Computer Networking: A Top-Down Approach Featuring the Internet, 5
thEdition
Solutions to Review Questions and Problems
Version Date: 1 September 2009
This document contains the solutions to review questions and problems for the 5th edition of Computer Networking: A Top-Down Approach Featuring the Internet by Jim Kurose and Keith Ross. These solutions are being made available to instructors ONLY.
Please do NOT copy or distribute this document to others (even other instructors). Please do not post any solutions on a publicly-available Web site. We’ll be happy to provide a copy (up-to-date) of this solution manual ourselves to anyone who asks.
Acknowledgments: Over the years, several students and colleagues have helped us prepare this solutions manual. Special thanks goes to Honggang Zhang, Rakesh Kumar, and Prithula Dunghel.
All material © copyright 1996-2009 by J.F. Kurose and K.W. Ross. All rights reserved
Chapter 1 Review Questions
1. There is no difference. Throughout this text, the words “host” and “end system” are used interchangeably. End systems include PCs, workstations, Web servers, mail servers, Internet-connected PDAs, WebTVs, etc.
2. Suppose Alice, an ambassador of country A wants to invite Bob, an ambassador of country B, over for dinner. Alice doesn’t simply just call Bob on the phone and say,
“come to our dinner table now”. Instead, she calls Bob and suggests a date and time.
Bob may respond by saying he’s not available that particular date, but he is available another date. Alice and Bob continue to send “messages” back and forth until they agree on a date and time. Bob then shows up at the embassy on the agreed date, hopefully not more than 15 minutes before or after the agreed time. Diplomatic protocols also allow for either Alice or Bob to politely cancel the engagement if they have reasonable excuses.
3. A networking program usually has two programs, each running on a different host, communicating with each other. The program that initiates the communication is the client. Typically, the client program requests and receives services from the server program.
4. 1. Dial-up modem over telephone line: residential; 2. DSL over telephone line:
residential or small office; 3. Cable to HFC: residential; 4. 100 Mbps switched Etherent: company; 5. Wireless LAN: mobile; 6. Cellular mobile access (for example, WAP): mobile
5. HFC bandwidth is shared among the users. On the downstream channel, all packets emanate from a single source, namely, the head end. Thus, there are no collisions in the downstream channel.
6. Current possibilities include: dial-up; DSL; cable modem; fiber-to-the-home.
7. Ethernet LANs have transmission rates of 10 Mbps, 100 Mbps, 1 Gbps and 10 Gbps.
For an X Mbps Ethernet (where X = 10, 100, 1,000 or 10,000), a user can continuously transmit at the rate X Mbps if that user is the only person sending data.
If there are more than one active user, then each user cannot continuously transmit at X Mbps.
8. Ethernet most commonly runs over twisted-pair copper wire and “thin” coaxial cable.
It also can run over fibers optic links and thick coaxial cable.
9. Dial up modems: up to 56 Kbps, bandwidth is dedicated; ISDN: up to 128 kbps, bandwidth is dedicated; ADSL: downstream channel is .5-8 Mbps, upstream channel is up to 1 Mbps, bandwidth is dedicated; HFC, downstream channel is 10-30 Mbps and upstream channel is usually less than a few Mbps, bandwidth is shared. FTTH: 2- 10Mbps upload; 10-20 Mbps download; bandwidth is not shared.
10. There are two most popular wireless Internet access technologies today:
a) Wireless LAN
In a wireless LAN, wireless users transmit/receive packets to/from a base station (wireless access point) within a radius of few tens of meters. The base station is typically connected to the wired Internet and thus serves to connect wireless users to the wired network.
b) Wide-area wireless access network
In these systems, packets are transmitted over the same wireless infrastructure used for cellular telephony, with the base station thus being managed by a telecommunications provider. This provides wireless access to users within a
radius of tens of kilometers of the base station.
11. A circuit-switched network can guarantee a certain amount of end-to-end bandwidth for the duration of a call. Most packet-switched networks today (including the Internet) cannot make any end-to-end guarantees for bandwidth.
12. In a packet switched network, the packets from different sources flowing on a link do not follow any fixed, pre-defined pattern. In TDM circuit switching, each host gets the same slot in a revolving TDM frame.
13. At time t0 the sending host begins to transmit. At time t1 = L/R1, the sending host completes transmission and the entire packet is received at the router (no propagation delay). Because the router has the entire packet at time t1, it can begin to transmit the packet to the receiving host at time t1. At time t2 = t1 + L/R2, the router completes transmission and the entire packet is received at the receiving host (again, no propagation delay). Thus, the end-to-end delay is L/R1 + L/R2.
14. A tier-1 ISP connects to all other tier-1 ISPs; a tier-2 ISP connects to only a few of the tier-1 ISPs. Also, a tier-2 ISP is a customer of one or more tier-1.
15. a) 2 users can be supported because each user requires half of the link bandwidth.
b) Since each user requires 1Mbps when transmitting, if two or fewer users transmit simultaneously, a maximum of 2Mbps will be required. Since the available bandwidth of the shared link is 2Mbps, there will be no queuing delay before the link. Whereas, if three users transmit simultaneously, the bandwidth required will be 3Mbps which is more than the available bandwidth of the shared link. In this case, there will be queuing delay before the link.
c) Probability that a given user is transmitting = 0.2
d) Probability that all three users are transmitting simultaneously = 3
(
1)
3 33
3 −
⎟⎟ −
⎠
⎜⎜ ⎞
⎝
⎛ p p
= (0.2)3 = 0.008. Since the queue grows when all the users are transmitting, the fraction of time during which the queue grows (which is equal to the probability that all three users are transmitting simultaneously) is 0.008.
16. The delay components are processing delays, transmission delays, propagation delays, and queuing delays. All of these delays are fixed, except for the queuing delays, which are variable.
17. Java Applet
18. 10msec; d/s; no; no 19. a) 500 kbps
b) 64 seconds
c) 100kbps; 320 seconds
20. End system A breaks the large file into chunks. To each chunk, it adds header generating multiple packets from the file. The header in each packet includes the address of the destination: end system B. The packet switch uses the destination address to determine the outgoing link. Asking which road to take is analogous to a packet asking which outgoing link it should be forwarded on, given the packet’s address.
21. Java Applet
22. Five generic tasks are error control, flow control, segmentation and reassembly, multiplexing, and connection setup. Yes, these tasks can be duplicated at different layers. For example, error control is often provided at more than one layer.
