transmits acknowledgement immediately upon receiving a data
packet
TCP may retransmit packets upon receiving duplicate acknowledgements
TCP uses cumulative acknowledgements
TCP may retransmit packets upon timer timeout events
Maximum marks: 1
TCP uses a single timer for the oldest unacknowledged segment.
The receiver employs a delayed ACK mechanism as noted in the lectures
The sender retransmits a segment when it receives triple duplicate ACKs as per the fast
retransmit mechanism
TCP uses cumulative acks
The expiration of a timer will result in the transmission of the oldest unacknowledged
segment
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18 GBN and SR
Two hosts A and B are using the go-back-N (GBN) protocol with a window size of 4. Host A
sends host B four segments back to back with sequence numbers 15, 16, 17 and 18. These
segments arrive at host B in the following order: 15, 18, 17, 16. When host B receives each of
these segments, it sends an acknowledgment segment to host A. What are the
acknowledgement numbers in the acknowledgement segments that host B sends to host A in the
order in which host B sends them? You may assume that all previous segments (14, 13, …) have
been correctly received by host B in the expected order.
Repeat the above if the two hosts are using the selective-repeat (SR) protocol.
Note: You are not required to provide any explanation. Simply note down the sequence of ack
numbers in the space provided below, first for GBN and then for SR.
Fill in your answer here
Format
Σ
Words: 0
Maximum marks: 2
Go-Back-N (1 mark):
15, 15, 15, 16
The second and third packets arrive out of order. These are thus discarded. The acknowledgments for
these packets will contain the last in-order received segment number, which is 15 (cumulative acks). The
last received packet is the next expected packet (sequence number 16) and thus the acknowledgment
will now contain the sequence number 16.
Selective Repeat (1 mark):
15, 18, 17, 16
In selective repeat, each packet is individually acknowledged and out of order packets are buffered, thus
resulting in the above-noted pattern.
GBN: 15, 15, 15, 16
SR: 15, 18, 17, 16
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19 UDP and Applications
Which of the following statements is corrrect?
Select one alternative:
BitTorrent, DNS and First Person Shooter Games typically use UDP.
DNS and First Person Shooter Games typically use UDP.
E-mail and DNS typically use UDP.
E-mail, DNS, BitTorrent and First Person Shooter Games use UDP.
DNS and BitTorrent typically use UDP.
Maximum marks: 0.75
E-mail and BitTorrent use TCP.
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20 TCP Sequence Number
Host A sends a 128-byte TCP segment carrying a sequence number of 100 to Host B. Host B
receives it correctly and sends an ACK to Host A. What is the acknowledgement number in the
ACK?
Select one alternative:
101
227
228
226
Maximum marks: 0.75
The segment contains bytes numbered from 100 to 227. The ACK number is always for the next
expected sequence number which is 228.
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Consider the network in the figure below. Host A can choose between two different
paths to communicate with host B. Host can choose to send packets via either
Router 1 or Router 2 to host B. The communication links are of two different types,
as indicated in the figure. The characteristics of these two types of links are:
Link type 1: Each link is of length 2000km, propagation speed is 2 x 10 m/s and
bandwidth is 100kbps.
Link type 2: Each link is of length 4000km, propagation speed is 2 x 10 m/s and
bandwidth is 50kbps.
Host A wishes to transmit a message of size 4Kbytes to host B. It breaks this
message into 4 packets of equal size. Neglect any packet headers. Remember that
routers work on the store-and-forward principle.
Assume that the processing delay and queuing delay in the routers are negligible.
You may also approximate file sizes to be an order of 10 (i.e. 4Kbytes = 4000 bytes
instead of 4096 bytes).
8
8
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21 Switching Q1
If host A chooses to send the packets via Router 1, determine the time it takes to move the
packets from host A to host B, i.e., beginning from the time that host A starts to send the first bit
of the first packet till the time that host B receives the last bit of the last packet.
You are encouraged to draw a timing diagram to help you visualise the delays. However, you are
NOT required to upload such a diagram with your answers.
Do not simply write the final answer. Show us your work (just type it in the space provided).
Fill in your answer here
Format
Σ
Words: 0
Maximum marks: 2
From the timing diagram below, the time required = 5dt1 + 2 dp1 = 5(0.08 ) + 2(0.01) = 0.42 second
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22 Switching Q2
Now assume that host A chooses to send the packets via Router 2 to host B. Determine the time
it takes to move the packets from host A to host B, i.e., beginning from the time that host A starts
to send the first bit of the first packet till the time that host B receives the last bit of the last
packet.
You are encouraged to draw a timing diagram to help you visualise the delays. However, you are
NOT required to upload such a diagram with your answers.
Do not simply write the final answer. Show us your work (just type it in the space provided).
Fill in your answer here
Format
Σ
Words: 0
Maximum marks: 2
From the timing diagram below, the time required = dp1 + dt1 + dp2 + 4dt2
= 0.01 + 0.08 + 0.02 + 4(0.16) = 0.75 s.
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Consider the picture below. Process P3 on the host with IP address X has set up a
TCP connection with process P4 on the server with IP address Y. Process P2 on the
host with IP address Z has set up a TCP connection with process P5 on the server
with IP address Y. There are no other TCP connections open at the server.
4 TCP segments S1 - S4 are shown in the picture. The source and destination IP
addresses and port numbers for S1, S3 and S4 are noted. S1 is sent by P3 to P4, S2
is sent by P4 to P3 and S3 is sent by P2 to P5.
Answer the 5 following questions.
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23 Socket Q1
What is the source IP address for TCP segment S2? No explanation needed.
(Y, y)
Maximum marks: 0.25
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24 Socket Q2
What is the source port number for TCP segment S2? No explanation needed.
(12004)
Maximum marks: 0.25
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25 Socket Q3
What is the destination IP address for TCP segment S2? No explanation needed.
(X, x)
Maximum marks: 0.25
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26 Socket Q4
What is the destination port number for TCP segment S2? No explanation needed
(4242)
Maximum marks: 0.25
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27 Socket Q5
Consider TCP segment S4 sent by process P1. Assume that S4 contains data. Describe what
happens to this segment and why? 2-3 sentences should be sufficient.
Fill in your answer here
Maximum marks: 1
S4 will be routed by the Internet to arrive at server Y. The server will try to match the
(source IP, source port, destination port) tuple, i.e. (Z, 4242, 12005) against all active
TCP sockets.
The two active TCP connections will have the following entries in the active TCP socket
table:
(X, 4242, 12004) -> For the TCP connection between P3 and P4.
(Z, 2543, 12005) -> For the TCP connection between P2 and P5.
Notice that, S4 does not generate an exact match with either. Thus this segment will be
dropped.
NOT REQUIRED FOR MARKING: The server would respond to P1 with a TCP RST
(Reset) segment.
![計算機作業系統（第四版）複習材料-完全 第一章 作業系統引論
第二章 程序的描述與控制
第三章 處理機排程與死鎖
第四章 存儲器管理
第五章 虛拟存儲器
第六章 輸入輸出系統
第七章 檔案管理
第八章 磁盤存儲器的管理[圖]](data:image/gif;base64,R0lGODlhAQABAIAAAP///wAAACwAAAAAAQABAAACAkQBADs=)