ccan 1 chapter 4

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1.Refer to the exhibit. What two pieces of information can be determined from the

output that is shown? (Choose two.)

The local host is using three client sessions.

The local host is using web sessions to a remote server.

The local host is listening for TCP connections using public addresses.

The local host is using well-known port numbers to identify the source ports.

The local host is performing the three-way handshake with 192.168.1.101:1037.

2. After a web browser makes a request to a web server that is listening to the standard

port, what will be the source port number in the TCP header of the response from the

server?

13

53

80

1024

1728

3. Which information is found in both the TCP and UDP header information?

sequencing

flow control

acknowledgments

source and destination

4. Which is an important characteristic of UDP?

acknowledgement of data delivery
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Refer to the exhibit. In line 7 of thisWiresharkcapture, what TCP operation is being

performed?

session establishment

segment retransmit

data transfer

session disconnect

7.

Refer to the exhibit. The initial TCP exchange of data between two hosts is shown in the

exhibit. Assuming an initial sequence number of 0, what sequence number will be

included in Acknowledgment 2 if Segment 6 is lost?

2

3

6

1850

3431

3475

8. Why is flow control used for TCP data transfer?

to synchronize equipment speed for sent data

to synchronize and order sequence numbers so data is sent in complete numerical order

to prevent the receiver from being overwhelmed by incoming data

to synchronize window size on the server

to simplify data transfer to multiple hosts
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13. What mechanism is used by TCP to provide flow control as segments travel from

source to destination?

sequence numbers


window size

acknowledgments

14. What is dynamically selected by the source host when forwarding data?

destination logical address

source physical address

default gateway address

source port

15. Why are port numbers included in the TCP header of a segment?

to indicate the correctrouterinterface that should be used to forward a segment

to identify which switch ports should receive or forward the segment

to determine which Layer 3 protocol should be used to encapsulate the data

to enable a receiving host to forward the data to the appropriate application

to allow the receiving host to assemble the packet in the proper order

16. Which two options represent Layer 4 addressing? (Choose two.)

identifies the destination network

identifies source and destination hosts

identifies the communicating applications

identifies multiple conversations between the hosts

identifies the devices communicating over the local media

17. Which three features allow TCP to reliably and accurately track the transmission of

data from source to destination?

encapsulation
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flow control

connectionless services


numbering and sequencing

best effort delivery

18. Which OSI model layer is responsible for regulating the flow of information from

source to destination, reliably and accurately?

application

presentation

session

transport

network

19.

Based on the transport layer header shown in the diagram, which of the following

statements describe the established session? (Choose two.)

This is a UDP header.

**This contains a Telnet request.

This contains a TFTP data transfer.

The return packet from this remote host will have an Acknowledgement Number of

43693.

**This is a TCP header.

20. Which event occurs during the transport layer three-way handshake?


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The two applications exchange data.

**TCP initializes the sequence numbers for the sessions.

UDP establishes the maximum number of bytes to be sent.

The server acknowledges the bytes of data received from the client.

21.

Refer to the exhibit. Host1 is in the process of setting up a TCP session with Host2.

Host1 has sent a SYN message to begin session establishment. What happens next?

Host1 sends a segment with the ACK flag = 0, SYN flag = 0 to Host2.






Bn ting vit
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Hy tham kho trin lm.Hai mu thng tin c th c xc nh t u ra c hin

th?(Chn hai.)

Cc my ch a phng ang s dng ba bui khch hng.

Cc my ch a phng ang s dng phin web n mt my ch t xa.

Cc my ch a phng ang lng nghe cho cc kt ni TCP s dng a ch cng cng.

Cc my ch a phng ang s dng s cng ni ting xc nh cng ngun.

Cc my ch a phng ang thc hin bt tay ba chiu vi 192.168.1.101:1037.

2 . Sau khi mt trnh duyt web lm cho mt yu cu n mt my ch web l lng

nghe cng tiu chun, nhng g s l cng ngun s trong tiu TCP ca cc phn ng

t my ch?

13

53

80

1024

1728

3 . Nhng thng tin c tm thy trong c TCP v UDP thng tin tiu ?

trnh t

kim sot dng chy

Li cm n

ngun v ch

4 . l mt c tnh quan trng ca UDP ?

xc nhn cung cp d liu

ti thiu chm tr trong vic cung cp d liu

cao tin cy ca d liu giao hng

cung cp d liu cng


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5 - Hy tham kho trin lm.Host A l s dng FTP ti v mt file ln t Server

1. Trong qu trnh ti v, Server 1 khng nhn c mt s tha nhn t Host A cho

mt vi byte d liu c chuyn.Hnh ng s Server 1 kt qu l?

to ra mt tn hiu lp mt 1

tc mt thi gian ch v gi li d liu cn phi c tha nhn

gi mt cht t li my ch

thay i kch thc ca s trong tiu 4 lp

6 -

Hy tham kho trin lm.Trong dng 7 caWiresharknm bt ny , nhng g hot

ng TCP ang c thc hin nh th no?

k hp thnh lp
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phn khc truyn li

truyn d liu

phin ngt kt ni

7 - Hy tham kho trin lm.TCP ban u trao i d liu gia hai my ch c th

hin trong trin lm.Gi s s th t ban u t 0, nhng g s th t s c bao gm

trong Xc Nhn 2 nu on 6 b mt ?