23. The five layers in the Internet protocol stack are – from top to bottom – the application layer, the transport layer, the network layer, the link layer, and the physical layer. The principal responsibilities are outlined in Section 1.5.1.
24. Application-layer message: data which an application wants to send and passed onto the transport layer; transport-layer segment: generated by the transport layer and encapsulates application-layer message with transport layer header; network-layer datagram: encapsulates transport-layer segment with a network-layer header; link- layer frame: encapsulates network-layer datagram with a link-layer header.
25. Routers process layers 1 through 3. (This is a little bit of a white lie, as modern routers sometimes act as firewalls or caching components, and process layer four as well.) Link layer switches process layers 1 through 2. Hosts process all five layers.
26. a) Virus
Requires some form of human interaction to spread. Classic example: E-mail viruses.
b)Worms
No user replication needed. Worm in infected host scans IP addresses and port numbers, looking for vulnerable processes to infect.
c) Trojan horse
Hidden, devious part of some otherwise useful software.
27. Creation of a botnet requires an attacker to find vulnerability in some application or system (e.g. exploiting the buffer overflow vulnerability that might exist in an application). After finding the vulnerability, the attacker needs to scan for hosts that are vulnerable. The target is basically to compromise a series of systems by exploiting that particular vulnerability. Any system that is part of the botnet can automatically scan its environment and propagate by exploiting the vulnerability. An important property of such botnets is that the originator of the botnet can remotely control and issue commands to all the nodes in the botnet. Hence, it becomes possible for the attacker to issue a command to all the nodes, that target a single node (for example, all nodes in the botnet might be commanded by the attacker to send a TCP SYN message to the target, which might result in a TCP SYN flood attack at the target).
28. Trudy can pretend to be Bob to Alice (and vice-versa) and partially or completely modify the message(s) being sent from Bob to Alice. For example, she can easily change the phrase “Alice, I owe you $1000” to “Alice, I owe you $10,000”.
Furthermore, Trudy can even drop the packets that are being sent by Bob to Alice (and vise-versa), even if the packets from Bob to Alice are encrypted.
Chapter 1 Problems
Problem 1
There is no single right answer to this question. Many protocols would do the trick.
Here's a simple answer below:
Messages from ATM machine to Server Msg name purpose
--- ---
HELO <userid> Let server know that there is a card in the ATM machine
ATM card transmits user ID to Server PASSWD <passwd> User enters PIN, which is sent to server BALANCE User requests balance
WITHDRAWL <amount> User asks to withdraw money BYE user all done
Messages from Server to ATM machine (display) Msg name purpose
--- ---
PASSWD Ask user for PIN (password)
OK last requested operation (PASSWD, WITHDRAWL) OK
ERR last requested operation (PASSWD, WITHDRAWL) in ERROR
AMOUNT <amt> sent in response to BALANCE request BYE user done, display welcome screen at ATM Correct operation:
client server
HELO (userid) ---> (check if valid userid) <--- PASSWD PASSWD <passwd> ---> (check password)
<--- OK (password is OK) BALANCE --->
<--- AMOUNT <amt>
WITHDRAWL <amt> ---> check if enough $ to cover withdrawl <--- OK
ATM dispenses $
BYE --->
<--- BYE In situation when there's not enough money:
HELO (userid) ---> (check if valid userid) <--- PASSWD PASSWD <passwd> ---> (check password)
<--- OK (password is OK) BALANCE --->
<--- AMOUNT <amt>
WITHDRAWL <amt> ---> check if enough $ to cover withdrawl
<--- ERR (not enough funds) error msg displayed
no $ given out
BYE --->
<--- BYE
Problem 2
a) A circuit-switched network would be well suited to the application described, because the application involves long sessions with predictable smooth bandwidth requirements. Since the transmission rate is known and not bursty, bandwidth can be
reserved for each application session circuit with no significant waste. In addition, we need not worry greatly about the overhead costs of setting up and tearing down a circuit connection, which are amortized over the lengthy duration of a typical application session.
b) Given such generous link capacities, the network needs no congestion control mechanism. In the worst (most potentially congested) case, all the applications simultaneously transmit over one or more particular network links. However, since each link offers sufficient bandwidth to handle the sum of all of the applications' data rates, no congestion (very little queuing) will occur.
Problem 3
a) We can have n connections between each of the four pairs of adjacent switches. This gives a maximum of 4n connections.
b) We can n connections passing through the switch in the upper-right-hand corner and another n connections passing through the switch in the lower-left-hand corner, giving a total of 2n connections.
Problem 4
Tollbooths are 75 km apart, and the cars propagate at 100km/hr. A tollbooth services a car at a rate of one car every 12 seconds.
a) There are ten cars. It takes 120 seconds, or 2 minutes, for the first tollbooth to service the 10 cars. Each of these cars has a propagation delay of 45 minutes (travel 75 km) before arriving at the second tollbooth. Thus, all the cars are lined up before the second tollbooth after 47 minutes. The whole process repeats itself for traveling between the second and third tollbooths. It also takes 2 minutes for the third tollbooth to service the 10 cars. Thus the total delay is 96 minutes.
b) Delay between tollbooths is 8*12 seconds plus 45 minutes, i.e., 46 minutes and 36 seconds. The total delay is twice this amount plus 8*12 seconds, i.e., 94 minutes and 48 seconds.
Problem 5
a) dprop =m/s seconds.
b) dtrans =L/R seconds.
c) dend−to−end =(m/s+L/R) seconds.
d) The bit is just leaving Host A.
e) The first bit is in the link and has not reached Host B.
f) The first bit has reached Host B.
g) Want
(
2.5 10)
53610 56
120 8
3 × =
= ×
= s R
m L km.
Problem 6
Consider the first bit in a packet. Before this bit can be transmitted, all of the bits in the packet must be generated. This requires
103
64 8 56
×
⋅ sec=7msec.
The time required to transmit the packet is
106
2 8 56
×
⋅ sec=224μsec.
Propagation delay = 10 msec.
The delay until decoding is
7msec +224μsec + 10msec = 17.224msec A similar analysis shows that all bits experience a delay of 17.224 msec.
