计算机网络自顶向下(第七版)课后答案—英文

2021年2月13日发(作者：blackguard)

Computer Networking: A Top-Down Approach,

7

th

Edition

Solutions to Review Questions and Problems

Version Date: December 2016

This

document

contains

the

solutions

to

review

questions

and

problems

for

the

7th

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 1996-2016 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. Ethernet LANs have transmission rates of 10 Mbps, 100 Mbps, 1 Gbps and 10 Gbps.

8. Today, Ethernet most commonly runs over twisted-pair copper wire. It also can run

over fibers optic links.

9.

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

t

0

the

sending

host

begins

to

transmit.

At

time

t

1

=

L/R

1

,

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

t

1

, it can begin to transmit the

packet to the receiving host at time

t

1

. At time

t

2

= t

1

+ L/R

2

, 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/R

1

+ L/R

2

.

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

?

3

3< /p>

?

3

?

d) Probability that all three users are transmitting simultaneously =

?

?

?

p

1

?

p

?

3

?

?

?

3

=

(0.2)

= 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

purpose

-------

Let

server

know

that

there

is

a

card

in

the

ATM machine

ATM card transmits user ID to Server

PASSWD

User enters PIN, which is sent to server

BALANCE

User requests balance

WITHDRAWL

User asks to withdraw money

BYE

user all done

Messages from Server to ATM machine (display)

Msg name

--------

PASSWD

OK

ERR

AMOUNT

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

Correct operation:

client server

HELO (userid)

-------------->

(check if valid userid)

<-------------

PASSWD

PASSWD

-------------->

(check password)

<-------------

OK (password is OK)

BALANCE

-------------->

<-------------

AMOUNT

WITHDRAWL

-------------->

check if enough $$ to cover

withdrawl

<-------------

OK

ATM dispenses $$

BYE

-------------->

<-------------

BYE

In situation when there's not enough money:

HELO (userid)

-------------->

(check if valid userid)

<-------------

PASSWD

PASSWD

-------------->

(check password)

<-------------

OK (password is OK)

BALANCE

-------------->

<-------------

AMOUNT

WITHDRAWL

-------------->

check

if

enough

$$

to

cover

withdrawl

<-------------

ERR (not enough funds)

error msg displayed

no $$ given out

BYE

-------------->

<-------------

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)

d

prop

?

m

/

s

seconds.

b)

d

tran s

?

L

/

R< /p>

seconds.

c)

d

end

?

to

?

end

?

(

m

/

s

?

L

< p>
/

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 < /p>

L

120

?

m< /p>

?

s

?

2

.

5

?

10

8

?

?

536

km.

3

R

56

?

10

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.

3

64

?

10

The time required to transmit the packet is

56

?

8

sec=

224

?

sec.

2

?

10

6

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

?< /p>

0

.

1

.

?

120

?

n

120

?

n

?

c)

?

.

?< /p>

?

p

1

?

p

?

n

?

< p>
?

?

20

120

?

?

n

120

?

n

?

d)

1

?

?

?

.

?

?

p

1

?

p

?

n

?

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”

?

?< /p>

1

?

P

?

X

?

21

?

?

j

?

?

j

?

1

?

< br>

?

?

120

< br>X

j

?

12

?

?

120

?

9

?

?

j

?

1

?

P

?< /p>

X

?

21

?

P

?

?

?

?

?

j

?

120

?

0

.

1

?

0

< br>.

9

?

?

120

?

0

.

1

?

0

.

9

?

j

?

1

?

?

?

9

?

?

?

P

?

Z

< p>
?

?

?

P

?

Z

?

2

.

74

?

3

.

286

?

?

?

0

.

997

when

Z

is a standard normal r.v. Thus

P

?

“

21

or more users”

?< /p>

?

0

.

003< /p>

.

Problem 9

a)

10,000

M

?

M

?

n

M

?

n

?

b)

?

?

?

?

p

1

?

p

?

n

?

n

?

N

?

< br>1

?

?

