程序代写代做 C go algorithm cache data structure graph chain clock html DHCP assembly database game FTP dns Computer Networking: A Top-Down Approach, 6th Edition

Computer Networking: A Top-Down Approach, 6th Edition
Solutions to Review Questions and Problems
Version Date: May 2012
This document contains the solutions to review questions and problems for the 5th edition of Computer Networking: A Top-Down Approach 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, Prithula Dhungel, and Vijay Annapureddy. Also thanks to all the readers who have made suggestions and corrected errors.
All material © copyright 1996-2012 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, PDAs, Internet-connected game consoles, etc.
2. From Wikipedia: Diplomatic protocol is commonly described as a set of international courtesy rules. These well-established and time-honored rules have made it easier for nations and people to live and work together. Part of protocol has always been the acknowledgment of the hierarchical standing of all present. Protocol rules are based on the principles of civility.
3. Standards are important for protocols so that people can create networking systems and products that interoperate.
4. 1. Dial-up modem over telephone line: home; 2. DSL over telephone line: home or small office; 3. Cable to HFC: home; 4. 100 Mbps switched Ethernet: enterprise; 5. Wifi (802.11): home and enterprise: 6. 3G and 4G: wide-area wireless.
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. In most American cities, the current possibilities include: dial-up; DSL; cable modem; fiber-to-the-home.
7. 8.
9.
Ethernet LANs have transmission rates of 10 Mbps, 100 Mbps, 1 Gbps and 10 Gbps.
Today, Ethernet most commonly runs over twisted-pair copper wire. It also can run over fibers optic links.
Dial up modems: up to 56 Kbps, bandwidth is dedicated; ADSL: up to 24 Mbps downstream and 2.5 Mbps upstream, bandwidth is dedicated; HFC, rates up to 42.8 Mbps and upstream rates of up to 30.7 Mbps, bandwidth is shared. FTTH: 2-10Mbps upload; 10-20 Mbps download; bandwidth is not shared.
10. There are two popular wireless Internet access technologies today:
a) Wifi (802.11) In a wireless LAN, wireless users transmit/receive packets to/from an base station (i.e., 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) 3G and 4G wide-area wireless access networks. 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. 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.
12. 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. FDM requires sophisticated analog hardware to shift signal into appropriate frequency bands.
13. 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
3
d) Probability that all three users are transmitting simultaneously = 3p3 1 p33

= (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.
14. If the two ISPs do not peer with each other, then when they send traffic to each other they have to send the traffic through a provider ISP (intermediary), to which they have to pay for carrying the traffic. By peering with each other directly, the two ISPs can reduce their payments to their provider ISPs. An Internet Exchange Points (IXP) (typically in a standalone building with its own switches) is a meeting point where multiple ISPs can connect and/or peer together. An ISP earns its money by charging each of the the ISPs that connect to the IXP a relatively small fee, which may depend on the amount of traffic sent to or received from the IXP.
15. Google’s private network connects together all its data centers, big and small. Traffic between the Google data centers passes over its private network rather than over the public Internet. Many of these data centers are located in, or close to, lower tier ISPs. Therefore, when Google delivers content to a user, it often can bypass higher tier ISPs. What motivates content providers to create these networks? First, the content provider has more control over the user experience, since it has to use few intermediary ISPs. Second, it can save money by sending less traffic into provider

networks. Third, if ISPs decide to charge more money to highly profitable content providers (in countries where net neutrality doesn’t apply), the content providers can avoid these extra payments.
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. a) 1000 km, 1 Mbps, 100 bytes b) 100 km, 1 Mbps, 100 bytes
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. It adds header to each chunk, thereby generating multiple packets from the file. The header in each packet includes the IP address of the destination (end system B). The packet switch uses the destination IP address in the packet 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 destination address.
21. The maximum emission rate is 500 packets/sec and the maximum transmission rate is 350 packets/sec. The corresponding traffic intensity is 500/350 =1.43 > 1. Loss will eventually occur for each experiment; but the time when loss first occurs will be different from one experiment to the next due to the randomness in the emission process.
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 network, link and physical layers (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 Transport layer as well.) Link layer switches process link and physical layers (layers 1 through2). 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.
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
——–
HELO
PASSWD BALANCE
WITHDRAWL
purpose
——-
Let server know that there is a card in the ATM machine
ATM card transmits user ID to Server
User enters PIN, which is sent to server
User requests balance
User asks to withdraw money

BYE user all done
Messages from Server to ATM machine (display)
Msg name
——–
PASSWD
OK
ERR
AMOUNT
BYE
Correct operation:
client
HELO (userid)
PASSWD BALANCE
WITHDRAWL
ATM dispenses $
BYE
purpose
——-
Ask user for PIN (password)
last requested operation (PASSWD, WITHDRAWL) OK
last requested operation (PASSWD, WITHDRAWL) in ERROR
sent in response to BALANCE request
user done, display welcome screen at ATM
server
(check if valid userid)
PASSWD
(check password)
OK (password is OK)
AMOUNT
check if enough $ to cover
withdrawl
OK
BYE
(check if valid userid)
PASSWD
(check password)
OK (password is OK)
AMOUNT
check if enough $ to cover
ERR (not enough funds)
BYE
In situation when there’s not enough money:
HELO (userid)
PASSWD BALANCE
WITHDRAWL
withdrawl
————–>
<------------- -------------->
<------------- -------------->
<------------- -------------->
<------------- error msg displayed no $ given out BYE -------------->
<------------- -------------->
<------------- -------------->
<------------- -------------->
<------------- -------------->
<------------- -------------->
<------------- Problem 2 At time N*(L/R) the first packet has reached the destination, the second packet is stored in the last router, the third packet is stored in the next-to-last router, etc. At time N*(L/R) + L/R, the second packet has reached the destination, the third packet is stored in the last router, etc. Continuing with this logic, we see that at time N*(L/R) + (P-1)*(L/R) = (N+P-1)*(L/R) all packets have reached the destination. Problem 3 a) A circuit-switched network would be well suited to the application, 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 without significant waste. In addition, the overhead costs of setting up and tearing down connections are amortized over the lengthy duration of a typical application session. b) In the worst case, all the applications simultaneously transmit over one or more network links. However, since each link has sufficient bandwidth to handle the sum of all of the applications' data rates, no congestion (very little queuing) will occur. Given such generous link capacities, the network does not need congestion control mechanisms. Problem 4 a) Between the switch in the upper left and the switch in the upper right we can have 4 connections. Similarly we can have four connections between each of the 3 other pairs of adjacent switches. Thus, this network can support up to 16 connections. b) We can 4 connections passing through the switch in the upper-right-hand corner and another 4 connections passing through the switch in the lower-left-hand corner, giving a total of 8 connections. c) Yes. For the connections between A and C, we route two connections through B and two connections through D. For the connections between B and D, we route two connections through A and two connections through C. In this manner, there are at most 4 connections passing through any link. Problem 5 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 6 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 mLs 120 2.5108536km. R 56103 Problem 7 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 56  8 sec=7msec. 64103 The time required to transmit the packet is 568 sec=224sec. 2106 Propagation delay = 10 msec. The delay until decoding is 7msec + 224 sec + 10msec = 17.224msec A similar analysis shows that all bits experience a delay of 17.224 msec. Problem 8 a) 20 users can be supported. b) p0.1. 120  nn 120n c) p1p .  20 120 nn 120n d)1 p 1p . n0   We use the central limit theorem to approximate this probability. Let X j be independent random variables such that PX j  1 p . 120  P “21 or more users” 1 PX  21 120  120X 12 PXj 21P j1 j    j1 j   9 1200.10.9   j1   1200.10.9  9  P  Z  3 . 2 8 6   P  Z  2 . 7 4   0.997 when Z is a standard normal r.v. Thus P “21 or more users”   0.003 . Problem 9 a) 10,000 M  M  b)  n pn1 pMn nN1  Problem 10 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 11 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 10, we get 6 + 20 + 16 + 4 = 46 msec. Problem 12 The arriving packet must first wait for the link to transmit 4.5 *1,500 bytes = 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 13 a) 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: 1 + 2 + ....... + N = N(N+1)/2 b) It takes LN / R seconds to transmit the N packets. Thus, the buffer is empty when a each batch of N packets arrive. Thus, the average delay of a packet across all batches is the average delay within one batch, i.e., (N-1)L/2R. Problem 14 a) The transmission delay is L / R . The total delay is IL  L  L/R R(1I) R 1I b) LetxL/R. Total delay = x 1ax For x=0, the total delay =0; as we increase x, total delay increases, approaching infinity as x approaches 1/a. Problem 15 Total delay  L / R  L / R  1/   1 . 1I 1aL/R 1a/ a Problem 16 The total number of packets in the system includes those in the buffer and the packet that 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 d q denote the proc processing delay at the q th node. Let Rq be the transmission rate of the q th link and let d q trans d b) Let d q queue d qqq Qd d d .  L / Rq . Let d q be the propagation delay across the q th link. Then prop endtoend q1 proc trans prop endtoend q1 proc trans prop queue denote the average queuing delay at node q . Then Q dq dq dq dq . Problem 18 On linux you can use the command traceroute www.targethost.com and in the Windows command prompt you can use tracert www.targethost.com In either case, you will get three delay measurements. For those three measurements you can calculate the mean and standard deviation. Repeat the experiment at different times of the day and comment on any changes. Here is an example solution: Traceroutes between San Diego Super Computer Center and www.poly.edu a) The average (mean) of the round-trip delays at each of the three hours is 71.18 ms, 71.38 ms and 71.55 ms, respectively. The standard deviations are 0.075 ms, 0.21 ms, 0.05 ms, respectively. b) Inthisexample,thetracerouteshave12routersinthepathateachofthethreehours. No, the paths didn’t change during any of the hours. c) Traceroute packets passed through four ISP networks from source to destination. Yes, in this experiment the largest delays occurred at peering interfaces between adjacent ISPs. Traceroutes from www.stella-net.net (France) to www.poly.edu (USA). d) The average round-trip delays at each of the three hours are 87.09 ms, 86.35 ms and 86.48 ms, respectively. The standard deviations are 0.53 ms, 0.18 ms, 0.23 ms, respectively. In this example, there are 11 routers in the path at each of the three hours. No, the paths didn’t change during any of the hours. Traceroute packets passed three ISP networks from source to destination. Yes, in this experiment the largest delays occurred at peering interfaces between adjacent ISPs. Problem 19 An example solution: Traceroutes from two different cities in France to New York City in United States a) In these traceroutes from two different cities in France to the same destination host in United States, seven links are in common including the transatlantic link. b) In this example of traceroutes from one city in France and from another city in Germany to the same host in United States, three links are in common including the transatlantic link. Traceroutes to two different cities in China from same host in United States c) Five links are common in the two traceroutes. The two traceroutes diverge before reaching China Problem 20 Throughput = min{Rs, Rc, R/M} Problem 21 If only use one path, the max throughput is given by: max{min{R1,R1,,R1 },min{R2,R2,,R2 },,min{RM ,RM ,,RM }}. 12N12N12N M If use all paths, the max throughput is given by min{Rk , Rk ,, Rk }. k1 Problem 22 Probability of successfully receiving a packet is: ps= (1-p)N. The number of transmissions needed to be performed until the packet is successfully 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/ps -1. Problem 23 Let’s call the first packet A and call the second packet B. a) If the bottleneck link is the first link, then packet B is queued at the first link waiting for the transmission of packet A. So the packet inter-arrival time at the destination is simply L/Rs. b) If the second link is the bottleneck link and both packets are sent back to back, it 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 12N The left hand side of the above inequality represents the time needed by the second 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 time needed by the first packet to finish its transmission onto the second link. If we send the second packet T seconds later, we will ensure that there is no queuing delay for the second packet at the second link if we have: L/Rs +L/Rs+dprop +T>=L/Rs +dprop +L/Rc Thus, the minimum value of T is L/Rc  L/Rs .
Problem 24
40 terabytes = 40 * 1012 * 8 bits. So, if using the dedicated link, it will take 40 * 1012 * 8 / (100 *106 ) =3200000 seconds = 37 days. But with FedEx overnight delivery, you can guarantee the data arrives in one day, and it should cost less than $100.
Problem 25
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 26
s/R=20000km, then R=s/20000km= 2.5*108/(2*107)= 12.5 bps Problem 27
a) 80,000,000 bits
b) 800,000bits,thisisbecausethatthemaximumnumberofbitsthatwillbeinthelink
at any given time = min(bandwidth delay product, packet size) = 800,000 bits.
c) .25 meters

Problem 28
a) ttrans + tprop = 400 msec + 80 msec = 480 msec.
b) 20 * (ttrans + 2 tprop) = 20*(20 msec + 80 msec) = 2 sec.
c) Breaking up a file takes longer to transmit because each data packet and its
corresponding acknowledgement packet add their own propagation delays.
Problem 29
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 30
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 31
8106
a) Time to send message from source host to first packet switch = 2106 sec  4sec
With store-and-forward switching, the total time to move message from source host
to destination host = 4sec3 hops 12sec
b) Time to send 1st packet from source host to first packet switch = .
1104 nd
2106 sec5msec.Timeatwhich2 packetisreceivedatthefirstswitch=time
at which 1st packet is received at the second switch = 2  5m sec  10 m sec
c) Time at which 1st packet is received at the destination host = 5 m sec 3 hops  15 m sec . After this, every 5msec one packet will be received; thus
time at which last (800th) packet is received = 15 m sec 799 * 5m sec  4.01sec . It
can be seen that delay in using message segmentation is significantly less (almost 1/3rd).

d)
i. Without message segmentation, if bit errors are not tolerated, if there is a
single bit error, the whole message has to be retransmitted (rather than a single
packet).
ii. Without message segmentation, huge packets (containing HD videos, for
example) are sent into the network. Routers have to accommodate these huge packets. Smaller packets have to queue behind enormous packets and suffer unfair delays.
e)
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 32
Yes, the delays in the applet correspond to the delays in the Problem 31.The propagation delays affect the overall end-to-end delays both for packet switching and message switching equally.
Problem 33
There are F/S packets. Each packet is S=80 bits. Time at which the last packet is received at the first router is S  80  F sec. At this time, the first F/S-2 packets are at the
RS
destination, and the F/S-1 packet is at the second router. The last packet must then be transmitted by the first router and the second router, with each transmission taking
S  80 sec. Thus delay in sending the whole file is delay  S  80  ( F  2) RRS
To calculate the value of S which leads to the minimum delay,
d delay0S 40F dS
Problem 34
The circuit-switched telephone networks and the Internet are connected together at “gateways”. When a Skype user (connected to the Internet) calls an ordinary telephone, a circuit is established between a gateway and the telephone user over the circuit switched network. The skype user’s voice is sent in packets over the Internet to the gateway. At the gateway, the voice signal is reconstructed and then sent over the circuit. In the other direction, the voice signal is sent over the circuit switched network to the gateway. The gateway packetizes the voice signal and sends the voice packets to the Skype user.

Chapter 2 Review Questions
1. The Web: HTTP; file transfer: FTP; remote login: Telnet; e-mail: SMTP; BitTorrent file sharing: BitTorrent protocol
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. 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 socket in the destination process.
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. One such example is remote word processing, for example, with Google docs. However, because Google docs runs over the Internet (using TCP), timing guarantees are not provided.
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) Confidentiality (via encryption) 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 server creates a unique identification number, creates an entry in its back-end database, and returns this identification number as 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, possibly 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. Telnet is not available in Windows 7 by default. to make it available, go to Control Panel, Programs and Features, Turn Windows Features On or Off, Check Telnet client. To start Telnet, in Windows command prompt, issue the following command
> telnet webserverver 80
where “webserver” is some webserver. After issuing the command, you have established a TCP connection between your client telnet program and the web server. Then type in an HTTP GET message. An example is given below:
Since the index.html page in this web server was not modified since Fri, 18 May 2007 09:23:34 GMT, and the above commands were issued on Sat, 19 May 2007, the server returned “304 Not Modified”. Note that the first 4 lines are the GET message and header lines inputed 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. The message is first 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:
Received:
Received: Message-ID: Received:
From:
To:
Bcc:
Subject:
Date:
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 messages from the forth SMTP server to the last SMTP server (i.e. the receiver’s mail server) in the chain.
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
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
from mail pickup service by hotmail.com with Microsoft SMTPSVC; Sat, 19 May 2007 16:52:41 -0700
from 65.55.135.123 by by130fd.bay130.hotmail.msn.com with HTTP; Sat, 19 May 2007 23:52:36 GMT
“prithula dhungel” prithula@yahoo.com
Test mail
Sat, 19 May 2007 23:52:36 +0000

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. You should be able to see the sender’s IP address for a user with an .edu email address. But you will not be able to see the sender’s IP address if the user uses a gmail account.
21. 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 provides chunks to Bob throughout a 30-second interval.

22. Recall that in BitTorrent, a peer picks a random peer and optimistically unchokes the peer for a short period of time. Therefore, Alice will eventually be optimistically unchoked by one of her neighbors, during which time she will receive chunks from that neighbor.
23. 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.
24. Mesh DHT: The advantage is in order to a route a message to the peer (with ID) that is closest to the key, only one hop is required; the disadvantage is that each peer must track all other peers 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 the peer that is closest to the key.
25.
a) File Distribution
b) Instant Messaging
c) Video Streaming
d) Distributed Computing
26. 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.
27. 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 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 : = (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
rirN 1 rju/(N1)(usu/(N1))/N(usu)/N ji
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 DP2P max{F/us,NF/(usu)}
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 host, let each client “free-ride,” and combine the collected chunks from the different hosts into a single file. He can even write a small scheduling program to make the different hosts ask for different chunks of the file. This is actually a kind of Sybil attack in P2P networks.
Problem 27
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). Peer 3 will then make peer 8 its second successor.
Problem 28
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.
Problem 29
For each key, we first calculate the distances (using d(k,p)) between itself and all peers, and then store the key in the peer that is closest to the key (that is, with smallest distance value).
Problem 30
Yes, randomly assigning keys to peers does not consider the underlying network at all, so it very likely causes mismatches.
Such mismatches may degrade the search performance. For example, consider a logical path p1 (consisting of only two logical links): ABC, where A and B are neighboring peers, and B and C are neighboring peers. Suppose that there is another logical path p2 from A to C (consisting of 3 logical links): ADEC.
It might be the case that A and B are very far away physically (and separated by many routers), and B and C are very far away physically (and separated by many routers). But

it may be the case that A, D, E, and C are all very close physically (and all separated by few routers). In other words, a shorter logical path may correspond to a much longer physical path.
Problem 31
a) If you run TCPClient first, then the client will attempt to make a TCP connection with a non-existent server process. A TCP connection will not be made.
b) UDPClient doesn’t establish a TCP connection with the server. Thus, everything should work fine if you first run UDPClient, then run UDPServer, and then type some input into the keyboard.
c) If you use different port numbers, then the client will attempt to establish a TCP connection with the wrong process or a non-existent process. Errors will occur.
Problem 32
In the original program, UDPClient does not specify a port number when it creates the socket. In this case, the code lets the underlying operating system choose a port number. With the additional line, when UDPClient is executed, a UDP socket is created with port number 5432 .
UDPServer needs to know the client port number so that it can send packets back to the correct client socket. Glancing at UDPServer, we see that the client port number is not “hard-wired” into the server code; instead, UDPServer determines the client port number by unraveling the datagram it receives from the client. Thus UDP server will work with any client port number, including 5432. UDPServer therefore does not need to be modified.
Before:
Client socket = x (chosen by OS) Server socket = 9876
After:
Client socket = 5432 Problem 33
Yes, you can configure many browsers to open multiple simultaneous connections to a Web site. The advantage is that you will you potentially download the file faster. The

disadvantage is that you may be hogging the bandwidth, thereby significantly slowing down the downloads of other users who are sharing the same physical links.
Problem 34
For an application such as remote login (telnet and ssh), a byte-stream oriented protocol is very natural since there is no notion of message boundaries in the application. When a user types a character, we simply drop the character into the TCP connection.
In other applications, we may be sending a series of messages that have inherent boundaries between them. For example, when one SMTP mail server sends another SMTP mail server several email messages back to back. Since TCP does not have a mechanism to indicate the boundaries, the application must add the indications itself, so that receiving side of the application can distinguish one message from the next. If each message were instead put into a distinct UDP segment, the receiving end would be able to distinguish the various messages without any indications added by the sending side of the application.
Problem 35
To create a web server, we need to run web server software on a host. Many vendors sell web server software. However, the most popular web server software today is Apache, which is open source and free. Over the years it has been highly optimized by the open- source community.
Problem 36
The key is the infohash, the value is an IP address that currently has the file designated by the infohash.