Problem 7
a) 20 users can be supported.
b) p =0.1.
c) pn
(
p)
n. n− −
⎟⎟⎠
⎜⎜ ⎞
⎝
⎛ 120
120 1
d)
∑ ( )
.=
− −
⎟⎟⎠
⎜⎜ ⎞
⎝
− 20 ⎛
0
1 120
1 120
n
n n
p n p
We use the central limit theorem to approximate this probability. Let be independent random variables such that
Xj
(
X)
pP j = 1 = .
(
P “21 or more users”
)
⎟⎟⎠
⎜⎜ ⎞
⎝
⎛ ≤
−
=
∑
=
21 1 120
1 j
Xj
P
⎟⎟
⎟
⎠
⎞
⎜⎜
⎜
⎝
⎛
⋅
≤ ⋅
⋅
⋅
= −
⎟⎟⎠
⎞
⎜⎜⎝
⎛
∑
≤∑
== 120 0.1 0.9
9 9
. 0 1 . 0 120 21 12
120 120 1
1
j j
j j
X P
X P
(
2.74)
286 . 3
9 ⎟= ≤
⎠
⎜ ⎞
⎝⎛ ≤
≈P Z P Z
=0.997
when Z is a standard normal r.v. Thus P
(
“21 or more users”)
≈0.003.Problem 8 a) 10,000
b)
∑ ( )
+
=
− −
⎟⎟⎠
⎜⎜ ⎞
⎝
⎛
M
N n
n n M
p n p
M
1
1
Problem 9
The first end system requires L/R1 to transmit the packet onto the first link; the packet propagates over the first link in d1/s1; the packet switch adds a processing delay of dproc; after receiving the entire packet, the packet switch connecting the first and the second link requires L/R2 to transmit the packet onto the second link; the packet propagates over the second link in d2/s2. Similarly, we can find the delay caused by the second switch and the third link: L/R3, dproc, and d3/s3.
Adding these five delays gives
dend-end = L/R1 + L/R2 + L/R3 + d1/s1 + d2/s2 + d3/s3+ dproc+ dproc
To answer the second question, we simply plug the values into the equation to get 6 + 6 + 6 + 20+16 + 4 + 3 + 3 = 64 msec.
Problem 10
Because bits are immediately transmitted, the packet switch does not introduce any delay;
in particular, it does not introduce a transmission delay. Thus, dend-end = L/R + d1/s1 + d2/s2+ d3/s3
For the values in Problem 9, we get 6 + 20 + 16 + 4 = 46 msec.
Problem 11
The arriving packet must first wait for the link to transmit 6,750 bytes or 54,000 bits.
Since these bits are transmitted at 2 Mbps, the queuing delay is 27 msec. Generally, the queuing delay is (nL + (L - x))/R.
Problem 12
The queuing delay is 0 for the first transmitted packet, L/R for the second transmitted packet, and generally, (n-1)L/R for the nth transmitted packet. Thus, the average delay for the N packets is
(L/R + 2L/R + ... + (N-1)L/R)/N
= L/(RN) * (1 + 2 + ... + (N-1))
= L/(RN) * N(N-1)/2
= LN(N-1)/(2RN)
= (N-1)L/(2R)
Note that here we used the well-known fact that 1 + 2 + ... + N = N(N+1)/2
Problem 13
It takes LN /R seconds to transmit the N packets. Thus, the buffer is empty when a batch of N packets arrive.
The first of the N packets has no queuing delay. The 2nd packet has a queuing delay of seconds. The th packet has a delay of
R
L / n (n−1)L/R seconds.
The average delay is
2 ) 1 ( 2
) 1 ( 1 / 1
) 1
1 ( 1
0 1
= −
= −
=
−
∑
∑
−=
=
N R L N N N R n L N R R L L N n
N
n N
n
.
Problem 14
a) The transmission delay is L /R. The total delay is I R L R L I R
IL
= −
− + 1
/ )
1 ( b) Let x=L/R.
Total delay = ax x
− 1
Problem 15
Total delay
a a
R aL
R L I R L
= −
= −
= −
= −
μ μ
μ 1
/ 1
/ 1 /
1 / 1
/ .
Problem 16
The total number of packets in the system includes those in the buffer and the pack is being transmitted. So, N=10+1.
Because N =a⋅d, so (10+1)=a*(queuing delay + transmission delay). That is, 11=a*(0.01+1/100)=a*(0.01+0.01), thus, a=550 packets/sec.
Problem 17
a) There are Q nodes (the source host and the Q−1 routers). Let denote the processing delay at the th node. Let
q
dproc
q R be the transmission rate of the th link and let q q
q q
trans L R
d = / . Let dpropq be the propagation delay across the th link. Then q
[ ]
∑
=−
− = Q + +
q
q prop q
trans q
proc end
to
end d d d
d
1
.
b) Let dqueueq denote the average queuing delay at node q. Then
[ ]
∑
=−
− = Q + + +
q
q queue q
prop q
trans q
proc end
to
end d d d d
d
1
.
Problem 18 The command:
traceroute -q 20 www.eurecom.fr
will get 20 delay measurements from the issuing host to the host, www.eurecom.fr. The average and standard deviation of these 20 measurements can then be collected. Do you see any differences in your answers as a function of time of day?
Problem 19
Throughput = min{Rs, Rc, R/M}
Problem 20
If only use one path, the max throughput is given by
. }}
, , , min{
, }, , , , min{
}, , , ,
max{min{R11 R12 K R1N R12 R22 K RN2 K R1M R2M K RNM
If use all paths, the max throughput is given by .
Problem 21
robability of successfully receiving a packet is:
of transmissions needs to be performed until the packet is successfully
s -1.
Problem 22
ets call the first packet A and call the second packet B.
queued at the first link waiting
). If the second link is the bottleneck link and since both packets are sent back to back, it
resents the time needed by the second
e needed by the first packet to finish its
e know that (1) is possible as Rc < Rs. And it is clear that (1) shows that the second re is no queuing
Problem 23
0 terabytes = 40 * 1012 * 8 bits.
ill take 40 * 1012 * 8 / (100 *106 ) =3200000 seconds =
∑
= Mk
k N k
k R R
R
1min{ 1, 2,K, }
P
ps= (1-p)N. The number
received by the client is a geometric random variable with success probability ps. Thus, the average number of transmissions needed is given by: 1/ps
Then, the average number of re-transmissions needed is given by: 1/p
L
a). If the bottleneck link is the first link, then, packet B is
for the transmission of packet A. So, the packet inter-arrival time at the destination is simply L/Rs.
b
must be true that the second packet arrives at the input queue of the second link before the second link finishes the transmission of the first packet. That is,
L/Rs + L/Rs + dprop < L/Rs + dprop + L/Rc (1) The left hand side of the above inequality rep
packet to arrive at the input queue of the second link (the second link has not started transmitting the second packet yet).