Problem 10

The first

end system requires

L/R

1

to transmit the packet onto the first link; the packet

propagates over the first link in

< p>
d

1

/s

1

; the packet switch adds a processing delay of

d

proc

;

after

receiving

the

entire

packet,

the

packet

switch

connecting

the

first

and

the

second

link requires

L/R

2

to transmit the packet onto the second link; the packet propagates over

the second link in

d

2

/s

2

< p>
. Similarly, we can find the delay caused by the second switch and

the third link:

L/R

3

,

d

proc

, and

d

3

/s

3

.

Adding these five delays gives

d

end- end

= L/R

1

+ L/R

2

+ L/R

3

+ d

1

/s

1

+ d

2

/s

2

+ d

3

/s

3

+ d

proc

+ d

proc

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,

d

end-end

= L/R + d

1

/s

1

+ d

2

/s

2

+ d

3

/s

3

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

n

th

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

?

< br>?

R

(

1

?

I

)

R

1

?

I

b)< /p>

Let

x

?

L

/

R

.

x

Total delay =

< /p>

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

< p>
/

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< /p>

, 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

q

a)

There

are

Q

nodes

(the

source

host

and

the

Q

?

1

routers).

Let

d

proc

denote

the

processing delay at the

q

th node. Let

R

q

be the transmission rate of the

q

th link and

let

q

q

be the propagation delay across the

q

th link. Then

d< /p>

trans

?

L

/

R

q

. Let

d

prop

q

q

q

d

en d

?

to

?

e nd

?

?

d

p roc

?

d

trans

?

d

prop

.

< p>
q

?

1

Q

?

?

q

b)

Let

d

queue

denote the average queuing delay at node

q

. Then

q

< br>q

q

q

d

end

?

to

?

end

?

?

d

proc

?

d

trans

?

d

prop

?

d

queue

.

q

?

1

Q

?

?

Problem 18

On linux you can use the command

traceroute

and in the Windows command prompt you can use

tracert

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

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)

In this example, the traceroutes have 12 routers in the path at each of the three hours.

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 (France) to (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{R

s

, R

c

, R/M}

Problem 21

If only use one path, the max throughput is given by:

1

1

2

2

M

M

m

ax

{m

in{

R

1

,

R

2

,

?

,

R

1

in{

R

1

2

,

R

2

,

< br>?

,

R

N

},

?

,

m

in{

R

1

M

,

R

2

,

?< /p>

,

R

N

}}

.

N

},

m

< /p>

k

k

,

?

,

R

N

}

< p>
.

If use all paths, the max throughput is given by

?

min{

R

1

k

,

R

2

k

?

1

M

Problem 22

Probability of successfully receiving a packet is: p

s

= (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 p

s

. Thus,

the average number of transmissions needed is given by: 1/p

s

. Then, the average number

of re-transmissions needed is given by: 1/p

s

-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/R

s

.

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/R

s

+ L/R

s

+ d

prop

< L/R

s

+ d

prop

+ L/R

c

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/R

s

+ L/R

s

+ d

prop

+ T >= L/R

s

+ d

prop

+ L/R

c

Thus, the minimum value of T is

L/R

c

?

L/R

s

.

Problem 24

40 terabytes = 40 * 10

12

* 8 bits. So, if using the dedicated link, it will take 40 * 10

12

* 8 /

(100 *10

6

) =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

=20000k m, then

R

=

s

/20000km= 2.5*10

8

/(2*10

7

)= 12.5 bps

Problem 27

a)

80,000,000 bits

b)

800,000 bits, this is because that the maximum number of bits that will be in the link

at any given time = min(bandwidth delay product, packet size) = 800,000 bits.

c)

.25 meters

Problem 28

a)

t

trans

+ t

prop

= 400 msec + 80 msec = 480 msec.

b)

20 * (

t

trans

+ 2 t

prop

) = 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

a)

8

?< /p>

10

6

sec

?

4

sec

Time

to

send

message

from

source

host

to

first

packet

switch

=

6

2

?

10

With store-and-forward switching, the total time to move message from source host

to destination host =

4

sec

< p>
?

3

hops

?

12

sec

Time

to

send

1

st

packet

from

source

host

to

first

packet

switch

=

.

1

?

10

4

sec

?