Chapter 3 Review Questions
1.
a) Call this protocol Simple Transport Protocol (STP). At the sender side, STP accepts
from the sending process a chunk of data not exceeding 1196 bytes, a destination host address, and a destination port number. STP adds a four-byte header to each chunk and puts the port number of the destination process in this header. STP then gives the destination host address and the resulting segment to the network layer. The network layer delivers the segment to STP at the destination host. STP then examines the port number in the segment, extracts the data from the segment, and passes the data to the process identified by the port number.
b) Thesegmentnowhastwoheaderfields:asourceportfieldanddestinationportfield. At the sender side, STP accepts a chunk of data not exceeding 1192 bytes, a destination host address, a source port number, and a destination port number. STP creates a segment which contains the application data, source port number, and destination port number. It then gives the segment and the destination host address to the network layer. After receiving the segment, STP at the receiving host gives the application process the application data and the source port number.
c) No, the transport layer does not have to do anything in the core; the transport layer “lives” in the end systems.
2.
1. For sending a letter, the family member is required to give the delegate the letter
itself, the address of the destination house, and the name of the recipient. The delegate clearly writes the recipient’s name on the top of the letter. The delegate then puts the letter in an envelope and writes the address of the destination house on the envelope. The delegate then gives the letter to the planet’s mail service. At the receiving side, the delegate receives the letter from the mail service, takes the letter out of the envelope, and takes note of the recipient name written at the top of the letter. The delegate then gives the letter to the family member with this name.
2. No, the mail service does not have to open the envelope; it only examines the address on the envelope.
3. Source port number y and destination port number x.
4. An application developer may not want its application to use TCP’s congestion control, which can throttle the application’s sending rate at times of congestion. Often, designers of IP telephony and IP videoconference applications choose to run their applications over UDP because they want to avoid TCP’s congestion control. Also, some applications do not need the reliable data transfer provided by TCP.
5. Since most firewalls are configured to block UDP traffic, using TCP for video and

voice traffic lets the traffic though the firewalls.
6. Yes. The application developer can put reliable data transfer into the application layer protocol. This would require a significant amount of work and debugging, however.
7. Yes, both segments will be directed to the same socket. For each received segment, at the socket interface, the operating system will provide the process with the IP addresses to determine the origins of the individual segments.
8. For each persistent connection, the Web server creates a separate “connection socket”. Each connection socket is identified with a four-tuple: (source IP address, source port number, destination IP address, destination port number). When host C receives and IP datagram, it examines these four fields in the datagram/segment to determine to which socket it should pass the payload of the TCP segment. Thus, the requests from A and B pass through different sockets. The identifier for both of these sockets has 80 for the destination port; however, the identifiers for these sockets have different values for source IP addresses. Unlike UDP, when the transport layer passes a TCP segment’s payload to the application process, it does not specify the source IP address, as this is implicitly specified by the socket identifier.
9. Sequence numbers are required for a receiver to find out whether an arriving packet contains new data or is a retransmission.
10. To handle losses in the channel. If the ACK for a transmitted packet is not received within the duration of the timer for the packet, the packet (or its ACK or NACK) is assumed to have been lost. Hence, the packet is retransmitted.
11. A timer would still be necessary in the protocol rdt 3.0. If the round trip time is known then the only advantage will be that, the sender knows for sure that either the packet or the ACK (or NACK) for the packet has been lost, as compared to the real scenario, where the ACK (or NACK) might still be on the way to the sender, after the timer expires. However, to detect the loss, for each packet, a timer of constant duration will still be necessary at the sender.
12.
a) The packet loss caused a time out after which all the five packets were retransmitted.
b) Loss of an ACK didn’t trigger any retransmission as Go-Back-N uses cumulative acknowledgements.
c) The sender was unable to send sixth packet as the send window size is fixed to 5.

13.
a) When the packet was lost, the received four packets were buffered the receiver. After
the timeout, sender retransmitted the lost packet and receiver delivered the buffered packets to application in correct order.
b) Duplicate ACK was sent by the receiver for the lost ACK.
c) The sender was unable to send sixth packet as the send window size is fixed to 5
When a packet was lost, GO-Back-N retransmitted all the packets whereas Selective Repeat retransmitted the lost packet only. In case of lost acknowledgement, selective repeat sent a duplicate ACK and as GO-Back-N used cumulative acknowledgment, so that duplicate ACK was unnecessary.
14. a) false b) false c) true d) false e) true f) false g) false
15. a) 20 bytes b) ack number = 90
16. 3 segments. First segment: seq = 43, ack =80; Second segment: seq = 80, ack = 44; Third segment; seq = 44, ack = 81
17. R/2
18. False, it is set to half of the current value of the congestion window.
19. Let X = RTTFE, Y = RTTBE and ST = Search time. Consider the following timing diagram.

TCP packet exchange diagram between a client and a server (Back End) with a proxy (Front End) between them.
From this diagram we see that the total time is 4X + Y+ ST = 4*RTTFE + RTTBE + Search time
Chapter 3 Problems
Problem 1
e) Yes. f) No.
Problem 2
a) AS b) BS c) SA d) SB
source port numbers
destination port numbers
467 23 513 23 23 467 23 513
Suppose the IP addresses of the hosts A, B, and C are a, b, c, respectively. (Note that a, b, c are distinct.)
To host A: Source port =80, source IP address = b, dest port = 26145, dest IP address = a
To host C, left process: Source port =80, source IP address = b, dest port = 7532, dest IP address = c
To host C, right process: Source port =80, source IP address = b, dest port = 26145, dest IP address = c
Problem 3
Note, wrap around if overflow.
01010011 01100110
10111001

10111001 01110100
00101110
One’s complement = 1 1 0 1 0 0 0 1.
To detect errors, the receiver adds the four words (the three original words and the checksum). If the sum contains a zero, the receiver knows there has been an error. All one-bit errors will be detected, but two-bit errors can be undetected (e.g., if the last digit of the first word is converted to a 0 and the last digit of the second word is converted to a 1).
Problem 4
a) Adding the two bytes gives 11000001. Taking the one’s complement gives 00111110. b) Adding the two bytes gives 01000000; the one’s complement gives 10111111.
c) First byte = 01010100; second byte = 01101101.
Problem 5
No, the receiver cannot be absolutely certain that no bit errors have occurred. This is because of the manner in which the checksum for the packet is calculated. If the corresponding bits (that would be added together) of two 16-bit words in the packet were 0 and 1 then even if these get flipped to 1 and 0 respectively, the sum still remains the same. Hence, the 1s complement the receiver calculates will also be the same. This means the checksum will verify even if there was transmission error.
Problem 6
Suppose the sender is in state “Wait for call 1 from above” and the receiver (the receiver shown in the homework problem) is in state “Wait for 1 from below.” The sender sends a packet with sequence number 1, and transitions to “Wait for ACK or NAK 1,” waiting for an ACK or NAK. Suppose now the receiver receives the packet with sequence number 1 correctly, sends an ACK, and transitions to state “Wait for 0 from below,” waiting for a data packet with sequence number 0. However, the ACK is corrupted. When the rdt2.1 sender gets the corrupted ACK, it resends the packet with sequence number 1. However, the receiver is waiting for a packet with sequence number 0 and (as shown in the home work problem) always sends a NAK when it doesn’t get a packet with sequence number 0. Hence the sender will always be sending a packet with sequence number 1, and the receiver will always be NAKing that packet. Neither will progress forward from that state.

Problem 7
To best answer this question, consider why we needed sequence numbers in the first place. We saw that the sender needs sequence numbers so that the receiver can tell if a data packet is a duplicate of an already received data packet. In the case of ACKs, the sender does not need this info (i.e., a sequence number on an ACK) to tell detect a duplicate ACK. A duplicate ACK is obvious to the rdt3.0 receiver, since when it has received the original ACK it transitioned to the next state. The duplicate ACK is not the ACK that the sender needs and hence is ignored by the rdt3.0 sender.
Problem 8
The sender side of protocol rdt3.0 differs from the sender side of protocol 2.2 in that timeouts have been added. We have seen that the introduction of timeouts adds the possibility of duplicate packets into the sender-to-receiver data stream. However, the receiver in protocol rdt.2.2 can already handle duplicate packets. (Receiver-side duplicates in rdt 2.2 would arise if the receiver sent an ACK that was lost, and the sender then retransmitted the old data). Hence the receiver in protocol rdt2.2 will also work as the receiver in protocol rdt 3.0.
Problem 9
Suppose the protocol has been in operation for some time. The sender is in state “Wait for call from above” (top left hand corner) and the receiver is in state “Wait for 0 from below”. The scenarios for corrupted data and corrupted ACK are shown in Figure 1.

Sender sends M0
M0 corrupted A1
Sender ignores A1
Timeout: sender M0 resends M0 A0
M1 A1
sender sends M0 M0
Packet garbled, receiver resends last ACK (A1)
Corrupted data
sender sends M1 Ignore ACK
A0 M1
Corrupted ACK
A1 corrupted Timeout: sender M1
resends M1
A1 M0
Figure 1: rdt 3.0 scenarios: corrupted data, corrupted ACK Problem 10
Here, we add a timer, whose value is greater than the known round-trip propagation delay. We add a timeout event to the “Wait for ACK or NAK0” and “Wait for ACK or NAK1” states. If the timeout event occurs, the most recently transmitted packet is retransmitted. Let us see why this protocol will still work with the rdt2.1 receiver.
 Suppose the timeout is caused by a lost data packet, i.e., a packet on the sender- to-receiver channel. In this case, the receiver never received the previous transmission and, from the receiver’s viewpoint, if the timeout retransmission is received, it looks exactly the same as if the original transmission is being received.
 Suppose now that an ACK is lost. The receiver will eventually retransmit the packet on a timeout. But a retransmission is exactly the same action that if an ACK is garbled. Thus the sender’s reaction is the same with a loss, as with a garbled ACK. The rdt 2.1 receiver can already handle the case of a garbled ACK.

Problem 11
If the sending of this message were removed, the sending and receiving sides would deadlock, waiting for an event that would never occur. Here’s a scenario:
 Sender sends pkt0, enter the “Wait for ACK0 state”, and waits for a packet back from the receiver
 Receiver is in the “Wait for 0 from below” state, and receives a corrupted packet from the sender. Suppose it does not send anything back, and simply re-enters the ‘wait for 0 from below” state.
Now, the ender is awaiting an ACK of some sort from the receiver, and the receiver is waiting for a data packet form the sender – a deadlock!
Problem 12
The protocol would still work, since a retransmission would be what would happen if the packet received with errors has actually been lost (and from the receiver standpoint, it never knows which of these events, if either, will occur).
To get at the more subtle issue behind this question, one has to allow for premature timeouts to occur. In this case, if each extra copy of the packet is ACKed and each received extra ACK causes another extra copy of the current packet to be sent, the number of times packet n is sent will increase without bound as n approaches infinity.
Problem 13
M0 A0
M1 A1
M0
M0 A0
M1 A1
old version of M0 accepted!

Problem 14
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.
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.
Problem 15
It takes 12 microseconds (or 0.012 milliseconds) to send a packet, as 1500*8/109=12 microseconds. In order for the sender to be busy 98 percent of the time, we must have
util  0.98  (0.012n) / 30.012
or n approximately 2451 packets.
Problem 16
Yes. This actually causes the sender to send a number of pipelined data into the channel. Yes. Here is one potential problem. If data segments are lost in the channel, then the sender of rdt 3.0 won’t re-send those segments, unless there are some additional mechanism in the application to recover from loss.

Problem 17
rdt_send(data) Rdt_unable_to_send(data)
Wait: send to A
rdt_send(data) Rdt_unable_to_send(data)
Problem 18
rdt_receive(packet)
extract(packet,data) deliver_data(data)
rdt_send(data)
packet=make_pkt(data) udt_send(packet)
rdt_send(data) rdt_unable_to_send(data)
Wait: send to B
rdt_send(data)
packet=make_pkt(data) udt_send(packet)
A Wait: receive from B
B Wait: receive from A
rdt_receive(packet)
extract(packet,data) deliver_data(data)
rdt_send(data) rdt_unable_to_send(data)
In our solution, the sender will wait until it receives an ACK for a pair of messages (seqnum and seqnum+1) before moving on to the next pair of messages. Data packets have a data field and carry a two-bit sequence number. That is, the valid sequence numbers are 0, 1, 2, and 3. (Note: you should think about why a 1-bit sequence number space of 0, 1 only would not work in the solution below.) ACK messages carry the sequence number of the data packet they are acknowledging.
The FSM for the sender and receiver are shown in Figure 2. Note that the sender state records whether (i) no ACKs have been received for the current pair, (ii) an ACK for seqnum (only) has been received, or an ACK for seqnum+1 (only) has been received. In this figure, we assume that the seqnum is initially 0, and that the sender has sent the first

two data messages (to get things going). A timeline trace for the sender and receiver recovering from a lost packet is shown below:
Figure 2: Sender and receiver for Problem (3.18)
Sender Receiver
make pair (0,1)
send packet 0

Packet 0 drops
send packet 1
receive ACK 1
(timeout)
resend packet 0
receive ACK 0
Problem 19
receive packet 1
buffer packet 1
send ACK 1
receive packet 0
deliver pair (0,1)
send ACK 0
This problem is a variation on the simple stop and wait protocol (rdt3.0). Because the channel may lose messages and because the sender may resend a message that one of the receivers has already received (either because of a premature timeout or because the other receiver has yet to receive the data correctly), sequence numbers are needed. As in rdt3.0, a 0-bit sequence number will suffice here.
The sender and receiver FSM are shown in Figure 3. In this problem, the sender state indicates whether the sender has received an ACK from B (only), from C (only) or from neither C nor B. The receiver state indicates which sequence number the receiver is waiting for.

Figure 3. Sender and receiver for Problem 3.19(Problem 19)

Problem 20
rdt_rcv(rcvpkt)&&from_B(rcvpkt) Λ
rdt_rcv(rcvpkt)&&from_A(rcvpkt) Λ
Wait for 0 from A
Wait for 0 from B
rdt_rcv(rcvpkt)&&(corrupt(rcvpkt) ||has_seq1(rcvpkt))&&from_A(rcvpkt)
sndpkt=make_pkt(ACK, 1, checksum) udt_send(A,sndpkt)
rdt_rcv(rcvpkt)&&not_corrupt(rcvpkt)&& has_seq0(rcvpkt)&&from_A(rcvpkt)
extract(rcvpkt,data)
deliver_data(data) sndpkt=make_pkt(ACK, 0, checksum) udt_send(A,sndpkt)
rdt_rcv(rcvpkt)&&not_corrupt(rcvpkt) &&has_seq1(rcvpkt)&&from_B(rcvpkt)
extract(rcvpkt,data) deliver_data(data) sndpkt=make_pkt(ACK,1,checksum) udt_send(B,sndpkt)
Wait for 1 from B
Wait
for 1 from A
rdt_rcv(rcvpkt)&&not_corrupt(rcvpkt)&&ha s_seq1(rcvpkt)&&from_A(rcvpkt)
extract(rcvpkt,data) deliver_data(data) sndpkt=make_pkt(ACK,1,checksum) udt_send(A,sndpkt)
rdt_rcv(rcvpkt)&&(corrupt(rcvpkt) ||has_seq1(rcvpkt))&&from_B(rcvpkt)
sndpkt=make_pkt(ACK, 1, checksum) udt_send(B,sndpkt)
rdt_rcv(rcvpkt)&&from_B(rcvpkt)
Λ
rdt_rcv(rcvpkt)&&(corrupt(rcvpkt) ||has_seq0(rcvpkt))&&from_B(rcvpkt)
sndpkt=make_pkt(ACK, 0, checksum) udt_send(B,sndpkt)
rdt_rcv(rcvpkt)&&from_A(rcvpkt) Λ
rdt_rcv(rcvpkt)&&(corrupt(rcvpkt) ||has_seq0(rcvpkt))&&from_A(rcvpkt)
sndpkt=make_pkt(ACK, 0, checksum) udt_send(A,sndpkt)
Figure 4: Receiver side FSM for 3.18
rdt_rcv(rcvpkt)&&not_corrupt(rcvpkt)&& has_seq0(rcvpkt)&&from_B(rcvpkt)
extract(rcvpkt,data) deliver_data(data) sndpkt=make_pkt(ACK,0,checksum) udt_send(B,sndpkt)

Sender
The sender side FSM is exactly same as given in Figure 3.15 in text
Problem 21
Because the A-to-B channel can lose request messages, A will need to timeout and retransmit its request messages (to be able to recover from loss). Because the channel delays are variable and unknown, it is possible that A will send duplicate requests (i.e., resend a request message that has already been received by B). To be able to detect duplicate request messages, the protocol will use sequence numbers. A 1-bit sequence number will suffice for a stop-and-wait type of request/response protocol.
A (the requestor) has 4 states:
 “Wait for Request 0 from above.” Here the requestor is waiting for a call from
above to request a unit of data. When it receives a request from above, it sends a request message, R0, to B, starts a timer and makes a transition to the “Wait for D0” state. When in the “Wait for Request 0 from above” state, A ignores anything it receives from B.
 “Wait for D0”. Here the requestor is waiting for a D0 data message from B. A timer is always running in this state. If the timer expires, A sends another R0 message, restarts the timer and remains in this state. If a D0 message is received from B, A stops the time and transits to the “Wait for Request 1 from above” state. If A receives a D1 data message while in this state, it is ignored.
 “Wait for Request 1 from above.” Here the requestor is again waiting for a call from above to request a unit of data. When it receives a request from above, it sends a request message, R1, to B, starts a timer and makes a transition to the “Wait for D1” state. When in the “Wait for Request 1 from above” state, A ignores anything it receives from B.
 “Wait for D1”. Here the requestor is waiting for a D1 data message from B. A timer is always running in this state. If the timer expires, A sends another R1 message, restarts the timer and remains in this state. If a D1 message is received from B, A stops the timer and transits to the “Wait for Request 0 from above” state. If A receives a D0 data message while in this state, it is ignored.
The data supplier (B) has only two states:
 “Send D0.” In this state, B continues to respond to received R0 messages by sending D0, and then remaining in this state. If B receives a R1 message, then it knows its D0 message has been received correctly. It thus discards this D0 data (since it has been received at the other side) and then transits to the “Send D1” state, where it will use D1 to send the next requested piece of data.