2

3

6

1850

3431

3475

8 . Ti sao l kim sot dng chy c s dng chuyn d liu TCP?

ng b ha tc trang thit b cho cc d liu c gi

ng b ha v th t s th t v vy d liu c gi theo th t s y

ngn chn ngi nhn khng b trn ngp bi cc d liu n

ng b ha kch thc ca s trn my ch

n gin ha d liu chuyn giao cho nhiu my ch

9 - Trong mt phin giao tip TCP, nu cc gi tin n ch trong trt t, iu g s xy

ra vi thng ip ban u?


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Cc gi tin s khng c chuyn giao.

Cc gi tin s c truyn li t ngun.

Cc gi tin s c cung cp v tp hp li ti im n.

Cc gi tin s c phn phi v khng tp hp li ti im n.

10 - Vi TCP / ng gi d liu IP, phm vi s cng xc nh tt c cc ng dng ni

ting ?

0-255

256-1022

0-1023

1024-2047

49153-65535

11 - Giao thc lp vn chuyn cung cp chi ph thp v s c s dng cho cc ng

dng khng yu cu d liu cung cp ng tin cy ?

TCP

IP

UDP

HTTP

DNS

12 . Hai tnh nng ca Hng Datagram Protocol (UDP) l g?(Chn hai.)

kim sot dng chy

ph thp

kt ni

hng kt ni

trnh t v li cm n

13 - C ch no c s dng bi TCP cung cp kim sot dng chy nh phn oni t ngun ti ch ?


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chui s

k hp thnh lp

kch thc ca s

Li cm n

14 - Nhng g l t ng la chn bi cc my ch ngun khi chuyn tip d liu?

im n hp l a ch

ngun a ch vt l

a ch gateway mc nh

cng ngun

15 - Ti sao s cng bao gm trong tiu TCP ca mt phn on ?

ch ragiao dinb nh tuynchnh xc nn c s dng chuyn tip mt on

xc nh cc cng chuyn i s nhn c hoc chuyn tip cc phn khc

xc nh giao thc lp 3 nn c s dng ng gi cc d liu

cho php mt my ch tip nhn chuyn tip d liu ng dng thch hp

cho php cc my ch nhn c lp rp cc gi tin theo th t thch hp

16 - Hai la chn i din cho lp 4 gii quyt? (Chn hai.)

xc nh mng ch

xc nh cc host ngun v ch

xc nh cc ng dng giao tip

xc nh nhiu cuc hi thoi gia cc my ch

xc nh cc thit b giao tip qua cc phng tin truyn thng a phng

17 . Ba tnh nng cho php TCP ng tin cy v chnh xc theo di vic truyn ti d liu

t ngun ti ch ?

ng gi

kim sot dng chy
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kt ni dch v

k hp thnh lp

nh s v sp xp

n lc phn phi tt nht

18 . Lp m hnh OSI chu trch nhim iu tit dng chy ca thng tin t ngun ti

ch, ng tin cy v chnh xc?

ng dng

trnh by

phin

giao thng vn ti

mng

19 -

Da trn tiu lp vn chuyn c th hin trong biu ca cc bo co sau y m

t cc session c thit lp?(Chn hai.)

y l mt tiu UDP.

** Ny c cha mt yu cu Telnet.

Ny c cha mt TFTP truyn d liu.

Cc gi tin tr v t my ch t xa ny s c mt s Li cm n ca 43693.

** y l mt tiu TCP.

20 . S kin xy ra trong qu trnh bt tay ba chiu lp truyn ti?

Hai ng dng trao i d liu.


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** TCP khi to s th t cho cc phin hp.

UDP thit lp s lng ti a ca byte c gi i.

Cc my ch tha nhn cc byte d liu nhn c t khch hng.

21Tham kho trin lm .Host1 l trong qu trnh thit lp mt phin TCP vi

host2. Host1 gi mt tin nhn SYN bt u thnh lp phin . iu g s xy ra tip

theo? host1 s gi mtphn khc vi c ACK = 0, SYN flag = 0 n host2.

Host1 s gi mt phn khc vi c ACK = 1, SYN flag = 0 n host2.

Host1 s gi mt phn khc vi c ACK = 1, SYN flag = 1 host2.

Host2 gi mt phn khc vi c ACK = 0, c SYN = 1 ti host1.

Host2 gi mt phn khc vi c ACK = 1, SYN flag = 0 n host1.

Host2 s gi mt phn khc vi c ACK = 1, SYN flag = 1 host1.

ccan 1 chapter 4

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