5

m

sec

. Time at which 2

nd

packet is received at the first switch = time

6

2

?

10

at which 1

st

packet is received at the second switch =

2

?

5

m

sec

?

10

m

sec

Time

at

which

1

st

packet

is

received

at

the

destination

host

=

5

m

s ec

?

3

hops

?

15

m

sec

< br>. After this, every 5msec one packet will be received; thus

time

at

which

last

(800

th

)

packet

is

received

=

15

m

sec

?

799

*

5

m

sec

?

4

.

01

sec

.

It

can

be

seen

that

delay

in

using

message

segmentation

is

significantly

less

(almost

1/3

rd

).

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

b)

c)

d)

packets. Smaller packets have to queue behind enormous packets and suffer

unfair delays.

e)

i.

ii.

Packets have to be put in sequence at the destination.

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 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

S

?

80

F

at

the

first

router

is

?

sec.

At

this

time,

the

first

F/S-2

packets

are

at

the

R

S

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

< br>S

?

80

S

?

80

F

sec. Thus delay in sending the whole file is

delay

?

?

(

?

2

)

R

R

S

To calculate the value of S which leads to the minimum delay,

d

delay

?

0

?

S

?

40

F

dS

Problem 34

The

circuit- switched

telephone

networks

and

the

Internet

are

connected

together

at

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, p

ossibly to the same

LAN to which the user’s hos

t 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

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 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

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.

A list of several popular messaging apps: WhatsApp, Facebook Messenger, WeChat,

and Snapchat. These apps use the different protocols than SMS.

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.

from

65.54.246.203

(EHLO

)

Received:

(65.54.246.203)

by

with

SMTP;

Sat,

19

May 2007 16:53:51 -0700

from

([65.55.135.106])

by

Received:

with Microsoft SMTPSVC(6.0.3790.2668); Sat, 19 May 2007 16:52:42 -

0700

from mail pickup service by with Microsoft SMTPSVC; Sat,

Received:

19 May 2007 16:52:41 -0700

Message-ID:

from

65.55.135.123

by

with

HTTP;

Received:

Sat, 19 May 2007 23:52:36 GMT

From:

To:

prithula@

Bcc:

Subject:

Test mail

Date:

Sat, 19 May 2007 23:52:36 +0000

Mime-Version:

1

.0

Content-Type: Text/html; format=flowed

Return-Path:

prithuladhungel@

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 .

The third “Received:” header indicates the mail message flow from the second SMTP

server in the chain to the third server, and so on.

Finally, the first “Received:” header indicates the flow of the mail message

s from the

forth

SMTP

server

to

the

last

SMTP

server

(i.e.

the

receiver’s

mail

server)

in

the

chain.

Message-id:

The

message

has

been

given

this

number

BAY130-

F26D9E35BF59E0D18A819 AFB9310@

(by

bay0-omc3-

. 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@”

To: This field indicates the email address of the receiver of the mail. In the example,

the receiver is “prithula@”

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 th

is 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@”.

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.

One

server

placement

philosophy

is

called

Enter

Deep,

which

enter

deep

into

the

access networks of Internet Service Providers, by deploying server clusters in access

ISPs

all

over

the

world.

The

goal

is

to

reduce

delays

and

increase

throughput

between end users and the CDN servers. Another philosophy is Bring Home, which

bring the

ISPs home by building

large CDN server clusters at

a smaller number of

sites

and typically placing these server clusters

in

IXPs

(Internet

Exchange Points).

This Bring Home design typically results in lower maintenance and management cost,

compared with the enter-deep design philosophy.

25.

Other than network-related factors, there are some important factors to consider, such

as load- balancing (clients should not be directed to overload clusters), diurnal effects,

variations

across

DNS

servers

within

a

network,

limited

availability

of

rarely

accessed video, and the need to alleviate hot-spots that may arise due to popular video

content.

Reference paper:

Torres,

Ruben,

et

al.

video

server

selection

strategies

in

the

YouTube

CDN.

(ICDCS), 2011.

Another factor to consider is 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.

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

SMS (Short Message Service) is a technology that allows the sending and receiving of

text

messages

between

mobile

phones

over

cellular

networks.