 “Send D1.” In this state, B continues to respond to received R1 messages by sending D1, and then remaining in this state. If B receives a R1 message, then it knows its D1 message has been received correctly and thus transits to the “Send D1” state.
Problem 22
a) Here we have a window size of N=3. Suppose the receiver has received packet k-1, and has ACKed that and all other preceding packets. If all of these ACK’s have been received by sender, then sender’s window is [k, k+N-1]. Suppose next that none of the ACKs have been received at the sender. In this second case, the sender’s window contains k-1 and the N packets up to and including k-1. The sender’s window is thus [k-N,k-1]. By these arguments, the senders window is of size 3 and begins somewhere in the range [k-N,k].
b) If the receiver is waiting for packet k, then it has received (and ACKed) packet k-1 and the N-1 packets before that. If none of those N ACKs have been yet received by the sender, then ACK messages with values of [k-N,k-1] may still be propagating back.Because the sender has sent packets [k-N, k-1], it must be the case that the sender has already received an ACK for k-N-1. Once the receiver has sent an ACK for k-N-1 it will never send an ACK that is less that k-N-1. Thus the range of in- flight ACK values can range from k-N-1 to k-1.
Problem 23
In order to avoid the scenario of Figure 3.27, we want to avoid having the leading edge of the receiver’s window (i.e., the one with the “highest” sequence number) wrap around in the sequence number space and overlap with the trailing edge (the one with the “lowest” sequence number in the sender’s window). That is, the sequence number space must be large enough to fit the entire receiver window and the entire sender window without this overlap condition. So – we need to determine how large a range of sequence numbers can be covered at any given time by the receiver and sender windows.
Suppose that the lowest-sequence number that the receiver is waiting for is packet m. In this case, it’s window is [m,m+w-1] and it has received (and ACKed) packet m-1 and the w-1 packets before that, where w is the size of the window. If none of those w ACKs have been yet received by the sender, then ACK messages with values of [m-w,m-1] may still be propagating back. If no ACKs with these ACK numbers have been received by the sender, then the sender’s window would be [m-w,m-1].
Thus, the lower edge of the sender’s window is m-w, and the leading edge of the receivers window is m+w-1. In order for the leading edge of the receiver’s window to not overlap with the trailing edge of the sender’s window, the sequence number space must

thus be big enough to accommodate 2w sequence numbers. That is, the sequence number space must be at least twice as large as the window size, k  2w .
Problem 24
a) True. Suppose the sender has a window size of 3 and sends packets 1, 2, 3 at t0. At t1 (t1  t0) the receiver ACKS 1, 2, 3. At t2 (t2  t1) the sender times out and
resends 1, 2, 3. At t3 the receiver receives the duplicates and re-acknowledges 1, 2, 3. At t4 the sender receives the ACKs that the receiver sent at t1 and advances its window to 4, 5, 6. At t5 the sender receives the ACKs 1, 2, 3 the receiver sent at t2 . These ACKs are outside its window.
b) True. By essentially the same scenario as in (a).
c) True.
d) True. Note that with a window size of 1, SR, GBN, and the alternating bit protocol are functionally equivalent. The window size of 1 precludes the possibility of out-of- order packets (within the window). A cumulative ACK is just an ordinary ACK in this situation, since it can only refer to the single packet within the window.
Problem 25
a) Consider sending an application message over a transport protocol. With TCP, the application writes data to the connection send buffer and TCP will grab bytes without necessarily putting a single message in the TCP segment; TCP may put more or less than a single message in a segment. UDP, on the other hand, encapsulates in a segment whatever the application gives it; so that, if the application gives UDP an application message, this message will be the payload of the UDP segment. Thus, with UDP, an application has more control of what data is sent in a segment.
b) WithTCP,duetoflowcontrolandcongestioncontrol,theremaybesignificantdelay from the time when an application writes data to its send buffer until when the data is given to the network layer. UDP does not have delays due to flow control and congestion control.
Problem 26
There are 232  4,294,967,296 possible sequence numbers.
a) The sequence number does not increment by one with each segment. Rather, it
increments by the number of bytes of data sent. So the size of the MSS is irrelevant — the maximum size file that can be sent from A to B is simply the number of bytes
representable by 232  4.19 Gbytes.

d)
 232   8,012,999
b) The number of segments is536 . 66 bytes of header get added to each segment giving a total of 528,857,934 bytes of header. The total number of bytes transmitted is 232  528,857,934  4.824 109 bytes.
Thus it would take 249 seconds to transmit the file over a 155~Mbps link.
Problem 27
a) In the second segment from Host A to B, the sequence number is 207, source port number is 302 and destination port number is 80.
b) If the first segment arrives before the second, in the acknowledgement of the first arriving segment, the acknowledgement number is 207, the source port number is 80 and the destination port number is 302.
c) If the second segment arrives before the first segment, in the acknowledgement of the first arriving segment, the acknowledgement number is 127, indicating that it is still waiting for bytes 127 and onwards.
Host A
Seq = 127, 80 bytes
Timeout interval
Seq = 207, 40 bytes Seq = 127, 80 bytes
Host B
Ack = 207 Ack = 247
Ack = 247
Problem 28
Timeout interval
Since the link capacity is only 100 Mbps, so host A’s sending rate can be at most 100Mbps. Still, host A sends data into the receive buffer faster than Host B can remove data from the buffer. The receive buffer fills up at a rate of roughly 40Mbps. When the buffer is full, Host B signals to Host A to stop sending data by setting RcvWindow = 0. Host A then stops sending until it receives a TCP segment with RcvWindow > 0. Host A will thus repeatedly stop and start sending as a function of the RcvWindow values it

receives from Host B. On average, the long-term rate at which Host A sends data to Host B as part of this connection is no more than 60Mbps.
Problem 29
a) The server uses special initial sequence number (that is obtained from the hash of source and destination IPs and ports) in order to defend itself against SYN FLOOD attack.
b) No,theattackercannotcreatehalf-openorfullyopenconnectionsbysimplysending and ACK packet to the target. Half-open connections are not possible since a server using SYN cookies does not maintain connection variables and buffers for any connection before full connections are established. For establishing fully open connections, an attacker should know the special initial sequence number corresponding to the (spoofed) source IP address from the attacker. This sequence number requires the “secret” number that each server uses. Since the attacker does not know this secret number, she cannot guess the initial sequence number.
c) No, the sever can simply add in a time stamp in computing those initial sequence numbers and choose a time to live value for those sequence numbers, and discard expired initial sequence numbers even if the attacker replay them.
Problem 30
a) If timeout values are fixed, then the senders may timeout prematurely. Thus, some packets are re-transmitted even they are not lost.
b) If timeout values are estimated (like what TCP does), then increasing the buffer size certainly helps to increase the throughput of that router. But there might be one potential problem. Queuing delay might be very large, similar to what is shown in Scenario 1.
Problem 31
EstimatedRTT  xSampleRTT  (1 x)EstimatedRTT DevRTT  y SampleRTT  EstimatedRTT  (1 y)DevRTT
TimeoutInterval  EstimatedRTT  4 * DevRTT
After obtaining first sampleRTT is EstimatedRTT  0.125*106  0.875*100
 100.75ms .
DevRTT 0.25106100.75 0.75*5  5.06ms .
TimeoutInterval  100.75  4 * 5.06

 120.99ms .
After obtaining second sampleRTT = 120ms: EstimatedRTT  0.125*120  0.875*100.75
 103.15ms .
DevRTT  0.25 120 103.15  0.75*5.06 8ms.
TimeoutInterval103.154*8  135.15ms .
After obtaining Third sampleRTT = 140ms: EstimatedRTT  0.125*140  0.875*103.15
 107.76ms .
DevRTT 0.25140107.76 0.75*8  14.06ms .
TimeoutInterval107.764*14.06  164ms .
After obtaining fourth sampleRTT = 90ms: EstimatedRTT  0.125*90  0.875*107.76
 105.54ms .
DevRTT  0.25 90 105.54  0.75*14.06  14.42ms .
TimeoutInterval105.544*14.42  163.22ms .
After obtaining fifth sampleRTT = 115ms: EstimatedRTT  0.125*115  0.875*105.54
 106.71ms .
DevRTT 0.25115106.710.75*14.42  12.88ms .
TimeoutInterval 106.71 4*12.88  158.23ms .
Problem 32
a)
Denote EstimatedRTT(n) for the estimate after the nth sample.
EstimatedRTT (4)  xSampleRTT  1
(1x)[xSampleRTT2  (1x)[xSampleRTT3 (1x)SampleRTT4]]

b)
c)
j
xSampleRTT1 (1x)xSampleRTT2 (1x)2xSampleRTT (1x)3SampleRTT
n 1
EstimatedRTT(n) x(1x)j1SampleRTT j1
 (1 x)n1 SampleRTTn
EstimatedRTT()  
x  (1 x)j SampleRTTj 1x j1
.9 j SampleRTT The weight given to past samples decays exponentially.
1  9 j1
34
j
Problem 33
Let’s look at what could wrong if TCP measures SampleRTT for a retransmitted segment. Suppose the source sends packet P1, the timer for P1 expires, and the source then sends P2, a new copy of the same packet. Further suppose the source measures SampleRTT for P2 (the retransmitted packet). Finally suppose that shortly after transmitting P2 an acknowledgment for P1 arrives. The source will mistakenly take this acknowledgment as an acknowledgment for P2 and calculate an incorrect value of SampleRTT.
Let’s look at what could be wrong if TCP measures SampleRTT for a retransmitted segment. Suppose the source sends packet P1, the timer for P1 expires, and the source then sends P2, a new copy of the same packet. Further suppose the source measures SampleRTT for P2 (the retransmitted packet). Finally suppose that shortly after transmitting P2 an acknowledgment for P1 arrives. The source will mistakenly take this acknowledgment as an acknowledgment for P2 and calculate an incorrect value of SampleRTT.

Problem 34
At any given time t, SendBase – 1 is the sequence number of the last byte that the sender knows has been received correctly, and in order, at the receiver. The actually last byte received (correctly and in order) at the receiver at time t may be greater if there are acknowledgements in the pipe. Thus
SendBase–1  LastByteRcvd Problem 35
When, at time t, the sender receives an acknowledgement with value y, the sender knows for sure that the receiver has received everything up through y-1. The actual last byte received (correctly and in order) at the receiver at time t may be greater if y  SendBase or if there are other acknowledgements in the pipe. Thus
y-1  LastByteRvcd Problem 36
Suppose packets n, n+1, and n+2 are sent, and that packet n is received and ACKed. If packets n+1 and n+2 are reordered along the end-to-end-path (i.e., are received in the order n+2, n+1) then the receipt of packet n+2 will generate a duplicate ack for n and would trigger a retransmission under a policy of waiting only for second duplicate ACK for retransmission. By waiting for a triple duplicate ACK, it must be the case that two packets after packet n are correctly received, while n+1 was not received. The designers of the triple duplicate ACK scheme probably felt that waiting for two packets (rather than 1) was the right tradeoff between triggering a quick retransmission when needed, but not retransmitting prematurely in the face of packet reordering.
Problem 37
a) GoBackN:
A sends 9 segments in total. They are initially sent segments 1, 2, 3, 4, 5 and later re- sent segments 2, 3, 4, and 5.
B sends 8 ACKs. They are 4 ACKS with sequence number 1, and 4 ACKS with sequence numbers 2, 3, 4, and 5.
Selective Repeat:
A sends 6 segments in total. They are initially sent segments 1, 2, 3, 4, 5 and later re- sent segments 2.
B sends 5 ACKs. They are 4 ACKS with sequence number 1, 3, 4, 5. And there is one ACK with sequence number 2.

TCP:
A sends 6 segments in total. They are initially sent segments 1, 2, 3, 4, 5 and later re- sent segments 2.
B sends 5 ACKs. They are 4 ACKS with sequence number 2. There is one ACK with sequence numbers 6. Note that TCP always send an ACK with expected sequence number.
b) TCP. This is because TCP uses fast retransmit without waiting until time out.
Problem 38
Yes, the sending rate is always roughly cwnd/RTT.
Problem 39
If the arrival rate increases beyond R/2 in Figure 3.46(b), then the total arrival rate to the queue exceeds the queue’s capacity, resulting in increasing loss as the arrival rate increases. When the arrival rate equals R/2, 1 out of every three packets that leaves the queue is a retransmission. With increased loss, even a larger fraction of the packets leaving the queue will be retransmissions. Given that the maximum departure rate from the queue for one of the sessions is R/2, and given that a third or more will be transmissions as the arrival rate increases, the throughput of successfully deliver data can not increase beyond out. Following similar reasoning, if half of the packets leaving the queue are retransmissions, and the maximum rate of output packets per session is R/2, then the maximum value of out is (R/2)/2 or R/4.
Problem 40
a) TCP slowstart is operating in the intervals [1,6] and [23,26]
b) TCP congestion avoidance is operating in the intervals [6,16] and [17,22]
c) After the 16th transmission round, packet loss is recognized by a triple duplicate
ACK. If there was a timeout, the congestion window size would have dropped to 1.
d) After the 22nd transmission round, segment loss is detected due to timeout, and hence
the congestion window size is set to 1.
e) The threshold is initially 32, since it is at this window size that slow start stops and
congestion avoidance begins.
f) The threshold is set to half the value of the congestion window when packet loss is
detected. When loss is detected during transmission round 16, the congestion
windows size is 42. Hence the threshold is 21 during the 18th transmission round.
g) The threshold is set to half the value of the congestion window when packet loss is detected. When loss is detected during transmission round 22, the congestion windows size is 29. Hence the threshold is 14 (taking lower floor of 14.5) during the
24th transmission round.
h) During the 1st transmission round, packet 1 is sent; packet 2-3 are sent in the 2nd
transmission round; packets 4-7 are sent in the 3rd transmission round; packets 8-15 are sent in the 4th transmission round; packets 16-31 are sent in the 5th transmission

round; packets 32-63 are sent in the 6th transmission round; packets 64 – 96 are sent
in the 7th transmission round. Thus packet 70 is sent in the 7th transmission round.
i) The threshold will be set to half the current value of the congestion window (8) when the loss occurred and congestion window will be set to the new threshold value + 3
MSS . Thus the new values of the threshold and window will be 4 and 7 respectively.
j) threshold is 21, and congestion window size is 1.
k) round 17, 1 packet; round 18, 2 packets; round 19, 4 packets; round 20, 8 packets;
round 21, 16 packets; round 22, 21 packets. So, the total number is 52.
Problem 41
Refer to Figure 5. In Figure 5(a), the ratio of the linear decrease on loss between connection 1 and connection 2 is the same – as ratio of the linear increases: unity. In this case, the throughputs never move off of the AB line segment. In Figure 5(b), the ratio of the linear decrease on loss between connection 1 and connection 2 is 2:1. That is, whenever there is a loss, connection 1 decreases its window by twice the amount of connection 2. We see that eventually, after enough losses, and subsequent increases, that connection 1’s throughput will go to 0, and the full link bandwidth will be allocated to connection 2.
Figure 5: Lack of TCP convergence with linear increase, linear decrease
Problem 42
If TCP were a stop-and-wait protocol, then the doubling of the time out interval would suffice as a congestion control mechanism. However, TCP uses pipelining (and is therefore not a stop-and-wait protocol), which allows the sender to have multiple outstanding unacknowledged segments. The doubling of the timeout interval does not prevent a TCP sender from sending a large number of first-time-transmitted packets into the network, even when the end-to-end path is highly congested. Therefore a congestion-

control mechanism is needed to stem the flow of “data received from the application above” when there are signs of network congestion.
Problem 43
In this problem, there is no danger in overflowing the receiver since the receiver’s receive buffer can hold the entire file. Also, because there is no loss and acknowledgements are returned before timers expire, TCP congestion control does not throttle the sender. However, the process in host A will not continuously pass data to the socket because the send buffer will quickly fill up. Once the send buffer becomes full, the process will pass data at an average rate or R << S. Problem 44 a) It takes 1 RTT to increase CongWin to 6 MSS; 2 RTTs to increase to 7 MSS; 3 RTTs to increase to 8 MSS; 4 RTTs to increase to 9 MSS; 5 RTTs to increase to 10 MSS; 6 RTTs to increase to 11 MSS; and 7 RTTs to increase to 12MSS. b) In the first RTT 5 MSS was sent; in the second RTT 6 MSS was sent; in the third RTT 7 MSS was sent; in the fourth RTT 8 MSS was sent; in the fifth RTT, 9 MSS was sent; and in the sixth RTT, 10 MSS was sent. Thus, up to time 6 RTT, 5+6+7+8+9+10 = 45 MSS were sent (and acknowledged). Thus, we can say that the average throughput up to time 6 RTT was (45 MSS)/(6 RTT) = 7.5 MSS/RTT. Problem 45 a) The loss rate, L , is the ratio of the number of packets lost over the number of packets sent. In a cycle, 1 packet is lost. The number of packets sent in a cycle is W W  W/2W 2  2 1W  ( 2 n)   n0 W W W/2   2 1 2  n   n0 W 1W W/2(W/21) 22 2 W2 W W2 W 4284 3W2 3W 84 Thus the loss rate is L1 3W2 3W 84 b)ForWlarge,3W23W.ThusL8/3W2 orW 8.Fromthetext,we therefore have Problem 46 84 3L average throughput  3 8  MSS 4 3L RTT  1.22  MSS RTT L a) 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. b) 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. c) 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. Problem 47 Let W denote max window size. Let S denote the buffer size. For simplicity, suppose TCP sender sends data packets in a round by round fashion, with each round corresponding to a RTT. If the window size reaches W, then a loss occurs. Then the sender will cut its congestion window size by half, and waits for the ACKs for W/2 outstanding packets before it starts sending data segments again. In order to make sure the link always busying sending data, we need to let the link busy sending data in the period W/(2*C) (this is the time interval where the sender is waiting for the ACKs for the W/2 outstanding packets). Thus, S/C must be no less than W/(2*C), that is, S>=W/2.
Let Tp denote the one-way propagation delay between the sender and the receiver.
When the window size reaches the minimum W/2 and the buffer is empty, we need to make sure the link is also busy sending data. Thus, we must have W/2/(2Tp)>=C, thus, W/2>=C*2Tp.
Thus, S>=C*2Tp.

Problem 48
a) Let W denote the max window size. Then, W*MSS/RTT = 10Gbps, as packets will be dropped if maximum sending rate reaches link capacity. Thus, we have W*1500*8/0.15=10*10^9, then W= 125000 segments.
b) As congestion window size varies from W/2 to W, then the average window size is 0.75W=93750 segments. Average throughput is 93750*1500*8/0.1=7.5Gbps.
c) 93750/2 *0.15 /60= 117 minutes. In order to speed up the window increase process, we can increase the window size by a much larger value, instead of increasing window size only by one in each RTT. Some protocols are proposed to solve this problem, such as ScalableTCP or HighSpeed TCP.
Problem 49
As TCP’s average throughput B is given by B  1.22  MSS , so we know that,
L= (1.22*MSS / (B*RTT) ) 2
Since between two consecutive packet losses, there are 1/L packets sent by the TCP sender, thus, T=(1/L)*MSS/B. Thus, we find that T=B*RTT2/(1.222*MSS), that is, T is a function of B.
Problem 50
a) The key difference between C1 and C2 is that C1’s RTT is only half of that of C2. Thus C1 adjusts its window size after 50 msec, but C2 adjusts its window size after 100 msec. Assume that whenever a loss event happens, C1 receives it after 50msec and C2 receives it after 100msec. We further have the following simplified model of TCP. After each RTT, a connection determines if it should increase window size or not. For C1, we compute the average total sending rate in the link in the previous 50 msec. If that rate exceeds the link capacity, then we assume that C1 detects loss and reduces its window size. But for C2, we compute the average total sending rate in the link in the previous 100msec. If that rate exceeds the link capacity, then we assume that C2 detects loss and reduces its window size. Note that it is possible that the average sending rate in last 50msec is higher than the link capacity, but the average sending rate in last 100msec is smaller than or equal to the link capacity, then in this case, we assume that C1 will experience loss event but C2 will not.
The following table describes the evolution of window sizes and sending rates based on the above assumptions.
RTT L
C1
C2
Time (msec)
Window Size (num. of segments sent in next 50msec)
Average data sending rate (segments per second, =Window/0.05)
Window Size(num. of segments sent in next 100msec)
Average data sending rate (segments per second, =Window/0.1)
0
10
200 (in [0-50]msec]
10
100 (in [0-50]msec)

50
5
(decreases window size as the avg. total sending rate to the link in last 50msec is 300= 200+100)
100 (in [50-100]msec]
100 (in [50-100]msec)
100
2
(decreases window size as the avg. total sending rate to the link in last 50msec is 200= 100+100)
40
5
(decreases window size as the avg. total sending rate to the link in last 100msec is 250= (200+100)/2 + (100+100)/2)
50
150
1
(decreases window size as the avg. total sending rate to the link in last 50msec is 90= (40+50)
20
50
200
1
(no further decrease, as window size is already 1)
20
2
(decreases window size as the avg. total sending rate to the link in last 100msec is 80= (40+20)/2 + (50+50)/2)
20
250
1
(no further decrease, as window size
20
20

is already 1)
300
1
(no further decrease, as window size is already 1)
20
1
(decreases window size as the avg. total sending rate to the link in last 100msec is 40= (20+20)/2 + (20+20)/2)
10
350
2
40
10
400
1
20
1
10
450
2
40
10
500
1
(decreases window size as the avg. total sending rate to the link in last 50msec is 50= (40+10)
20
1
10
550
2
40
10
600
1
20
1
10
650
2
40
10
700
1
20
1
10
750
2
40
10
800
1
20
1
10
850
2
40
10
900
1
20
1
10
950
2
40
10
1000
1
20
1
10
Based on the above table, we find that after 1000 msec, C1’s and C2’s window sizes are 1 segment each.
b) No. In the long run, C1’s bandwidth share is roughly twice as that of C2’s, because C1 has shorter RTT, only half of that of C2, so C1 can adjust its window size twice as fast as C2. If we look at the above table, we can see a cycle every 200msec, e.g. from 850msec to 1000msec, inclusive. Within a cycle, the sending rate of C1 is (40+20+40+20) = 120, which is thrice as large as the sending of C2 given by (10+10+10+10) = 40.