The right hand side represents the tim transmission onto the second link.
W
packet must have queuing delay at the input queue of the second link.
If we send the second packet T seconds later, then we can ensure the the delay for the second packet at the second link if we have,
L/Rs + L/Rs + dprop + T >= L/Rs + dprop + L/Rc
Thus, T must be L/Rc − L/Rs .
4
So, if using the dedicated link, it w 37 days.
But with FedEx overnight delivery, you can guarantee the data arrives in one day, and it only costs you no more than $100.
Problem 24 a) 160,000 bits b) 160,000 bits
c) The bandwidth-delay product of a link is the maximum number of bits that can be in the link.
d) the width of a bit = length of link / bandwidth-delay product, so 1 bit is 125 meters long, which is longer than a football field
e) s/R
Problem 25
s/R=20000km, then R=s/20000km= 2.5*108/(2*107)= 12.5 bps
Problem 26 a) 80,000,000 bits
b) 800,000 bits, this is because that the maximum number of bits that will be in the link at any given time = min(bandwidth delay product, packet size) = 800,000 bits.
c) .25 meters
Problem 27
a) ttrans + tprop = 400 msec + 80 msec = 480 msec
b) 20 * (ttrans + 2 tprop) = 20*(20 msec + 80 msec) = 2 sec
Problem 28
Recall geostationary satellite is 36,000 kilometers away from earth surface.
a) 150 msec b) 1,500,000 bits c) 600,000,000 bits
Problem 29
Let’s suppose the passenger and his/her bags correspond to the data unit arriving to the top of the protocol stack. When the passenger checks in, his/her bags are checked, and a
tag is attached to the bags and ticket. This is additional information added in the Baggage layer if Figure 1.20 that allows the Baggage layer to implement the service or separating the passengers and baggage on the sending side, and then reuniting them (hopefully!) on the destination side. When a passenger then passes through security, and additional stamp is often added to his/her ticket, indicating that the passenger has passed through a security check. This information is used to ensure (e.g., by later checks for the security information) secure transfer of people.
Problem 30
a) Time to send message from source host to first packet switch = sec
4 10 sec
2 10 8
6
6 =
×
× . With store-and-forward switching, the total time to move message from source host to destination host = 4sec× hops3 =12sec
b) Time to send 1st packet from source host to first packet switch = . sec
1 10 sec
2 10 2
6 3
= m
×
× . Time at which 2nd packet is received at the first switch = time at which 1st packet is received at the second switch = 2×1msec=2msec c) Time at which 1st packet is received at the destination host = .
sec
3 . After this, every 1msec one packet will be received;
thus time at which last (4000th) packet is received = sec
002
. . It can be seen that delay in using message segmentation is significantly less (almost 1/3rd).
3 sec
1m × hops= m
4 sec 1
* 3999 sec
3m + m =
d) Drawbacks:
i. Packets have to be put in sequence at the destination.
ii. Message segmentation results in many smaller packets. Since header size is usually the same for all packets regardless of their size, with message segmentation the total amount of header bytes is more.
Problem 31 Java Applet
Problem 32
Time at which the 1st packet is received at the destination = +80 ×3 R
S sec. After this, one packet is received at destination every
R S+80
sec. Thus delay in sending the whole file = )
2 80 (
80) ( ) 1 (
80×3+ − × + = + × +
= +
S F R
S R
S S
F R
delay S
To calculate the value of S which leads to the minimum delay, F
S delay
dS
d =0⇒ = 40
Chapter 2 Review Questions
1. The Web: HTTP; file transfer: FTP; remote login: Telnet; Network News: NNTP;
e-mail: SMTP.
2. Network architecture refers to the organization of the communication process into layers (e.g., the five-layer Internet architecture). Application architecture, on the other hand, is designed by an application developer and dictates the broad structure of the application (e.g., client-server or P2P)
3. The process which initiates the communication is the client; the process that waits to be contacted is the server.
4. No. As stated in the text, all communication sessions have a client side and a server side. In a P2P file-sharing application, the peer that is receiving a file is typically the client and the peer that is sending the file is typically the server.
5. The IP address of the destination host and the port number of the destination socket.
6. You would use UDP. With UDP, the transaction can be completed in one roundtrip time (RTT) - the client sends the transaction request into a UDP socket, and the server sends the reply back to the client's UDP socket. With TCP, a minimum of two RTTs are needed - one to set-up the TCP connection, and another for the client to send the request, and for the server to send back the reply.
7. There are no good examples of an application that requires no data loss and timing. If you know of one, send an e-mail to the authors.
8. a) Reliable data transfer
TCP provides a reliable byte-stream between client and server but UDP does not.
b) A guarantee that a certain value for throughput will be maintained Neither
c) A guarantee that data will be delivered within a specified amount of time Neither
d) Security Neither
9. SSL operates at the application layer. The SSL socket takes unencrypted data from the application layer, encrypts it and then passes it to the TCP socket. If the
application developer wants TCP to be enhanced with SSL, she has to include the SSL code in the application.
10. A protocol uses handshaking if the two communicating entities first exchange control packets before sending data to each other. SMTP uses handshaking at the application layer whereas HTTP does not.
11. The applications associated with those protocols require that all application data be received in the correct order and without gaps. TCP provides this service whereas UDP does not.
12. When the user first visits the site, the site returns a cookie number. This cookie number is stored on the user’s host and is managed by the browser. During each subsequent visit (and purchase), the browser sends the cookie number back to the site. Thus the site knows when this user (more precisely, this browser) is visiting
the site.
13. Web caching can bring the desired content “closer” to the user, perhaps to the same LAN to which the user’s host is connected. Web caching can reduce the delay for all objects, even objects that are not cached, since caching reduces the traffic on links.
14. Issued the following command (in Windows command prompt) followed by the HTTP GET message to the “utopia.poly.edu” web server:
> telnet utopia.poly.edu 80
Since the index.html page in this web server was not modified since Fri, 18 May 2007 09:23:34 GMT, the following output was displayed when the above commands were issued on Sat, 19 May 2007. Note that the first 4 lines are the GET message and header lines input by the user and the next 4 lines (starting from HTTP/1.1 304 Not Modified) is the response from the web server.
