CENG/BILM 362

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CENG/BILM 362
Midterm Review
Spring 2014
Question # 1
• How many types of delay are present in an
end-to-end transfer path? Explain each briefly
(Uçtan uca transfer yolunda kaç çeşit gecikme
vardır? Herbirini kısaca açıklayınız)
Answer
• Processing Delay(düğüm işlem gecikmesi) : Time to
examine the packet’s header , decide route of the
packet , do error checks
• Queuing Delay(kuyrukta gecikme) : Time spent in the
queue of the node waiting to be transmitted onto the
link
• Transmission Delay(iletim gecikmesi): Time required to
push all of the packet’s bits into the link
• Propagation Delay(yayınım gecikmesi): Time required
for a bit to travel from one end to the other
Question #2
• How many types of access networks are
there? (Kaç çeşit giriş ağı vardır?)
Answer
• Home access (Ev ağı): DSL, Cable, FTTH, Dialup, Satellite
• Enterprise access(Kurumsal ağ) :Ethernet, WiFi
• Wide-area wireless access(Geniş alan kablosuz
ağ): 3G, LTE
Question #3
• List the types of physical media(Fiziksel ortam
türlerini listeleyiniz)?
• Answer:
– Twisted pair Cooper Wire(Çift sarımlı bakır kablo)
– Coaxial Cable(koaksiyel kablo)
– Fiber Optics
– Terrestrial Radio (3G, Wifi, microwave)
– Satellite Radio (uydu)
Question #4
•
This elementary problem begins to explore propagation delay and transmission
delay, two central concepts in data networking. Consider two hosts, Hosts A and B,
connected by a single link of rate R bps. Suppose that the two hosts are separated
by m meters, and suppose the propagation speed along the link is s meters/sec.
Host A is to send a packet of size L bits to Host B. (Host A ve Host B, R bps iletim
oranına sahip bir linkle birbirine bağlıdır ve m metre kadar birbirinden uzaktadır ve
yayınım hızı ise s metre/saniyedir. Host A, L bit boyutundaki paketi Host B’ye
göndermeye çalışmaktadır )
•
a) Express the propagation delay, dprop in terms of m and s. (Buna göre yayınım
gecikmesini (dprop) m ve s cinsinden yazınız)
•
b) Determine the transmission time of the packet, dtrans in terms of L and R(İletim
hızını(dtrans) L ve R cinsinden yazınız).
•
c) Ignoring processing and queuing delays, obtain an expression for the end-to-end
delay (işlem ve kuyruk gecikmelerini hesaba katmayarak uçtan-uca toplam gecikmeyi
hesaplayınız).
Question #4
•
This elementary problem begins to explore propagation delay and
transmission delay, two central concepts in data networking. Consider
two hosts, Hosts A and B, connected by a single link of rate R bps.
Suppose that the two hosts are separated by m meters, and suppose
the propagation speed along the link is s meters/sec. Host A is to send
a packet of size L bits to Host B.
•
d) Suppose Host A begins to transmit the packet at time t=0. At time t=dtrans, where
is the last bit of the packet? (Host A paketi t=0. saniyede göndermeye başlıyor. t=
dtrans eşit olduğunda paketin son biti nerededir)
•
e) Suppose dprop is greater than dtrans . At time t=dtrans, where is the first bit of the
packet? (dprop, dtrans’dan büyük ise t= dtrans ‘da paketin ilk biti nerededir?)
•
f) Suppose dprop is less than dtrans . At time t=dtrans, where is the first bit of the
packet?((dprop, dtrans’dan küçük ise t= dtrans ‘da paketin ilk biti nerededir?)
•
g) Suppose s=2.5*108, L=100bits and R=28 kbps. Find the distance m so that dprop
equals dtrans. (s = 2.5*108, L = 100bit ve R=28kbps ise dprop ve dtrans’ın birbirine eşit
olduğu m değerini bulunuz)
Answer
Question #5
• What are cookies used for in HTTP ? Is the cookie
stored at the client or the server?(HTTP
protokolünde Çerezler ne için kullanılır? Çerez
istemci mi yoksa sunucu tarafında mı depolanır?)