Problem 51
a) Similarly as in last problem, we can compute their window sizes over time in the following table. Both C1 and C2 have the same window size 2 after 2200msec.
100 (in [0-100]msec) 50
20
10
20
10
20
10
20
10
20
10
20
10
20
10
20
10
20
10
20
10
20
b) Yes, this is due to the AIMD algorithm of TCP and that both connections have the same RTT.
c) Yes, this can be seen clearly from the above table. Their max window size is 2.
d) No, this synchronization won’t help to improve link utilization, as these two connections act as a single connection oscillating between min and max window size. Thus, the link is not fully utilized (recall we assume this link has no buffer). One possible way to break the synchronization is to add a finite buffer to the link and randomly drop packets in the buffer before buffer overflow. This will cause different connections cut their window sizes at different times. There are many AQM (Active Queue Management) techniques to do that, such as RED (Random Early Detect), PI (Proportional and Integral AQM), AVQ (Adaptive Virtual Queue), and REM
(Random Exponential Marking), etc.
C1
C2
Time (msec)
Window Size (num. of segments sent in next 100msec)
Data sending speed (segments per second, =Window/0.1)
Window Size(num. of segments sent in next
Data sending speed (segments per second, =Window/0.1)
100msec)
0
15
150 (in [0-100]msec]
10
100
7
70
5
200
3
30
2
300
1
10
1
400
2
20
2
500
1
10
1
600
2
20
2
700
1
10
1
800
2
20
2
900
1
10
1
1000
2
20
20
2
1100
1
10
1
1200
2
20
2
1300
1
10
1
1400
2
20
2
1500
1
10
1
1600
2
2
1700
1
10
1
1800
2
20
2
1900
1
10
1
2000
2
20
2
2100
1
10
1
2200
2
20
2

Problem 52
Note that W represents the maximum window size.
First we can find the total number of segments sent out during the interval when TCP changes its window size from W/2 up to and include W. This is given by:
S= W/2 + (W/2)*(1+) + (W/2)*(1+)2 + (W/2)*(1+)3 + … + (W/2)*(1+)k
We find k=log(1+)2, then S=W*(2+1)/(2).
Loss rate L is given by:
L= 1/S = (2) / (W*(2+1) ).
The time that TCP takes to increase its window size from W/2 to W is given by: k*RTT= (log(1+)2) * RTT,
which is clearly independent of TCP’s average throughput.
Note, TCP’s average throughput is given by: B=MSS * S/((k+1)*RTT) = MSS / (L*(k+1)*RTT).
Note that this is different from TCP which has average throughput: B1.22MSS , RTT L
where the square root of L appears in the denominator.
Problem 53
Let’s assume 1500-byte packets and a 100 ms round-trip time. From the TCP throughput equation B  1.22  MSS , we have
RTT L
10 Gbps = 1.22 * (1500*8 bits) / (.1 sec * srqt(L)), or
sqrt(L) = 14640 bits / (10^9 bits) = 0.00001464, or L = 2.14 * 10^(-10)
Problem 54
An advantage of using the earlier values of cwnd and ssthresh at t2 is that TCP would not have to go through slow start and congestion avoidance to ramp up to the throughput value obtained at t1. A disadvantage of using these values is that they may be no longer accurate. In particular, if the path has become more congested between t1 and t2, the sender will send a large window’s worth of segments into an already (more) congested path.
Problem 55
a) The server will send its response to Y.

b) The server can be certain that the client is indeed at Y. If it were at some other address spoofing Y, the SYNACK would have been sent to the address Y, and the TCP in that host would not send the TCP ACK segment back. Even if the attacker were to send an appropriately timed TCP ACK segment, it would not know the correct server sequence number (since the server uses random initial sequence numbers.)
Problem 56
a) Referring to the figure below, we see that the total delay is
RTT + RTT + S/R + RTT + S/R + RTT + 12S/R = 4RTT + 14 S/R
b) Similarly, the delay in this case is:
RTT+RTT + S/R + RTT + S/R + RTT + S/R + RTT + 8S/R = 5RTT +11 S/R
c) Similarly, the delay in this case is:
RTT + RTT + S/R + RTT + 14 S/R = 3 RTT + 15 S/R
initiate TCP connection
request object
RTT
first window = S/R
second window = 2S/R
third window = 4S/R
object delivered
fourth window = 8S/R
complete transmission
time at client
time at server

Chapter 4 Review Questions
1. A network-layer packet is a datagram. A router forwards a packet based on the packet’s IP (layer 3) address. A link-layer switch forwards a packet based on the packet’s MAC (layer 2) address.
2. Datagram-based network layer: forwarding; routing. Additional function of VC-based network layer: call setup.
3. Forwarding is about moving a packet from a router’s input port to the appropriate output port. Routing is about determining the end-to-routes between sources and destinations.
4. Yes, both use forwarding tables. For descriptions of the tables, see Section 4.2.
5. Single packet: guaranteed delivery; guaranteed delivery with guaranteed maximum delay. Flow of packets: in-order packet delivery; guaranteed minimal bandwidth; guaranteed maximum jitter. None of these services is provided by the Internet’s network layer. ATM’s CBR service provides both guaranteed delivery and timing. ABR does not provide any of these services.
6. Interactive live multimedia applications, such as IP telephony and video conference, could benefit from ATM CBR’s service, which maintains timing.
7. With the shadow copy, the forwarding lookup is made locally, at each input port, without invoking the centralized routing processor. Such a decentralized approach avoids creating a lookup processing bottleneck at a single point within the router.
8. Switching via memory; switching via a bus; switching via an interconnection network. An interconnection network can forward packets in parallel as long as all the packets are being forwarded to different output ports.
9. If the rate at which packets arrive to the fabric exceeds switching fabric rate, then packets will need to queue at the input ports. If this rate mismatch persists, the queues will get larger and larger and eventually overflow the input port buffers, causing packet loss. Packet loss can be eliminated if the switching fabric speed is at least n times as fast as the input line speed, where n is the number of input ports.
10. Assuming input and output line speeds are the same, packet loss can still occur if the rate at which packets arrive to a single output port exceeds the line speed. If this rate mismatch persists, the queues will get larger and larger and eventually overflow the output port buffers, causing packet loss. Note that increasing switch fabric speed cannot prevent this problem from occurring.

11. HOL blocking: Sometimes the a packet that is first in line at an input port queue must wait because there is no available buffer space at the output port to which it wants to be forwarded. When this occurs, all the packets behind the first packet are blocked, even if their output queues have room to accommodate them. HOL blocking occurs at the input port.
12. Yes. They have one address for each interface.
13. 11011111 00000001 00000011 00011100.
14. Students will get different correct answers for this question.
15. 8 interfaces; 3 forwarding tables.
16. 50% overhead.
17. The 8-bit protocol field in the IP datagram contains information about which transport layer protocol the destination host should pass the segment to.
18. Typically the wireless router includes a DHCP server. DHCP is used to assign IP addresses to the 5 PCs and to the router interface. Yes, the wireless router also uses NAT as it obtains only one IP address from the ISP.
19.
20. Yes, because the entire IPv6 datagram (including header fields) is encapsulated in an IPv4 datagram.
21. Link state algorithms: Computes the least-cost path between source and destination using complete, global knowledge about the network. Distance-vector routing: The calculation of the least-cost path is carried out in an iterative, distributed manner. A node only knows the neighbor to which it should forward a packet in order to reach given destination along the least-cost path, and the cost of that path from itself to the destination.
22. Routers are organized into autonomous systems (ASs). Within an AS, all routers run the same intra-AS routing protocol. The problem of scale is solved since an router in an AS need only know about routers within its AS and the subnets that attach to the AS. To route across ASes, the inter-AS protocol is based on the AS graph and does not take individual routers into account.
IPv6 has a fixed length header, which does not include most of the options an IPv4 header can include. Even though the IPv6 header contains two 128 bit addresses (source and destination IP address) the whole header has a fixed length of 40 bytes only. Several of the fields are similar in spirit. Traffic class, payload length, next header and hop limit in IPv6 are respectively similar to type of service, datagram
length, upper-layer protocol and time to live in IPv4.

23. No. Each AS has administrative autonomy for routing within an AS.
24. No. The advertisement tells D that it can get to z in 11 hops by way of A. However, D can already get to z by way of B in 7 hops. Therefore, there is no need to modify the entry for z in the table. If, on the other hand, the advertisement said that A were only 4 hops away from z by way of C, then D would indeed modify its forwarding table.
25. With OSPF, a router periodically broadcasts routing information to all other routers in the AS, not just to its neighboring routers. This routing information sent by a router has one entry for each of the router’s neighbors; the entry gives the distance from the router to the neighbor. A RIP advertisement sent by a router contains information about all the networks in the AS, although this information is only sent to its neighboring routers.
26. “sequence of ASs on the routes”
27. Policy: Among ASs, policy issues dominate. It may well be important that traffic originating in a given AS not be able to pass through another specific AS. Similarly, a given AS may want to control what transit traffic it carries between other ASs. Within an AS, everything is nominally under the same administrative control and thus policy issues a much less important role in choosing routes with in AS.
Scale: The ability of a routing algorithm and its data structures to scale to handle routing to/among large numbers of networks is a critical issue in inter-AS routing. Within an AS, scalability is less of a concern. For one thing, if a single administrative domain becomes too large, it is always possible to divide it into two ASs and perform inter-AS routing between the two new ASs.
Performance: Because inter-AS routing is so policy oriented, the quality (for example, performance) of the routes used is often of secondary concern (that is, a longer or more costly route that satisfies certain policy criteria may well be taken over a route that is shorter but does not meet that criteria). Indeed, we saw that among ASs, there is not even the notion of cost (other than AS hop count) associated with routes. Within a single AS, however, such policy concerns are of less importance, allowing routing to focus more on the level of performance realized on a route.
28. ISP C can use the BGP Multi-Exit Descriptor to suggest to ISP B that the preferred route to ISP D is through the east coast peering point. For example, the east coast BGP router in ISP C can advertise a route to D with an MED value of 5. The west coast router in ISP C can advertise a route to D with an MED value of 10. Since a lower value is preferred, ISP B knows that ISP C wants to receive traffic on the east coast. In practice, a router can ignore the MED value, and so ISP B can still use hot potato routing to pass traffic to ISP C destined to ISP D via the west coast peering point.

29. A subnet is a portion of a larger network; a subnet does not contain a router; its boundaries are defined by the router and host interfaces. A prefix is the network portion of a CDIRized address; it is written in the form a.b.c.d/x ; A prefix covers one or more subnets. When a router advertises a prefix across a BGP session, it includes with the prefix a number of BGP attributes. In BGP jargon, a prefix along with its attributes is a BGP route (or simply a route).
30. Routers use the AS-PATH attribute to detect and prevent looping advertisements; they also use it in choosing among multiple paths to the same prefix. The NEXT- HOP attribute indicates the IP address of the first router along an advertised path (outside of the AS receiving the advertisement) to a given prefix. When configuring its forwarding table, a router uses the NEXT-HOP attribute.
31. A tier-1 ISP B may not to carry transit traffic between two other tier-1 ISPs, say A and C, with which B has peering agreements. To implement this policy, ISP B would not advertise to A routes that pass through C; and would not advertise to C routes that pass through A.
32. N-way unicast has a number of drawbacks, including:
 Efficiency: multiple copies of the same packet are sent over the same link for potentially many links; source must generate multiple copies of same packet
 Addressing: the source must discover the address of all the recipients
33. a) uncontrolled flooding: T; controlled flooding: T; spanning-tree: F b) uncontrolled flooding: T; controlled flooding: F; spanning-tree: F
34. False
35. IGMP is a protocol run only between the host and its first-hop multicast router. IGMP allows a host to specify (to the first-hop multicast router) the multicast group it wants to join. It is then up to the multicast router to work with other multicast routers (i.e., run a multicast routing protocol) to ensure that the data for the host-joined multicast group is routed to the appropriate last-hop router and from there to the host.
36. In a group-shared tree, all senders send their multicast traffic using the same routing tree. With source-based tree, the multicast datagrams from a given source are routed over s specific routing tree constructed for that source; thus each source may have a different source-based tree and a router may have to keep track of several source- based trees for a given multicast group.

Chapter 4 Problems
Problem 1
a) With a connection-oriented network, every router failure will involve the routing of that connection. At a minimum, this will require the router that is “upstream” from the failed router to establish a new downstream part of the path to the destination node, with all of the requisite signaling involved in setting up a path. Moreover, all of the routers on the initial path that are downstream from the failed node must take down the failed connection, with all of the requisite signaling involved to do this.
With a connectionless datagram network, no signaling is required to either set up a new downstream path or take down the old downstream path. We have seen, however, that routing tables will need to be updated (e.g., either via a distance vector algorithm or a link state algorithm) to take the failed router into account. We have seen that with distance vector algorithms, this routing table change can sometimes be localized to the area near the failed router. Thus, a datagram network would be preferable. Interestingly, the design criteria that the initial ARPAnet be able to function under stressful conditions was one of the reasons that datagram architecture was chosen for this Internet ancestor.
b) In order for a router to maintain an available fixed amount of capacity on the path between the source and destination node for that source-destination pair, it would need to know the characteristics of the traffic from all sessions passing through that link. That is, the router must have per-session state in the router. This is possible in a connection-oriented network, but not with a connectionless network. Thus, a connection-oriented VC network would be preferable.
c) In this scenario, datagram architecture has more control traffic overhead. This is due to the various packet headers needed to route the datagrams through the network. But in VC architecture, once all circuits are set up, they will never change. Thus, the signaling overhead is negligible over the long run.
Problem 2
a) Maximum number of VCs over a link = 28 = 256.
b) The centralized node could pick any VC number which is free from the set
{0,1,…,28-1}. In this manner, it is not possible that there are fewer VCs in progress
than 256 without there being any common free VC number.
c) Each of the links can independently allocate VC numbers from the set {0,1,…,28-1}.
Thus, a VC will likely have a different VC number for each link along its path. Each router in the VC’s path must replace the VC number of each arriving packet with the VC number associated with the outbound link.

Problem 3
For a VC forwarding table, the columns are : Incoming Interface, Incoming VC Number, Outgoing Interface, Outgoing VC Number. For a datagram forwarding table, the columns are: Destination Address, Outgoing Interface.
Problem 4
a) Data destined to host H3 is forwarded through interface 3
Destination Address Link Interface H3 3
b) No, because forwarding rule is only based on destination address.
c) One possible configuration is:
Incoming interface Incoming VC# Outgoing Interface 1 12 3 22 2 63 4 18
Note, that the two flows could actually have the same VC numbers.
d) One possible configuration is:
Router B.
Incoming interface Incoming VC# Outgoing Interface 1 22 2 24
Router C.
Incoming interface Incoming VC# Outgoing Interface 1 18 2 50
Router D.
Incoming interface Incoming VC# Outgoing Interface 1 24 3 70
2 50 3 76
Problem 5
Outgoing VC#
Outgoing VC#
Outgoing VC#
Outgoing VC#
a) No VC number can be assigned to the new VC; thus the new VC cannot be established in the network.

b) Each link has two available VC numbers. There are four links. So the number of combinations is 24 = 16. One example combination is (10,00,00,10).
Problem 6
In a virtual circuit network, there is an end-to-end connection in the sense that each router along the path must maintain state for the connection; hence the terminology connection service. In a connection-oriented transport service over a connectionless network layer, such as TCP over IP, the end systems maintain connection state; however the routers have no notion of any connections; hence the terminology connection-oriented service.
Problem 7
a) No, you can only transmit one packet at a time over a shared bus.
b) Yes, as discussed in the text, as long as the two packets use different input busses and different output busses, they can be forwarded in parallel.
c) No, in this case the two packets would have to be sent over the same output bus at the same time, which is not possible.
Problem 8
a) (n-1)D b) (n-1)D c) 0
Problem 9
The minimal number of time slots needed is 3. The scheduling is as follows. Slot 1: send X in top input queue, send Y in middle input queue.
Slot 2: send X in middle input queue, send Y in bottom input queue
Slot 3: send Z in bottom input queue.
Largest number of slots is still 3. Actually, based on the assumption that a non-empty input queue is never idle, we see that the first time slot always consists of sending X in the top input queue and Y in either middle or bottom input queue, and in the second time slot, we can always send two more datagram, and the last datagram can be sent in third time slot.
NOTE: Actually, if the first datagram in the bottom input queue is X, then the worst case would require 4 time slots.

Problem 10
a)
b)
Prefix Match
11100000 00
11100000 01000000
1110000
11100001 1
otherwise
Link Interface
0 1 2 3 3
Prefix match for first address is 5th entry: link interface 3 Prefix match for second address is 3nd entry: link interface 2
Prefix match for third address is 4th
entry: link interface 3
Link Interface
Problem 11
Destination Address Range
00000000 through 00111111
01000000 through 01011111
01100000 through 01111111
10000000 through
10111111
11000000 through
11111111
number of number of number of number of
0
1
2
2
3
addresses for interface 0 = 26 addresses for interface 1 = 25  32 addresses for interface 2 = 26  25 addresses for interface 3 = 26  64
 64  32  96
 64

Problem 12
Destination Address Range
11000000
through (32 addresses)
11011111
10000000
through(64 addresses)
10111111
11100000
through (32 addresses)
11111111
00000000
through (128 addresses) 01111111
Problem 13
223.1.17.0/26 223.1.17.128/25 223.1.17.192/28
Problem 14
Destination Address
200.23.16/21 200.23.24/24 200.23.24/21 otherwise
Problem 15
Destination Address
11100000 00 (224.0/10) 11100000 01000000 (224.64/16) 1110000 (224/8)
11100001 1 (225.128/9)
otherwise
Link Interface
0
1
2
3
Link Interface
0 1 2 3
Link Interface
0 1 2 3 3

Problem 16
Any IP address in range 128.119.40.128 to 128.119.40.191
Four equal size subnets: 128.119.40.64/28, 128.119.40.80/28, 128.119.40.96/28, 128.119.40.112/28
Problem 17
From a)
b)
Subnet Subnet Subnet
Subnet Subnet Subnet
214.97.254/23, possible assignments are
A: 214.97.255/24 (256 addresses)
B: 214.97.254.0/25 – 214.97.254.0/29 (128-8 = 120 addresses) C: 214.97.254.128/25 (128 addresses)
D: 214.97.254.0/31 (2 addresses) E: 214.97.254.2/31 (2 addresses) F: 214.97.254.4/30 (4 addresses)
To simplify the solution, assume that no datagrams have router interfaces as ultimate destinations. Also, label D, E, F for the upper-right, bottom, and upper- left interior subnets, respectively.
Router 1
Longest Prefix Match
11010110 01100001 11111111 11010110 01100001 11111110 11010110 01100001 11111110
Router 2
Longest Prefix Match
11010110 01100001 11111111 11010110 01100001 11111110 11010110 01100001 11111110
Router 3
Longest Prefix Match
0000000
000001
0000000
0
0000001
Outgoing Interface
Subnet A Subnet D Subnet F
Outgoing Interface
Subnet D Subnet B Subnet E
Outgoing Interface

11010110 01100001 11111111 000001 11010110 01100001 11111110 0000001 11010110 01100001 11111110 1
Problem 18
Subnet F Subnet E Subnet C
The IP address blocks of Polytechnic Institute of New York University are: NetRange: 128.238.0.0 – 128.238.255.255
CIDR: 128.238.0.0/16
The IP address blocks Stanford University are: NetRange: 171.64.0.0 – 171.67.255.255 CIDR: 171.64.0.0/14
The IP address blocks University of Washington are: NetRange: 140.142.0.0 – 140.142.255.255
CIDR: 140.142.0.0/16
determine with certainty the geographical www.maxmind.com is used to determine the locations of the Web servers at Polytechnic
Institute of New York University, Stanford University and University of Washington. Locations of the Web server at Polytechnic Institute of New York University is
Locations of the Web server Stanford University is
Locations of the Web server at University of Massachusetts is
No, the whois services cannot be used to
location of a specific IP address.

Problem 19
The maximum size of data field in each fragment = 680 (because there are 20 bytes IP header). Thus the number of required fragments   2400  20   4
 680 
Each fragment will have Identification number 422. Each fragment except the last one
will be of size 700 bytes (including IP header). The last datagram will be of size 360 bytes (including IP header). The offsets of the 4 fragments will be 0, 85, 170, 255. Each of the first 3 fragments will have flag=1; the last fragment will have flag=0.
Problem 20
MP3 file size = 5 million bytes. Assume the data is carried in TCP segments, with each TCP segment also having 20 bytes of header. Then each datagram can carry 1500- 40=1460 bytes of the MP3 file
Number of datagrams required  5106   3425 . All but the last datagram will be 1,500  1460 
bytes; the last datagram will be 960+40 = 1000 bytes. Note that here there is no fragmentation – the source host does not create datagrams larger than 1500 bytes, and these datagrams are smaller than the MTUs of the links.
Problem 21
a) Home addresses: 192.168.1.1, 192.168.1.2, 192.168.1.3 with the router interface being 192.168.1.4
b)
WAN Side 24.34.112.235, 4000 24.34.112.235, 4001 24.34.112.235, 4002 24.34.112.235, 4003 24.34.112.235, 4004 24.34.112.235, 4005
192.168.1.1, 3345 192.168.1.1, 3346 192.168.1.2, 3445 192.168.1.2, 3446 192.168.1.3, 3545 192.168.1.3, 3546
Problem 22
NAT Translation Table
LAN Side
a) Since all IP packets are sent outside, so we can use a packet sniffer to record all IP packets generated by the hosts behind a NAT. As each host generates a sequence of IP packets with sequential numbers and a distinct (very likely, as they are randomly chosen from a large space) initial identification number (ID), we can group IP packets with consecutive IDs into a cluster. The number of clusters is the number of hosts behind the NAT.