15. FTP uses two parallel TCP connections, one connection for sending control information (such as a request to transfer a file) and another connection for actually transferring the file. Because the control information is not sent over the
same connection that the file is sent over, FTP sends control information out of band.
16. Message is sent from Alice’s host to her mail server over HTTP. Alice’s mail server then sends the message to Bob’s mail server over SMTP. Bob then transfers the message from his mail server to his host over POP3.
17.
Received:
from 65.54.246.203 (EHLO bay0-omc3-s3.bay0.hotmail.com)
(65.54.246.203) by mta419.mail.mud.yahoo.com with SMTP; Sat, 19 May 2007 16:53:51 -0700
Received:
from hotmail.com ([65.55.135.106]) by bay0-omc3-s3.bay0.hotmail.com with Microsoft SMTPSVC(6.0.3790.2668); Sat, 19 May 2007 16:52:42 - 0700
Received: from mail pickup service by hotmail.com with Microsoft SMTPSVC; Sat, 19 May 2007 16:52:41 -0700
Message-ID: <BAY130-F26D9E35BF59E0D18A819AFB9310@phx.gbl>
Received: from 65.55.135.123 by by130fd.bay130.hotmail.msn.com with HTTP;
Sat, 19 May 2007 23:52:36 GMT
From: "prithula dhungel" <prithuladhungel@hotmail.com>
To: prithula@yahoo.com Bcc:
Subject: Test mail
Date: Sat, 19 May 2007 23:52:36 +0000 Mime-Version:1.0
Content-Type: Text/html; format=flowed Return-Path: prithuladhungel@hotmail.com
Figure: A sample mail message header
Received: This header field indicates the sequence in which the SMTP servers send and receive the mail message including the respective timestamps.
In this example there are 4 “Received:” header lines. This means the mail message passed through 5 different SMTP servers before being delivered to the receiver’s mail box. The last (forth) “Received:” header indicates the mail message flow from the SMTP server of the sender to the second SMTP server in the chain of servers. The sender’s SMTP server is at address 65.55.135.123 and the second SMTP server in the chain is by130fd.bay130.hotmail.msn.com.
The third “Received:” header indicates the mail message flow from the second SMTP server in the chain to the third server, and so on.
Finally, the first “Received:” header indicates the flow of the mail message from the forth SMTP server to the last SMTP server (i.e. the receiver’s mail server) in the chain.
Message-id: The message has been given this number BAY130- F26D9E35BF59E0D18A819AFB9310@phx.gbl (by bay0-omc3- s3.bay0.hotmail.com. Message-id is a unique string assigned by the mail system when the message is first created.
From: This indicates the email address of the sender of the mail. In the given example, the sender is “prithuladhungel@hotmail.com”
To: This field indicates the email address of the receiver of the mail. In the example, the receiver is “prithula@yahoo.com”
Subject: This gives the subject of the mail (if any specified by the sender). In the example, the subject specified by the sender is “Test mail”
Date: The date and time when the mail was sent by the sender. In the example, the sender sent the mail on 19th May 2007, at time 23:52:36 GMT.
Mime-version: MIME version used for the mail. In the example, it is 1.0.
Content-type: The type of content in the body of the mail message. In the example, it is “text/html”.
Return-Path: This specifies the email address to which the mail will be sent if the receiver of this mail wants to reply to the sender. This is also used by the sender’s mail server for bouncing back undeliverable mail messages of mailer-daemon error messages. In the example, the return path is
“prithuladhungel@hotmail.com”.
18. With download and delete, after a user retrieves its messages from a POP server, the messages are deleted. This poses a problem for the nomadic user, who may want to access the messages from many different machines (office PC, home PC, etc.). In the download and keep configuration, messages are not deleted after the user retrieves the messages. This can also be inconvenient, as each time the user retrieves the stored messages from a new machine, all of non-deleted messages will be transferred to the new machine (including very old messages).
19. Yes an organization’s mail server and Web server can have the same alias for a host name. The MX record is used to map the mail server’s host name to its IP address.
20. It is not necessary that Bob will also provide chunks to Alice. Alice has to be in the top 4 neighbors of Bob for Bob to send out chunks to her; this might not occur even if Alice is provides chunks to Bob throughout a 30-second interval.
21. Alice will get her first chunk as a result of she being selected by one of her neighbors as a result of an “optimistic unchoke,” for sending out chunks to her.
22. The overlay network in a P2P file sharing system consists of the nodes participating in the file sharing system and the logical links between the nodes.
There is a logical link (an “edge” in graph theory terms) from node A to node B if there is a semi-permanent TCP connection between A and B. An overlay network does not include routers. With Gnutella, when a node wants to join the Gnutella network, it first discovers (“out of band”) the IP address of one or more nodes already in the network. It then sends join messages to these nodes. When the node receives confirmations, it becomes a member of the of Gnutella network. Nodes maintain their logical links with periodic refresh messages.
23. It is a hybrid of client server and P2P architectures:
a) There is a centralized component (the index) like in the case of a client server system.
b) Other functions (except the indexing) do not use any kind of central server. This is similar to what exists in a P2P system.
24. Mesh DHT: The advantage is to a route a message to the peer closest to the key, only one hop is required; the disadvantage is that each peer must track all other peers in the in the DHT. Circular DHT: the advantage is that each peer needs to track only a few other peers; the disadvantage is that O(N) hops are needed to route a message to a peer responsible for the key.
25. a) User location b) NAT traversal 26. a) File Distribution
b) Instant Messaging c) Video Streaming d) Distributed Computing
27. With the UDP server, there is no welcoming socket, and all data from different clients enters the server through this one socket. With the TCP server, there is a welcoming socket, and each time a client initiates a connection to the server, a new socket is created. Thus, to support n simultaneous connections, the server would need n+1 sockets.
28. For the TCP application, as soon as the client is executed, it attempts to initiate a TCP connection with the server. If the TCP server is not running, then the client will fail to make a connection. For the UDP application, the client does not initiate connections (or attempt to communicate with the UDP server) immediately upon execution
Chapter 2 Problems
Problem 1 a) F
b) T c) F d) F e) F
Problem 2
Access control commands:
USER, PASS, ACT, CWD, CDUP, SMNT, REIN, QUIT.
Transfer parameter commands:
PORT, PASV, TYPE STRU, MODE.
