Translate Team Solution for Computer Networking 5th Edition

Computer Networking: A Top-Down Approach Featuring the Internet, 5th Edition Translate by K55CC

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Computer Networking - A Top-Down Approach Featuring the Internet, 5th edition

Solutions to Review Questions and Problems (Vietnamese version)

Author : K55CC translate team.(K55CC-University of engineering and technology-Vietnam national university)

This document belong to K55CC.All copies or sharing must be allowed by K55CC.

Chapter1: Introduction-Review Question

Cu 1: S kh c nhau gia 1 host v 1 end system. Lit k c c kiu end system. Web server c phi l 1

end system khng?

Khng c s khc nhau. host v end system l 2 t c th thay th cho nhau.

End system gm c: PC, my trm, web server, mail server, web TV, cc PDA c kt ni internet...

Cu 2: V d v giao thc ngo i giao (diplomatic protocol)

Gi s Alice, mt i s ca quc gia A mun mi Bob, mt i s nc B, n ti. Alice khng ch n

gin l ch cn gi Bob trn in thoi v ni, "n bn n ca chng ti by gi". Thay vo ,

c gi Bob v cho thy mt ngy v thi gian. Bob c th p ng bng cch ni rng ng khng phi c

sn m c th ngy, nhng anh c sn mt ngy khc. Alice v Bob tip tc gi "thng ip" qua li cho

n khi h ng vo mt ngy v thi gian. Bob sau cho thy ti i s qun vo ngy tho thun,

hy vng khng qu 15 pht trc khi hoc sau khi thi gian tho thun. ngoi giao giao thc cng cho

php hoc Alice hoc Bob lch s hy b s tham gia nu h c l do hp l.

Cu 3: Ch ng trnh client l g? Ch ng trnh server l g? Mt ch ng trnh server yu cu v nhn

dch v t 1 ch ng trnh client ng k?

Mt chng trnh mng thng c 2 chng trnh, mi ci chy trn 1 host khc nhau, c ni vs

nhau. Chng trnh m khi u vic trao i thng tin l client. Thng thng, chng trnh client yu

cu v nhn dch v t chng trnh server.

Cu 4: Lit k 6 cng ngh truy cp.

1. Dial-up modem over telephone line: residential;

2. DSL over telephone line: residential or small office;

3. Cable to HFC: residential;

4. 100 Mbps switched Ethernet: company;

5. Wireless LAN: mobile;

6. Cellular mobile access (for example, WAP): mobile

Cu 7: Tc truy n ca LAN Ethernet l g? Vi 1 tc truy n mi user trong m ng LAN c th

truy n lin tc vi cng tc c k?



Ethernet LANs c tc truyn l 10 Mbps, 100 Mbps, 1 Gbps and 10 Gbps. VD vi X Mbps Ethernet (

X = 10, 100, 1,000 or 10,000), mt user c th truyn lien tc vi tc X Mbps nu ch c user

ang gi d liu. Nu c nhiu user hot ng th mi user khng th truyn lin tc vi tc X Mbps.

Cu 8: Mi tr ng vt l m Ethernet c th truy n qua l g?

Ethernet c th truyn qua cp ng trc mnh v dy ng xon i. N cng c th truyn qua si

quang v cp ng trc dy.

Cu 9: Cho bit khong tc truy n ca Dial up modems, ISDN, ADSL, HFC, FTTH. Chng thuc

kiu chia s hay ring bit?

Dial up modems: up to 56 Kbps, bandwidth is dedicated; ISDN: up to 128 kbps, bandwidth is

dedicated; ADSL: downstream channel is .5-8 Mbps, upstream channel is up to 1 Mbps, bandwidth is

dedicated; HFC, downstream channel is 10-30 Mbps and upstream channel is usually less than a few

Mbps, bandwidth is shared. FTTH: 2-10Mbps upload; 10-20 Mbps download; bandwidth is not shared.

Cu 10: M t c c cng ngh truy cp Internet khng dy ph bin hin nay. So s nh v nu im kh c

bit ca chng

Hin nay c 2 cng ngh truy cp Internet khng dy ph bin.

- Mng LAN khng dy (Wireless LAN). Trong WLAN, ngi dng khng dy truyn v cc nhn gi

tin n 1 trm c s (im truy cp khng dy) trong bn knh khong vi chc mt. Trm c s thng

c kt ni Internet bng dy.

- Mng truy cp khng dy din rng: Trong cc h thng ny, cc gi tin c truyn qua cng c s h

tng khng dy c dng cho h thng in thoi, trm c s ny c qun l bi 1 nh cung cp

dch v vin thng. N cung cp s truy cp khng dy cho ngi dung trong vng bn knh vi chc

kilomet t my trm c s.

Cu 11: Thun li ca m ng chuyn m ch in so vi chuyn m ch gi l g?

Mng chuyn mch in c th m bo 1 lng c nh bng thng gia 2 im ni duy tr 1 cuc

gi. Hu ht cc mng chuyn mch gi hin nay (bao gm c Internet) khng to s m bo bng

thng gia 2 im ni.

Cu 13: Gi s c 1 gi tin c chuyn t host gi n host nhn. Tc truy n gia host gi v switch

v gia switch v host nhn ln l t l R1 v R2. Gi s switch s dng chuyn m ch gi l u tr v

chuyn tip. T nh tng thi gian delay gia 2 im ni gi 1 packet c di L (b qua hng i

tr khi truy n v khi x l)

Ti thi im t0, host gi bt u truyn. Ti thi im t1 = L/R1, host gi hon thnh vic truyn v

ton b gi tin c nhn ti router (khng c tr khi truyn). V router c ton b gi tin ti thi im t1

nn n c th truyn gi tin n host nhn ti thi im t1. Ti thi im t2 = t1 + L/R2, router hon

thnh vic truyn v ton b gi tin c nhn ti host nhn. V vy, tng thi gian tr l L/R1 + L/R2.

Cu 15: Gi s nhiu user chia s 1 link 2Mbps. V cng gi s rng mi user truyn lien tc vs tc

1Mbps khi truyn, nhng mi user ch truyn 20% thi gian.



a. Nu dng chuyn mch in th c bao nhiu user c s dng?

b. Gi s dng chuyn mch gi. Ti sao nu c t hn hoc bng 2 user cng truyn th khng phi

i? v nu nhiu hn 2 user cng truyn th phi c hng i?

c. Tnh xc sut mi user c truyn.

d. Gi s c 3 user. Tnh xc sut ti bt k thi im no, c 3 user cng truyn 1 lc. Tnh khong

thi gian hnh thnh hng i.

TL:

a. C 2 user c truyn v mi user cn 1 na bng thng ng truyn.

b. Theo gi thit, mi user cn 1Mbps khi truyn, nu c 2 hoc t hn 2 user cng truyn th ti a

cn 2Mbps bng thng. M ng truyn cho c bng thng l 2Mbps nn khng c hng i

no c.

Ngc li, nu 3 user cng truyn th bng thng cn s l 3Mbps, nhiu hn bng thng hin c.

Trong trng hp ny s c hng i trc khi vo ng truyn.

c. Xc sut = 0.2

d. Xc sut 3 user cng truyn cng lc = C(3,3) . p3 . (1-p)3-3 = (0.2)3 = 0.008. Theo gt, hng i

tng khi tt c user ang truyn, nn khong tgian m hng i tng (= xc sut tt c 3 user cng

truyn) l 0.008.

Cu 16: Gi s ang gi 1 gi tin t 1 host ngun n 1 host ch qua 1 ng c nh. Lit k c c

khong tr. C i no l c nh, c i no thay i?

Cc thnh phn tr gm: tr khi x l, tr khi lan ta (propagation delay), tr khi truyn (transmission

delay) v tr khi i. Tr tr khi i l thay i, tt c nhng khong tr cn li u l c nh.

Cu 18: T nh khong thi gian truy n 1 gi tin di L qua 1 ng truy n khong c ch d, tc

truy n l s, tc chuyn giao l R bps? N c ph thuc vo L v R khng?

Khong thi gian = d/s.

N khng ph thuc vo di gi tin (L) v tc chuyn R.

Cu 19: Gi s host A mun gi 1 file ln n host B. C 3 ng i t A n B, tc ln l t l R1 =

500 kbps, R2 = 2 Mbps v R3 = 1 Mbps.

a. Gi s khng c tc nghn trong m ng. T nh thng l ng truy n file?

b. Gi s file c dung l ng 4 triu byte. Chia k ch th c file bng thng l ng. T nh thi gian

xp x truy n file n host B?

c. Lm cu a, b vi R2 gim cn 100 kbps.

TL:

a. 500 kbps. t 1

b. 4.000.000 byte = 4.000 KB = 32.000 Kbit. Tgian = 32000/500 = 64 s



c. thng lng = 100 kbps. Tgian = 320 s.

Cu 22: Lit k 5 nhim v m 1 tng c th thc hin c. C th xy ra tr ng hp mt hay nhi u

nhim v c thc hin bi nhi u tng c k?

5 nhim v chung l: kim sot li, iu khin lung, phn on v lp rp li, dn knh v ci t kt

ni. C th xy ra trng hp 1 hay nhiu nhim v c thc hin bi nhiu tng. VD: kim sot li

thng c cung cp nhiu tng.

Cu 24: Mt message tng ng dng l g? Mt segment tng giao vn? Mt datagram tng

network? Mt frame tng lin kt?

- Message: d liu m 1 ng dng mun gi v truyn xung tng giao vn.

- Segment: c to ra bi tng giao vn bng cch ng gi message ca tng ng dng vi header ca

tng giao vn.

- Datagram: ng gi segment ca tng giao vn vi header ca tng mng.

- Frame: ng gi datagram ca tng mng vi header ca tng lin kt.

Cu 25: Mt router x l tng no? Mt switch x l tng no? V mt host x l tng no?

Cc router x l tng 1, 2, 3 tc l tng vt l, tng lin kt v tng network. Cc switch x l tng 1 v

2 tc l tng vt l v tng lin kt. Cc host x l c 5 tng.

Chapter1: : Introduction- Problem

P5: Gi s c 2 host A v B kt ni vi nhau bi 1 ng c tc R bps. 2 host c ch nhau m mt v tc

lan ta (propagation speed) l s (m/s). Host A gi 1 packet k ch th c L n host B.

a. Biu din tr lan ta (dprop) thng qua m v s.

b. X c nh thi gian truy n ca gi tin dtrans thng qua L v R.

c. B qua tr x l v hng i. Vit biu thc t nh tr gia 2 im kt thc.

d. Gi s Host A bt u truy n gi tin t i thi im t = 0. T i thi im t = dtrans, bit cui cng

ca gi tin u?

e. Gi s dprop > dtrans. T i thi im t = dtrans, bit u tin ca gi tin u?

f. Gi s dprop < dtrans. T i thi im t = dtrans, bit u tin ca gi tin u?

g. Gi s s = 2,5.108, L = 120 bits v R = 56 kbps. T nh khong c ch m dprop = dtrans

TL:

a. dprop = m/s (s) l tr lan ta (thi gian gi tin i ht khong cch m)

b. dtrans = L/R (s) l tr truyn (thi gian t lc bit u tin ca gi tin ra khi host n khi bit cui

cng ca gi tin ra khi host).

c. dA-B = dprop + dtrans = m/s + L/R (s)

d. Ti t = dtrans, bit cui cng ca gi tin va ri khi host A.

e. Ti t = dtrans, bit u tin ca gi tin vn ang trn ng truyn, cha n c host B.

f. Ti t = dtrans, bit u tin ca gi tin n host B.



g. dprop = dtrans m/s = L/R m = Ls/R = 523158 (m)

P6: Xt vic gi ging ni thi gian thc t host A n host B qua 1 m ng chuyn m ch gi (VoIP). Host

A chuyn ging ni thnh lung bit t n hiu s 64 kbps. Sau host A nhm c c bit thnh tng gi tin 56

bytes. Ch c 1 ng duy nht t A n B tc truy n l 2 Mbps v tr lan ta l 10 ms. Ngay khi

host A t o c 1 gi tin n gi gi tin cho B. Ngay khi host B nhn c gi ton b gi tin n chuyn

c c bit ca gi tin thnh t n hiu t ng t. T nh thi gian k t khi bit u tin c t o (t t n hiu gc

host A) n khi bit cui cng c gii m ( host B)?