One

SMS

message

can

contain data of 140 bytes and it supports languages internationally. The maximum size of

a message can be 160 7-bit characters, 140 8-bit characters, or 70 16-bit characters. SMS

is

realized

through

the

Mobile

Application

Part

(MAP)

of

the

SS#7

protocol,

and

the

Short Message protocol is defined by 3GPP TS 23.040 and 3GPP TS 23.041. In addition,

MMS (Multimedia Messaging Service) extends the capability of original text messages,

and support sending photos, longer text messages, and other content.

iMessage is an instant messenger service developed by Apple. iMessage supports texts,

photos, audios or videos that we send to iOS devices and Macs over cellular data network

or WiFi. Apple’s i

Message is based on a proprietary, binary protocol APNs (Apple Push

Notification Service).

WhatsApp

Messenger

is

an

instant

messenger

service

that

supports

many

mobile

platforms

such

as

iOS,

Android,

Mobile

Phone,

and

Blackberry.

WhatsApp

users

can

send each other unlimited images, texts, audios, or videos over cellular data network or

WiFi. WhatsApp uses the XMPP protocol (Extensible Messaging and Presence Protocol).

iMessage

and

WhatsApp

are

different

than

SMS

because

they

use

data

plan

to

send

messages and they work on TCP/IP networks, but SMS use the text messaging plan we

purchase from our wireless carrier. Moreover, iMessage and WhatsApp support sending

photos, videos, files,

etc., while the original

SMS can only send text

message. Finally,

iMessage and WhatsApp can work via WiFi, but SMS cannot.

Problem 3

Application layer protocols: DNS and HTTP

Transport layer protocols: UDP for DNS; TCP for HTTP

Problem 4

a)

The

document

request

was

/cs453/.

The

Host

:

field indicates the server's name and /cs453/ 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 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

:

The

server

agreed

to

a

persistent connection, as indicated by the Connection: Keep-Alive field

Problem 6

a)

Persistent connections are discussed in section 8 of RFC 2616 (the real goal of this

question was to get you to retrieve and read an RFC). Sections 8.1.2 and 8.1.2.1 of

the RFC indicate that either the client or the server can indicate to the other that it is

going to close the persistent connection. It does so by including the connection-token

b)

HTTP does not provide any encryption services.

c)

(From R

FC 2616) “Clients that use persistent connections should limit the number of

simultaneous

connections

that

they

maintain

to

a

given

server.

A

single-user

client

SHOULD NOT maintain more than 2 connections with any server or proxy.”

d)

Yes. (From RFC 2616) “A

client might have started to send a new request at the same

time that the server has decided to close the

of view, the connection is being closed while it was idle, but from the client's point of

view, a request i

s in progress.”

Problem 7

The total amount of time to get the IP address is

RTT

1

?

RTT

2

?

?

?

RTT

n

.

Once the IP address is known,

RTT

O

elapses to set up the TCP connection and another

RTT

O

elapses to request and receive the small object. The total response time is

2

RTT

o

?

RTT

1

?

RTT

2

?

?

?

RTT

n

Problem 8

a)

RT T

1

?

?

?< /p>

RTT

n

?

2< /p>

RTT

o

?

8< /p>

?

2

RTT

o< /p>

?

18

RTT

o

?

RTT

1

?

?

?

RTT

n

.

b)

RTT

1

?

?

?

RTT

n

?

2

RTT

o

?

2

?

2

RTT

o

?

6

RTT

o

?

RTT

1

?

?

?

RTT

n

c)

Persistent connection with pipelining. This is the default mode of HTTP.

RTT

1

?

?

?

RTT

n

?

2

RTT

o

?

RTT

o

?

3

RTT

o

?

RTT

1

?

?

?

RTT

n

.

Persistent connection without pipelining, without parallel connections.

RTT

1

?

?

?

RTT

n

?

2

RTT

o

?

8

RTT

o

?

10

RT T

o

?

RTT

1

?

?

?

RT T

n

.

Problem 9

a)

The time to transmit an object of size

L

over a link or rate

R

is

L/R

. The average time

is the average size of the object divided by

R

:

?