Service commands:
RETR, STOR, STOU, APPE, ALLO, REST, RNFR, RNTO, ABOR, DELE, RMD, MRD, PWD, LIST, NLST, SITE, SYST, STAT, HELP, NOOP.
Problem 3
Application layer protocols: DNS and HTTP
Transport layer protocols: UDP for DNS; TCP for HTTP
Problem 4
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.
Problem 5
a) The status code of 200 and the phrase OK indicate that the server was able to locate the document successfully. The reply was provided on Tuesday, 07 Mar 2008 12:39:45 Greenwich Mean Time.
b) The document index.html was last modified on Saturday 10 Dec 2005 18:27:46 GMT.
c) There are 3874 bytes in the document being returned.
d) The first five bytes of the returned document are : <!doc. The server agreed to a persistent connection, as indicated by the Connection: Keep-Alive field
Problem 6
a) Persistent connections are discussed in section 8 of RFC 2616 (the real goal of this question was to get you to retrieve and read an RFC). Sections 8.1.2 and 8.1.2.1 of the RFC indicate that either the client or the server can indicate to the other that it is going to close the persistent connection. It does so by including the including the connection-token "close" in the Connection-header field of the http request/reply.
b) HTTP does not provide any encryption services.
c) (From RFC 2616) “Clients that use persistent connections should limit the number of simultaneous connections that they maintain to a given server. A single-user client SHOULD NOT maintain more than 2 connections with any server or proxy.”
d) Yes. (From RFC 2616) “A client might have started to send a new request at the same time that the server has decided to close the "idle" connection. From the server's point of view, the connection is being closed while it was idle, but from the client's point of view, a request is in progress.”
Problem 7
The total amount of time to get the IP address is
RTTn
RTT
RTT1+ 2 +L+ .
Once the IP address is known, elapses to set up the TCP connection and another elapses to request and receive the small object. The total response time is
RTTO
RTTO
n
o RTT RTT RTT
RTT + 1+ 2 +L+ 2
Problem 8 a)
o o
n RTT RTT
RTT
RTT1+ L+ +2 +8⋅2 =18RTTo +RTT1 +L+RTTn. b)
o o
n RTT RTT
RTT
RTT1+L+ +2 +2⋅2 =6RTTo +RTT1 +L+RTTn c) RTT1+L+RTTn +2RTTo +RTTo
=3RTTo+RTT1+L+RTTn.
Problem 9
a) The time to transmit an object of size L over a link or rate R is L/R. The average time is the average size of the object divided by R:
Δ= (850,000 bits)/(15,000,000 bits/sec) = .0567 sec
The traffic intensity on the link is given by βΔ=(16 requests/sec)(.0567 sec/request) = 0.907. Thus, the average access delay is (.0567 sec)/(1 - .907) ≈ .6 seconds. The total average response time is therefore .6 sec + 3 sec = 3.6 sec.
b) The traffic intensity on the access link is reduced by 60% since the 60% of the requests are satisfied within the institutional network. Thus the average access delay is (.0567 sec)/[1 – (.4)(.907)] = .089 seconds. The response time is approximately zero if the request is satisfied by the cache (which happens with probability .6); the average response time is .089 sec + 3 sec = 3.089 sec for cache misses (which happens 40% of the time). So the average response time is (.6)(0 sec) + (.4)(3.089 sec) = 1.24 seconds. Thus the average response time is reduced from 3.6 sec to 1.24 sec.
Problem 10
Note that each downloaded object can be completely put into one data packet. Let Tp denote the one-way propagation delay between the client and the server.
First consider parallel downloads via non-persistent connections. Parallel download would allow 10 connections share the 150 bits/sec bandwidth, thus each gets just 15 bits/sec. Thus, the total time needed to receive all objects is given by:
(200/150+Tp + 200/150 +Tp + 200/150+Tp + 100,000/150+ Tp )
+ (200/(150/10)+Tp + 200/(150/10) +Tp + 200/(150/10)+Tp + 100,000/(150/10)+ Tp )
= 7377 + 8*Tp (seconds)
Then consider persistent HTTP connection. The total time needed is give by:
(200/150+Tp + 200/150 +Tp + 200/150+Tp + 100,000/150+ Tp ) + 10*(200/150+Tp + 100,000/150+ Tp )
=7351 + 24*Tp (seconds)
Assume the speed of light is 300*106 m/sec, then Tp=10/(300*106)=0.03 microsec. Tp is negligible compared with transmission delay.
Thus, we see that the persistent HTTP does not have significant gain (less than 1 percent) over the non-persistent case with parallel download.
Problem 11
a). Yes, because Bob has more connections, so he can proportionally get more aggregate bandwidth share out of the total link bandwidth.
b) Yes, Bob still needs to perform parallel download, otherwise he will get less bandwidth share than other four users. In fact, all users might tend to open more connections in order to gain more bandwidth share.
Problem 12 TCPServer.java import java.io.*;
import java.net.*;
class TCPServer {
public static void main(String argv[]) throws Exception {
String clientSentence;
ServerSocket welcomeSocket = new ServerSocket(6789);
while(true) {
Socket connectionSocket = welcomeSocket.accept();
BufferedReader inFromClient = new BufferedReader(new InputStreamReader(connectionSocket.getInputStream( ) ) );
clientSentence = inFromClient.readLine();
System.out.println(“RECEIVED FROM CLIENT : ” +
clientSentence + “\n”);
} }
}
Problem 13
The MAIL FROM: in SMTP is a message from the SMTP client that identifies the sender of the mail message to the SMTP server. The From: on the mail message itself is NOT an SMTP message, but rather is just a line in the body of the mail message.
Problem 14
SMTP uses a line containing only a period to mark the end of a message body.
HTTP uses “Content-Length header field” to indicate the length of a message body.
No, HTTP cannot use the method used by SMTP, because HTTP message could be binary data, whereas in SMTP, the message body must be in 7-bit ASCII format.
Problem 15
MTA stands for Mail Transfer Agents. A mail is forwarded by a source to a MTA and then it follows a sequence of MTAs to reach the receiver’s mail reader.
We see that this spam email follows a chain of MTAs. An honest MTA should report where it receives the message. Notice that in this email, “asusus-4b96
([58.88.21.177])” does not report where it receives the email. As we assume that the only the originator is dishonest, so “asusus-4b96 ([58.88.21.177])” must be the originator.