For more practical algorithms, see the following papers.
“A Technique for Counting NATted Hosts”, by Steven M. Bellovin, appeared in IMW’02, Nov. 6-8, 2002, Marseille, France.
“Exploiting the IPID field to infer network path and end-system characteristics.” Weifeng Chen, Yong Huang, Bruno F. Ribeiro, Kyoungwon Suh, Honggang Zhang, Edmundo de Souza e Silva, Jim Kurose, and Don Towsley.
PAM’05 Workshop, March 31 – April 01, 2005. Boston, MA, USA.
b) However, if those identification numbers are not sequentially assigned but randomly assigned, the technique suggested in part (a) won’t work, as there won’t be clusters in sniffed data.
Problem 23
It is not possible to devise such a technique. In order to establish a direct TCP connection between Arnold and Bernard, either Arnold or Bob must initiate a connection to the other. But the NATs covering Arnold and Bob drop SYN packets arriving from the WAN side. Thus neither Arnold nor Bob can initiate a TCP connection to the other if they are both behind NATs.
Problem 24
y-x-u, y-x-v-u, y-x-w-u, y-x-w-v-u,
y-w-u, y-w-v-u, y-w-x-u, y-w-x-v-u, y-w-v-x-u,
y-z-w-u, y-z-w-v-u, y-z-w-x-u, y-z-w-x-v-u, y-z-w-v-x-u,
Problem 25
x to z:
x-y-z, x-y-w-z,
x-w-z, x-w-y-z, x-v-w-z, x-v-w-y-z, x-u-w-z, x-u-w-y-z, x-u-v-w-z, x-u-v-w-y-z
z to u:
z-w-u,
z-w-v-u, z-w-x-u, z-w-v-x-u, z-w-x-v-u, z-w-y-x-u, z-w-y-x-v-u,
z-y-x-u, z-y-x-v-u, z-y-x-w-u, z-y-x-w-y-u, z-y-x-v-w-u,
z-y-w-v-u, z-y-w-x-u, z-y-w-v-x-u, z-y-w-x-v-u, z-y-w-y-x-u, z-y-w-y-x-v-u
z to w:
z-w, z-y-w, z-y-x-w, z-y-x-v-w, z-y-x-u-w, z-y-x-u-v-w, z-y-x-v-u-w

Problem 26
Step N’
0 x
1 xv
2 xvu
3 xvuw
4 xvuwy
5 xvuwyt
6 xvuwytz
Problem 27
a)
Step N’
0 t
1 tu
2 tuv
3 tuvw
4 tuvwx
5 tuvwxy
6 tuvwxyz
b)
Step N’
D(t),p(t) D(u),p(u) D(v),p(v)
∞ ∞ 3,x 7,v 6,v 3,x 7,v 6,v 3,x 7,v 6,v 3,x 7,v 6,v 3,x 7,v 6,v 3,x 7,v 6,v 3,x
D(x), p(x) D(u),p(u) D(v),p(v)
∞ 2,t 4,t ∞ 2,t 4,t 7,v 2,t 4,t 7,v 2,t 4,t 7,v 2,t 4,t 7,v 2,t 4,t 7,v 2,t 4,t
D(x), p(x) D(t),p(t) D(v),p(v)
D(w),p(w) D(y),p(y) D(z),p(z)
6,x 6,x 8,x 6,x 6,x 8,x 6,x 6,x 8,x 6,x 6,x 8,x 6,x 6,x 8,x 6,x 6,x 8,x 6,x 6,x 8,x
D(w),p(w) D(y),p(y) D(z),p(z)
∞ 7,t ∞ 5,u 7,t ∞ 5,u 7,t ∞ 5,u 7,t ∞ 5,u 7,t 15,x 5,u 7,t 15,x 5,u 7,t 15,x
D(w),p(w) D(y),p(y) D(z),p(z)
3,u ∞ ∞ 3,u 9,t ∞ 3,u 9,t ∞ 3,u 9,t ∞ 3,u 9,t 14,x 3,u 9,t 14,x 3,u 9,t 14,x
u ∞ 2,u 3,u ut ∞ 2,u 3,u utv 6,v 2,u 3,u utvw 6,v 2,u 3,u utvwx 6,v 2,u 3,u utvwxy 6,v 2,u 3,u utvwxyz 6,v 2,u 3,u

c)
Step N’
D(x), p(x) D(u),p(u) D(t),pt)
D(w),p(w) D(y),p(y) D(z),p(z)
4,v 8,v ∞ 4,v 8,v 11,x 4,v 8,v 11,x 4,v 8,v 11,x 4,v 8,v 11,x 4,v 8,v 11,x 4,v 8,v 11,x
D(t),p(t) D(y),p(y) D(z),p(z)
∞ ∞ ∞ 5,u ∞ ∞ 5,u 12,v ∞ 5,u 12,v ∞ 5,u 12,v 14,x 5,u 12,v 14,x 5,u 12,v 14,x
D(w),p(w) D(t),p(t) D(z),p(z)
∞ 7,y 12,y 12,x 7,y 12,y 12,x 7,y 12,y 12,x 7,y 12,y 12,x 7,y 12,y 12,x 7,y 12,y 12,x 7,y 12,y
D(w),p(w) D(y),p(y) D(t),p(t)
∞ 12,z ∞ 14,x 12,z ∞ 14,x 12,z 15,v 14,x 12,z 15,v 14,x 12,z 15,v
d)
Step N’
D(x), p(x) D(u),p(u) D(v),p(v)
e)
Step N’
D(x), p(x) D(u),p(u) D(v),p(v)
f)
Step N’
D(x), p(x) D(u),p(u) D(v),p(v)
v 3,v 3,v 4,v vx 3,v 3,v 4,v vxu 3,v 3,v 4,v vxut 3,v 3,v 4,v vxutw 3,v 3,v 4,v vxutwy 3,v 3,v 4,v vxutwyz 3,v 3,v 4,v
w 6,w 3,w 4,w wu 6,w 3,w 4,w wuv 6,w 3,w 4,w wuvt 6,w 3,w 4,w wuvtx 6,w 3,w 4,w wuvtxy 6,w 3,w 4,w wuvtxyz 6,w 3,w 4,w
y 6,y ∞ 8,y yx 6,y ∞ 8,y yxt 6,y 9,t 8,y yxtv 6,y 9,t 8,y yxtvu 6,y 9,t 8,y yxtvuw 6,y 9,t 8,y yxtvuwz 6,y 9,t 8,y
z 8,z ∞ ∞ zx 8,z ∞ 11,x zxv 8,z 14,v 11,x zxvy 8,z 14,v 11,x zxvyu 8,z 14,v 11,x

zxvyuw 8,z 14,v 11,x zxvyuwt 8,z 14,v 11,x
Problem 28
Cost to uvxyz
v∞∞∞∞∞ Fromx ∞ ∞ ∞ ∞ ∞
z∞62∞0
Cost to uvxyz
v103∞6 Fromx ∞ 3 0 3 2 z75250
Cost to uvxyz
v10335 Fromx 4 3 0 3 2 z65250
Cost to uvxyz
v10335 Fromx 4 3 0 3 2 z65250
Problem 29
14,x 12,z 15,v 14,x 12,z 15,v
The wording of this question was a bit ambiguous. We meant this to mean, “the number of iterations from when the algorithm is run for the first time” (that is, assuming the only information the nodes initially have is the cost to their nearest neighbors). We assume

that the algorithm runs synchronously (that is, in one step, all nodes compute their distance tables at the same time and then exchange tables).
At each iteration, a node exchanges distance tables with its neighbors. Thus, if you are node A, and your neighbor is B, all of B’s neighbors (which will all be one or two hops from you) will know the shortest cost path of one or two hops to you after one iteration (i.e., after B tells them its cost to you).
Let d be the “diameter” of the network – the length of the longest path without loops between any two nodes in the network. Using the reasoning above, after d 1 iterations, all nodes will know the shortest path cost of d or fewer hops to all other nodes. Since any path with greater than d hops will have loops (and thus have a greater cost than that path with the loops removed), the algorithm will converge in at most d 1 iterations.
ASIDE: if the DV algorithm is run as a result of a change in link costs, there is no a priori bound on the number of iterations required until convergence unless one also specifies a bound on link costs.
Problem 30
a) Dx(w) = 2, Dx(y) = 4, Dx(u) = 7
b) First consider what happens if c(x,y) changes. If c(x,y) becomes larger or smaller (as long as c(x,y) >=1) , the least cost path from x to u will still have cost at least 7. Thus a change in c(x,y) (if c(x,y)>=1) will not cause x to inform its neighbors of any changes.
If c(x,y)= <1, then the least cost path now passes through y and has cost +6. Now consider if c(x,w) changes. If c(x,w) =   1, then the least-cost path to u continues to pass through w and its cost changes to 5 + ; x will inform its neighbors of this new cost. If c(x,w) =  > 6, then the least cost path now passes through y and has cost 11; again x will inform its neighbors of this new cost.
c) Any change in link cost c(x,y) (and as long as c(x,y) >=1) will not cause x to inform its neighbors of a new minimum-cost path to u .
Problem 31
Node x table
Cost to
xyz x034

Fromy ∞ ∞ ∞ z∞∞∞
Cost to xyz
x034 Fromy 3 0 6 z460
Node y table
Cost to
xyz x∞∞∞
Fromy 3 0 6 z∞∞∞
Cost to xyz
x034 Fromy 3 0 6 z460
Node z table
Cost to
xyz x∞∞∞ Fromy ∞ ∞ ∞
z460
Cost to xyz
x034 Fromy 3 0 6 z460
Problem 32
NO, this is because that decreasing link cost won’t cause a loop (caused by the next-hop relation of between two nodes of that link). Connecting two nodes with a link is equivalent to decreasing the link weight from infinite to the finite weight.
Problem 33

At each step, each updating of a node’s distance vectors is based on the Bellman-Ford equation, i.e., only decreasing those values in its distance vector. There is no increasing in values. If no updating, then no message will be sent out. Thus, D(x) is non-increasing. Since those costs are finite, then eventually distance vectors will be stabilized in finite steps.
Problem 34
a)
b) Yes,therewillbeacount-to-infinityproblem.Thefollowingtableshowstherouting converging process. Assume that at time t0, link cost change happens. At time t1, y updates its distance vector and informs neighbors w and z. In the following table, “” stands for “informs”.
Router z
Informs w, Dz(x)=
Informs y, Dz(x)=6
Router w
Informs y, Dw(x)=
Informs z, Dw(x)=5
Router y
Informs w, Dy(x)=4
Informs z, Dy(x)=4
time
t0
t1
t2
t3
t4
Z
w, Dz(x)= y, Dz(x)=6
No change
w, Dz(x)= y, Dz(x)=11
W
y, Dw(x)= z, Dw(x)=5
y, Dw(x)= z, Dw(x)=10
No change
Y
w, Dy(x)=4 z, Dy(x)=4
w, Dy(x)=9 z, Dy(x)= 
No change
w, Dy(x)=14 z, Dy(x)= 
We see that w, y, z form a loop in their computation of the costs to router x. If we continue the iterations shown in the above table, then we will see that, at t27, z detects that its least cost to x is 50, via its direct link with x. At t29, w learns its least cost to x is 51 via z. At t30, y updates its least cost to x to be 52 (via w). Finally, at time t31, no updating, and the routing is stabilized.
time
t27
t28
t29
t30
t31
Z
w, Dz(x)=50 y, Dz(x)=50
via w,  via y, 55 via z, 50
W
y, Dw(x)= z, Dw(x)=50
y, Dw(x)=51 z, Dw(x)= 
via w,  via y,  via z, 51
Y
w, Dy(x)=53 z, Dy(x)= 
w, Dy(x)=  z, Dy(x)= 52
via w, 52 via y, 60 via z, 53

c) cut the link between y and z.
Problem 35
Since full AS path information is available from an AS to a destination in BGP, loop detection is simple – if a BGP peer receives a route that contains its own AS number in the AS path, then using that route would result in a loop.
Problem 36
The chosen path is not necessarily the shortest AS-path. Recall that there are many issues to be considered in the route selection process. It is very likely that a longer loop-free path is preferred over a shorter loop-free path due to economic reason. For example, an AS might prefer to send traffic to one neighbor instead of another neighbor with shorter AS distance.
Problem 37
a) eBGP b) iBGP c) eBGP d) iBGP
Problem 38
a) I1 because this interface begins the least cost path from 1d towards the gateway router 1c.
b) I2. Both routes have equal AS-PATH length but I2 begins the path that has the closest NEXT-HOP router.
c) I1. I1 begins the path that has the shortest AS-PATH.
Problem 39
One way for C to force B to hand over all of B’s traffic to D on the east coast is for C to only advertise its route to D via its east coast peering point with C.

Problem 40
Bx Cy Cy
wBxwA A
X’s view of the topology W’s view of the topology
In the above solution, X does not know about the AC link since X does not receive an advertised route to w or to y that contain the AC link (i.e., X receives no advertisement containing both AS A and AS C on the path to a destination.
Problem 41
BitTorrent file sharing and Skype P2P applications.
Consider a BitTorrent file sharing network in which peer 1, 2, and 3 are in stub networks W, X, and Y respectively. Due the mechanism of BitTorrent’s file sharing, it is quire possible that peer 2 gets data chunks from peer 1 and then forwards those data chunks to 3. This is equivalent to B forwarding data that is finally destined to stub network Y.
Problem 42
A should advise to B two routes, AS-paths A-W and A-V. A should advise to C only one route, A-V.
C receives AS paths: B-A-W, B-A-V, A-V.
Problem 43
Since Z wants to transit Y’s traffic, Z will send route advertisements to Y. In this manner, when Y has a datagram that is destined to an IP that can be reached through Z, Y will have the option of sending the datagram through Z. However, if Z advertizes routes to Y, Y can re-advertize those routes to X. Therefore, in this case, there is nothing Z can do from preventing traffic from X to transit through Z.
Problem 44
The minimal spanning tree has z connected to y via x at a cost of 14(=8+6). z connected to v via x at a cost of 11(=8+3);
z connected to u via x and v, at a cost of 14(=8+3+3);
z connected to w via x, v, and u, at a cost of 17(=8+3+3+3).

This can be obtained by Prim’s algorithm to grow a minimum spanning tree.
Problem 45
The 32 receives are shown connected to the sender in the binary tree configuration shown above. With network-layer broadcast, a copy of the message is forwarded over each link exactly once. There are thus 62 link crossings (2+4+8+16+32). With unicast emulation, the sender unicasts a copy to each receiver over a path with5 hops. There are thus 160 link crossings (5*32).
A topology in which all receivers are in a line, with the sender at one end of the line, will have the largest disparity between the cost of network-layer broadcast and unicast emulation.
Problem 46
A
B
FE
G
The thicker shaded lines represent The shortest path tree from A to all destination. Other solutions are possible, but in these solutions, B can not route to either C or D from A.
c
D
Problem 47

A
B
c
Problem 48
Z
x
w
Problem 49
F
D E
y
v
G
t
u
The center-based tree for the topology shown in the original figure connects A to C; B to C; E to C; and F to C (all directly). D connects to C via E, and G connects to C via D, E. This center-based tree is different from the minimal spanning tree shown in the figure.
Problem 50
The center-based tree for the topology shown in the original figure connects t to v; u to v; w to v; x to v; and y to v (all directly). And z connected to v via x. This center-based tree is different from the minimal spanning tree.
Problem 51
Dijkstra’s algorithm for the network below, with node A as the source, results in a least- unicast-cost path tree of links AC, AB, and BD, with an overall free cost of 20. The minimum spanning tree contains links AB, BD, and DC, at a cost of 11.

c
B
After 1 step 3 copies are transmitted, after 2 steps 6 copies are transmitted. After 3 steps, 12 copies are transmitted, and so on. After k steps, 3*2k-1 copies will be transmitted in that step.
1 A91D
10
Problem 52
The protocol must be built at the application layer. For example, an application may periodically multicast its identity to all other group members in an application-layer message.
Problem 54
A simple application-layer protocol that will allow all members to know the identity of all other members in the group is for each instance of the application to send a multicast message containing its identity to all other members. This protocol sends message in- band, since the multicast channel is used to distribute the identification messages as well as multicast data from the application itself. The use of the in-band signaling makes use of the existing multicast distribution mechanism, leading to a very simple design.
Problem 55
32  4  28 bits are available for multicast addresses. Thus, the size of the multicast address space is N  228 .
The probability that two groups choose the same address is
3 6 12
24 48
…
Problem 53

1 228 3.73109 N
The probability that 1000 groups all have different addresses is
N  ( N  1)  ( N  2 )  ( N  9 9 9 )  1   2   9 9 9  N1000  1 N 1 N 1 N 
Ignoring cross-product terms, this is approximately equal to
12999 9991000
1   N   1  2 N  0 . 9 9 8

Chapter 5 Review Questions
1. The transportation mode, e.g., car, bus, train, car.
2. Although each link guarantees that an IP datagram sent over the link will be received at the other end of the link without errors, it is not guaranteed that IP datagrams will arrive at the ultimate destination in the proper order. With IP, datagrams in the same TCP connection can take different routes in the network, and therefore arrive out of order. TCP is still needed to provide the receiving end of the application the byte stream in the correct order. Also, IP can lose packets due to routing loops or equipment failures.
3. Framing: there is also framing in IP and TCP; link access; reliable delivery: there is also reliable delivery in TCP; flow control: there is also flow control in TCP; error detection: there is also error detection in IP and TCP; error correction; full duplex: TCP is also full duplex.
4. There will be a collision in the sense that while a node is transmitting it will start to receive a packet from the other node.
5. Slotted Aloha: 1, 2 and 4 (slotted ALOHA is only partially decentralized, since it requires the clocks in all nodes to be synchronized). Token ring: 1, 2, 3, 4.
6. After the 5th collision, the adapter chooses from {0, 1, 2,…, 31}. The probability that it chooses 4 is 1/32. It waits 204.8 microseconds.
7. In polling, a discussion leader allows only one participant to talk at a time, with each participant getting a chance to talk in a round-robin fashion. For token ring, there isn’t a discussion leader, but there is wine glass that the participants take turns holding. A participant is only allowed to talk if the participant is holding the wine glass.
8. When a node transmits a frame, the node has to wait for the frame to propagate around the entire ring before the node can release the token. Thus, if L/R is small as compared to tprop, then the protocol will be inefficient.
9. 248 MAC addresses; 232 IPv4 addresses; 2128 IPv6 addresses.
10. C’s adapter will process the frames, but the adapter will not pass the datagrams up the protocol stack. If the LAN broadcast address is used, then C’s adapter will both process the frames and pass the datagrams up the protocol stack.
11. An ARP query is sent in a broadcast frame because the querying host does not which adapter address corresponds to the IP address in question. For the response, the sending node knows the adapter address to which the response should be sent, so

there is no need to send a broadcast frame (which would have to be processed by all the other nodes on the LAN).
12. No it is not possible. Each LAN has its own distinct set of adapters attached to it, with each adapter having a unique LAN address.
13. The three Ethernet technologies have identical frame structures.
14. 2 (the internal subnet and the external internet)
15. In 802.1Q there is a 12- bit VLAN identifier. Thus 212 = 4,096 VLANs can be supported.
16. We can string the N switches together. The first and last switch would use one port for trunking; the middle N-2 switches would use two ports. So the total number of ports is 2+ 2(N-2) = 2N-2 ports.
Chapter 5 Problems
Problem 1
1110 1 0110 0 1001 0 1101 1 1100 0
Problem 2
Suppose we begin with the initial two-dimensional parity matrix:
000 0 111 1 010 1 101 0
With a bit error in row 2, column 3, the parity of row 2 and column 3 is now wrong in the matrix below:
000 0 110 1 010 1 101 0

Now suppose there is a bit error in row 2, column 2 and column 3. The parity of row 2 is now correct! The parity of columns 2 and 3 is wrong, but we can’t detect in which rows the error occurred!
000 0 100 1 010 1 101 0
The above example shows that a double bit error can be detected (if not corrected).
Problem 3
01001100 01101001 + 01101110 01101011 ——————————
10111010 11010100 + 00100000 01001100 ——————————
11011011 00100000 + 01100001 01111001 —————————–
00111100 10011010 (overflow, then wrap around) + 01100101 01110010
——————————
10100010 00001100
The one’s complement of the sum is 01011101 11110011
Problem 4
a) To compute the Internet checksum, we add up the values at 16-bit quantities:
00000001 00000010 00000011 00000100 00000101 00000110 00000111 00001000 00001001 00001010 ————————- 00011001 00011110
The one’s complement of the sum is 11100110 11100001.
b) To compute the Internet checksum, we add up the values at 16-bit quantities:

01000010 01000011 01000100 01000101 01000110 01000111 01001000 01001001 01001010 01001011 ————————- 10011111 10100100
The one’s complement of the sum is 01100000 01011011
c) To compute the Internet checksum, we add up the values at 16-bit quantities:
01100010 01100011
01100100 01100101
01100110 01100111
01101000 01101001
01101010 01101011
————————-
00000000 00000101
The one’s complement of the sum is 11111111 11111010.
Problem 5
If we divide 10011 into 1010101010 0000, we get 1011011100, with a remainder of R=0100. Note that, G=10011 is CRC-4-ITU standard.
Problem 6
a) we get 1000100011, with a remainder of R=0101.
b) we get 1011111111, with a remainder of R=0001.
c) we get 0101101110, with a remainder of R=0010.
Problem 7
a) Without loss of generality, suppose ith bit is flipped, where 0<= i <= d+r-1 and assume that the least significant bit is 0th bit. A single bit error means that the received data is K=D*2r XOR R + 2i. It is clear that if we divide K by G, then the reminder is not zero. In general, if G contains at least two 1’s, then a single bit error can always be detected. b) ThekeyinsighthereisthatGcanbedividedby11(binarynumber),butanynumber of odd-number of 1’s cannot be divided by 11. Thus, a sequence (not necessarily contiguous) of odd-number bit errors cannot be divided by 11, thus it cannot be divided by G. Problem 8 a) E(p)  Np(1 p)N1 E'(p)  N(1 p)N1  Np(N 1)(1 p)N2  N(1 p)N2((1 p) p(N 1)) E'(p)0p* 1 N b) Thus lim E(p*)  1 N e Problem 9 (1 1)N E(p*)N1(11)N1 (11)N1  N N N N 11 N lim(11)1 N N lim(11)N 1 N N e E(p)  Np(1 p)2(N1) E'(p)  N(1 p)2(N2)  Np2(N 1)(1 p)2(N3)  N(1 p)2(N3)((1 p) p2(N 1)) E'(p)0 p* 1 E(p*) N (1 1 )2(N1) 2N 1 2N 1 2N 1 limE(p*)11 1 N 2 e 2e Problem 10 a) A’s average throughput is given by pA(1-pB). Total efficiency is pA(1-pB) + pB(1-pA). b) A’s throughput is pA(1-pB)=2pB(1-pB)= 2pB- 2(pB)2. B’s throughput is pB(1-pA)=pB(1-2pB)= pB- 2(pB)2. Clearly, A’s throughput is not twice as large as B’s. In order to make pA(1-pB)= 2 pB(1-pA), we need that pA= 2 – (pA / pB). c) A’s throughput is 2p(1-p)N-1, and any other node has throughput p(1-p)N-2(1-2p). Problem 11 a) (1 – p(A))4 p(A) where, p(A) = probability that A succeeds in a slot p(A) = p(A transmits and B does not and C does not and D does not) = p(A transmits) p(B does not transmit) p(C does not transmit) p(D does not transmit) = p(1 – p) (1 – p)(1-p) = p(1 – p)3 Hence, p(A succeeds for first time in slot 5) = (1 – p(A))4 p(A) = (1 – p(1 – p)3)4 p(1 – p)3 b) p(A succeeds in slot 4) = p(1-p)3 p(B succeeds in slot 4) = p(1-p)3 p(C succeeds in slot 4) = p(1-p)3 p(D succeeds in slot 4) = p(1-p)3 p(either A or B or C or D succeeds in slot 4) = 4 p(1-p)3 (because these events are mutually exclusive) c) p(some node succeeds in a slot) = 4 p(1-p)3 p(no node succeeds in a slot) = 1 - 4 p(1-p)3 Hence, p(first success occurs in slot 3) = p(no node succeeds in first 2 slots) p(some node succeeds in 3rd slot) = (1 - 4 p(1-p)3)2 4 p(1-p)3 d) efficiency = p(success in a slot) =4 p(1-p)3 Problem 12 Problem 13 The length of a polling round is N(Q/Rdpoll). The number of bits transmitted in a polling round is NQ . The maximum throughput therefore is NQ  R N(Q/Rdpoll) 1dpollR Problem 14 a), b) See figure below. Q A 192.168.1.001 00-00-00-00-00-00 192.168.2.001 44-44-44-44-44-44 LAN LAN B c) 1. Forwarding table in E determines that the datagram should be routed to interface 192.168.3.002. 2. The adapter in E creates and Ethernet packet with Ethernet destination address 88- 88-88-88-88-88. 3. Router 2 receives the packet and extracts the datagram. The forwarding table in this router indicates that the datagram is to be routed to 198.162.2.002. 4. Router 2 then sends the Ethernet packet with the destination address of 33-33-33- 33-33-33 and source address of 55-55-55-55-55-55 via its interface with IP address of 198.162.2.003. 5. The process continues until the packet has reached Host B. LAN 192.168.1.002 22-22-22-22-22-22 192.168.2.002 33-33-33-33-33-33 Router 1 C Router 2 E 192.168.3.001 77-77-77-77-77-77 192.168.2.003 55-55-55-55-55 192.168.2.004 66-66-66-66-66 192.168.3.003 99-99-99-99-99-99 192.168.3.002 88-88-88-88-88-88 192.168.1.003 11-11-11-11-11-11 D F d) ARPinEmustnowdeterminetheMACaddressof198.162.3.002.HostEsendsout an ARP query packet within a broadcast Ethernet frame. Router 2 receives the query packet and sends to Host E an ARP response packet. This ARP response packet is carried by an Ethernet frame with Ethernet destination address 77-77-77-77-77-77. Problem 15 a) No. E can check the subnet prefix of Host F’s IP address, and then learn that F is on the same LAN. Thus, E will not send the packet to the default router R1. Ethernet frame from E to F: Source IP = E’s IP address Destination IP = F’s IP address Source MAC = E’s MAC address Destination MAC = F’s MAC address b) No, because they are not on the same LAN. E can find this out by checking B’s IP address. Ethernet frame from E to R1: Source IP = E’s IP address Destination IP = B’s IP address Source MAC = E’s MAC address Destination MAC = The MAC address of R1’s interface connecting to Subnet 3. c) Switch S1 will broadcast the Ethernet frame via both its interfaces as the received ARP frame’s destination address is a broadcast address. And it learns that A resides on Subnet 1 which is connected to S1 at the interface connecting to Subnet 1. And, S1 will update its forwarding table to include an entry for Host A. Yes, router R1 also receives this ARP request message, but R1 won’t forward the message to Subnet 3. B won’t send ARP query message asking for A’s MAC address, as this address can be obtained from A’s query message. Once switch S1 receives B’s response message, it will add an entry for host B in its forwarding table, and then drop the received frame as destination host A is on the same interface as host B (i.e., A and B are on the same LAN segment). Problem 16 Lets call the switch between subnets 2 and 3 S2. That is, router R1 between subnets 2 and 3 is now replaced with switch S2. a) No. E can check the subnet prefix of Host F’s IP address, and then learn that F is on the same LAN segment. Thus, E will not send the packet to S2. Ethernet frame from E to F: Source IP = E’s IP address Destination IP = F’s IP address Source MAC = E’s MAC address Destination MAC = F’s MAC address b) Yes,becauseEwouldliketofindB’sMACaddress.Inthiscase,EwillsendanARP query packet with destination MAC address being the broadcast address. This query packet will be re-broadcast by switch 1, and eventually received by Host B. Ethernet frame from E to S2: Source IP = E’s IP address Destination IP = B’s IP address Source MAC = E’s MAC address Destination MAC = broadcast MAC address: FF-FF-FF-FF-FF-FF. c) Switch S1 will broadcast the Ethernet frame via both its interfaces as the received ARP frame’s destination address is a broadcast address. And it learns that A resides on Subnet 1 which is connected to S1 at the interface connecting to Subnet 1. And, S1 will update its forwarding table to include an entry for Host A. Yes, router S2 also receives this ARP request message, and S2 will broadcast this query packet to all its interfaces. B won’t send ARP query message asking for A’s MAC address, as this address can be obtained from A’s query message. Once switch S1 receives B’s response message, it will add an entry for host B in its forwarding table, and then drop the received frame as destination host A is on the same interface as host B (i.e., A and B are on the same LAN segment). Problem 17 Wait for 51,200 bit times. For 10 Mbps, this wait is 51.2103bits 5.12msec 10106bps For 100 Mbps, the wait is 512  sec. Problem 18 At t  0 A transmits. At t  576 , A would finish transmitting. In the worst case, B begins transmitting at time t=324, which is the time right before the first bit of A’s frame arrives at B. At time t=324+325=649 B 's first bit arrives at A . Because 649> 576, A finishes transmitting before it detects that B has transmitted. So A incorrectly thinks that its frame was successfully transmitted without a collision.
Problem 19
Time, t
0
245
293
293+245 = 538 538+96=634
293+512 = 805 634+245=879
Event
A and B begin transmission
A and B detect collision
A and B finish transmitting jam signal
B ‘s last bit arrives at A ; A detects an idle channel
A starts transmitting
B returns to Step2
B must sense idle channel for 96 bit times before it transmits
A’s transmission reaches B
Because A ‘s retransmission reaches B before B ‘s scheduled retransmission time (805+96), B refrains from transmitting while A retransmits. Thus A and B do not collide. Thus the factor 512 appearing in the exponential backoff algorithm is sufficiently large.
Problem 20
a) Let Y be a random variable denoting the number of slots until a success: P(Y m)(1)m1,
where  is the probability of a success.
This is a geometric distribution, which has mean 1/  . The number of consecutive
wasted slots is X  Y  1 that x  E[ X ]  E[Y ]  1  1  
  Np(1 p)N1


x  1 Np(1 p)N1 Np(1 p)N1
k k
kx k1Np(1p)N1
efficiency Np(1 p)N1 b)
Maximizing efficiency is equivalent to minimizing x , which is equivalent to maximizing
 . We know from the text that  is maximized at p  1 . N
c)
efficiency  k
1(1 1)N1 k 1N
(1  N ) N 1
k  k
d) Clearly, k
k  e 1
Problem 21
LAN
B
lim efficiency  N
k 11/e 1/e
k  e 1 approaches 1 as k   .
C
A
111.111.111.001 00-00-00-00-00-00
122.222.222.001 44-44-44-44-44-44
133.333.333.001 77-77-77-77-77-77
E
Router 1
LAN
LAN
Router 2
111.111.111.002 22-22-22-22-22-22
122.222.222.002 33-33-33-33-33-33
122.222.222.003 55-55-55-55-55
122.222.222.004 66-66-66-66-66
133.333.333.002 88-88-88-88-88-88
111.111.111.003 11-11-11-11-11-11
D
133.333.333.003 99-99-99-99-99-99
F

i) from A to left router: Source MAC address: 00-00-00-00-00-00 Destination MAC address: 22-22-22-22-22-22
Source IP: 111.111.111.001 Destination IP: 133.333.333.003
ii) from the left router to the right router: Source MAC address: 33-33-33-33-33-33 Destination MAC address: 55-55-55-55-55-55
Source IP: 111.111.111.001 Destination IP: 133.333.333.003
iii) from the right router to F: Source MAC address: 88-88-88-88-88-88 Destination MAC address: 99-99-99-99-99-99
Source IP: 111.111.111.001 Destination IP: 133.333.333.003
Problem 22
i) from A to switch: Source MAC address: 00-00-00-00-00-00 Destination MAC address: 55-55-55-55-55-55
Source IP: 111.111.111.001
Destination IP: 133.333.333.003
ii) from switch to right router: Source MAC address: 00-00-00-00-00-00
Destination MAC address: 55-55-55-55-55-55 Source IP: 111.111.111.001
Destination IP: 133.333.333.003
iii) from right router to F: Source MAC address: 88-88-88-88-88-88 Destination MAC address: 99-99-99-99-99-99
Source IP: 111.111.111.001 Destination IP: 133.333.333.003
Problem 23
If all the 11=9+2 nodes send out data at the maximum possible rate of 100 Mbps, a total aggregate throughput of 11*100 = 1100 Mbps is possible.
Problem 24
Each departmental hub is a single collision domain that can have a maximum throughput of 100 Mbps. The links connecting the web server and the mail server has a maximum

throughput of 100 Mbps. Hence, if the three collision domains and the web server and mail server send out data at their maximum possible rates of 100 Mbps each, a maximum total aggregate throughput of 500 Mbps can be achieved among the 11 end systems.
Problem 25
All of the 11 end systems will lie in the same collision domain. In this case, the maximum total aggregate throughput of 100 Mbps is possible among the 11 end sytems.
Problem 26
Action
Switch Table State
Link(s) packet is forwarded to
Explanation
B sends a frame to E
Switch learns interface corresponding to MAC address of B
A, C, D, E, and F
Since switch table is empty, so switch does not know the interface corresponding to MAC address of E
E replies with a frame to B
Switch learns interface corresponding to MAC address of E
B
Since switch already knows interface corresponding to MAC address of B
A sends a frame to B
Switch learns the interface corresponding to MAC address of A
B
Since switch already knows the interface corresponding to MAC address of B
B replies with a frame to A
Switch table state remains the same as before
A
Since switch already knows the interface corresponding to MAC address of A
Problem 27
a) The time required to fill L  8 bits is L8 secLmsec.
128103 16
b) For L  1,500, the packetization delay is
1500 msec  93.75msec. 16

For L  50, the packetization delay is 50 msec  3.125msec.
16
c) Store-and-forward delay  L  8  40
For L  1,500 , the delay is 1500  8  40 sec  19.4 sec
R
622106
For L  50, store-and-forward delay  1 sec .
d) Store-and-forward delay is small for both cases for typical link speeds. However, packetization delay for L  1500 is too large for real-time voice applications.
Problem 28
The IP addresses for those three computers (from left to right) in EE department are: 111.111.1.1, 111.111.1.2, 111.111.1.3. The subnet mask is 111.111.1/24.
The IP addresses for those three computers (from left to right) in CS department are: 111.111.2.1, 111.111.2.2, 111.111.2.3. The subnet mask is 111.111.2/24.
The router’s interface card that connects to port 1 can be configured to contain two sub- interface IP addresses: 111.111.1.0 and 111.111.2.0. The first one is for the subnet of EE department, and the second one is for the subnet of CS department. Each IP address is associated with a VLAN ID. Suppose 111.111.1.0 is associated with VLAN 11, and 111.111.2.0 is associated with VLAN 12. This means that each frame that comes from subnet 111.111.1/24 will be added an 802.1q tag with VLAN ID 11, and each frame that comes from 111.111.2/24 will be added an 802.1q tag with VLAN ID 12.
Suppose that host A in EE department with IP address 111.111.1.1 would like to send an IP datagram to host B (111.111.2.1) in CS department. Host A first encapsulates the IP datagram (destined to 111.111.2.1) into a frame with a destination MAC address equal to the MAC address of the router’s interface card that connects to port 1 of the switch. Once the router receives the frame, then it passes it up to IP layer, which decides that the IP datagram should be forwarded to subnet 111.111.2/24 via sub-interface 111.111.2.0. Then the router encapsulates the IP datagram into a frame and sends it to port 1. Note that this frame has an 802.1q tag VLAN ID 12. Once the switch receives the frame port 1, it knows that this frame is destined to VLAN with ID 12, so the switch will send the frame

to Host B which is in CS department. Once Host B receives this frame, it will remove the 802.1q tag.
Problem 29
in label
out label
dest
out interf.
7 5
10
12 8
A
D A
0
0 1
in label
out label
dest
out interf.
7
A
0
in label
out label
dest
out interf.
10 12
6 9
A D
1 0
R6 0 0
R5
00
R4 1
D
R3 1 00
in label
out label
dest
out interf.
5
A
0
R2
A R1
in label
out label
dest
out interf.
8
6
A
0
in label
out label
dest
out inter.
6
–
A
0
Problem 30

in label
out label
dest
out interf.
3 2
12 4
D D
0 1
in label
out label
dest
out interf.
3
D
0
in label
out label
dest
out interf.
12
–
D
0
R6 0
0
00 R4 1 R3 1
D
R1
R5 1
0
A
in label
out label
dest
out interf.
2
D
0
R2
in label
out label
dest
out interf.
4
1
D
0
in label
out label
dest
out inter.
1
12
D
1
Problem 31
(The following description is short, but contains all major key steps and key protocols involved.)
Your computer first uses DHCP to obtain an IP address. You computer first creates a special IP datagram destined to 255.255.255.255 in the DHCP server discovery step, and puts it in a Ethernet frame and broadcast it in the Ethernet. Then following the steps in the DHCP protocol, you computer is able to get an IP address with a given lease time.
A DHCP server on the Ethernet also gives your computer a list of IP addresses of first- hop routers, the subnet mask of the subnet where your computer resides, and the addresses of local DNS servers (if they exist).
Since your computer’s ARP cache is initially empty, your computer will use ARP protocol to get the MAC addresses of the first-hop router and the local DNS server.
Your computer first will get the IP address of the Web page you would like to download. If the local DNS server does not have the IP address, then your computer will use DNS protocol to find the IP address of the Web page.
Once your computer has the IP address of the Web page, then it will send out the HTTP request via the first-hop router if the Web page does not reside in a local Web server. The

HTTP request message will be segmented and encapsulated into TCP packets, and then further encapsulated into IP packets, and finally encapsulated into Ethernet frames. Your computer sends the Ethernet frames destined to the first-hop router. Once the router receives the frames, it passes them up into IP layer, checks its routing table, and then sends the packets to the right interface out of all of its interfaces.
Then your IP packets will be routed through the Internet until they reach the Web server.
The server hosting the Web page will send back the Web page to your computer via HTTP response messages. Those messages will be encapsulated into TCP packets and then further into IP packets. Those IP packets follow IP routes and finally reach your first-hop router, and then the router will forward those IP packets to your computer by encapsulating them into Ethernet frames.
Problem 32
a) Each flow evenly shares a link’s capacity with other flows traversing that link, then the 80 flows crossing the B to access-router 10 Gbps links (as well as the access router to border router links) will each only receive 10 Gbps / 80 = 125 Mbps
b) In Topology of Figure 5.31, there are four distinct paths between the first and third tier-2 switches, together providing 40 Gbps for the traffic from racks 1-4 to racks 9-12. Similarly, there are four links between second and fourth tier-2 switches, together providing 40 Gbps for the traffic from racks 5-8 to 13-16. Thus the total aggregate bandwidth is 80 Gbps, and the value per flow rate is 1 Gbps.
c) Now 20 flows will need to share each 1 Gbps bandwidth between pairs of TOR switches. So the host-to-host bit rate will be 0.5 Gbps.
Problem 33
a) Both email and video application uses the fourth rack for 0.1 percent of the time.
b) Probability that both applications need fourth rack is 0.001*0.001 = 10-6.
c) Suppose the first three racks are for video, the next rack is a shared rack for both video and email, and the next three racks are for email. Let’s assume that the fourth rack has all the data and software needed for both the email and video applications. With the topology of Figure 5.31, both applications will have enough intra-bandwidth as long as both are not simultaneously using the fourth rack. From part b, both are using the fourth rack for no more than .00001 % of time, which is within the .0001% requirement.

Chapter 6 Review Questions
1. In infrastructure mode of operation, each wireless host is connected to the larger network via a base station (access point). If not operating in infrastructure mode, a network operates in ad-hoc mode. In ad-hoc mode, wireless hosts have no infrastructure with which to connect. In the absence of such infrastructure, the hosts themselves must provide for services such as routing, address assignment, DNS-like name translation, and more.
2. a) Single hop, infrastructure-based b) Single hop, infrastructure-less c) Multi-hop, infrastructure-based d) Multi-hop, infrastructure-less
3. Path loss is due to the attenuation of the electromagnetic signal when it travels through matter. Multipath propagation results in blurring of the received signal at the receiver and occurs when portions of the electromagnetic wave reflect off objects and ground, taking paths of different lengths between a sender and receiver. Interference from other sources occurs when the other source is also transmitting in the same frequency range as the wireless network.
4. a) Increasing the transmission power b) Reducing the transmission rate
5. APs transmit beacon frames. An AP’s beacon frames will be transmitted over one of the 11 channels. The beacon frames permit nearby wireless stations to discover and identify the AP.
6. False
7. APs transmit beacon frames. An AP’s beacon frames will be transmitted over one of the 11 channels. The beacon frames permit nearby wireless stations to discover and identify the AP.
8. False
9. Each wireless station can set an RTS threshold such that the RTS/CTS sequence is used only when the data frame to be transmitted is longer than the threshold. This ensures that RTS/CTS mechanism is used only for large frames.
10. No, there wouldn’t be any advantage. Suppose there are two stations that want to transmit at the same time, and they both use RTS/CTS. If the RTS frame is as long as a DATA frames, the channel would be wasted for as long as it would have been wasted for two colliding DATA frames. Thus, the RTS/CTS exchange is only useful when the RTS/CTS frames are significantly smaller than the DATA frames.

11. Initially the switch has an entry in its forwarding table which associates the wireless station with the earlier AP. When the wireless station associates with the new AP, the new AP creates a frame with the wireless station’s MAC address and broadcasts the frame. The frame is received by the switch. This forces the switch to update its forwarding table, so that frames destined to the wireless station are sent via the new AP.
12. Any ordinary Bluetooth node can be a master node whereas access points in 802.11 networks are special devices (normal wireless devices like laptops cannot be used as access points).
13. False
14. “Opportunistic Scheduling” refers to matching the physical layer protocol to channel conditions between the sender and the receiver, and choosing the receivers to which packets will be sent based on channel condition. This allows the base station to make best use of the wireless medium.
15. UMTS to GSM and CDMA-2000 to IS-95.
16. No. A node can remain connected to the same access point throughout its connection to the Internet (hence, not be mobile). A mobile node is the one that changes its point of attachment into the network over time. Since the user is always accessing the Internet through the same access point, she is not mobile.
17. A permanent address for a mobile node is its IP address when it is at its home network. A care-of-address is the one its gets when it is visiting a foreign network. The COA is assigned by the foreign agent (which can be the edge router in the foreign network or the mobile node itself).
18. False
19. The home network in GSM maintains a database called the home location register (HLR), which contains the permanent cell phone number and subscriber profile information about each of its subscribers. The HLR also contains information about the current locations of these subscribers. The visited network maintains a database known as the visitor location register (VLR) that contains an entry for each mobile user that is currently in the portion of the network served by the VLR. VLR entries thus come and go as mobile users enter and leave the network.
The edge router in home network in mobile IP is similar to the HLR in GSM and the edge router in foreign network is similar to the VLR in GSM.