Xt bit u tin trong 1 gi tin. Trc khi bit ny c truyn i, tt c cc bit trong gi tin phi c to

ra. V vy cn: 310.64

8.56 (s) = 7 (ms).

Thi gian cn truyn gi tin l: 610.2

8.56 (s) = 224 s

tr lan ta l 10 ms.

tr cho n khi gii m xong l: 7 ms + 224 s + 10 ms = 17,224 ms

P12: Gi s N gi tin cng n 1 ng truy n v hin t i khng c gi tin no c truy n i hoc phi

xp hng. Mi gi tin c di L v tc ng truy n l R. T nh tr hng i trung bnh ca N gi

tin.

tr hng i l 0 khi gi tin u tin truyn i, l L/R khi gi tin th 2 truyn i

v tng qut, l (n 1) L/R khi gi tin th n truyn i.

V vy, tr trung bnh ca N gi tin l:

(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)

P24: Gi s c 2 host A v B c ch nhau 20.000 km, c ni trc tip vs nhau bng 1 ng tc R = 2

Mbps. Tc lan ta trn ng truy n l 2,5.108 m/s.

a. T nh t ch ca bng thng v tr (R.dprop).

b. Xt vic gi 1 file di 800.000 bits t A n B. Gi s file c truy n lin tc. S bit ti a

c truy n i trn ng truy n t i bt k thi im no l bao nhiu?

c. Gii th ch ngha t ch R.dprop.

d. rng ca 1 bit (t nh theo mt) trn ng truy n bng bao nhiu?

e. Vit cng thc tng qu t t nh rng ca 1 bit trn ng truy n. Cho bit tc lan ta l s, tc

truy n l R v di ng truy n l m.

TL:



a. R.dprop = R. m/s = 2. 20.106 / (2,5.10

8) = 160.000 bits.

b. S bit ti a c truyn i trn ng truyn ti bt k thi im = R.dprop = 160.000 bits.

c. Tch s gia bng thng(R) v tr(dprop) ca ng truyn l s bit ti a c th trn ng

truyn

d. rng ca 1 bit = di ng truyn / (R.dprop) = 20.106 / 160.000 = 125 m

e. rng ca 1 bit = m / (R.dprop) = s/R.

Chapter2: Application Layer -Review Question

Cu 1: Lit k 5 ng dng internet khng c quy n v giao thc m chng s dng t i tng ng dng

- The Web - HTTP; - File transfer (bittorrent) - FTP; - Remote login - Telnet; - Network News - NNTP; - E-mail - SMTP;

Cu 2: S kh c bit gia Network architecture v Application architecture?

- Network architecture l mt h thng cc tin trnh giao tip vi nhau thng qua cc tng

- Application architecture c hiu theo ngha khc, n c thit k bi nh pht trin ng dng v n

iu khin hot ng ca cc application.

Cu 4: Cho 1 ng dng chia s file P2P. B n c ng vi nhn xt khng c nh ngha bn client and

server trong 1 giao tip gia chng? T i sao?

Khng, v trong mi giao dch u c 1 bn l client v 1 bn l server. Trong ng dng chia s file P2P

th bn nhn c file l client cn bn gi file l server.

Cu 6: Gi s b n mun lm mt giao dch gia 1 remote client v 1 server vi tc cao, b n nn chn

UDP hay TCP?

Nn chn UDP, v s dng UDP khng cn thit lp ng chuyn nn nhanh hn bn ch mt 1 RTT

l client gi yu cu ti UDP socket. Cn nu s dng TCP bn mt ti thiu 2 RTT: 1 cho vic thit

lp kt ni TCP, cn li cho client gi yu cu v server tr phn hi.

Cu 15: V sao ni FTP gi thng tin i u khin l out-of-band?

FTP s dng 2 kt ni TCP song song. Mt kt ni kim sot thng tin (chng hn nh 1 yu cu

chuyn giao 1 tp tin) v mt kt ni khc dng chuyn giao cc tp tin . Bi v cc thng tin kim

sot khng c gi qua cng 1 kt ni vi tp tin nn c th ni rng FTP gi thng tin iu khin

"out-of-band" .

Cu 18 : Nu s kh c bit gia download-and-delete mode and the download-and-keep mode in POP3?



- Vi ch download-and-delete sau khi ngi dng ly tin nhn t mt my ch POP, cc tin nhn s

b xa. iu ny t ra mt vn cn gi quyt l khi nhiu ngi cng truy cp cc tin nhn t

nhiu my khc nhau (vn phng my tnh, my tnh gia nh,vv.)

- Trong ch download-and-keep, tin nhn s khng b xa sau khi ngi s dng ly cc tin nhn.

iu ny cng c th l bt tin l mi ln ngi s dng ly cc tin nhn, tt c cc tin nhn khng b

xa s c chuyn vo my tnh (bao gm cc thng bo rt c).

Cu 19 : C hay khng mt t chc Mail server and Web server no c cng 1 b danh(alias) cho 1

hostname khng? C i g s l u tr host name of the mail server?

C. The MX record c s dng nh x tn ca mail server vi a ch IP.

Cu 22: Overlay network (m ng bao ph) l g? N c cha router khng? The edges(ra) ca n l g?

N c t o ra v duy tr nh th no?

- Overlay network l mt h thng gm cc nt tham gia vo chia s tp tin v cc lin kt gia chng.

- N khng cha router.

- The edges ca n l logical link (lin kt logic).

- Cch to ra n l: khi 1 node mi mun tham gia vo h thng n cn bit a ch IP ca 1 hay nhiu

node ca h thng, sau n s gi thng ip cho cc node ny, cc nt ny nhn v xc nhn, v n

s tr thnh 1 phn ca h thng.

Cu 25: Skype cng ngh P2P cho 2 giao thc quan trng no?

User location v NAT traversal.

Cu 26: 4 ng dng quan trng ph hp vi kin trc P2P?

a) File Distribution

b) Instant Messaging

c) Video Streaming

d) Distributed Computing

Chapter2: Application Layer -Problem

Cu 1: ng / Sai

a. Mt ngi dng yu cu t 1 trang web 1 text v 3 hnh nh, client s gi 1 tin nhn yu cu v nhn 4

tin nhn phn hi? - Sai: gi 4 nhn 4

b. 2 trang web khc bit c th gi qua cng 1 kt ni kin tr. - ng

c. Vi mt kt ni khng kin tr gia trnh duyt v my ch c th cho 1 gi tin TCP thc hin 2 request

HTTP khc nhau. - Sai

d. Thng ip yu cu ca giao thc HTTP khng th rng.



Cu 3: Gi s mt HTTP client mun ly 1 ti liu ca trang web m bit URL, a ch IP ch a bit.

Th tng ng dng v tng giao vn cn giao thc no

- Application layer protocols: DNS and HTTP

- Transport layer protocols: UDP for DNS; TCP for HTTP

Cu 7: Gi s trong trnh duyt web b n click vo mt link cha 1 trang web. B n cn ly 1 a ch IP

ca trang web ch a c trong cache. B n phi i qua n DNS sau mi c c n. Mi ln thm 1

DNS b n mt RTT thi gian ln l t l RTT1, RTT2 RTTn. Trong trang web c 1 i t ng text.

B n mt RTT0 i t host ti server cha i t ng . T nh thi gian t khi click n khi nhn c

i t ng.

Tng thi gian ly c a ch IP l: RTT1++RTTn;

Sau khi bit a ch IP bn mt RTT0 kt ni TCP v RTT0 na gi thng ip v nhn i

tng.

Nn tng thi gian cn thit l: 2.RTT0 + RTT1 + + RTTn;

Cu 8: Vi bi cu 7, gi s HTML c thm 8 i t ng na trn cng server. Mt bao lu thi gian

vi:

a. Khng kin tr v khng c kt ni song song:

Khi trang web thm 8 i tng ta cn 8 ln thit lp v 8 ln gi thng ip v nhn i tng nn mt

thi gian l: 8.2.RTT0 = 16 RTT0

Nn tng thi gian l: 2RTT0 + RTT1 + + RTTn + 16 RTT0

b. Khng kin tr c 5 kt ni song song:

1 ln kt ni gi nhn c 5 i tng nn 8 i tng cn 2 ln kt ni nn mt thi gian l: 4RTT0

Nn tng thi gian l: 2.RTT0 + RTT1 + + RTTn + 4 RTT0

c. Kt ni kin tr

Do khi to kt ni ly i tng text nn 8 i tng ny khng cn khi to kt ni na.

S dng kt ni kin tr nn cn 1 RTT0 gi yu cu v nhn i tng

Nn tng thi gian l: 2.RTT0 + RTT1 + + RTTn + RTT0 = 3 RTT0 + RTT1 + + RTTn

Cu 14: SMTP kt thc thn mail nh th no? HTTP th sao? HTTP c th s dng ph ng thc ging

SMTP kt thc mail c khng?

- SMTP kt thc th bng mt dng ch cha du chm.

- HTTP qun l th bng trng di trong header.

- HTTP khng th s dng phng thc ging SMTP c. v HTTP message c th dng m nh

phn cn SMTP th phi dng ASCII

Cu 17: Gi s truy cp mail ca b n bng POP3



a. Gi s bn nh dng cho th ca bn to lm vic ch down xong xa. Hon thnh giao dch bn

di:

C: dele 1

C: retr 2

S: (blah blah

S: ..blah)

S: .

C: dele 2

C: quit

S: +OK POP3 server signing of

b. Ch down xong gi

C: retr 2

S: blah blah

S: ..blah

S: .

C: quit

S: +OK POP3 server signing off

c. Ban u ch down xong gi c th 1 xong tt i sau 5 pht c th 2. a ra bn ghi cho trng

hp ny?

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

Cu 20: Suppose you can access the caches in the local DNS servers of your depart ment. Can you

propose a way to roughly determine the Web servers (outside your department) that are most popular

among the users in your department?

Chng ta c th nh k lu nhanh cc bn ca DNS caches trong nhng my ch DNS a phng .

Web server xut hin thng xuyn nht trong DNS caches chnh l server ph bin nht. iu ny l

bi nu nhiu ngi dng quan tm n 1 Web server th DNS requests n Web server c gi

thng xuyn hn. Do , Web server s c xut hin trong DNS caches nhiu hn local DNS cache,

v vy thi gian truy vn s l 0 msec. Cc trng hp khc, thi gian truy vn ln hn.