• Answer: Cookies are used to hold state
information in HTTP. They are stored both at the
client and the server (Durum bilgisi tutmak için
kullanılır. Her iki tarafta da depolanır)
Question #6
•
Consider the following string of ASCII characters that were captured by Wireshark when the
browser sent an HTTP GET message (i.e., this is the actual content of an HTTP GET message). The
characters <cr><lf> are carriage return and line-feed characters (that is, the italized character string
<cr> in the text below represents the single carriage-return character that was contained at that
point in the HTTP header). Answer the following questions, indicating where in the HTTP GET
message below you find the answer. (Aşağıdaki HTTP GET bağlığına bakarak soruları cevaplayın)
GET /cs453/index.html HTTP/1.1<cr><lf>Host: gaia.cs.umass.edu<cr><lf>User-Agent: Mozilla/5.0
(Windows;U; Windows NT 5.1; en-US; rv:1.7.2) Gecko/20040804 Netscape/7.2 (ax) <cr><lf>
Accept:ext/xml, application/xml, application/xhtml+xml, text/html;q=0.9,
text/plain;q=0.8,image/png,*/*;q=0.5<cr><lf>Accept-Language: enus, en;q=0.5<cr><lf>AcceptEncoding: zip,deflate<cr><lf>Accept-Charset: ISO-8859-1,utf-8;q=0.7,*;q=0.7<cr><lf>Keep-Alive
300<cr><lf> Connection:keepalive<cr><lf><cr><lf>
•
a. What is the URL of the document requested by the browser? (Tarayıcı tarafından istenen
dokümanın URL’i nedir?)
b. What version of HTTP is the browser running? (Tarayıcı hangi versiyon HTTP kullanmaktadır?)
c. Does the browser request a non-persistent or a persistent connection? (Tarayıcı kalıcı mı yoksa
kalıcı olmayan bağlantı mı istemektedir?)
d. What is the IP address of the host on which the browser is running? (Tarayıcının çalıştığı
makinenin IP adresi nedir?)
e. What type of browser initiates this message? Why is the browser type needed in an HTTP
request message? (Ne tip bir tarayıcı bu mesajı oluşturmuştur? HTTP mesajında tarayıcı tipine
neden ihtiyaç duyulmaktadır?)
•
•
•
•
Answer
• a. The document request was
http://gaia.cs.umass.edu/cs453/index.html. The Host : field
indicates the server's name and /cs453/index.html indicates the file
name.
• b. The browser is running HTTP version 1.1, as indicated just before
the first <cr><lf> pair.
• c. The browser is requesting a persistent connection, as indicated by
the Connection: keep-alive.
• d. This is a trick question. This information is not contained in an
HTTP message anywhere. So there is no way to tell this from
looking at the exchange of HTTP messages alone. One would need
information from the IP datagrams (that carried the TCP segment
that carried the HTTP GET request) to answer this question.
• e. Mozilla/5.0. The browser type information is needed by the
server to send different versions of the same object to different
types of browsers.
Question #7
• With the TCP connection established POP3 progresses through
three phases. List them. (TCP bağlantısı oluşturulduktan sonra POP3
3 aşamadan geçer. Bunları listeleyin)
• Answer:
– Authorization(Yetki verme): The user agent sends a username and a
password to authenticate the user(kullanıcı, kullanıcı ismi ve parolayı
gönderir)
– Transaction(İşlem yapma) : User agent retrieves messages, mark
messages for deletion and obtain statistics (email mesajları indirilir,
silinmek için işaretlenir, istatistik bilgisi alınır)
– Update(Güncelleme): After the client issues a QUIT ending the
session, the mail server deleted messages marked for deletion (QUIT
komutundan sonra silinmek için işaretlenen mesajlar sunucu
tarafından silinir.)