Problem 16
UIDL abbreviates “unique-ID listing”. When a POP3 client issues the UIDL command, the server responds with the unique message ID for all of the messages present in the users mailbox. This command is useful for “download and keep”. By keeping a file that lists the messages retrieved in earlier sessions, the client can use the UIDL command to determine which messages on the server have already been seen.
Problem 17 a) C: dele 1 C: retr 2
S: (blah blah … S: ………..blah) S: .
C: dele 2 C: quit
S: +OK POP3 server signing off
b) C: retr 2 S: blah blah … S: ………..blah S: .
C: quit
S: +OK POP3 server signing off c) C: list
S: 1 498 S: 2 912 S: . C: retr 1 S: blah …..
S: ….blah S: . C: retr 2 S: blah blah … S: ………..blah S: .
C: quit
S: +OK POP3 server signing off
Problem 18
a) For a given input of domain name (such as ccn.com), IP address or network administrator name, whois database can be used to locate the corresponding registrar, whois server, DNS server, and so on.
b) NS4.YAHOO.COM from www.register.com; NS1.MSFT.NET from ww.register.com c) Local Domain: www.mindspring.com
Web servers : www.mindspring.com
207.69.189.21, 207.69.189.22, 207.69.189.23, 207.69.189.24,
207.69.189.25, 207.69.189.26, 207.69.189.27, 207.69.189.28
Mail Servers : mx1.mindspring.com (207.69.189.217) mx2.mindspring.com (207.69.189.218) mx3.mindspring.com (207.69.189.219) mx4.mindspring.com (207.69.189.220) Name Servers: itchy.earthlink.net (207.69.188.196)
scratchy.earthlink.net (207.69.188.197) www.yahoo.com
Web Servers: www.yahoo.com (216.109.112.135, 66.94.234.13)
Mail Servers: a.mx.mail.yahoo.com (209.191.118.103) b.mx.mail.yahoo.com (66.196.97.250)
c.mx.mail.yahoo.com (68.142.237.182, 216.39.53.3) d.mx.mail.yahoo.com (216.39.53.2)
e.mx.mail.yahoo.com (216.39.53.1)
f.mx.mail.yahoo.com (209.191.88.247, 68.142.202.247) g.mx.mail.yahoo.com (209.191.88.239, 206.190.53.191) Name Servers: ns1.yahoo.com (66.218.71.63)
ns2.yahoo.com (68.142.255.16) ns3.yahoo.com (217.12.4.104) ns4.yahoo.com (68.142.196.63) ns5.yahoo.com (216.109.116.17) ns8.yahoo.com (202.165.104.22) ns9.yahoo.com (202.160.176.146) www.hotmail.com
Web Servers: www.hotmail.com (64.4.33.7, 64.4.32.7)
Mail Servers: mx1.hotmail.com (65.54.245.8, 65.54.244.8, 65.54.244.136) mx2.hotmail.com (65.54.244.40, 65.54.244.168, 65.54.245.40) mx3.hotmail.com (65.54.244.72, 65.54.244.200, 65.54.245.72) mx4.hotmail.com (65.54.244.232, 65.54.245.104, 65.54.244.104) Name Servers: ns1.msft.net (207.68.160.190)
ns2.msft.net (65.54.240.126) ns3.msft.net (213.199.161.77) ns4.msft.net (207.46.66.126) ns5.msft.net (65.55.238.126) d) The yahoo web server has multiple IP addresses
www.yahoo.com (216.109.112.135, 66.94.234.13)
e) The address range for Polytechnic University: 128.238.0.0 – 128.238.255.255
f) An attacker can use the whois database and nslookup tool to determine the IP address ranges, DNS server addresses, etc., for the target institution.
g) By analyzing the source address of attack packets, the victim can use whois to obtain information about domain from which the attack is coming and possibly inform the administrators of the origin domain.
Problem 19 a)
The following delegation chain is used for gaia.cs.umass.edu
a.root-servers.net
E.GTLD-SERVERS.NET ns1.umass.edu(authoritative) First command:
dig +norecurse @a.root-servers.net any gaia.cs.umass.edu
;; AUTHORITY SECTION:
edu. 172800 IN NS E.GTLD-SERVERS.NET.
edu. 172800 IN NS A.GTLD-SERVERS.NET.
edu. 172800 IN NS G3.NSTLD.COM.
edu. 172800 IN NS D.GTLD-SERVERS.NET.
edu. 172800 IN NS H3.NSTLD.COM.
edu. 172800 IN NS L3.NSTLD.COM.
edu. 172800 IN NS M3.NSTLD.COM.
edu. 172800 IN NS C.GTLD-SERVERS.NET.
Among all returned edu DNS servers, we send a query to the first one.
dig +norecurse @E.GTLD-SERVERS.NET any gaia.cs.umass.edu umass.edu. 172800 IN NS ns1.umass.edu.
umass.edu. 172800 IN NS ns2.umass.edu.
umass.edu. 172800 IN NS ns3.umass.edu.
Among all three returned authoritative DNS servers, we send a query to the first one.
dig +norecurse @ns1.umass.edu any gaia.cs.umass.edu gaia.cs.umass.edu. 21600 IN A 128.119.245.12 b) The answer for google.com could be:
a.root-servers.net
E.GTLD-SERVERS.NET ns1.google.com(authoritative)
Problem 20
We can periodically take a snapshot of the DNS caches in those local DNS servers. The Web server that appears most frequently in the DNS caches is the most popular server.
This is because if more users are interested in a Web server, then DNS requests for that server are more frequently sent by users. Thus, that Web server will appear in the DNS caches more frequently.
For a complete measurement study, see:
Craig E. Wills, Mikhail Mikhailov, Hao Shang
“Inferring Relative Popularity of Internet Applications by Actively Querying DNS Caches”, in IMC'03, October 27-29, 2003, Miami Beach, Florida, USA
Problem 21
Yes, we can use dig to query that Web site in the local DNS server.
For example, “dig cnn.com” will return the query time for finding cnn.com. If cnn.com is just accessed a couple of seconds ago, an entry for cnn.com is cached in the local DNS cache, so the query time is 0 msec. Otherwise, the query time is large.