Cu 22: Consider distributing a file of F = 15 Gbits to N peers. The server has an upload rate of us = 30

Mbps, and each peer has a download rate of d i = 2 Mbps and an upload rate of u. For N = 10, 100, and

1,000 and u = 300 Kbps, 700 Kbps, and 2 Mbps, prepare a chart giving the minimum distribution time

for each of the combinations of Nand u for both client-server distribution and P2P distribution.

tnh ton thi gian iu phi ti thiu (minimum distribution time) cho iu phi client-server, ta s

dng cng thc:

Dcs = max {NF/us, F/dmin}

Tng t, tnh ton thi gian iu phi ti thiu cho iu phi P2P, ta s dng cng thc:

y, F = 15 Gbits = 15 * 1024 Mbits

us = 30 Mbps

dmin = di = 2 Mbps

Note, 300Kbps = 300/1024 Mbps

Cu 23. Consider distributing a file of F bits to N peers using a client-server architecture. Assume a fluid

model where the server can simultaneously transmit to multiple peers, transmitting to each peer at

different rates, as long as the combined rate does not exceed us

a. Suppose that us/N



nhn file hon chnh l F/(us/N) = NF/us. Tt c cc client u nhn hon chnh file trong thi gian NF/us,

nn tng thi gian iu phi cng l NF/us.

Cu 34

a. Gi s bn m FTPclient trc khi m FTPserver. C vn g xy ra khng?

Khi client s c gng kt ni vi server trong khi server cha m=> kt ni li.

b. Bn chy UDP client trc khi chy UDP server?

Khng c vn g. V client khng cn khi to to kt ni ti server.

c. S dng port khc nhau gia client v server.

Khi client s c gng kt ni TCP vi 1 tin trnh khng ng. Nn s c li

Chapter3: Transport Layer -Review Question

Cu 1. Xem xt mt kt ni TCP t host A n B.Gi s TCP segment i t A -> B c port ngun l x v

port ch l y.Vy port ngun v port ch ca segment i t B -> A l j?

Source Port: y, Dest Port: x

Question 2

Gii th ch t i sao nh ph t trin ng dng l i chn UDP h n TCP.

Solution :

Nh pht trin ng dng chn UDP v h khng mun ng dng ca h phi s dng c ch kim sot tc

nghn(c ch iu tit tc truyn d liu ca ng dng lc b tc nghn).C ch ny c th lm gim tc

truyn,iu ny c th nh hng ti ng dng,nht l nhng ng dng chat voice hay hi tho trc

tuyn).Nhng ng dng k cn tin cy ca d liu c truyn,m quan trng l thi gian.

Question 3. Liu c ng dng no c c kh nng truy n d liu ng tin cy ngay c khi n ch y trn

giao thc UDP?

Solution :

C. Nh pht trin ng dng c th thm giao thc truyn d liu ng tin cy vo giao thc ca lp ng

dng.Tt nhin,n s i hi mt lng cng vic ng k cho vic pht hin li.

Question 4 True or False

a. Gi s host A truyn mt tp tin ln ti host B trn giao thc TCP,v host B khng c d liu j gi

ti host A. Host B s khng gi ACK ti host A v n khng th ng gi ACK vo d liu c.(

unlogical )

b. Kch thc ca ca s bn nhn (TCP Rcv window) khng bao h thay i trong sut qu trnh kt ni.

c. Gi s host A truyn mt tp tin ln ti host B trn giao thc TCP,s byte d liu NAK khng th vt

qu b nh m ca pha nhn.



d. Gi s host A truyn mt tp tin ln ti host B trn giao thc TCP,S th t ca segment cho ln gi

ny l m,th s th t cho segment tip theo phi l m+1.( Sai v nu qu trnh gi gi trc m b li th

phi gi li gi m).

e. TCP segment c 1 trng trong tiu cho ca s bn nhn.

f. Gi s thi gian RTT gn nht trong kt ni TCP l 1 s.Vy phi ci t thi gian timeout >=1 cho ln

truyn ny.(Sai v cn sd nhiu RTT trong qu kh ch k sd RTT gn nht).

g. Gi s host A gi 1 segment ti B vs STT l 38 v 4-bytes d liu.Vy ACK cho segment ny phi ny

42.(sai.ACK l 38)

Solution : a. F b. F c.T d. F e.T f. F g. F

Question 5

Gi s A gi 2 gi TCP li n nhau ti B.Segment th nht c STT l 90, segment th 2 c STT l 110.

a. Segment u tin c dung l ng l bao nhiu.

b. Gi s segment u tin b mt,nh ng segment th 2 vn n c B. Vy stt ca ACK c gi t B v A

l j?

Solution :

a. 110-90 = 20 bytes.

b. V gi 1 mt nn ACK vn l 90

Question 7: Hin t i c 2 kt ni TCP vi 1 nt c chai c tc l R bps. C 2 kt ni u c 1 file ln

cn gi (n cng 1 h ng qua nt c chai ). Vic truy n d liu bt u cng mt thi im. Vy

tc truy n m TCP s phn cho mi kt ni l bao nhiu?

Ch : Qua t nh cng bng ca giao thc TCP ta bit c, nu N kt ni cng nhau chia s mt knh

truy n tc nghn th mi kt ni s nhn c 1/N bng thng. (Tc l nhn c bng thng bng nhau).

Solution :

Qua ch th p n l : R/2

Question 8

Xem xt c ch kim sot tc nghn trong TCP. Nu xy ra timeout bn gi th threshold s c t bng

mt na ca gi tr threshold trc .

Sai v nu xy ra mt gi tin, ngng threshold s c t bng mt na ca Congwin (Congrestion

window)

Note : Tc truyn b gii hn bi ca s tc nghn Congwin. Threshold l ngng gia 2 pha



Question 9 (P1 trong sch) Gi s Network Layercung cp dch v sau. Network layer host ngun chp

nhn gi 1 segment c max size l 1200 bytes v a ch ch t transport layer.Network layer m bo gi

thnh cng segment ti transport layer ti host ch.

Gi s c nhiu tin trnh cng chy ti host ch.

a.Thit k mt giao thc n gin cho transport layer nhn c d liu ph hp cho mi tin trnh ti

host ch.Gi s HH host ch c th cp 4 bytes port number cho mi tin trnh ang chy.

b.iu chnh giao thc n c th tr v a ch (return address) cho host ch.

c.Trong giao thc ca bn liu transport layer c phi lm j trong li ca mng my tnh?

Solution :

a. Gi giao thc ny l STP (Simple transfer protocol).Bn pha gi,STP chp nhn cho tin trnh gi 1

chunk kch thc k vt qu 1196bytes d liu ,mt a ch host ch v mt cng nhn.STP thm 4

bytes vo trg header ca mi chunk,4 bytes lu port number ca tin trnh bn nhn.V network

layer s chuyn gi tin ti transport layer ca bn nhn.Sau STP s gii nn segment v kim tra

segment nhn c c port number ng vi tin trnh no ri gi segment ti ng tin trnh cn nhn.

b. By h mi segment s c 2 trg tiu (header).Mt trg cho port ngun v 1 trg cho port ch.Bn pha

gi chp nhn gi 1 chunk kch thc k vt qu 1192 bytes, mt a ch host ch ,mt port ngun v

mt port ch.

STP to mt segment bao gm d liu ca application,port ngun v port ch.Sau n s chuyn

segment v a ch host ch cho tng network gi sang bn nhnSau khi bn ch nhn c

segment,STP s chuyn ti ng dng d liu cn nhn v c port ngun.

c. N k phi lm j.N ch nm trn thit b u cui.

Question 10.(P2) Gi s c 1 hnh tinh c dch v gi th.Mi gia nh c 1 a ch(1 hm th ring).Mi

thnh vin trong gia nh u c mt tn ring.Dch v gi th c th gi t nh ny sang nh khc.N yu

cu phi cho bc th v a ch ca nh cn gi vo trong phong b.Mi gia nh u c 1ng i din,ng

ny s nhn th v phn pht th cho cc thnh vin khc trong gia nh.



a.S dng giao thc Question 9 gi th.M t cch ng i din nhn v phn pht th.

b.Trong giao thc ny,liu mail server c phi m phong b kim tra bc th bn trong hay khng?

Solution :

a) Ng gi s a cho ng i din ca gia nh h bc th cng vi tn ng nhn v a ch ca ng .Sau

ng i din s vit tn ng nhn vo u bc th,nht n vo phong b.Bn ngoi phong b ghi a ch ca

ngi nh pha ng nhn.Sau a cho bu in(chnh l dch v gi th ca hnh tinh ) Bn kia,ng

i din nhn c phong b s ly bc th ra,xem tn ng nhn,v gi cho ng ng .

(Gi s l th tnh m ng i din l b or m th e k bik What will happen? lol)

b) Mail server k cn m phong b,n ch kim tra a ch trn phong b .

Question 11(P5) Ti sao by h cc ng dng chat voice hay video li dng TCP hn UPD,trong khi TCP c

c ch kim sot tc nghn gy tr ng truyn?

Sol : v by gi firewall hay chn cc lung truyn ti data ca UDP.Trong khi TCP c th vt qua c.

Question 12.(P7) Gi s mt tin trnh host C c mt UDP socket vi port l 6789.Gi s c 2 host A v B

u c th gi segment ti C vi cng mt port ch l 6789.Liu c 2 segment u cng hng n cng

mt socket ch host C hay khng?Nu c th lm sao host C c th bit c s khc nhau gia 2

segment (n c ngun gc t u)?

Solution: C.Mi segment nhn c du c th h iu hnh s cung cp 1 tin trnh vi a ch IP xc nh

ngun gc ca segment .

Question 13.(P8) Gi s Web server chy trn host C ti port 80 v dng kt ni kin tr.N nhn c 2 yu

cu t host A v B.Liu c hai yu cu c c gi trn cng mt socket ca C.Nu 2 yu cu c tr li qua 2

socket khc nhau th liu c 2 socket u c port l 80?

Solution : Trong kt ni kin tr, Web server to ra tng kt ni socket ring bit . Mi socket kt ni c xc

nh vi mt b bn : (a ch IP ngun, port ngun , a ch IP ch ,port ch ) .Khi host C nhn c IP

datagram/segment , n kim tra 4 trg ny xc nh socket m n phi truyn segment qua . V vy , cc yu

cu t A v B c truyn thng qua cc socket khc nhau .C 2 u c port number l 80 , tuy nhin , vic nhn

din cc socket ny da vo a ch IP ngun.Khng ging nh UDP, khi lp transport gi mt segment TCP

ln tin trnh tng ng dng , n khng ch ra a ch IP ngun ,m n ngm quy nh bi cc trnh nh danh

socket.

Question 14(P9) Trong giao thc rdt,ti sao chng ta cn s dng STT(sequence number)?

Solution : STT c s dng cho bn nhn,gip phn bit c khi no d liu nhn c l b trng.

Question 15(P10) Trong rdt protocol,ti sao phi s dng b m thi gian(timer)?



Solution : Nu nh thi gian bn gi ch ACK (or NAK) vt qu thi gian quy nh ca b m,th bn gi

kt lun gi tin c th b mt.Do ,n s truyn li gi tin .

Question 16 (P11) Gi s roundtrip delay ( tr phn hi) gia bn nhn v bn gi l hng s,v bn gi

bik tr ny.Vy ta c cn timer cho rdt 3.0 na hay khng?Gi s gi tin c th b mt.

Solution : Timer vn cn thit cho rdt 3.0 v nu bn gi bik c RTT(RTT l thi gian t lc gi n lc nhn

c ACK) th c 1 u im duy nht l bn gi bik c packet hoc ACK (NAK) c th b mt.Hy t vo tnh

hung c th,nu ACK k mt m ch n tr,lc bn gi vn cn timer bit c gi tin mt hay cha.

Chapter3: Transport Layer - Problem

1. Gi s ti cng 1 thi im, client A v client B u mun khi to 1 phin Telnet ti server S.

Cung cp s hiu cng ngun v ch c th c cho:

a. Cc segments gi t A n S

b. Cc segments gi t B n S

c. Cc segments gi t S n A

d. Cc segments gi t S n B

e. Nu A, B l 2 host khc nhau, th s hiu cng ngun trong segments t A ti S c c ging nh

l t B ti S khng?

f. Nu chng l cng 1 host th sao?