= (850,000 bits)/(15,000,000 bits/sec) = .0567 sec

The traffic intensity on the link is given by

?

?

=(16 requests/sec)(.0567 sec/request) =

0.907. Thus, the average access delay is (.0567 sec)/(1 - .907)

?

.6 seconds. The total

average response time is therefore .6 sec + 3 sec = 3.6 sec.

b)

The

traffic

intensity

on

the

access

link

is

reduced

by

60%

since

the

60%

of

the

requests are satisfied within the institutional network. Thus the average access delay

is

(.0567

sec)/[1

–

(.4)(.907)]

= .089

seconds.

The

response

time

is

approximately

zero if the request is satisfied by the cache (which happens with probability .6); the

average

response

time

is .089

sec

+

3

sec

=

3.089

sec

for

cache

misses

(which

happens 40% of the time). So the average response time is (.6)(0 sec) + (.4)(3.089 sec)

= 1.24 seconds. Thus the average response time is reduced from 3.6 sec to 1.24 sec.

Problem 10

Note

that

each

downloaded

object

can

be

completely

put

into

one

data

packet.

Let

Tp

denote the one-way propagation delay between the client and the server.

First

consider

parallel

downloads

using

non-persistent

connections.

Parallel

downloads

would

allow

10

connections

to

share

the

150

bits/sec

bandwidth,

giving

each

just

15

bits/sec. Thus, the total time needed to receive all objects is given by:

(200/150+

T

p + 200/150 +

T

p + 200/150+

T

p + 100,000/150+

T

p )

+ (200/(150/10)+

T

p + 200/(150/10) +

T

p + 200/(150/10)+

T

p + 100,000/(150/10)+

T

p )

= 7377 + 8*

T

p (seconds)

Now consider a persistent HTTP connection. The total time needed is given by:

(200/150+

T

p + 200/150 +

T

p + 200/150+

T

p + 100,000/150+

T

p )

+ 10*(200/150+

T

p + 100,000/150+

T

p )

=7351 + 24*

T

p (seconds)

Assuming the speed of light is 300*10

6

m/sec, then Tp=10/(300*10

6

)=0.03 microsec. Tp

is therefore negligible compared with transmission delay.

Thus, we see that persistent HTTP is not significantly faster (less than 1 percent) than the

non-persistent case with parallel download.

Problem 11

a)

Yes,

because

Bob

has

more

connections,

he

can

get

a

larger

share

of

the

link

bandwidth.

b)

Yes,

Bob

still

needs

to

perform

parallel

downloads;

otherwise

he

will

get

less

bandwidth than the other four users.

Problem 12

from socket import *

serverPort=12000

serverSocke t=socket(AF_INET,SOCK_STREAM)

(('',serverPort))

(1)

connectionSocket, addr = ()

while 1:

sentence = (1024)

print

'From

Server:',

sentence,

()

'n'

Problem 13

The MAIL FROM: in SMTP is a message from the SMTP client that identifies the sender

of the mail message to the SMTP server. The From: on the mail message itself is NOT an

SMTP message, but rather is just a line in the body of the mail message.

Problem 14

SMTP uses a line containing only a period to mark the end of a message body.

HTTP uses “

Content-Length header field

” to indicate the length of a message body.

No,

HTTP

cannot

use

the

method

used

by

SMTP,

because

HTTP

message

could

be

binary data, whereas in SMTP, the message body must be in 7-bit ASCII format.

Problem 15

MTA stands for Mail Transfer Agent. A host sends the message to an MTA. The message

then

follows a sequence of MTAs to

reach the receiver’s mail

reader.

We see that this

spam message follows a chain of MTAs. An honest MTA should report where it receives

the message. Notice that in this message,

“asu

sus-

4b96 ([58.88.21.177])”

does not

report from where it received the email. Since we assume only the originator is dishonest,

so

“asusus

-

4b96 ([58.88.21.177])”

must be the originator.