20. Anchor MSC is the MSC visited by the mobile when a call first begins; anchor MSC thus remains unchanged during the call. Throughout the call’s duration and regardless of the number of inter-MSC transfers performed by the mobile, the call is routed from the home MSC to the anchor MSC, and then from the anchor MSC to the visited MSC where the mobile is currently located.
21. a) Local recovery
b) TCP sender awareness of wireless links c) Split-connection approaches
Chapter 6 Problems
Problem 1
Output corresponding to bit d1 = [-1,1,-1,1,-1,1,-1,1] Output corresponding to bit d0 = [1,-1,1,-1,1,-1,1,-1]
Problem 2
Sender 2 output = [1,-1,1,1,1,-1,1,1]; [ 1,-1,1,1,1,-1,1,1]
Problem 3
d21 11(1)(1)111111(1)(1)11111 8
d2 11(1)(1)111111(1)(1)11111 8
Problem 4
Sender 1: (1, 1, 1, -1, 1, -1, -1, -1) Sender 2: (1, -1, 1, 1, 1, 1, 1, 1)
Problem 5
a) The two APs will typically have different SSIDs and MAC addresses. A wireless station arriving to the café will associate with one of the SSIDs (that is, one of the APs). After association, there is a virtual link between the new station and the AP. Label the APs AP1 and AP2. Suppose the new station associates with AP1. When the new station sends a frame, it will be addressed to AP1. Although AP2 will also receive the frame, it will not process the frame because the frame is not addressed to it. Thus, the two ISPs can work in parallel over the same channel. However, the two ISPs will be sharing the same wireless bandwidth. If wireless stations in different

ISPs transmit at the same time, there will be a collision. For 802.11b, the maximum aggregate transmission rate for the two ISPs is 11 Mbps.
b) NowiftwowirelessstationsindifferentISPs(andhencedifferentchannels)transmit at the same time, there will not be a collision. Thus, the maximum aggregate transmission rate for the two ISPs is 22 Mbps for 802.11b.
Problem 6
Suppose that wireless station H1 has 1000 long frames to transmit. (H1 may be an AP that is forwarding an MP3 to some other wireless station.) Suppose initially H1 is the only station that wants to transmit, but that while half-way through transmitting its first frame, H2 wants to transmit a frame. For simplicity, also suppose every station can hear every other station’s signal (that is, no hidden terminals). Before transmitting, H2 will sense that the channel is busy, and therefore choose a random backoff value.
Now suppose that after sending its first frame, H1 returns to step 1; that is, it waits a short period of times (DIFS) and then starts to transmit the second frame. H1’s second frame will then be transmitted while H2 is stuck in backoff, waiting for an idle channel. Thus, H1 should get to transmit all of its 1000 frames before H2 has a chance to access the channel. On the other hand, if H1 goes to step 2 after transmitting a frame, then it too chooses a random backoff value, thereby giving a fair chance to H2. Thus, fairness was the rationale behind this design choice.
Problem 7
A frame without data is 32 bytes long. Assuming a transmission rate of 11 Mbps, the time to transmit a control frame (such as an RTS frame, a CTS frame, or an ACK frame) is (256 bits)/(11 Mbps) = 23 usec. The time required to transmit the data frame is (8256 bits)/(11 Mbps) = 751
DIFS + RTS + SIFS + CTS + SIFS + FRAME + SIFS + ACK
= DIFS + 3SIFS + (3*23 + 751) usec = DIFS + 3SIFS + 820 usec
Problem 8
a) 1 message/ 2 slots
b) 2 messages/slot
c) 1 message/slot

d) i) 1 message/slot ii) 2 messages/slot iii) 2 messages/slot
e) i) 1 message/4 slots
ii) slot 1: Message AB, message DC
slot 2: Ack BA slot 3: Ack CD
= 2 messages/ 3 slots iii)
slot 1: Message C D
slot 2: Ack DC, message A B slot 3: Ack BA
= 2 messages/3 slots
Problem 10
Repeat
a) 10 Mbps if it only transmits to node A. This solution is not fair since only A is getting served. By “fair” it means that each of the four nodes should be allotted equal number of slots.
b) For the fairness requirement such that each node receives an equal amount of data during each downstream sub-frame, let n1, n2, n3, and n4 respectively represent the number of slots that A, B, C and D get.
Now,
data transmitted to A in 1 slot = 10t Mbits
(assuming the duration of each slot to be t)
Hence,
Total amount of data transmitted to A (in n1 slots) = 10t n1
Similarly total amounts of data transmitted to B, C, and D equal to 5t n2, 2.5t n3, and t n4 respectively.
Now, to fulfill the given fairness requirement, we have the following condition:
10t n1 = 5t n2 = 2.5t n3 = t n4
Hence,
n2 = 2 n1
n3 = 4 n1 n4 = 10 n1

Now, the total number of slots is N. Hence, n1+ n2+ n3+ n4 = N
i.e. n1+ 2 n1 + 4 n1 + 10 n1 = N
i.e. n1 = N/17
Hence,
n2 = 2N/17
n3 = 4N/17
n4 = 10N/17
The average transmission rate is given by:
(10t n1+5t n2+ 2.5t n3+t n4)/tN
= (10N/17 + 5 * 2N/17 + 2.5 * 4N/17 + 1 * 10N/17)/N = 40/17 = 2.35 Mbps
c) Let node A receives twice as much data as nodes B, C, and D during the sub-frame. Hence,
10tn1 = 2 * 5tn2 = 2 * 2.5tn3 = 2 * tn4 i.e. n2 = n1
n3 = 2n1
n4 = 5n1
Again,
n1 + n2 + n3 + n4 = N
i.e. n 1+ n1 + 2n1 + 5n1 = N i.e. n1 = N/9
Now, average transmission rate is given by: (10t n1+5t n2+ 2.5t n3+t n4)/tN
= 25/9 = 2.78 Mbps
Similarly, considering nodes B, C, or D receive twice as much data as any other nodes, different values for the average transmission rate can be calculated.
Problem 11
a) No. All the routers might not be able to route the datagram immediately. This is because the Distance Vector algorithm (as well as the inter-AS routing protocols like BGP) is decentralized and takes some time to terminate. So, during the time when the algorithm is still running as a result of advertisements from the new foreign network, some of the routers may not be able to route datagrams destined to the mobile node.
b) Yes. This might happen when one of the nodes has just left a foreign network and joined a new foreign network. In this situation, the routing entries from the old foreign network might not have been completely withdrawn when the entries from the new network are being propagated.

c) The time it takes for a router to learn a path to the mobile node depends on the number of hops between the router and the edge router of the foreign network for the node.

Problem 12
If the correspondent is mobile, then any datagrams destined to the correspondent would have to pass through the correspondent’s home agent. The foreign agent in the network being visited would also need to be involved, since it is this foreign agent that notifies the correspondent’s home agent of the location of the correspondent. Datagrams received by the correspondent’s home agent would need to be encapsulated/tunneled between the correspondent’s home agent and foreign agent, (as in the case of the encapsulated diagram at the top of Figure 6.23.
Problem 13
Because datagrams must be first forward to the home agent, and from there to the mobile, the delays will generally be longer than via direct routing. Note that it is possible, however, that the direct delay from the correspondent to the mobile (i.e., if the datagram is not routed through the home agent) could actually be smaller than the sum of the delay

from the correspondent to the home agent and from there to the mobile. It would depend on the delays on these various path segments. Note that indirect routing also adds a home agent processing (e.g., encapsulation) delay.
Problem 14
First, we note that chaining was discussed at the end of section 6.5. In the case of chaining using indirect routing through a home agent, the following events would happen:
 The mobile node arrives at A, A notifies the home agent that the mobile is now visiting A and that datagrams to the mobile should now be forwarded to the specified care-of-address (COA) in A.
 The mobile node moves to B. The foreign agent at B must notify the foreign agent at A that the mobile is no longer resident in A but in fact is resident in B and has the specified COA in B. From then on, the foreign agent in A will forward datagrams it receives that are addressed to the mobile’s COA in A to the mobile’s COA in B.
 The mobile node moves to C. The foreign agent at C must notify the foreign agent at B that the mobile is no longer resident in B but in fact is resident in C and has the specified COA in C. From then on, the foreign agent in B will forward datagrams it receives (from the foreign agent in A) that are addressed to the mobile’s COA in B to the mobile’s COA in C.
Note that when the mobile goes offline (i.e., has no address) or returns to its home network, the datagram-forwarding state maintained by the foreign agents in A, B and C must be removed. This teardown must also be done through signaling messages. Note that the home agent is not aware of the mobile’s mobility beyond A, and that the correspondent is not at all aware of the mobile’s mobility.
In the case that chaining is not used, the following events would happen:
 

The mobile node arrives at A, A notifies the home agent that the mobile is now visiting A and that datagrams to the mobile should now be forwarded to the specified care-of-address (COA) in A.
The mobile node moves to B. The foreign agent at B must notify the foreign agent at A and the home agent that the mobile is no longer resident in A but in fact is resident in B and has the specified COA in B. The foreign agent in A can remove its state about the mobile, since it is no longer in A. From then on, the home agent will forward datagrams it receives that are addressed to the mobile’s COA in B.
The mobile node moves to C. The foreign agent at C must notify the foreign agent at B and the home agent that the mobile is no longer resident in B but in fact is resident in C and has the specified COA in C. The foreign agent in B can remove its state about the mobile, since it is no longer in B. From then on, the home agent will forward datagrams it receives that are addressed to the mobile’s COA in C.

When the mobile goes offline or returns to its home network, the datagram-forwarding state maintained by the foreign agent in C must be removed. This teardown must also be done through signaling messages. Note that the home agent is always aware of the mobile’s current foreign network. However, the correspondent is still blissfully unaware of the mobile’s mobility.
Problem 15
Two mobiles could certainly have the same care-of-address in the same visited network. Indeed, if the care-of-address is the address of the foreign agent, then this address would be the same. Once the foreign agent decapsulates the tunneled datagram and determines the address of the mobile, then separate addresses would need to be used to send the datagrams separately to their different destinations (mobiles) within the visited network.
Problem 16
If the MSRN is provided to the HLR, then the value of the MSRN must be updated in the HLR whenever the MSRN changes (e.g., when there is a handoff that requires the MSRN to change). The advantage of having the MSRN in the HLR is that the value can be provided quickly, without querying the VLR. By providing the address of the VLR Rather than the MSRN), there is no need to be refreshing the MSRN in the HLR.

Chapter 7 Review Questions
1.
2. Spatial Redundancy: It is the redundancy within a given image. Intuitively, an image consists of mostly white space has a high degree of redundancy and can be efficiently compressed without significantly sacrificing image quality.
Temporal Redundancy reflects repetition from image to subsequent image. If, for example, an image and the subsequent image are exactly the same, there is no reason to re-encode the subsequent image; it is instead more efficient simply to indicate during encoding that the subsequent image is exactly the same. If the two images are very similar, it may be not efficient to indicate how the second image differs from the first, rather than re-encode the second image.
3. Quantizing a sample into 1024 levels means 10 bits per sample. The resulting rate of the PCM digital audio signal is 160 Kbps.
4. Streaming stored audio/video: In this class of applications, the underlying medium is prerecorded video, such as a movie, a television show, or a prerecorded sporting event. These prerecorded videos are played on servers, and users send requests to the servers to view the videos on demand. Many internet companies today provide streaming video, including YouTube, Netflix, and Hulu.
Conversational Voice- and Video-over-IP: Real-time conversational voice over the Internet is often referred to as Internet telephony, since, from the user’s perspective, it is similar to the traditional circuit-switched telephone service. It is also commonly called Voice-over-IP (VOIP).Conversational video is similar except that it includes the video of the participants as well as their voices. Conversational voice and video are widely used in the Internet today, with the Internet companies like Skype and Google Talk boasting hundreds of millions of daily users.
Streaming Live Audio and Video: These applications allow users to receive a live radio or television transmission over the Internet. Today, thousands of radio and television stations around the world are broadcasting content over the internet.
5. UDP Streaming: With UDP streaming, the server transmits video at a rate that matches the client’s video consumption rate by clocking out the video chunks over UDP at a steady rate.
HTTP Streaming: In HTTP streaming, the video simply stored in an HTTP server as ordinary file with a specific URL. When a user wants to see the video, the client
Bit rate
Bytes transferred in 67 mins
Facebook Frank
40 kbps
20 Mbytes
Martha Music
200 kbps
100 Mbytes
Victor Video
4 Mbps
2 Gbytes

establishes a TCP connection with the server and issues an HTTP GET request for that URL. The server then sends the video file, within an HTTP response message, as quickly as possible, that is, as quickly as TCP congestion control and flow control will allow.
Adaptive HTTP Streaming (DASH): In Dynamic Adaptive Streaming over HTTP, the video is encoded several different versions, with each version having a different bit rate and, correspondingly, a different quality level. The client dynamically requests the chunks of video segments of a few seconds in length from the different versions. When the amount of available bandwidth is high, the client naturally selects chunks from a high-rate version; and when the available bandwidth is low, it naturally selects from a low-rate version.
6. The three significant drawbacks of UDP Streaming are:
1. Due to unpredictable and varying amount of available bandwidth between server
and client, constant-rate UDP streaming can fail to provide continuous play out.
2. It requires a media control server, such as an RTSP server, to process client-to- server interactivity requests and to track client state for each ongoing client
session.
3. Many firewalls are configured to block UDP traffic, preventing users behind
these firewalls from receiving UDP video.
7. No. On the client side, the client application reads bytes from the TCP receive buffer and places the bytes in the client application buffer.
8. The initial buffering delay is tp = Q/x = 4 seconds.
9. Enter Deep: This philosophy is to enter deep into the access networks of Internet Service Providers, by deploying server clusters in access ISPs all over the world.
Bring Home: A second design philosophy is to bring the ISP’s home by building large clusters at a smaller number of key locations and connecting these clusters using a private high-speed network.
10. Geographically closest cluster selection and real time measurements selection can find a good cluster with respect to LDNS. IP anycast chooses good cluster with respect to client itself.
11. Load on the cluster – clients should not be directed to overload clusters.
ISP delivery cost – the clusters may be chosen so that specific ISPs are used to carry CDN-to-client traffic, taking into account the different cost structures in the contractual relationships between ISPs and cluster operators.
12. End-to-end delay is the time it takes a packet to travel across the network from source to destination. Delay jitter is the fluctuation of end-to-end delay from packet to the next packet.

13. A packet that arrives after its scheduled play out time cannot be played out. Therefore, from the perspective of the application, the packet has been lost.
14. First scheme: send a redundant encoded chunk after every n chunks; the redundant chunk is obtained by exclusive OR-ing the n original chunks. Second scheme: send a lower-resolution low-bit rate scheme along with the original stream. Interleaving does not increase the bandwidth requirements of a stream.
15. RTP streams in different sessions: different multicast addresses; RTP streams in the same session: SSRC field; RTP packets are distinguished from RTCP packets by using distinct port numbers.
16. The role of a SIP registrar is to keep track of the users and their corresponding IP addresses which they are currently using. Each SIP registrar keeps track of the users that belong to its domain. It also forwards INVITE messages (for users in its domain) to the IP address which the user is currently using. In this regard, its role is similar to that of an authoritative name server in DNS.
17. In non-preemptive priority queuing, the transmission of a packet is not interrupted once it has begun. In preemptive priority queuing, the transmission of a packet will be interrupted if a higher priority packet arrives before transmission completes. This would mean that portions of the packet would be sent into the network as separate chunks; these chunks would no longer all have the appropriate header fields. For this reason, preemptive priority queuing is not used.
18. A scheduling discipline that is not work conserving is time division multiplexing, whereby a rotating frame is partitioned into slots, with each slot exclusively available to a particular class.
19. FIFO: line at Starbucks. RR: merging traffic (taking 1 vehicle from first lane then 1 from the second, then 1 from the first, and so on). WFQ: ticket counter at airport providing service to 2 people from first class and 1 from economy, and again 2 from first class, and so on.

Chapter 7 Problems
Problem 1
a) Client begins playout as soon as first block arrives at t1 and video blocks are to be played out over the fixed amount of time, d. So it follows that second video block should be arrived before time t1 + d to be played at right time, third block at
t1 + 2d and so on. We can see from figure that only blocks numbered 1,4,5,6 arrive at receiver before their playout times.
b) Client begins playout at time t1 + d and video blocks are to be played out over the fixed amount of time, d. So it follows that second video block should be arrived before time t1 +2d to be played at right time, third block at t1 + 3d and so on. We can see from figure that video blocks numbered from 1 to 6 except 7 arrive at receiver before their playout times.
c) Maximum two video blocks are ever stored in the client buffer. Video blocks numbered 3 and 4 arrive before t1 + 3d and after t1 + 2d, hence these two blocks are stored in the client buffer. Video block numbered 5 arrives before time t1 + 4d and after t1 + 3d, which is stored in the client buffer along with already stored video block numbered 4.
d) The smallest playout at the client should be t1 + 3d to ensure that every block has arrived in time.
Problem 2
a) During a playout period, the buffer starts with Q bits and decreases at rate r – x. Thus, after Q/(r – x) seconds after starting playback the buffer becomes empty. Thus, the continuous playout period is Q/(r – x) seconds. Once the buffer becomes empty, it fills at rate x for Q/x seconds, at which time it has Q bits and playback begins. Therefore, the freezing period is Q/x seconds.
b) Time until buffer has Q bits is Q/x seconds. Time to add additional B – Q bits is (B – Q)/(x – r) seconds. Thus the time until the application buffer becomes full is
seconds.
Problem 3
a) The server’s average send rate is .
b) This part (b) is an odd question and will be removed from the next edition. After playing out the first frame, because x(t) < r, the next frame will arrive after the scheduled playout time of the next frame. Thus playback will freeze after displaying the first frame. c) Let q(t) denote the number of bits in the buffer at time t. Playout begins when q(t) = Q. Let’s assume throughout this problem that HT/2 ≥ Q, so that q(t) = Q by the end of the first cycle for x(t). We have ()∫. Therefore, q (t) = Q when t = √ = tp. d) At time t = T, q (t) = HT/2 = Q, so that playout begins. If subsequently there is no freezing, we need q(t + T) > 0 for all t ≥ T, we have
()∫() >()∫()
With t = nT + ∆, with 0 < ∆ < T, we have from above ()( ) =() Which is easily seen to be possible for all 0 < ∆ < T. e) First consider the [0, T]. We have ( ) ( ) for tp ≤ t ≤ T q (t) is minimized at t = rT/H. It can then be shown that q (rT/H) ≥ 0 if and only if tp ≥ rT/2H. Furthermore, if tp = rT/2H, proof can be extended to show q (t) > 0 for all t ≥ T thus, tp < rT/2H and Q = r2T/8H. f) This is a very challenging problem. Assuming that B is reached before time T, then tf is solution to ( √ ) = B. Problem 4 a) Buffer grows at rate x – r. At time E, (x - r)*E bits are in buffer and are wasted. b) Let S be the time when the server has transmitted the entire video. If S > E, buffer grows at rate x – r until time E, so the waste is again (x – r)*E. If S < E, then at time E there are T – E seconds of video still to be played in buffer. So the waste is r* (T - E). Problem 5 a) N*N = N2. b) N+N=2N Problem 6 a) 160  h bytes are sent every 20 msec. Thus the transmission rate is (160  h)  8 Kbps  (64  .4h)Kbps 20 b) IP header: 20 bytes UDP header: 8 bytes RTP header: 12 bytes c) h=40 bytes (a 25% increase in the transmission rate!) Problem 7 a) Denote d ( n ) for the estimate after the nth sample. d(1) r t d(2) u(r t )(1u)(r t ) 33 44 d(3) u(r t )(1u)u(r t )(1u)(r t ) 223344 u(r t )(1u)u(r t )(1u)2(r t ) 223344 d(4) u(rt)(1u)d(3) 11 u(r t )(1u)u(r t )(1u)2u(r t )(1u)3(r t ) 11223344 b) d ( n )  u n  1 ( 1  u ) j ( r  t )  ( 1  u ) n ( r  t ) j1 c) 44 jj nn d()  u  (1u)j(rj tj) 1u j1  1  . 9 j ( r  t ) Problem 8 a) Denote v(n) for the estimate after the nth sample. Let   r  t . jjj 9 j1 The weight given to past samples decays exponentially. jj v(1)   d(1) (=0) 4 v(2) u  d(2) (1u)  d(1) 34 v(3) u  d(3) (1u)v(2) 2 u  d(3) u(1u)  d(2) (1u)2  d(1) 234 v(4) u  d(4) (1u)v(3) 1 u  d(4) (1u)u  d(3) u(1u)2  d(2) 123 (1u)3  d(1) 4 u  d(4) (1u)  d(3) (1u)2  d(2)  123 (1u)3  d(1) 4 b) v(n) un1(1u)j1 j d(nj1) (1u)n n d(1) j1 Problem 9 a) r1 – t1 + r2 - t2 + ...+rn-1-tn-1 = (n-1)dn-1 Substituting this into the expression for dn gives dn1d rntn n nn1 n b) The delay estimate in part (a) is an average of the delays. It gives equal weight to recent delays and to “old” delays. The delay estimate in Section 6.3 gives more weight to recent delays; delays in the distant past have relatively little impact on the estimate. Problem 10 The two procedures are very similar. They both use the same formula, thereby resulting in exponentially decreasing weights for past samples. One difference is that for estimating average RTT, the time when the data is sent and when the acknowledgement is received is recorded on the same machine. For the delay estimate, the two values are recorded on different machines. Thus the sample delay can actually be negative. Problem 11 a) The delay of packet 2 is 7 slots. The delay of packet 3 is 9 slots. The delay of packet 4 is 8 slots. The delay of packet 5 is 7 slots. The delay of packet 6 is 9 slots. The delay of packet 7 is 8 slots. The delay of packet 8 is > 8 slots.
b) Packets 3, 4, 6, 7, and 8 will not be received in time for their playout if playout begins at t-8.
c) Packets 3 and 6 will not be received in time for their playout if playout begins at t=9.
d) No packets will arrive after their playout time if playout time begins at t=10.
Problem 12
The answers to parts a and b are in the table below:
Packet Number
ri – ti
di
vi
1
7
7
0
2
8
7.10
0.09
3
8
7.19
0.162
4
7
7.17
0.163
5
9
7.35
0.311
6
9
7.52
0.428
7
8
7.57
0.429
8
8
7.61
0.425