Answer:

source port numbers destination port numbers

a) AS 467 23

b) BS 513 23

c) SA 23 467

d) SB 23 513

e) Yes

f) No

2. UDP v TCP s dng b 1 cho trng checksum.

Gi s ban u bn c 3 byte 8 bit: 01010011, 01010100, 01110100. B 1 ca tng ca cc byte 8 bit l

g? Ti sao UDP li ly b 1 ca tng? Ti sao khng ch s dng tng? Vi vic s dng b 1, lm th

no ngi nhn pht hin li? C th no li 1 bit s i khng b pht hin? Li 2 bit th sao?

Answer:



0 1 0 1 0 0 1 1

+ 0 1 0 1 0 1 0 0

-----------------------

1 0 1 0 0 1 1 1

1 0 1 0 0 1 1 1

+ 0 1 1 1 0 1 0 0

-----------------------

0 0 0 1 1 1 0 0

B 1 l: 1 1 1 0 0 0 1 1

- t s t t 4

t 3 t 1 t su

- N u t 1 s 0 !

- t 1 t u t 2

t t t s t v

s u t u t u

t 0 s u t t 2

u t 1

3. Gi s ngi nhn UDP tnh checksum cho gi tin UDP segment v nhn thy rng n trng khp gi tr

truyn ti trong trng checksum ca gi tin. Ngi nhn c th hon ton chc chn rng li bit khng

xy ra hay khng? Gii thch.

Answer:

Khng, ngi nhn khng th hon ton chc chn li bit khng xy ra

Bi: nu cc bit tng ng ca 2 byte trong gi tin l 0 v 1 th khi chng i gi tr cho nhau (01, 10)

th tng ca chng vn ging nhau b 1 ca tng cng s ging nhau

4. So snh pha gi v pha nhn ca giao thc rdt2.2 v rdt3.0

Answer:

Pha gi dng giao thc rdt3.0 khc pha gi dng giao thc rdt2.2, trong c thm thi gian timeouts.

Chng ta thy rng vic thm thi gian timeouts lm tng kh nng trng lp cc gi tin. Tuy nhin, pha

nhn dng giao thc rdt.2.2 c th x l cc gi d liu trng lp. (Gi tin trng lp pha nhn trong

rdt 2.2 s pht sinh nu ACK b li, v ngi gi sau truyn li d liu c). Do pha nhn trong giao

thc rdt2.2 lm vic nh l pha nhn trong giao thc rdt3.0

5. Hy xem xt mt giao thc truyn d liu ng tin cy m ch s dng NAK.

Gi s pha gi s gi d liu khng thng xuyn. Giao thc ch dng NAK c thch hp hn giao thc

s dng ACKs hay khng? Ti sao?

By gi gi s phia gi c rt nhiu d liu gi v kt ni end-to-end c rt t mt mt. Trong trng

hp ny, giao thc ch dng NAK c thch hp hn giao thc s dng ACKs khng? Ti sao?

Answer:

Trong giao thc ch dng NAK, mt gi x ch c th pht hin c bi pha nhn khi pha nhn nhn c

gi x-1 v ngay sau l x+1.

Nu pha gi gi khng thng xuyn, v c 1 khong thi gian delay di gia 2 gi tin x v x+1 th ch

n khi no gi x+1 c gi i v n c pha nhn th pha nhn mi pht hin c l gi x b mt =>

s mt 1 khong thi gian di khi phc gi x theo giao thc ch dng NAK



Mt khc, nu d liu c gi thng xuyn th vic khi phc theo giao thc ch dng NAK s xy ra rt

nhanh chng. Hn na, nu li khng thng xuyn xy ra th NAK ch thnh thong mi c gi (khi

cn thit) cn ACK th khng cn gi. Do d lm gim ng k trong thng tin phn hi

Giao thc ch dng NAK ti u hn giao thc s dng ACK

6. Xem xt giao thc GBN vi kch thc windows pha gi l 3 v phm vi ca seq num l 1024.

Gi s ti thi im t, th t tip theo trong gi tin m pha nhn ang mong i l 1 sequence number ca

k. Cho rng, mi trng khng sp xp li th t ca cc msg. Tr li cc cu hi

a. Cc tp seq num c th c bn trong windows pha gi ti thi im t l g ?

b. Tt c cc gi tr c th c ca trng ACK trong tt c cc msg c th hin ang truyn quay

tr li pha ngi gi ti thi im t l g ?

Answer:

a. y chng ta c 1 windows vi kch thc N=3.

Gi s pha nhn nhn c gi tin k-1, v gi ACK cho n v tt c cc gi tin khc. Nu tt

c ACK u c nhn pha gi th sau windows pha ngi gi s l k, k+N-1].

Gi s th 2 l khng c ACK no c nhn ti pha gi, khi windows pha gi s cha N gi

tin v phi bao gm gi k-1, nn windows phi gi s l k-N, k-1]

Vi nhng lp lun trn th windows pha gi s c kch thc l 3 v s bt u ti 1 ni no

trong phm vi k-N, k]

b. Nu pha nhn ang ch gi tin th k, v sau n nhn c gi tin th k-1 v N-1 gi tin trc

=> P19

7. Tr li ng hoc sai cho cc cu hi sau y v gii thch cho cu tr li ca bn:

a. Vi giao thc selective repeat, n c th cho pha gi nhn c mt ACK cho mt gi tin nm bn

ngoi windows hin ti ca n.

b. Vi GBN, n c th cho pha gi nhn c mt ACK cho mt gi tin nm bn ngoi windows hin

ti ca n.

Answer:

a. ng.

Gi s pha gi c kch thc windows l 3 v gi cc gi tin 1,2,3 ti t0.

Ti t1 (t1>t0) pha nhn gi ACK 1,2,3.

Ti t2 (t2 > t1), pha gi timeouts v gi li gi tin 1,2,3.

Ti t3, pha nhn nhn cc gi tin trng lp, v gi li CK 1,2,3.

Ti t4 pha gi nhn ACK 1,2,3 m pha nhn gi ti thi im t1, v chuyn windows ln 4,5,6.

Ti t5 pha gi nhn ACK 1,2,3 m pha nhn gi ti t3. Khi ny ACK ngoi khong ca windows

b. ng.



8. Xem xt vic chuyn mt tp tin rt ln ca L byte t Host A n Host B. Gi s 1 MSS = 536 byte.

a. Gi tr ti a ca L TCP seq num khng b cn kit l g? Ttrng TCP seq num c 4 byte.

b. Vi gi tr L c c trong cu a), phi mt bao lu truyn ti cc tp tin.

Gi s rng c tng s 66 byte trng tiu ca tng giao vn, mng, lin kt c thm vo

segment trc khi gi d liu c gi qua mt lin kt 155 Mbps. B qua kim sot lu lng v

kim sot tc nghn v th c th gi ra cc segment lin tc.

Answer:

C tt c 232 = 4294967296 sequence number

a. Sequence number khng tng ln 1 mi segment. Thay vo , n gia tng s lng cc byte d

liu c gi. V vy kch c ca MSS l khng lin quan kch thc tp tin ti a c th c gi

t A n B ch n gin l s lng byte biu din bi 232 = 4.19 Gbytes

b. S lng segment l

= 8012999 (ly nguyn, lm trn ln)

66byte tiu c thm vo cho mi segment

=> tng c

= 66 * 8012999 = 528,857,934 bytes ca tiu

=> tng ln phi truyn i l s = 232 +

= 4,824*109 bytes

=> thi gian truyn ti l (s * 8) / (155*1000) = 249 (s)

9. Host A v B giao tip qua kt ni TCP v host B nhn c t A tt c cc byte thng qua byte 126. Gi

s rng host A sau gi thm 2 segment ti host B back-to-back. Segment u tin v th 2 cha tng

ng 70 v 50 byte d liu. Trong segment u tin sequence number l 127, source port l 302, destination

port l 80. Host B gi ACK ngay khi n nhn c segment t A

a. Trong segment th 2 c gi t host A n host B th sequence number, source port, destination

port l bao nhiu?

b. Nu segment u tin n trc segment th 2, trong acknowledgement ca segment u tin n

th acknowledgement number, source port, destination port l bao nhiu?

c. Nu segment th 2 n trc segment u tin, trong acknowledgement ca segment u tin n

thi acknowledgement number l bao nhiu?

d. Gi s 2 segment c gi bi A n theo th t ti B, acknowledgement u tin l b mt,

acknowledgement th 2 n sau khong thi gian time out. V s thi gian hin th cc segment

v acknowledgements gi.

Answer:

a. Trong segment th 2:

- sequence number l 197

- source port number l 302

- destination port number l 80



b. Nu segment u tin n trc segment th 2:

- acknowledgement number l 197

- source port number l 80

- destination port number l 302.

c. Nu segment th 2 n trc segment u tin:

- acknowledgement number l 127 ch ra rng n vn cn i cc bytes 127 tr i

d.

10. Host A v B c kt ni trc tip vi lin kt 100 Mbps.C mt kt ni TCP gia hai host, v host A gi

n host B mt tp tin c kch thc ln qua kt ni ny. Host A c th gi d liu ng dng ca n vo

socket TCP ca n vi tc cao l 120 Mbps nhng Host B c th c nhn c b m vi tc ti

a l 60 Mbps. M t nh hng ca kim sot lu lng TCP.

Answer:

Host A c th gi vi tc 100Mbps. Tuy nhin host A gi vo b m nhanh hn so vi host B c th

c. B m y ln vi tc 40Mbps. Khi b m y, host B bo hiu ti host A dng gi d liu bng

cch thit lp RcvWindow = 0. Host A s dng vic gi d liu cho n tn khi nhn c segment vi

RcvWindow > 0. Do host A s lin tc dng li v bt u gi. Trung bnh trong thi gian di, host A

gi d liu n host B nh l 1 phn ca kt ni ny v khng ln hn 60Mbps

11. Hy xem xt cc th tc TCP c lng RTT. Gi s = 0,1.

SampleRTT1 l SampleRTT gn y nht. SampleRTT2 l RTT gn y nht tip theo, v nh vy.

a. i vi mt kt ni TCP nht nh, gi s c bn acknowledgments c tr li vi

SampleRTTs tng ng: SampleRTT4, SampleRTT3, SampleRTT2, v SampleRTT1, Th hin

EstimatedRTT vi bn SampleRTTs.

b. Khi qut cng thc cho n SampleRTTs

c. i vi cng thc trong phn (b) p dng cho n.

Answer:



a. K hiu EstimatedRTT(n) l c lng sau n sample

EstimatedRTT(1)

= SampleRTT1

EstimatedRTT(2)

= .SampleRTT1 + (1-) .SampleRTT2

EstimatedRTT(3)

= .SampleRTT1 + (1-) .SampleRTT2 + (1-).SampleRTT3 ]

= .SampleRTT1 + (1-). .SampleRTT2 + (1-)2.SampleRTT3

EstimatedRTT(4)

= .SampleRTT1 + (1-). .SampleRTT2 + (1-)2. .SampleRTT3 + (1-)

3.SampleRTT3

b. EstimatedRTT(n) = . ( )

.SampleRTTj + ( )

c.