Question #8
• Compare the sizes of UDP and TCP headers (TCP ve UDP
başlıklarının büyüklüklerini karşılaştırın)
• Answer: 20 bytes of overhead for TCP and 8 bytes of overhead for
UDP
• What is receiver window in TCP header used for? (TCP başlığındaki
alıcı penceresi ne için kullanılır)
• Answer : Flow Control (Akış kontrolü)
• Both congestion control and flow control throttle the sender, then
how are they different? (Sıkışıklık ve akış kontrolü hangi yönleriyle
birbirinden farklıdır?)
• Answer: Flow control handles congestion at the receiver while
congestion control handles congestion at the network core
(Sıkışıklık kontrolü ağ merkezinde sıkışıklık meydana gelmesini
engellerken, akış kontrolü alıcının bufferlarının dolmasını engeller)
Question # 9
•
Consider a reliable data transfer protocol that uses only negative
acknowledgements. Suppose the sender sends data only
infrequently. Would a NAK-only protocol be preferable to a protocol
that uses ACKs? Why? (Sadece negatif ACK kullanan bir güvenilir
transfer protokolü düşünün. Gönderici sık olmayan aralıklarla paket
gönderyorsa sadece-NAK kullanan bir protokol sadece ACK kullanan
bir protokole tercih edilebilir mi? Neden?)
• Now suppose the sender has a lot of data to send and the end-toend connection experiences few losses. In this second case, would a
NAK-only protocol be preferable to a protocol that uses ACKs?
Why?(Eğer gönderici sık olarak paket gönderiyor ve kayıp çok az
yaşanıyorsa, sadece NAK kullanan bir protokol sadece ACK kullanan
bir protokole tercih edilebilir mi?)
Answer
• In a NAK only protocol, the loss of packet x is only detected by the
receiver when packet x+1 is received. That is, the receivers receives
x-1 and then x+1, only when x+1 is received does the receiver
realize that x was missed. If there is a long delay between the
transmission of x and the transmission of x+1, then it will be a long
time until x can be recovered, under a NAK only protocol. (NAK
tercih edilmez çünkü kaybolan x paketi ancak x+1. paket yerine
ulaştığında farkedilir)
• On the other hand, if data is being sent often, then recovery under
a NAK-only scheme could happen quickly. Moreover, if errors are
infrequent, then NAKs are only occasionally sent (when needed),
and ACK are never sent – a significant reduction in feedback in the
NAK-only case over the ACK-only case. (Sık paket gönderilen bir
durumda kayıplar çabuk farkedilir ve kayıp ta çok az yaşanıyorsa az
sayıda NAK gönderilerek gönderilen geri dönüşüm paketlerinin
sayısında büyük bir azalma olur.)
Question # 10
• )
Question # 10
• 5 kez ölçüm sonucunda SampleRTT değerleri şu
şekildedir: 106ms, 120ms, 140ms, 90ms,115ms.
EstimatedRTT değerini her ölçümden sonra
hesaplayınız (α = 0.125 ve ölçümlerden önceki
Estimated RTT değeri 100ms’dir
• DevRTT değerini her bir ölçümden sonra
hesaplayınız. Ölçümlerden öndeki DevRTT değeri
5ms’dir ve β=0.25
• En son timeoutInterval’ı hesaplayın
Answer
EstimatedRTT
DevRTT
 xSampleRTT
 y SampleRTT
TimeoutInterval
 (1  x )EstimatedRTT
 EstimatedRTT
 EstimatedRTT
 (1  y )DevRTT
 4 * DevRTT
Answer
Answer
Answer
Question #11
TCP Congestion Control
35
30
25
cwnd
20
Cwnd
15
10
5
0
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
transmission round
Question #11
• Identify the intervals SS and CA are working? (SS ve CA’in çalıştığı
aralıkları belirtin)
• Answer: SS-> 1-5, 25-30 CA-> 5-24, 31-36
• What is the value of ssthresh at round 21 (21.turda ssthresh değeri
nedir?)
• Answer: 18/2
• Which round packet 80 is sent? (80.paket hangi turda gönderilir)
• Answer: 1+2+4+8+16+17+18+19 = 85 at round 8
• At which rounds congestion is detected and what is the cause of
congestion for each? (Hangi turlarda sıkışıklık tespit edilir ve
sıkışıklığın sebebi nedir?)