Problem 22
For calculating the minimum distribution time for client-server distribution, we use the following formula:
Dcs = max {NF/us, F/dmin}
Similarly, for calculating the minimum distribution time for P2P distribution, we use the following formula:
)}
u , NF/(u
, F/d max{F/u D
N
1 i
i s
min P s
P
∑
=
+
2 =
Where, F = 15 Gbits = 15 * 1024 Mbits us = 30 Mbps
dmin = di = 2 Mbps
Note, 300Kbps = 300/1024 Mbps.
Client Server
N
10 100 1000
300 Kbps 7680 51200 512000
700 Kbps 7680 51200 512000
u
2 Mbps 7680 51200 512000
Peer to Peer
N
10 100 1000
300 Kbps 7680 25904 47559
700 Kbps 7680 15616 21525
u
2 Mbps 7680 7680 7680
Problem 23
a) Consider a distribution scheme in which the server sends the file to each client, in parallel, at a rate of a rate of us/N. Note that this rate is less than each of the client’s download rate, since by assumption us/N ≤ dmin. Thus each client can also receive at rate
us/N. Since each client receives at rate us/N, the time for each client to receive the entire file is F/( us/N) = NF/ us. Since all the clients receive the file in NF/ us, the overall distribution time is also NF/ us.
b) Consider a distribution scheme in which the server sends the file to each client, in parallel, at a rate of dmin. Note that the aggregate rate, N dmin, is less than the server’s link rate us, since by assumption us/N ≥ dmin. Since each client receives at rate dmin, the time for each client to receive the entire file is F/ dmin. Since all the clients receive the file in this time, the overall distribution time is also F/ dmin.
c) From Section 2.6 we know that DCS ≥ max {NF/us, F/dmin} (Equation 1)
Suppose that us/N ≤ dmin. Then from Equation 1 we have DCS ≥ NF/us . But from (a) we have DCS ≤ NF/us . Combining these two gives:
DCS = NF/us when us/N ≤ dmin. (Equation 2) We can similarly show that:
DCS =F/dmin when us/N ≥ dmin (Equation 3).
Combining Equation 2 and Equation 3 gives the desired result.
Problem 24
a) Define u = u1 + u2 + ….. + uN. By assumption us <= (us + u)/N Equation 1
Divide the file into N parts, with the ith part having size (ui/u)F. The server transmits the ith part to peer i at rate ri = (ui/u)us. Note that r1 + r2 + ….. + rN = us, so that the aggregate server rate does not exceed the link rate of the server. Also have each peer i forward the bits it receives to each of the N-1 peers at rate ri. The aggregate
forwarding rate by peer i is (N-1)ri. We have (N-1)ri = (N-1)(usui)/u <= ui,
where the last inequality follows from Equation 1. Thus the aggregate forwarding rate of peer i is less than its link rate ui.
In this distribution scheme, peer i receives bits at an aggregate rate of
s j i j
i r u
r +
∑
=<>
Thus each peer receives the file in F/us.
b) Again define u = u1 + u2 + ….. + uN. By assumption us >= (us + u)/N Equation 2 Let ri = ui/(N-1) and
rN+1 = (us – u/(N-1))/N
In this distribution scheme, the file is broken into N+1 parts. The server sends bits from the ith part to the ith peer (i = 1, …., N) at rate ri. Each peer i forwards the bits arriving at rate ri to each of the other N-1 peers. Additionally, the server sends bits from the (N+1) st part at rate rN+1 to each of the N peers. The peers do not forward the bits from the (N+1)st part.
The aggregate send rate of the server is
r1+ …. + rN + N rN+1 = u/(N-1) + us – u/(N-1) = us
Thus, the server’s send rate does not exceed its link rate. The aggregate send rate of peer i is
(N-1)ri = ui
Thus, each peer’s send rate does not exceed its link rate.
In this distribution scheme, peer i receives bits at an aggregate rate of
N u u N N
u u N
u j r r
r s s
i j N
i+ 1+ ∑ = /( −1)+( − /( −1))/ =( + )/
<>
+
Thus each peer receives the file in NF/(us+u).
(For simplicity, we neglected to specify the size of the file part for i = 1, …., N+1. We now provide that here. Let Δ = (us+u)/N be the distribution time. For i = 1, …, N, the ith file part is Fi = ri Δ bits. The (N+1)st file part is FN+1 = rN+1 Δ bits. It is straightforward to show that F1+ ….. + FN+1 = F.)
c) The solution to this part is similar to that of 17 (c). We know from section 2.6 that u)}
, NF/(u max{F/u
DP2P >= s s+
Combining this with (a) and (b) gives the desired result.
Problem 25
There are N nodes in the overlay network. There are N(N-1)/2 edges.
Problem 26
Yes. His first claim is possible, as long as there are enough peers staying in the swarm for a long enough time. Bob can always receive data through optimistic unchoking by other peers.
His second claim is also true. He can run a client on each machine, and let each client do
“free-riding”, and combine those collected chunks from different machines into a single file. He can even write a small scheduling program to let different machines only asking for different chunks of the file. This is actually a kind of Sybil attack in P2P networks.
Problem 27 a).
Note that we assume nb>=na. )
, (
) ,
(
b a b a
n N C
n n n N
C − −
, where C(N, n) is the notation for combination, which means the number of ways of choosing n out of N.
b). p(na)=
) , (
) ,
( 1
1
b a b a N
n
n
C N n
n n n N C
a
N
b
−
∑
−−
= .
c). prob= 5
1 0
)) 1 (
(
1
N an
n N p
a
∑
−=
−
.For a complete analysis, see:
Donyu Qiu and R. Srikant.
Modeling and Performance Analsysis of BitTorrent-Like Peer-to-Peer Networks.
ACM Sigcomm 2004, Portland, Oregon, USA
Problem 28
Peer 3 learns that peer 5 has just left the system, so Peer 3 asks its first successor (peer 4) for the identifier of its immediate successor (peer 8). Then peer 3 will make peer 8 as its second successor.
Note that peer 3 knows that peer 5 was originally the first successor of peer 4, so peer 3 would wait until peer 4 finishes updating its first successor.
Problem 29
Peer 6 would first send peer 15 a message, saying “what will be peer 6’s predecessor and successor?” This message gets forwarded through the DHT until it reaches peer 5, who realizes that it will be 6’s predecessor and that its current successor, peer 8, will become 6’s successor. Next, peer 5 sends this predecessor and successor information back to 6.
Peer 6 can now join the DHT by making peer 8 its successor and by notifying peer 5 that it should change its immediate successor to 6.