12. Host A gi mt tp tin rt ln n host B qua mt kt ni TCP. Qua kt ni ny l khng bao gi c bt k

mt mt gi d liu v khng c timeout. K hiu tc truyn dn ca lin kt kt ni host A n Internet

l R bps. Gi s rng qu trnh host A c kh nng gi d liu vo TCP socket vi tc S bps, trong S

= 10 * R. Hn na gi s rng b m nhn l ln cha ton b tp tin, v b m gi ch c th gi

mt phn trm ca tp tin. iu g s ngn chn qu trnh trong my ch lin tc truyn d liu n socket

TCP ca n vi tc S bps ?

Answer:

Trong vn ny, khng c nguy him trong trn b nh pha nhn nhn t khi vng m nhn pha nhn

c th gi ton b tp tin. Hn na, v khng c mt mt v acknowledgements c gi tr trc khi

timeout, TCP kim sot nghn khng iu tit pha gi. Tuy nhin, qu trnh trn host A s khng lin tc

chuyn d liu cho socket v vng m gi s nhanh chng c lp y. Mt khi vng m gi tr thnh

y, qu trnh s chuyn d liu t l trung bnh hoc R


cng mng my tnh Page 21

b. Trong RTT u tin: 5 MSS c gi;

Trong RTT th 2: 6 MSS c gi;




Trong RTT th 6: 10 MSS c gi.

V th trong khong 6 RTT, 5+6+7+8+9+10 = 45 MSS c gi (v acknowledged).

Do , chng ta c th ni rng thng lng trung bnh trong khong 6RTT l:

14. Trong giao thc rdt3.0, gi tin ACK t pha nhn n pha gi khng phi nh s, mc d chng c

trng trong ACK m cha stt ca packet m chng bit. Ti sao cc packet ACK khng i hi nh

s (sequence number)

Answer

Chng ta thy rng, pha gi cn seq num pha nhn c th bit packet c trng vi 1 packet nhn

t trc khng. i vi ACKs, th bn gi khng cn thng tin (VD: stt ca ACK) bit 1 ACK trng

lp. Mt ACK trng lp l r rng i vi pha nhn ca rdt3.0, k t khi n nhn c ACK gc th n

chuyn sang trng thi mi. ACK trng lp khng phi l ci ACK ngi gi cn nn n b b qua bi

rdt3.0

15. Suppose an application uses rdt3.0 as its transport layer protocol. As the stop-and-wait protocol has very

low channel utilization (shown in the cross-country example), the designers of this application let the

receiver keep sending back a number (more than two) of alternating ACK 0 and ACK 1 even if the

corresponding data have not arrived at the receiver. Would this application design increase the channel

utilization? Why? Are there any potential problems with this approach? Explain

Answer

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 wont

re-send those segments, unless there are some additional mechanism in the application to recover from loss



Chapter4: Network Layer Review Question

R1: a ra mt vi thut ng c s dng trong cun s ch. Tn gi packet ca tng transport l segment

v tn gi ca tng link l frame. Tn packet ca tng network? Routers v link layer switches c gi l

packet switches. Nguyn tc kh c nhau c bn ca routers v link layer switches?

Tr li:

Packet tng network c gi l datagram. Mt router mun chuyn packet phi da trn a ch IP ca packet (tng 3). A link-layer switch mun chuyn packet phi da trn a ch MAC (tng 2).

R2: Hai chc nng quan trng ca tng network trong chuyn m ch gi l g? Ba chc nng quan trng ca tng network trong mt chuyn m ch o?

Tr li: Hai chc nng quan trng ca tng network trong datagram network: chuyn v nh tuyn. Ba chc nng quan trng ca tng network trong virtual circuit: chuyn, nh tuyn, v thit lp tuyn ng.

R3: S kh c nhau gia routing v forwarding?

Tr li: Forwarding l chuyn mt packet t mt lin kt u vo ca mt router, ra mt lin kt u ra ca router sao cho thch hp. Routing l xc nh cc router trn ng truyn t ngun ti ch.

R4: Routers trong chuyn m ch gi v chuyn m ch in c c s dng trong bng nh tuyn? Nu vy, hy m t bng nh tuyn cho hai m hnh trn?

Tr li: c s dng cho c hai.

R7: T i sao mi cng vo trong vng nh ca router c tc cao (Discuss why each input port in a high-speed router stores a shadow copy of the forwarding table)

With the shadow copy, the forwarding decision is made locally, at each input port, without invoking the

centralized routing processor. Such decentralized forwarding avoids creating a forwarding processing bottleneck at

a single point within the router

R9: Miu t vic mt m t gi tin c th xy ra t i cng vo. Miu t vic mt m t gi tin t i cng vo do c th b lo i b. (tr tr ng hp s dng b nh m)

Tr li: Mt mt gi tin xy ra nu kch c hng i ti cng vo tng ln cao bi v tc forwarding chm do b nh router s b y. iu c th b loi tr nu tc forwarding ti thiu l n ln tc ca cng vo, n chnh l s cng vo.

R10: Miu t vic mt m t gi tin c th xy ra t i cng ra.

Tr li: Vic mt mt gi tin c th xy ra nu kch c hng i ti cng ra (s gi tin cn chuyn i ra khi buffer ca router) tng cao trong khi tc ra ca chm



R11: HOL blocking l g? N c th xy ra trong cng vo hoc cng ra khng?

Tr li: HOL blocking cc datagram xp hang ti trc hang i ngn khng cho datagram khc chuyn tip. N c th xy ra ti cng vo.

R12: Router c c a ch IP khng? Nu c, di ?

Tr li: C. N c 1 a ch IP cho mi interface.

R13: Chuyn sang h c s nh phn di a ch 223.1.3.27

Tr li: 11011111 00000001 00000011 00011100

R15: Gi s c 3 router gia host ngun v ch. B qua phn mnh, mt IP datagram gi t host ngun n ch s i qua bao nhiu interfaces? C bao nhiu bng chuyn tip chuyn gi t ngun n ch?

Tr li: 8 interfaces; 3 forwarding tables

R16: Gi s mt ng dng t o ra 40bytes d liu trong khong thi gian 20msec v mi d liu c ng gi trong gi TCP segment v sau thnh gi IP datagram. Ti l ca mi datagram s overhead l g? T l d liu ng dng l bao nhiu?

Tr li: Overhead: 20 byte ca TCP header, 20 byte ca IP header = 40 byte 50% overhead

R17: Gi s host A gi host B mt gi TCP segment c ng gi trong IP datagram. Khi host B nhn c gi datagram. Tng network trong host B s lm nh th no bit n nn truy n bng TCP tt h n UDP hay mt c i g kh c?

Tr li: 8 bit trong trng giao thc ca IP datagram bao gm thng tin v giao thc tng transport m my ch nn gi segment ln lp transport

R20: C nhn nh rng: Khi IPv6 c ng ng qua IPv4, IPv6 nm trong ng ng IPv4 nh mt giao thc t i tng lien kt. B n c ng vi ph t biu trn. T i sao v t i sao khng? Tr li: C. Bi v ton b gi tin IPv6 c ng gi trong gi tin IPv4.

R21. So s nh s kh c nhau gia hai thut to n linkstate v distance vector.

Thut ton linkstate: tnh ton con ng i c chi ph ngn nht t ngun ti ch bng cch s dng nhng kin

thc hon chnh, tng qut v mng.

nh tuyn theo distance vector: : Vic tnh ton chi ph ng i t nht c thc hin lp i lp li, phn phi

theo cch thc. Mt nt ch bit hng xm m n phi chuyn tip mt gi tin i qua t c ch n theo

con ng c chi ph l t nht , v chi ph ca con ng t n n ch.



R22, Tho lun lm th no m internet c th t chc sp xp c th phn chia cho hng triu ng i

dung?

Router c tng hp vo cc h thng t iu khin (cc AS). Trong mt h thng AS, tt c cc router chy

cng giao thc nh tuyn trong ni b AS. c bit gateway router trong cc AS khc nhau chy cc giao thc

nh tuyn lin h thng t iu khin xc nh ng dn nh tuyn gia cc AS.

Vn ca scale(phn chia) c gii quyt khi mt router trong AS ch cn bit v cc router v gateway router

trong AS ca n

R23, C cn thit vic mi h thng t tr (AS) s dng cng thut to n nh tuyn intra- AS khng?

Khng. Mi AS c quyn t ch trong qun l nh tuyn trong AS.

R26, I n vo trng sau: RIP advertisements typically announce the number of hops to various

destinations. BGP updates, on the other hand, announce the _____ to the various destinations.

sequence of ASs on the routes - Trnh t cc Ass trn cc tuyn ng.

R29, nh ngha v so s nh c c thut ng sau: Subnet,prefix v BGP route.

Tr li

Subnet l mt phn ca mng li ln hn, mt subnet khng cha 1 router. Ranh gii ca n uc xc nh bi

giao din ca router v host.

Prefix l mt phn ca a ch CDIRIzed c vit di dng abcd/x. Mt prefix c th gm 1 hay nhiu subnet.

Khi mt router qung b mt prefix qua mt phin BGP, n bao gm mt s thuc tnh BGP.

Trong thut ng BGP , mt prefix cng vi cc thuc tnh ca n l mt tuyn ng BGP (hoc ch n gin l

mttuyn ng).



Chapter4: Network Layer Problem

P1: Trong cu hi ny, chng ta xt 1 vi u nh c im ca virtual-circuit v datagram netwok.

a. Gi s routers l i t ng chu i u kin kh th ng xuyn b li. i u ny c gy tranh ci trong

vic n c li cho kin trc Vc hay kin trc datagram? T i sao

b. Gi s rng c 1 node ngun v 1 node ich, yu cu c nh cng sut lun c sn t i c c router trn

ng dn t ngun n ch, cho vic s dng c quy n ca c c trafic gia ngun v ch. S

dng VC hay kin trc datagram? T i sao?

c. Gi s c c links v router trong m ng l i khng bg b li v nh tuyn gia tt c c c router

ngun/ ch vn khng thay i. trong tnh hung ny, VC hay kin trc datagram c nhi u i u

khin lung giao thng overheadd h n? T i sao

Tr li

a. Vi 1 mng li kt ni mng, mi router b li s lin quan n phn nh tuyn ca kt ni . mc ti

thiu, n s yu cu router li ngc dng t router tht bi n xy dng 1 ng dn mi n node

ch, vi tt c cc du hiu xy dng lin quan n set up ng dn mi. Hn na, tt c cc router trong

ng dn ban u t node sai phi c i xung kt ni sai, vi tt c du hiu xy dng lin quan n

n.

Vi 1 mng ko kt ni datagram, khng du hiu thit lp set up cho c ng dn mi hay i xung

ng dn c, chng ta c th thy, tuy nhin, bng nh tuyn s cn update li( v d dng thut ton

distance vector hoc link state) a cc router khng thnh cng vo ti khon. chng ra c th thy vs

tt distance vector, bng nh tuyn ny c th thay i thi thong bng i phng ha tng vng gn

router tht bi. do , 1 datagram netwok s thch hp hn. Tht th v, cc tiu chun thit k trong

arpanet ban u c th s dng di iu kin cng thng l 1 trong nhng l do kin trc datagram c

la chn cho t tin Internet

b. cho 1 router c th duy tr s lng cng sut c nh trn ng dn gia node ngun v node ch, n

cn bit cc ct nh ca cc truy cp t tt c cc phin truyn thng qua cc lin kt trong router. iu

ny c th c trong mng li lin kt nh hng, nhng ko c trong mng li ko kt ni. v vy 1 kt

ni mng nh hng VC s thch hp hn.

c. trong tnh hung ny, kin trc datagram c kim sot lung trn khng hn. iu ny l do cc tiu gi

tin khc nhau cn thit nh tuyn cc gi tin thng qua mng. Nhng kin trc VC, 11 khi tt c cc

circuit c thit lp. n s khng bao h thay i. nh vy overhead l khng ng k v lu di

P2: Xt 1 virtual circuit netwok, gi s s VC l 1 tr ng 8 bits.

a. S ti a VC c th mang trong1 link?

b. Gi s node gia x c nh ng dn v VC number thit lp kt ni. gi d c 1 c i VC number

ging vy c s dng trong mi link cng VCs path. Miu t lm th no node trung tm c th

x c nh VC number thit lp lin kt. c th c t VC trong tin trnh h n s ti a x c nh?

c. Gi s s kh c nhau c cho php trong mi link dc theo ng dn ca VC. Trong qu trnh

thit lp kt ni, sau khi 1 ng dn end to end c x c nh. Miu t lm th no links c th

chn VC number ca n v cu hnh n theo c ch phn cp, m khng c s ph thuc vo 1 node

trung tm?