Problem 16

UIDL abbreviates “unique

-

ID listing”. When a POP3 client

issues the UIDL command,

the

server

responds

with

the

unique

message

ID

for

all

of

the

messages

present

in

the

user'

s mailbox. This command is useful for “download and keep”. By

maintaining a file

that

lists

the

messages

retrieved

during

earlier

sessions,

the

client

can

use

the

UIDL

command to determine which messages on the server have already been seen.

Problem 17

a)

C:

dele 1

C:

retr 2

S:

(blah blah …

S:

………..blah)

S:

.

C:

dele 2

C:

quit

S:

+OK POP3 server signing off

b)

C:

retr 2

S:

blah blah …

S:

………..blah

S:

.

C:

quit

S:

+OK POP3 server signing off

c)

C:

list

S:

1 498

S:

2 912

S:

.

C:

retr 1

S:

blah …..

S:

….blah

S:

.

C:

retr 2

S:

blah blah …

S:

………..blah

S:

.

C:

quit

S:

+OK POP3 server signing off

Problem 18

a)

For

a

given

input

of

domain

name

(such

as

),

IP

address

or

network

administrator

name,

the

whois

database

can

be

used

to

locate

the

corresponding

registrar, whois server, DNS server, and so on.

b)

from from

c)

Local Domain:

Web servers :

207.69.189.21, 207.69.189.22,

207.69.189.23, 207.69.189.24,

207.69.189.25, 207.69.189.26, 207.69.189.27,

207.69.189.28

Mail Servers : (207.69.189.217)

(207.69.189.218)

(207.69.189.219)

(207.69.189.220)

Name Servers: (207.69.188.196)

(207.69.188.197)

Web Servers: (216.109.112.135, 66.94.234.13)

Mail Servers: (209.191.118.103)

(66.196.97.250)

(68.142.237.182, 216.39.53.3)

(216.39.53.2)

(216.39.53.1)

(209.191.88.247, 68.142.202.247)

(209.191.88.239, 206.190.53.191)

Name Servers: (66.218.71.63)

(68.142.255.16)

(217.12.4.104)

(68.142.196.63)

(216.109.116.17)

(202.165.104.22)

(202.160.176.146)

Web Servers: (64.4.33.7, 64.4.32.7)

Mail Servers: (65.54.245.8, 65.54.244.8, 65.54.244.136)

(65.54.244.40, 65.54.244.168, 65.54.245.40)

(65.54.244.72, 65.54.244.200, 65.54.245.72)

(65.54.244.232, 65.54.245.104, 65.54.244.104)

Name Servers: (207.68.160.190)

(65.54.240.126)

(213.199.161.77)

(207.46.66.126)

(65.55.238.126)

d) The yahoo web server has multiple IP addresses

(216.109.112.135, 66.94.234.13)

e) The address range for Polytechnic University: 128.238.0.0

–

128.238.255.255

f) An attacker can use the

whois

database and nslookup tool to determine the IP address

ranges, DNS server addresses, etc., for the target institution.

g)

By analyzing the source address of attack packets, the victim can use whois to obtain

information about domain from which the attack is coming and possibly inform the

administrators of the origin domain.

Problem 19

a)

The following delegation chain is used for

(authoritative)

First command:

dig +norecurse @ any

;; AUTHORITY SECTION:

edu. 172800 IN NS .

edu. 172800 IN NS .

edu. 172800 IN NS .

edu. 172800 IN NS .

edu. 172800 IN NS .

edu. 172800 IN NS .

edu. 172800 IN NS .

edu. 172800 IN NS .

Among all returned edu DNS servers, we send a query to the first one.

dig +norecurse @ any

. 172800 IN NS .

. 172800 IN NS .

. 172800 IN NS .

Among all three returned authoritative DNS servers, we send a query to the first one.

dig +norecurse @ any

. 21600 IN A 128.119.245.12

b)

The answer for could be:

(authoritative)

Problem 20

We can periodically take a snapshot of the DNS caches in

the

local

DNS

servers. The

Web server that appears most frequently in the DNS caches is the most popular server.

This is because if more users are interested in a Web server, then DNS requests for that

server are more frequently sent by users. Thus, that Web server will appear in the DNS

caches more frequently.