Problem 13
a) Both schemes require 25% more bandwidth. The first scheme has a playback delay of 5 packets. The second scheme has a delay of 2 packets.
b) Thefirstschemewillbeabletoreconstructtheoriginalhigh-qualityaudioencoding. The second scheme will use the low quality audio encoding for the lost packets and will therefore have lower overall quality.
c) For the first scheme, many of the original packets will be lost and audio quality will be very poor. For the second scheme, every audio chunk will be available at the receiver, although only the low quality version will be available for every other chunk. Audio quality will be acceptable.
Problem 14
a) Each of the other N – 1 participants sends a single audio stream of rate r bps to the initiator. The initiator combines this stream with its own outgoing stream to create a stream of rate r. It then sends a copy of the combined stream to each of the N -1 other participants. The call initiator therefore sends at a total rate of (N-1)r bps, and the total rate aggregated over all participants is 2(N-1)r bps.
b) As before, each of the other N – 1 participants sends a single video stream of rate r bps to the initiator. But because the streams are now video, the initiator can no longer combine them into a single stream. The initiator instead must send each stream it receives to N – 2 participants. The call initiator therefore sends at a total rate of (N- 1)*(N-1)r bps, and the total rate aggregated over all participants is (N-1)r + (N-1)*(N- 1)r = N (N-1) r bps.
c) N *(N-1)r bps
Problem 15
a) As discussed in Chapter 2, UDP sockets are identified by the two-tuple consisting of destination IP address and destination port number. So the two packets will indeed pass through the same socket.
b) Yes, Alice only needs one socket. Bob and Claire will choose different SSRC’s, so Alice will be able distinguish between the two streams. Another question we could have asked is: How does Alice’s software know which stream (i.e. SSRC) belongs to Bob and which stream belongs to Alice? Indeed, Alice’s software may want to display the sender’s name when the sender is talking. Alice’s software gets the SSRC to name mapping from the RTCP source description reports.

Problem 16
a) True
b) True
c) No, RTP streams can be sent to/from any port number. See the SIP example in
Section 6.4.3
d) No, typically they are assigned different SSRC values.
e) True
f) False, she is indicating that she wishes to receive GSM audio
g) False, she is indicating that she wishes to receive audio on port 48753
h) True, 5060 for both source and destination port numbers
i) True
j) False, this is a requirement of H.323 and not SIP.
Problem 17
a) One possible sequence is 1 2 1 3 1 2 1 3 1 2 1 3 …
Another possible sequence is 1 1 2 1 1 3 1 1 2 1 1 3 1 1 2 1 1 3 …
b) 1 1 3 1 1 3 1 1 3 1 1 3 …
Problem 18
a)
Packet
Time leaving the queue
Delay
1
0
0
2
1
1
3
2
1
4
3
2
5
5
2
6
4
2
7
6
3
8
7
2
9
8
3
10
9
2
11
10
2
12
11
3
Average Delay
1.91
b)
Packet
Time leaving the queue
Delay
1
0
0
2
2
2
3
1
0

4
6
5
5
4
1
6
7
5
7
3
0
8
9
4
9
5
0
10
10
3
11
8
0
12
11
3
Average Delay
1.91
c)
Packet
Time leaving the queue
Delay
1
0
0
2
2
2
3
4
3
4
1
0
5
3
0
6
6
4
7
5
2
8
7
2
9
9
4
10
11
4
11
8
0
12
10
2
Average Delay
1.91
d)
Packet
Time leaving the queue
Delay
Note
1
0
0
WFQ
2
2
2
WFQ
3
1
0
WFQ
4
5
4
WFQ
5
3
0
WFQ
6
7
5
Idealized WFQ scheduling
7
4
1
WFQ
8
9
4
WFQ
9
6
1
Idealized WFQ scheduling
10
10
3
WFQ
11
8
0
WFQ
12
11
3
WFQ
Average Delay
1.91

e) It can be noticed that the average delay for all four cases is the same (1.91 seconds).
Problem 19
a)
Packet
Time leaving the queue
Delay
1
0
0
2
4
4
3
5
4
4
1
0
5
3
0
6
2
0
7
6
3
8
9
4
9
7
2
10
10
3
11
8
0
12
11
3
Average Delay
1.91
b)
Packet
Time leaving the queue
Delay
1
0
0
2
1
1
3
3
2
4
2
1
5
6
3
6
4
2
7
5
2
8
9
4
9
7
2
10
10
3
11
8
0
12
11
3
Average Delay
1.91
c)
Packet
Time leaving the queue
Delay
1
0
0
2
1
1
3
2
1
4
3
2

5
9
6
6
6
4
7
4
1
8
7
2
9
5
0
10
8
1
11
11
3
12
10
2
Average Delay
1.91
Problem 20
Time Slot
Packets in the queue
Number of tokens in bucket
0
1, 2, 3
2
1
3, 4
1
2
4,5
1
3
5,6
1
4
6
1
5
–
1
6
7, 8
2
7
9, 10
1
8
10
1
Time Slot
Packets in output buffer
0
1, 2
1
3
2
4
3
5
4
6
5
–
6
7, 8
7
9
8
10
Problem 21
Time Slot
Packets in the queue
Number of tokens in bucket
0
1, 2, 3
2
1
3, 4
2
2
5
2
3
6
2
4
–
2
5
–
2
6
7, 8
2
7
9, 10
2

8
–
2
Time Slot
Packets in output buffer
0
1, 2
1
3, 4
2
5
3
6
4
–
5
–
6
7, 8
7
9, 10
8
–
Problem 22
No. The answer still remains the same as in Problem 21.
Problem 23
See figure below. For the second leaky bucket, r  p, b  1.
Figure: Solution to problem 26
Problem 24
No.
Problem 25
Let  be a time at which flow 1 traffic starts to accumulate in the queue. We refer to 
as the beginning of a flow-1 busy period. Let t   be another time in the same flow-1
busy period. Let T (,t) be the amount of flow-1 traffic transmitted in the interval [,t]. 1
Clearly,

T(,t) W1 R(t) 1 Wj
Let Q1(t) be the amount of flow-1 traffic in the queue at time t. Clearly, Q(t)b r(t)T(,t)
1111
b r(t) W1 R(t) 11 Wj
b(t)r W1 R 1 1 W 
Since r  W1 R , Q (t)  b . Thus the maximum amount of flow-1 traffic in the queue 1W11
j 
j
is b1 . The minimal rate at which this traffic is served is W1R . Wj
Thus, the maximum delay for a flow-1 bit is
1
b d .
W1R/ W max j

Chapter 8 Review Questions
1. Confidentiality is the property that the original plaintext message can not be determined by an attacker who intercepts the ciphertext-encryption of the original plaintext message. Message integrity is the property that the receiver can detect whether the message sent (whether encrypted or not) was altered in transit. The two are thus different concepts, and one can have one without the other. An encrypted message that is altered in transmit may still be confidential (the attacker can not determine the original plaintext) but will not have message integrity if the error is undetected. Similarly, a message that is altered in transit (and detected) could have been sent in plaintext and thus would not be confidential.
2. User’s laptop and a web server; (ii) two routers; (iii) two DNS name servers.
3. One important difference between symmetric and public key systems is that in symmetric key systems both the sender and receiver must know the same (secret) key. In public key systems, the encryption and decryption keys are distinct. The encryption key is known by the entire world (including the sender), but the decryption key is known only by the receiver.
4. In this case, a known plaintext attack is performed. If, somehow, the message encrypted by the sender was chosen by the attacker, then this would be a chosen- plaintext attack.
5. An 8-block cipher has 28 possible input blocks. Each mapping is a permutation of the 28 input blocks; so there are 28! possible mappings; so there are 28! possible keys.
6. If each user wants to communicate with N other users, then each pair of users must have a shared symmetric key. There are N*(N-1)/2 such pairs and thus there are N*(N-1)/2 keys. With a public key system, each user has a public key which is known to all, and a private key (which is secret and only known by the user). There are thus 2N keys in the public key system.
7. a mod n = 23 , b mod n = 4. So (a*b) mod n = 23*4=92
8. 175
9. One requirement of a message digest is that given a message M, it is very difficult to find another message M’ that has the same message digest and, as a corollary, that given a message digest value it is difficult to find a message M’’ that has that given message digest value. We have “message integrity” in the sense that we have reasonable confidence that given a message M and its signed message digest that the message was not altered since the message digest was computed and signed. This is

not true of the Internet checksum, where we saw in Figure 7.18 that it easy to find two messages with the same Internet checksum.
10. No. This is because a hash function is a one-way function. That is, given any hash value, the original message cannot be recovered (given h such that h=H(m), one cannot recover m from h).
11. This is scheme is clearly flawed. Trudy, an attacker, can first sniff the communication and obtain the shared secret s by extracting the last portion of digits from H(m)+s. Trudy can then masquerade as the sender by creating her own message t and send (t, H(t)+s).
12. Suppose Bob sends an encrypted document to Alice. To be verifiable, Alice must be able to convince herself that Bob sent the encrypted document. To be non-forgeable, Alice must be able to convince herself that only Bob could have sent the encrypted document (e.g.,, no one else could have guessed a key and encrypted/sent the document) To be non-reputable, Alice must be able to convince someone else that only Bob could have sent the document. To illustrate the latter distinction, suppose Bob and Alice share a secret key, and they are the only ones in the world who know the key. If Alice receives a document that was encrypted with the key, and knows that she did not encrypt the document herself, then the document is known to be verifiable and non-forgeable (assuming a suitably strong encryption system was used). However, Alice cannot convince someone else that Bob must have sent the document, since in fact Alice knew the key herself and could have encrypted/sent the document.
13. A public-key signed message digest is “better” in that one need only encrypt (using the private key) a short message digest, rather than the entire message. Since public key encryption with a technique like RSA is expensive, it’s desirable to have to sign (encrypt) a smaller amount of data than a larger amount of data.
14. This is false. To create the certificate, certifier.com would include a digital signature, which is a hash of foo.com’s information (including its public key), and signed with certifier.com’s private key.
15. For a MAC-based scheme, Alice would have to establish a shared key with each potential recipient. With digital signatures, she uses the same digital signature for each recipient; the digital signature is created by signing the hash of the message with her private key. Digital signatures are clearly a better choice here.
16. The purpose of the nonce is to defend against the replay attack.
17. Once in a lifetimes means that the entity sending the nonce will never again use that value to check whether another entity is “live”.

18. In a man-in-the-middle attack, the attacker puts himself between Alice and Bob, altering the data sent between them. If Bob and Alice share a secret authentication key, then any alterations will be detected.
19. Alice provides a digital signature, from which Bob can verify that message came from Alice. PGP uses digital signatures, not MACs, for message integrity.
20. False. SSL uses implicit sequence numbers.
21. The purpose of the random nonces in the handshake is to defend against the connection replay attack.
22. True. The IV is always sent in the clear. In SSL, it is sent during the SSL handshake.
23. After the client will generate a pre-master secret (PMS), it will encrypt it with Alice’s public key, and then send the encrypted PMS to Trudy. Trudy will not be able to decrypt the PMS, since she does not have Alice’s private key. Thus Trudy will not be able to determine the shared authentication key. She may instead guess one by choosing a random key. During the last step of the handshake, she sends to Bob a MAC of all the handshake messages, using the guessed authentication key. When Bob receives the MAC, the MAC test will fail, and Bob will end the TCP connection.
24. False. Typically an IPsec SA is first established between Host A and Host B. Then all packets in the stream use the SA.
25. False. IPsec will increment the sequence number for every packet it sends.
26. False. An IKE SA is used to establish one or more IPsec SAs.
27. 01011100
28. True
29. Filter table and connection table. The connection table keeps track of connections, allowing for a finer degree of packet filtering.
30. True
31. True
32. If there isn’t a packet filter, than users inside the institution’s network will still be able to make direct connections to hosts outside the institution’s network. The filter forces the users to first connect to the application gateway.

33. True
Chapter 8 Problems
Problem 1
The encoding of “This is an easy problem” is “uasi si my cmiw lokngch”. The decoding of “rmij’u uamu xyj” is “wasn’t that fun”.
Problem 2
If Trudy knew that the words “bob” and “alice” appeared in the text, then she would know the ciphertext for b,o,a,l,i,c,e (since “bob” is the only palindrome in the message, and “alice” is the only 5-letter word. If Trudy knows the ciphertext for 7 of the letters, then she only needs to try 19!, rather than 26!, plaintext-ciphertext pairs. The difference between 19! and 26! is 26*25*24…*20, which is 3315312000, or approximately 109.
Problem 3
Every letter in the alphabet appears in the phrase “The quick fox jumps over the lazy brown dog.” Given this phrase in a chosen plaintext attack (where the attacker has both the plain text, and the ciphertext), the Caesar cipher would be broken – the intruder would know the ciphertext character for every plaintext character. However, the Vigenere cipher does not alway translate a given plaintext character to the same ciphertext character each time, and hence a Vigenere cipher would not be immediately broken by this chosen plaintext attack.
Problem 4
a) The output is equal to 00000101 repeated eight times.
b) The output is equal to 00000101 repeated seven times + 10000101.
c) We have (ARBRCR)R = CBA, where A, B, C are strings, and R means inverse
operation. Thus:
1. For (a), the output is 10100000 repeated eight times;
2. For (b), the output is 10100001 + 10100000 repeated seven times.
Problem 5
a) There are 8 tables. Each table has 28 entries. Each entry has 8 bits.
number of tables * size of each table * size of each entry = 8*28* 8= 214 bits
b) There are 264 entries. Each entry has 64 bits. 271 bits
Problem 6
a) 100100100 ==> 011011011
b) Trudy will know the three block plaintexts are the same.
c) c(i) = KS(m(i) XOR c(i-1))
c(1) = KS(100 XOR 111) = KS (011) = 100 c(2) = KS(100 XOR 100) = KS (000) = 110 c(1) = KS(100 XOR 110) = KS (010) = 101

Problem 7
a) Wearegiven p3andq11.Wethushaven33andq11.Choosee9(it might be a good idea to give students a hint that 9 is a good value to choose, since the resulting calculations are less likely to run into numerical stability problems than
other choices for e. ) since 3 and (p 1)*(q 1)  20 have no common factors.
Choose d 9 also so that e*d 81 and thus e*d 180 is exactly divisible by
20. We can now perform the RSA encryption and decryption using n  33, e  9 and d  9.
letter m d 4
o15 g7
ciphertext
25
m**e
262144
38443359375
40353607
c**d
38146972265625
ciphertext = m**e mod 33
25
3 19
m = c**d mod n letter 4d
3 19683 15 o 19 322687697779 7g
We first consider each letter as a 5-bit number: 00100, 01111, 00111. Now we concatenate each letter to get 001000111100111 and encrypt the resulting decimal number m=4583. The concatenated decimal number m (= 4583) is larger than current n(=33).Weneedm 1023
23
any
allow outside of 222.22/16 222.22/16
TCP
23
> 1023
ACK
x
Allow outside of 222.22.0.12 222.22/16
TCP
>1023
80
Any
Allow 222.22.0.12 outsideof 222.22/16
TCP
80
>1023
Any
deny All all
all
all
all
All
Connection Table:
Source address
Dest address
Source port
Dest port
222.22.1.7
37.96.87.123
12699
23
222.22.93.2
199.1.205.23
37654
23
222.22.65.143
203.77.240.43
48712
23

Problem 26
a)
Alice
b)
Alice
Proxy1
K1+(S1)
Proxy1
Proxy2
S1( K2+(S2))
K2+(S2)
Proxy2
c) Alice
S1(S2(req))
S1(S2(page))
)
Proxy1
Activist.com
S2(req)
S2(page)
req page

Chapter 9 Review Questions
1. A network manager would like to have network management capabilities when (a) a component of the network fails, (b) a component of the network is about to fail, and is acting “flaky” (c) a component of the network has been compromised from a security standpoint and is attacking the network, e.g., by launching a DOS attack by flooding the network with packets, (d) traffic levels exceed a certain threshold on a link, causing packets to be dropped, (e) everything is running smoothly (in order to know that everything is running smoothly and there are no problems). There are many additional reasons as well.
2. Performance management, fault management, configuration management, accounting management, security management.
3. Network management is more narrowly defined, as it focuses on the resources in the network – monitoring their functions and controlling their operation. These resources are combined (used) in various ways to implement services. Note that while the network resources may all be functioning as they should, they may not be sufficient to implement a service with a given level of performance; this latter concern is an aspect of service management.
4.
 Managing entity: control the collection, processing, analysis, display of network management information, and is used by the network manager to control the devices in the network.
 Managed device: a piece of network equipment that is under the control of the managing entity.
 Management agent: a software process running on a managed device that communicated with the managing entity and takes action on the managed device under the control of the managing entity.
 MIB: pieces of information associated with all of the managed objects in a device.
 Network management protocol: runs between the managing entity of the management agents on the managed devices, allowing the agents to alert the managing entity to potential problems, and allowing the managing entity to send commands to the management agents.
5. The SMI is a data-definition language used to defined the pieces of information in an SNMP MIB.

6. The trap message is sent by the management agent to the managing entity (and requires no response from the managing entity). A request-response message is sent by the managing entity, with the response coming back from the management agent.
7. GetRequest, GetNextRequest, GetBulkRequest, SetRequest, InformRequest, Response, Trap
8. The SNMP engine is the part of an SNMP implementation that handles the dispatching, processing, authentication, access control, and timeliness of he SNMP messages. See Figure 8.5.
9. The ASN.1 object identifier tree provides a standard way to name objects.
10. The role of the presentation layer is to allow the sending and receiving of data in a machine-independent format (i.e., without regard to the particular storage and architectural conventions of the sender and receiver).
11. The Internet does not have presentation layer. It is up to the implementers of protocols to make sure that data is sent and received in the desired format.
12. In TLV encoding, each piece of data is tagged with its type, length, and value.
Chapter 9 Problems
Problem 1
Request response mode will generally have more overhead (measured in terms of the number of messages exchanged) for several reasons. First, each piece of information received by the manager requires two messages: the poll and the response. Trapping generates only a single message to the sender. If the manager really only wants to be notified when a condition occurs, polling has more overhead, since many of the polling messages may indicate that the waited-for condition has not yet occurred. Trapping generates a message only when the condition occurs.
Trapping will also immediately notify the manager when an event occurs. With polling, the manager needs will need to wait for half a polling cycle (on average) between when the event occurs and the manager discovers (via its poll message) that the event has occurred.
If a trap message is lost, the managed device will not send another copy. If a poll message, or its response, is lost the manager would know there has been a lost message (since the reply never arrives). Hence the manager could repoll, if needed.

Problem 2
Often, the time when network management is most needed is in times of stress, when the network may be severely congested and packets are being lost. With SNMP running over TCP, TCP’s congestion control would cause SNMP to back-off and stop sending messages at precisely the time when the network manager needs to send SNMP messages.
Problem 3
1.3.6.1.2.1.5
Problem 4
Microsoft file formats are under 1.2.840.113556.4 (see section 8.3.2)
Problem 6
If value precedes the length, then there is a chance of misinterpreting length to be part of the value.
Problem 7
47Michael21 ‘10100101’
Problem 8
47Sridhar 21 ‘10010001’