Tr li

a. S ti a VC c th mang trong 1 link = 2^8 = 256



b. cc node trung tm c th chn bt k s lng VC free t {0, 1, , 2^8 1}. Theo cch ny, n khng

th c t VC hn 256 trong tin trnh m khng c bt k VC ph bin free

c. mi cp ca cc lin kt c lp c th phn b VC number t tp {0, 1, 2^8 1} do 1 s VC c kh

nng s c 1 s VC khc nhau cho mi lin kt dc theo con ng ca VC phi thay th VVC number ca

mi gi tin n VV number lin kt vs cc lk ra bn ngoi.

P3: Bng chuyn tip bare-bones trong VC netwok c 4 ct. ngha ca tng gi tr trong mi ct? Bng

chuyn tip bare-bones trong datagram network c 2 ct, ngha?

Tr li:

i vs bng chuyn tip VC, cc ct l: Incoming Interface, Incoming VC Number, Outgoing Interface,

Outgoing VC Number

i vs bng chuyn tip datagram, cc ct l: Destination Address, Outgoing Interface

P5: Xem nh 1 VC network c 2 b t cho tr ng VC number. Gi s network mun thit lp 1 ng truy n

chuyn o qua 4 links A B C D. Gi s mi link c cn thc hin 2 ng chuyn o kh c nhau. Vi s

VC nh hnh v:

Hy nh rng 1 VC ch truyn trn 1 trong 4 link.

a. Nu 1 VC mun dng 1 number VC ging nhau cho c 4 link. Th VC number c th gn cho new VC.

b. Nu mi VC c php c nhiu VC number trong cc link khc nhau th c bao nhiu t hp cho 4 VC

number c th s dng?

Tr li:

a. Khng c VC number no c gn cho new VC. Nn cng khng c VC no c thit lp.

b. 2^4 ( TRR)

P9: xem nh 1 datagram s dng 32 bit nh a ch ca host. Gi s 1 router c 4 links t 0 n 3, v

gi tin c th chuyn qua link interface nh hnh sau:



a. Cung cp 1 bng chuyn s dng prefix matching di nht v chuy n gi tin qua link interface ng.

b. B n x c nh bng chuyn nh th no vi c c a ch ch nh d i y

Gii:

a.

Prefix Match Link Interface

11100000 00 0

11100000 01000000 1

1110000 2

11100001 1 3

otherwise 3

b. 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

P12: gi s 1 router c kt ni ti 3 subnet 1,2,3. Mi subnet c d ng 223.1.17/24. Gi s rng sub1 c th h

tr cho 63 interface, sub 2 h tr cho 95 interface, sub 3 h tr cho 16 interface. Cung cp 3 a ch m ng

tha mn c c yu cu trn.

Tr li:

223.1.17.0/26 v 63 < 2^6

223.1.17.128/25 v 95< 2^7

223.1.17.192/28 v 16 =2^4

P15: Hy xem xt 1 subnet 128.119.40.128/26. a ra 1 v d v 1 a ch IP d ng xxx.xxx.xxx.xxx c th g n

cho m ng ny. Gi s 1 ISP s hu khi a ch c d ng 128.119.40.64/25. Gi s mun t o ra 4 m ng con

t khi ny, vi mi khi c cng s a ch IP. Subnet ca 4 m ng con l g?

Tr li:

- a ch IP nm trong mng c subnet 128.119.40.128/26:

Ta c 10000000 01110111 00101000 10000000

Cc IP t 128.119.40.128 10000000 01110111 00101000 10000000

n 128.119.40.191 10000000 01110111 00101000 10111111

- Vi khi a ch 128.119.40.64/25 => 10000000 01110111 00101000 01000000

25 bit u l c nh

V bit th 26 = 1 nn ta cn 6 bit na phn bit a ch IP (khng t tin cho cu ni ny)

c tt c 26 = 64

Mi mng con c s IP bng nhau nn mi mng c 64/4 = 16 = 24

Ta dng 4 bit cui phn bit a ch IP, 2 bit tip theo ta phn bit 4 mng con vi nhau

4 mng con c subnet l:

128.119.40.64/28 - 10000000 01110111 00101000 01000000

128.119.40.80/28 - 10000000 01110111 00101000 01010000

128.119.40.96/28 - 10000000 01110111 00101000 01100000

128.119.40.112/28 - 10000000 01110111 00101000 01110000



P17: Gi s gi 2400 byte datagram vo trong mt link, link ny c MTU ca 700bytes. Cho rng

datagram gc c nh du vi nh danh s 422. C bao nhiu c c t o ra? Gi tr ca mi tr ng

kh c nhau trong datagram c t o ra c lin quan g n phn mnh?

Tr li:

Kch c ti a ca trng d liu trong mi phn on = 680 byte (700-20 byte IP header)

Do s phn on cn n = (2400-20)/680=4

Mi phn on s c mt nh danh s 422 v loi tr ra ci cui cng c kch c 700byte (bao gm IP header).

Gi cui cng c kch c 360byte(bao gm IP header). V tr ca 4 phn on s l 0, 85, 170, 255. Ba ci phn

on u tin c c bng 1, phn on cui cng c c = 0.

P18: Gi s datagram c gii h n l 1500byte (bao gm header) gia host ngun A v host ch B. Tha

nhn rng c 20byte IP header. Hi c bao nhiu datagram cn n gi 1 MP3 bao gm 5 triu byte ?.

gii th ch cho t nh to n ca b n.

Tr li:

1 file MP3 = 5000000byte

Tha nhn d liu mang theo TCP segment, vi mi TCP segment c 20byte ca header. Mi datagram c th

mang 1500-40=1460 byte ca file MP3

S datagram cn n = 5.106/1460 = 3425.

Datagram cui cng 960 +40 = 1000byte. Ch rng, ti khng c phn mnh host ngun khng th to mt

datagram ln hn 1500 v ci datagram s nh hn MTUs the links.

P20: Gi s b n va kh m ph ra s ca host ng sau mt NAT. B n theo di thy rng lp IP c nh

du vi mt s l nh danh, c c s lin tip trn mi IP packet. S nh danh ca IP packet u tin

c t o ra bi host l mt s ngu nhin, v s nh danh ca nhng IP packet tip theo l tip tc. Tha

nhn tt c IP packet t o ra bi hosts ng sau NAT c gi ra bn ngoi.

a. Da vo nhng theo di v tha nhn, b n c th ph t hin ra tt packet c gi bi NAT ra bn

ngoi. B n c th a ta mt vi k thut n gin- k thut ph t hin ra s duy nht ca host ng

sau NAT ? Bo cha cho cu tr li ca b n.

b. Nu s nh danh khng c ch ra lin tc nh ng ngu nhin, k thut ca b n nn c l g?

Tr li:

a. K t khi tt c IP packet c gi ra ngoi, vy chng ta c th s dng mt packet pht hin ra bn ghi

tt c IP packet to ra bi host ng sau mt NAT. Khi mi host to ra mt chui ca IP packet vi s lin

tip v mt nh danh khc hn nhau lc ban u, chng c th to nhm IP packet vi IDs lin tc trong

mt khi (m).S ca khi l s ca hosts ng sau NAT.

Tuy nhin, nu s nh danh khng ch ra lin tip m c ch ra ngu nhin, k thut c gi phn (a) s

khng lm c g, khi s khng th to nhm cho d liu c pht hin ra



Chapter5: Link Layer Review Question

R2. Nu tt c c c ng kt ni trn Internet u cung cp dch v truy n thng tin cy th giao thc TCP

c cn thit na hay khng ??? T i sao c v t i sao khng?

M t r t t IP qu t s u t

r t IP u tr t t . V IP

tr t k t CP t s tu u tr v t

t. CP v t t u t t t g tr t t . N r IP

t t t v tu .

R3. Nhng dch v no tng lin kt d liu c th cung cp cho tng m ng ? Nhng dch v no trong

t ng t vi IP, vi TCP ???

Framing ( IP and TCP), link access,reliable delivery (TCP), flow control (TCP), error detection ( IP and TCP),

error correction, full duplex( TCP)

R4. Gi s 2 nt bt u truy n d liu cng lc vi k ch th c packet l L bit(broadcast khng phi p2p)

vi tc R. K hiu s chm tr phin truy n gia 2 nt l d. Hi c xung t hay khng nu d



R10. T i sao gi tin truy vn ARP l i c gi broadcast ? t i sao gi tin tr li ARP l i ch gi n 1 a

ch MAC duy nht ?

V i ARP query thi t b t a ch MAC c t rf n. Khi gi l t tr l t

a ch MAC c a thi t b g tr A P qu r n ph i gi broadcast n a.

R11. Vi m ng Figure 5.19, router c 2 ARP modules, mi modules c bng ARP ca n. C th hay

khng 1 a ch MAC xut hin cng 1 bng ARP

K t v i m L 1 t p adapter g vs vi m t r t 1 a ch MAC duy nh t.

R14. Trong CSMA/CD sau ln xung t th 5. X c sut mt nt chn K=4 l bao nhiu? T ng ng vi

K=4 th tr l bnhieu vi m ng 10Mbps.

Sau l n th 5 xu t t s ch n K trong {0 1 2. . . 31} x su t 1/32.

Th i gian ch : 4*512*0,1 = 204,8 microseconds

R16. S VLANS maximum c th c cu hnh trn switch c cung cp giao thc 802.1 . Why??

v c 12 bit VLANS c nh danh nn 212 = 4096 VLANS



Chapter5: Link Layer Problem

P1: Gi s ni dung thng tin ca mt packet l 1 mu bit 1100 1011 1001 1101 v t nh chn l cng c

s dng. Gi tr ca tr ng cha c c bit chn l i vi tr ng hp s dng bit chn l 2 chi u l g? Cu

tr li ca b n nn l tr ng checksum c di nh nht c s dng.

Tr li:

1 1 1 0 1

1 0 1 1 1

1 0 0 1 0

1 1 0 1 1

0 0 0 1 1

P2: Cho mt v d cho thy kim tra bit chn l 2 chi u c th sa v ph t hin c 1 li bit. Cho mt v d

cho thy 2 li bit c th c ph t hin nh ng khng c sa.

Tr li:

Gi s chng ta bt u vi ma trn chn l 2 chiu nh phn:

0 0 0 0

1 1 1 1

0 1 0 1

1 0 1 0

Vi 1 li bit hng 2, ct 3, tnh chn l ca hng 2 v ct 3 gi b sai, th hin trong ma trn sau:

0 0 0 0

1 1 0 1

0 1 0 1

1 0 1 0

By gi gi s c 1 li bit hng 2, ct 2 v ct 3. Tnh chn l ca hng 2 gi ng. Tnh chn l ca ct 2 v

3 sai, nhng ta khng th pht hin ra li sai xy ra hng no.