For a complete measurement study, see:

Craig E. Wills, Mikhail Mikhailov, Hao Shang

“

Inferring

Relative

Popularity

of

Internet

Applications

by

Actively

Querying

DNS

Caches

”, in

IMC'03, October 27-

29, 2003, Miami Beach, Florida, USA

Problem 21

Yes, we can use dig to query that Web site in the local DNS server.

For example, “dig ” will return the query time f

or finding . If

was just accessed a couple of seconds ago, an entry for is cached in the local

DNS cache, so the query time is 0 msec. Otherwise, the query time is large.

Problem 22

For calculating the minimum distribution time for client-server distribution, we use the

following formula:

D

cs

= max {NF/u

s

, F/d

min

}

Similarly, for calculating the minimum distribution time for P2P distribution, we use the

following formula:

D

P

2

P

?

max{F/u

s

, F/d

min

, NF/(u

s

?

?

u

i

)}

i

?

1

N

Where,

F

= 15 Gbits = 15 * 1024 Mbits

u

s

= 30 Mbps

d

min

=

d

i

= 2 Mbps

Note, 300Kbps = 300/1024 Mbps.

Client Server

N

10

100

300 Kbps

7680

51200

u

700 Kbps

7680

51200

2 Mbps

7680

51200

1000

512000

512000

512000

Peer to Peer

u

Problem 23

a)

Consider

a

distribution

scheme

in

which

the

server

sends

the

file

to

each

client,

in

parallel, at a rate of a rate of

u

s

/

N

< br>. Note that this rate is less than each of the client’s

download rate, since by assumption

u

s

/

N

≤

d

min

. Thus each client can also receive at

rate

u

s

/

N

. Since each client receives at rate

u

s

/

N

, the time for each client to receive the

entire

file

is

F/(

u

s

/

N

)

=

NF/

u

s

.

Since

all

the

clients

receive

the

file

in

NF/

u

s

,

the

overall distribution time is also

NF/ u

s

.

b)

Consider

a

distribution

scheme

in

which

the

server

sends

the

file

to

each

client,

in

parallel, at a rate of

d

min

. Note that the aggregate rate,

N

d

min

, is less than the server’s

link rate

u

s

, since by assumption

u

s

/

N

≥

d

min

. Since each client receives at rate

d

min

,

the time for each client to receive the entire file is F/

d

min

.

Since all the clients receive

the file in this time, the overall distribution time is also F/

d

min

.

c)

From Section 2.6 we know that

D

CS

≥ ma

x {

NF/u

s

, F/d

min

} (Equation 1)

Suppose that

u

s

/

N

≤

d

min

. Then from Equation 1 we have

D

CS

≥

NF/u

s

.

But from (a)

we have

D

CS

≤

NF/u

s

.

Combining these two gives:

D

CS

=

NF/u

s

when

u

s

/

N

≤

d

min

. (Equation 2)

We can similarly show that:

D

CS

=

F/d

min

when

u

s

/

N

≥

d

min

(Equation 3).

Combining Equation 2 and Equation 3 gives the desired result.

N

10

300 Kbps

7680

700 Kbps

7680

2 Mbps

7680

100

25904

15616

7680

1000

47559

21525

7680

Problem 24

a)

Define u = u1 + u2

+ ….. + u

N. By assumption

u

s

<= (u

s

+ u)/N

Equation 1

Divide the file into N parts, with the i

th

part having size (u

i

/u)F. The server transmits

the i

th

part to peer i at rate

r

i

= (

u

i

/

u

)< /p>

u

s

. Note that

r

1

+

r

2

+ ….. +

r

N

=

u

s

, so that the

aggregate server rate does not exceed the link rate of the server. Also have each peer i

forward

the

bits

it

receives

to

each

of

the

N-1

peers

at

rate

r

i

.

The

aggregate

forwarding rate by peer i is (N-1)r

i

. We have

(

N-1

)

r

i

= (

N-1

)(

u

s

u

i

)/

u

<=

u

i

,

where the last inequality follows from Equation 1. Thus the aggregate forwarding rate

of peer i is less than its link rate u

i

.