• Answer: 11 (3DUPACK), 16(3DUPACK), 24 (timeout), 30 (3DUPACK)
Question #12
• Compare GBN and SR. Assume that the timeout values for all three
protocols are sufficiently long such that 5 consecutive data
segments and their ACKs can be received if not lost in the channel
by the receiving host and the sending host respectively. Suppose
Host A sends 5 data segments to Host B and the ACK for the 2nd
segment is lost. In the end, all 5 data segments have been received
by Host B. How many segments has Host A sent in total and how
many ACKs has Host B sent in total? What are their sequence
numbers ? (Go-back-N, Selective Repeat protokollerini
karşılaştırınız. Üç protocol için de time out süresinin 5 tane veri
segmentinin gönderilip ACKlerinin alınmasına imkan tanıyacak
kadar uzun olduğunu varsayın(Eğer bağlantı üzerinde
kaybolmazlarsa). Host A Host B ye 5 veri segmenti gönderir fakat 2
nolu segmentin ACK paketi kaybolur.A ve B toplamda kaç segment
ve ACK göndermiştir ve sıra numaraları nedir?)
Answer
• GBN
– 5 segments sent by A: 1,2,3,4,5
– 5 ACKs sent by B: 1,2,3,4,5
• SR
– 6 segments sent by A: 1,2,3,4,5,2
– 6ACKs sent by B: 1,2,3,4,5,2
Question 13
• Consider that only a single TCP connection uses one 10Mbps link which
does not buffer any data. Suppose that this link is the only congested link
between the sending and receiving hosts. Assume that the TCP sender has
a huge file to send and the receiver’s buffer is much larger than the
congestion window. Each TCP segment is 1500 bytes and propagation
delay is 150 msec and this tcp connection is always in congestion
avoidance phase. (Tek bir TCP bağlantısı 10Mbps lik bir linki
kullanmaktadır. TCP bağlantısı büyük bir dosyayı göndermekte ve TCP
alıcı bufferları sıkışıklık penceresinden büyüktür. Her segment 1500 byte
ve aradaki gecikme 150msdir. Bu bağlantı hep CA fazında çalışmaktadır.)
– What is the maximum window size in segments that this TCP connection can
achieve?(Bu bağlantının ulaşabileceği en büyük pencere büyüklüğü nedir?)
– What is the average window size and average throughput? (Ortalama pencere
büyüklüğü ve ortalama throughput nedir?)
– How long would it take for this TCP connection to reach its maximum window
size again after recovering from a packet loss?(Bağlantının paket kaybından
sonra en yüksek pencere büyüklüğüne ulaşması ne kadar vakit alır?)
Answer
• Let W denote the max window size measured in segments. Then,
W*MSS/RTT = 10Mbps, as packets will be dropped if the maximum
sending rate exceeds link capacity. Thus, we have
W*1500*8/0.15=10*10^6, then W is about 125 segments. (W maximum
pencere büyüklüğü olarak verilirse ve segment cinsinden ise W*MSS/RTT
link gönderim hızını geçmemeli yoksa paket kaybına sebep olabilir.
Öyleyse W*1500*8/0.15=10*10^6 dan W yaklaşık 125 segment olur.)
• As congestion window size varies from W/2 to W, then the average
window size is 0.75W=94 (ceiling of 93.75) segments. Average throughput
is 94*1500*8/0.15 =7.52Mbps.(Pencere büyüklüğü W ve W/2 arasında
değiştiği için ortalama büyüklük 0.75W=94 segmenttir)
• 94/2 *0.15 = 7.05 seconds, as the number of RTTs (that this TCP
connections needs in order to increase its window size from W/2 to W) is
given by W/2. Recall the window size increases by one in each RTT. (Her
paket kaybında pencere yarılandığından ve daha sonra her RTT de 1
arttığında dolayı ortalama pencere büyüklüğüne ulaşmak için 94/2*0.15=
7.05 saniye gereklidir.)

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