0 0 0 0

1 0 0 1

0 1 0 1

1 0 1 0

V d trn cho thy l 2 li bit c th c pht hin (nu khng c sa).

P3: Gi s phn thng tin ca 1 packet (D trong phn 5.4) cha 10 bytes gm 8 bit nh phn khng du

ASCII miu t xu Link Layer. T nh Internet Checksum cho d liu ny.

Tr li:

01001100 01101001

+ 01101110 01101011

------------------ ------------

10111010 11010100

+ 00100000 01001100

------------------ ------------

11011011 00100000



+ 01100001 01111001

-----------------------------

00111100 10011010 (trn, sau wrap around) + 01100101 01110010

------------------ ------------

10100010 00001100

B ca tng ny l 01011101 11110011 P4: Xt vn tr c, nh ng thay v cha m nh phn ca c c s t 0 ti 9, gi s 10 bytes ny cha:

a. M nh phn ca c c s t 0 ti 10 b. M ASCII ca c c k t t A ti J (vit hoa) c. M ASCII ca c c k t t a ti j (vit th ng)

T nh Internet Checksum cho c c d liu trn.

Tr li:

a) tnh Internet Checksum, chng ta cng thm

cc gi tr di 16 bit.

00000001 00000010

00000011 00000100

00000101 00000110

00000111 00001000

00001001 00001010

-------------------------

00011001 00011110

B ca tng ny l 11100110 11100001.

b) tnh Internet Checksum, chng ta cng thm

cc gi tr di 16 bit:

01000001 01000010

01000011 01000100

01000101 01000110

01000111 01001000

01001001 01001010

-------------------------

01011000 01011111

B ca tng ny l 10100111 10100000

c) tnh Internet Checksum, chng ta cng thm cc gi tr di 16 bit:

01100001 01100010

01100011 01100100

01100101 01100110

01100111 01100111

01101000 01101001

-------------------------

11111001 11111101

B ca tng ny l 00000110 00000010.

P5: Xt b sinh 7 bit, G = 10011, gi s D c gi tr 1010101010. Gi tr ca R l g?

Tr li:

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.

Nu ta chia 10011 cho 1010101010 0000, ta c 1011011100, vi d l 0100. Ch rng, G = 10011 l chun

CRC-4-ITU.

P6: Xt nh bi trn, nh ng gi s D c gi tr :

a. 1001000101.

b. 1010001111 .

c. 0101010101.

Tr li :



a) ta c 1000100011, vi s d R=0101.

b) ta c 1011111111, vi s d R=0001.

c) ta c 0101101110, vi s d R=0010.

P7 : Chng ta tm hiu 1 s thuc t nh ca CRC. Vi b sinh G (= 1001) c cho trong Section 5.2.3, tr li

c c cu hi sau:

a. T i sao n c th ph t hin bt k li n no trong d liu D?

b. G trn c th ph t hin bt k 1 s l c c li bit khng? t i sao?

Tr li:

a. Khng mt tnh tng qut, gi s bit th I c lt, ni 0



P9: Chng minh l hiu sut ln nht ca ALOHA thun ty l 1/(2e). Ch : vn ny s rt d nu bn tr

li xong cu hi trn.

Tr li:

P10: Gi s c 2 node, A v B, s dng giao thc ALOHA chia khe. Gi s node A c nhi u d liu

chuyn h n node B, v x c sut truy n l i ca A l pA, ln h n ca B l pb.

a. Tm cng thc thng l ng trung bnh ca node A. Tng hiu sut ca giao thc vi 2 node l g?

b. Nu pA = 2pB th thng l ng trung bnh ca A c gp 2 ln ca B khng? T i sao? Nu khng, c th

chn pA v pb th no i u xy ra.

c. Tng qu t, gi s c N nodes, trong node A c x c sut truy n l i l 2p v c c node kh c c x c

sut truy n l i l p. T nh thng l ng trung bnh ca node A v ca c c node kh c.

Tr li:

a) Thng lng trung bnh ca A l pA(1-pB)

Tng hiu sut pA(1-pB) + pB(1-pA)

b) Thng lng ca A l pA(1-pB) = 2pB(1-pB) = 2pB 2(pB)2

Thng lng ca B l pB(1-pA) = pB(1-2pB) = pB 2(pB)2

Thng lng ca A khng gp 2 ln ca B

thng lng ca A gp 2 ln ca B th

pA(1-pB) = 2pB(1-pA)

pA = 2 (pA/pB)

c) Thng lng ca A l 2p(1-p)N-1, v cc node khc l p(1-p)N-2(1-2p)



P11: Gi s c 4 node ho t ng l A,B,C,D ang c nh tranh truy cp vo knh s dng ALOHA chia khe.

Gi s mi node c v h n packet gi. Mi node c gng gi vo mi slot vi x c sut p. Slot u c

nh s 1, slot 2 nh s 2, c th.

a. X c sut node A gi thnh cng ngay ln u tin t i slot 5 l bao nhiu?

b. X c sut vi node (c th l A, B, C hoc D) gi thnh cng t i slot 4?

c. X c sut ln gi thnh cng u tin xy ra l i slot 3?

d. Hiu sut ca h 4 node ny l bao nhiu?

Tr li:

a) (1-p(A))4p(A)

Trong :

p(A) = xc sut A thnh cng ti 1 slot

p(A) = p(A gi v B,C,D khng gi)

= p(A gi).p(B khng gi).p(C khng gi).p(D khng gi)

= p (1-p)(1-p)(1-p) = p(1-p)3

Do :

p(A thnh cng trong ln gi u ti slot 5)

= (1-p(A))4.p(A)

= (1-p(1-p)3)4.p(1-p)

3

b) p(A thnh cng slot 4) = p(1-p)3

p(B thnh cng slot 4) = p(1-p)3

p(C thnh cng slot 4) = p(1-p)3

p(D thnh cng slot 4) = p(1-p)3

p(A,B,C hoc D thnh cng ti slot 4) = 4 p(1-p)3

(bi v cc s kin ny loi tr ln nhau)

c) p(vi node thnh cng ti 1 slot) = 4p(1-p)3

p(khng c node no thnh cng ti 1 slot) = 1-4p(1-p)3

Do :

p(ln thnh cng u tin xy ra slot 3) = p(p(khng c node no thnh cng ti 2 slot u).p(vi node

thnh cng ti slot 3) = (1 - 4 p(1-p)3)24p(1-p)3

d) hiu sut = p(thnh cng ti 1 slot) = 4 p(1-p)3

P13: Gi s mt knh broadcast(qung b ) vi N node v transmission rate l R bps. Gi s knh

broadcast s dng giao thc hi vng (polling) (vi 1 node polling c thm vo) cho a truy cp. Gi s

khong thi gian t khi mt node hon thnh vic truy n n khi node tip theo c php chuyn (tr hi

vng polling delay) l dpoll. Gi s trong 1 vng hi, 1 node c php truy n ti a Q bit. Thng l ng ti

a ca knh broadcast ny l bao nhiu?

Tr li:

Thi gian 1 vng hi l

N(Q/R + dpoll)

S bit c truyn trong 1 vng hi l NQ. Thng lng ti a v th s l:

(

)

=



P14: gi s 3 m ng LANs kt ni vi nhau bi 2 router, xem hnh 5.8

a. t a ch IP cho tt c c c interface ny. Vi Subnet 1 th ta dng a ch c d ng 192.168.1.xxx,

Vi Subnet 2 th ta dng a ch c d ng 192.168.2.xxx, Vi Subnet 3 th ta dng a ch c d ng

192.168.3.xxx

b. t a ch MAC cho c c adapter.

c. Gi 1 IP datagram t host E ti host B.

Gi s tt c c c bng ARP u c cp

nht. Lit k c c b c, ging nh v d 1

router Section 5.4.2.

d. Lm l i cu c, gi gi s bng ARP host

gi ch a c d kin g, c c bng kh c u

c cp nht.

Tr li:

a,b: xem hnh:

c)

1. Bng Forwading E cho thy l datagram c nh tuyn ti interface 192.168.3.002.

2. Adapter E to 1 Ethernet packet vi a ch ch l 88-88-88-88-88-88.

3. Router 2 nhn packet v gii nn datagram. Bng forwarding router ny ch ra l datagram ny phi c nh

tuyn ti 198.162.2.002.

4. Router 2 gi Ethernet packet vi a ch ch 33-33-33-33-33-33 v a ch ngun 55-55-55-55-55-55 thng qua

interface ca n vi a ch IP l 198.162.2.003.

5. Tin trnh tip tc n khi packet n c host B.

d) ARP E lc ny phi xc nh a ch MAC ca 198.162.3.002. Host E gi 1 packet truy vn ARP trong 1

frame Ethernet qung b @@ Router 2 nhn packet truye vn v gi ti Host E mt packer ARP phn hi. Packet

ARP phn hi ny c mang i bi 1 frame Ethernet vi a ch ch Ethernet l 77-77-77-77-77-77.



P15: Xt hnh 5.38, gi chng ta thay v tr ca router gia subnet 1 v 2 bng 1 switch S1, v t tn router

gia subnet 2 v 3 l R1.

a. Gi 1 IP datagram t host E ti host F. Host E c phi hi router R1 gip forward datagram i

khng? T i sao? Trong Ethernet frame cha IP datagram, a ch IP ch, ngun v a ch MAC l g?

b. Gi s E mun gi 1 IP datagram ti B, v gi s b nh cache ARP ca E khng cha a ch MAC

ca B. B c phi gi 1 ARP truy vn tm a ch MAC ca B khng? T i sao? Trong Ethernet frame

(cha IP datagram gi ti B) c chuyn ti router 1, a ch ch, ngun v a ch MAC l g?

c. Gi s host A mun gi 1 IP datagram ti host B, b nh cache ARP ca A khng cha a ch MAC

ca B v b nh cache ARP ca B khng cha a ch MAC ca A. Gi s l bng forwarding ca switch S1

ch cha entries cho host B v router 1. V th, A s broadcast mt ARP request message. Switch S1 s thc

hin hnh ng no khi n nhn c ARP message? Router R1 c cng nhn c ARP request message

khng? Nu nhn c, R1 c forward message ti Subnet 3 khng? Ngay khi host B nhn c ARP

request message, n c gi l i host A 1 ARP message phn hi khng? T i sao? Switch S1 s lm g ngay khi

n nhn c ARP message phn hi t host B?

Tr li:

a) Khng. E c th kim tra tin t subnet (subnet prefix) ca a ch IP ca host F, sau nhn ra l F cng

1 mng LAN. Do , E s khng gi packet ti router mc nh R1.

Ethernet frame t E ti F:

IP ngun = a ch IP ca E

IP ch = a ch IP ca F

a ch MAC ngun = a ch MAC ca E

a ch MAC ch = a ch MAC ca F

b) Khng, v chng khng trong cng 1 mng LAN. E c th nhn ra nhiu ny bng cch kim tra a ch

IP ca B.

Ethernet frame t E ti R1:


IP ch = a ch IP ca B


a ch MAC ch = a ch MAC ca interface ca R1 kt ni ti Subnet 3

c) Switch S1 s broadcast Ethernet frame thng qua c 2 interface ca n as a ch ch ca ARP frame nhn

c l 1 a ch broadcast. V n nhn ra l A nm trong Subnet 1 ang kt ni ti S1 ti interface kt ni ti

Subnet 1. V S1 s cp nht bng forwarding ca n, bao gm c entry cho host A.

C, router R1 cng nhn c ARP request message, nhng R1 khng forward message ti Subnet 3.

B khng gi ARP message truy vn hi a ch MAC ca A, a ch ny c th nhn c t message truy vn

ca A.