In this distribution scheme, peer i receives bits at an aggregate rate of

r

i

?

?

r

j

?

u

s

j

??

i

Thus each peer receives the file in F/u

s

.

b)

Again define

u

=

u

1

+

u

2

+ ….. +

u

N

. By assumption

u

s

>= (

u

s

+

u

)/

N

Equation 2

Let

r

i

=

u

i

/(

N-1

) and

r

N+1

= (

u

s

–

u

/(< /p>

N-1

))/

N

In this distribution scheme, the file is broken into

N+1

parts. The server sends bits

from the i

th

part to the i

th

peer (i =

1, …., N

) at rate r

i

. Each peer i forwards the bits

arriving at rate r

i

to each of the other N-1 peers. Additionally, the server sends bits

from the (

N+1

)

st

part at rate r

N+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

r

1

+ …. +

r

N

+

N

r

N+1

=

u

/(

N-1

) +

u

s

–

u

/(< /p>

N-1

) =

u

s

Thus, the server’s send rate does not exceed its link rate. The aggregate send rate of

peer i is

(

< p>
N-1

)

r

i

< p>
=

u

i

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

r

i

?

r

N

?

1

?

< br>?

r

j

?

u

/(

N

?

1

)

?

(

u< /p>

s

?

u

/(

N

?

1

))

/

N

?

(

< p>
u

s

?

u

)

/

N

j

??

i

Thus each peer receives the file in NF/(u

s

+u).

(For simplicity, we neglected to specify the size of the file part for i = 1, …., N+1.

We now provide th

at here. Let Δ = (u

s

+u)/N be the distribution time. For i = 1, …, N,

the

i

th

file

part

is

F

i

=

r

i

Δ

bits.

The

(N+1)

st

file

part

is

F

N+1

=

r

N+1

Δ

bits.

It

is

straightforward to show that F

1

+ ….. + F

N+1

= F.)

c)

The solution to this part is similar to that of 17 (c). We know from section 2.6 that

D

P

2

P

< p>
??

max{F/u

s

, NF/(u

s

?

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

a.

N files, under the assumption that we do a one-to-one matching by pairing video

versions with audio versions in a decreasing order of quality and rate.

b.

2N files.

Problem 28

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 29

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 30

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 31

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 32

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.

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)

The segment now has two header fields: a source port field and destination port field.

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.

S

TP

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

=

RTT

FE

,

Y

=

RTT

BE

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

a) A

?

S

b) B

?

S

c) S

?

A

d) S

?

B

e) Yes.

f) No.

source port

numbers

467

513

23

23

destination port

numbers

23

23

467

513

Problem 2

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.

0< /p>

1

0

1

0

0

1

1

?

< p>
0

1

1

0

0

1

1

0

1

0

1

1

1

0

0

1

1

0

1

1

1

0

0

1

?

0

1

1

1

< br>0

1

0

0

0

0

1

0

1

1

1

0

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

Sender ignores A1

M0 corrupted

A1

Packet garbled, receiver

resends last ACK (A1)

Timeout: sender

resends M0

M0

A0

M1

A1

Corrupted

data

sender sends M0

sender sends M1

Ignore ACK

Timeout: sender

resends M1

M0

A0

M1

A1 corrupted

Corrupted

ACK

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*8/10

9

=12

microseconds. In order for the sender to be busy 98 percent of the time, we must have

util

?

0

.< /p>

98

?

(

0

.

012

n

)

/

30

.

012< /p>

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)

packet=make_pkt(data)

udt_send(packet)

Wait: send

to B

rdt_send(data)

Rdt_unable_to_send(data)

A

Wait: receive

from B

rdt_send(data)

rdt_unable_to_send(data)

rdt_receive(packet)

extract(packet,data)

deliver_data(data)

rdt_send(data)

packet=make_pkt(data)

udt_send(packet)

Wait: send

to A

rdt_send(data)

Rdt_unable_to_send(data)

B

Wait: receive

from A

rdt_send(data)

rdt_unable_to_send(data)

rdt_receive(packet)

extract(packet,data)

deliver_data(data)

Problem 18

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

make pair (0,1)

send packet 0

Receiver