Khi Switch S1 nhn message phn hi t B, n s thm 1 entry cho host B vo bng forwarding, sau th

frame va nhn c vi a ch ch l host A trong cng 1 interface nh host B (v d: A v B cng nm trong 1

mng LAN).

P16: Vn vn nh trn, nh ng gi s by gi router gia subnet 2 v 3 c thay bng 1 switch. Tr li

cu hi t a n c vi gi thit mi.

Tr li:



Gi switch gia subnet 2 v 3 l S2. Theo bi, router R1 gia subnet 2 v 3 gi c thay th bi switch S2.

a) Khng. E c th kim tra tin t subnet (subnet prefix) ca a ch IP ca host F, v nhn ra l F trong

cng 1 mng LAN. V th, E s khng gi packet ti S2.

Ethernet frame t E ti F:


IP ch = a ch IP ca F


a ch MAC ch = a ch MAC ca F

b) C, bi v E mun bit a ch MAC ca B. V th, E s gi 1 packet ARP truy vn vi a ch MAC ch

l a ch broadcast.

Packet truy vn ny s c broadcast li bi switch 1, v cui cng c host B nhn.

Ethernet frame t E ti S2:


IP ch = a ch IP ca B

MAC ngun = a ch MAC ca E

MAC ch = a ch MAC broadcast: FF-FF-FF-FF-FF-FF.

c)

Switch S1 s broadcast frame Ethernet thng kia c 2 interface ca n as a ch ch ca ARP frame nhn

c l 1 a ch broadcast. V n nhn ra l A nm trong Subnet 1 ang kt ni ti S1 ti interface kt ni ti

Subnet 1. V S1 s cp nht bng forwarding ca n, bao gm c entry cho host A.

C, router S2 cng nhn c ARP request message ny, v S2 s broadcast packet truy vn ny ti tt c cc

interface ca n.

B khng gi message truy vn hi a ch MAC ca A, a ch ny c th nhn c t message truy vn ca

A.

Khi S1 nhn c tin nhn phn hi ca B, n s thm 1 entry cho host B vo bng forwarding, sau th

frame va nhn c vi a ch ch l host A trong cng 1 interface nh host B (v d: A v B cng nm trong 1

mng LAN).

P17: Nh l i v giao thc CSMA/CD, adapter i K*512 bit times sau khi c xung t, K c rt ngu

nhin. Vi K = 100, adapter phi i bao lu ti khi tr l i b c 2 vi 10Mbps Ethernet? vi 100Mbps

Ethernet?

Tr li:



i 51200 bit times. Vi 10Mbps, thi gian i l:

Vi 100Mbps, thi gian i l 512 sec.

P18: Gi s node A v B trn cng 1 bus 10Mbps Ethernet, v tr lan ta gia 2 node l 325 bit times. Gi

s node A bt u gi 1 frame v, tr c khi kt thc, node B bt u gi 1 frame. A c kt thc vic gi

tr c khi n ph t hin l B gi khng? T i sao? Nu cu tr li l c, th A hiu nhm l frame ca

n c chuyn thnh cng m khng c xung t. Gi : gi s t i thi im t = 0 bit times, A bt u gi

1 frame. Trong tr ng hp xu nht, A gi 1 frame c k ch th c nh nht l 512 + 64 bit times. Nn A s

kt thc vic gi t i t = 512 + 64 bit times. Th nn, cu tr li l khng, nu t n hiu ca B ti c A tr c

bit time t = 512 + 64 bits. Trong tr ng hp xu nht, khi no th du hiu ca B ti c A?

Tr li:

Ti thi im t = 0, A bt u gi. Ti t = 576, A s kt thc vic gi. Trong trng hp xu nht, B bt u gi ti

thi im t = 324, l thi im ngay trc khi bit u tin ca frame ca A ti c B. Ti thi im t = 324 + 325

= 649, bit u tin ca B ti c A. V 649 > 576, A kt thc vic gi trc khi pht hin ra l B gi. Nn A

hiu nhm l frame ca n c gi thnh cng m khng c xung t.

P19: Gii th ch t i sao l i i hi k ch th c frame l nh nht vi Ethernet. V d, 10Base Ethernet yu

cu k ch th c nh nht ca frame bt buc l 64 bytes (nu b n lm vn tr c, b n c th hiu c

l do). Gi gi s khong c ch gia 2 ends of Ethernet LAN l d. B n c th xy dng 1 cng thc tm

k ch th c frame nh nht m 1 Ethernet packet cn khng? Da vo l do ca b n, k ch th c ti thiu

ca frame cn cho 1 Ethernet ko di 2 km l bao nhiu?

Tr li:

Theo l do vn trc, chng ta cn m bo l that one end of Ethernet c th pht hin xung t trc khi n

hon thnh vic chuyn frame. V th, i hi kch thc frame phi l nh nht.

BW biu th bng thng ca Ethernet. Xt trng hp xu nht cho vic pht hin xung t ca Ethernet:

1. Ti t = 0 (bit times): A gi 1 frame

2. Ti t = dprop -1(bit times): B gi 1 frame ngay trc khi n cm nhn c bit u tin ca A.

3. Ti t = 2dprop-2(bit times): nu A kt thc vic truyn bit cui cng ca n ngay trc khi frame ca B n

A, th A s khng th pht hin xung t trc khi kt thc vic truyn frame ca n. Do , A c th pht hin

xung t trc khi kt thc vic truyn, i hi kch thc ti thiu ca frame >= 2dprop-1 (bit times).

Gi s tc lan ta tn hiu trong 10BASE-T Ethernet l 1.8*108m/sec. Vi dprop = d/(1.8*108)*BW ( y,

chng ta cn convert tr lan truyn t giy sang bit times ph hp vi c trng ca link Ethernet). Hoc ta

chn 2*d/(1.8*108)*BW(bits). Nu d = 2km, th cn kch thc ti thiu ca frame l 222 bits.

P20 : Gi s b n c th tng tc link ca c p Ethernet ca b n, s nng cp ny nh h ng th no ti

k ch th c yu cu ti thiu ca packet ? Nu b n nng cp c p ca b n ti 1 tc cao h n v nhn ra l

b n k th thay i k ch th c packet, b n s lm g gi cho qu trnh ho t ng ng ?

Tr li :



Da vo gii php vn trc, bn bit l link c tc cao hn th i hi kch thc yu cu ti thiu ca

packet ln hn. Nu bn khng th thay i kch thc packet, th bn c th thm cc switch hoc router phn

vng mng LAN ca bn, m bo l kch thc mi vng mng LAN l nh so vi kch thc nh ca

frame.

P21 : Gi s node A v B cng 1 Ethernet bus 10Mbps, v tr lan ta gia 2 node l 245 bit times. Gi s

A v B gi frame cng lc, frame xung t, sau A v B chn gi tr kh c nhau ca K trong thut to n

CSMA/CD. Gi s khng c node kh c ho t ng, ln gi l i ca A v B c b xung t khng ? Gi s A v

B bt u gi t i t = 0 bit times. Chng cng ph t hin xung t t i = 245 bit times. Chng gi xong t n

hiu tc nghn t i t = 245+48 = 293 bit times. Gi s KA = 0 v KB = 1. T i thi im no th B sp xp gi

l i ? T i thi im no A bt u gi ? (Ch : node phi i 1 knh rnh ri sau khi quay tr l i b c 2

giao thc nhn.) T i thi im no th t n hiu ca A ti c B ? B c tr hon vic gi vo thi im

c sp xp ca n ?

Tr li :

Time,t Event

0 A v B bt u gi

245 A v B pht hin xung t

293 A v B kt thc vic gi tn hiu tc nghn

293+245=538 Bit cui cng ca B ti A; A pht hin 1 knh ri

538+96=634 A bt u gi

293+512=805 B quay li bc 2

B phi cm nhn c knh ri for 96 bit times trc khi n gi

634+246=879 frame A gi ti c B

V frame gi li ca A ti B trc thi im B sp xp gi li (805+96), B tr hon gi trong khi A gi li. V

th A v B khng xung t. Th nn nhn t 512 xut hin trong thut ton exponential backoff l ln.

P22: Xt 100BASE-T Ethernet 100 Mbps vi tt c c c node kt ni trc tip ti 1 hub. c hiu sut 0.50,

khong c ch ln nht gia 1 node v hub nn l bao nhiu? Gi s chi u di ca 1 frame l 1000 bytes v

khng c lp l i (repeater). Khong c ch ln nht ny c m bo l node A ang gi c th ph t hin l

c c node kh c gi khi A ang gi khng? T i sao? Khong c ch ln nht so vi chun 100 Mbps thc

t th th no? Gi s tc lan truy n t n hiu trong 100BASE-T Ethernet l 1.8*108 m/sec.

Tr li :

Kch c 1 frame l 1000*8 + 64 = 8064 bit, v phn preamble 64 bit c thm vo frame.

Chng ta cn 1/(1+5a) = 0.5 hoc tng ng, a = 0.2 = tprop/ttrans.tprop = d/(1.8*108) m/sec v ttrans = (8064

bit)/(108 bit/sec) = 80.64 sec. Tnh ra ta c d = 2903 mt.

Vi ni gi l A, pht hin xem c ni khc gi trong thi gian A ang gi, ttrans phi ln hn 2tprop = 2*2903

m/1.8*108 m/sec = 32.36 sec. V 32.26 < 80.64, A s pht hin tn hiu ca B trc khi qu trnh gi ca n

hon tt.

P23 : Gi s 4 node, A, B, C v D cng kt ni vo 1 hub thng qua c p Ethernet 10Mbps. Khong c ch

gia hub v 4 node ln l t l 300m, 400m, 500m v 700m. Giao thc CSMA/CD c s dng cho

Ethernet ny. Gi s tc lan truy n t n hiu l 2*108 m/sec.



a. Chi u di ti thiu yu cu ca frame l bao nhiu ? Ti a l bao nhiu ?

b. Nu tt c frame u c chi u di 1500 bit, tm hiu sut ca Ethernet ny.

Tr li :

a) Chiu di ti thiu yu cu ca frame l :

2*dprop*BW = 2*(500+700)/(2*108)*10*10

6 = 120 bit

Khng c chiu di ti a yu cu ca frame.

b) Hiu sut:

1/(1+5*dprop/dtrans) = 1/(1+5*120/2/1500) = 0.83

P25: Gi s c 2 node A,B hai u ca 1 cable di 800 m. v c 1 mt frame 1500 bit cn chuyn qua cable

. C 2 node u truy n t i thi im t=0. Gi s c 4 ln lp gia Av B. mi ln chn c 20 bit tr. Gi s

ng truy n l 100 Mbps v CSMA/CD with backoff intervals of multiples of 512 bits is used. Sau ln

xung t u tin, A cho kt qu K=0, B cho kt qu K=1 in the exponential backoff protocol. Khng c t n

hiu no b nhiu,t he 96-bit time delay.

a. tr lan ta t A ti B l? Gi s tc lan ta l 2.108 m/s.

b. Gi tin ca A chuyn xong khi no?

c. Gi s ch A cn gi gi tin, v repeaters thay thnh switch. Gi s mi switch c 20-bit processing

delay in addition to a store-and-forward delay.

Tr li:

a.

b. First note, the transmission time of a single frame is give by 1500/(100Mbps) =15 micro sec, longer than

the propagation delay of a bit.

- At time t=0, both A and B transmit.

- Vo thi im 4.8 micro giy A, B pht hin xung t, v kt thc.

- Vo thi im 9.6 micro giy bit cui cng ca gi tin li t B ti A

- Vo thi im 14,4 micro giy bt u tin ca frame c gi li n B.

- Vo thi im 14,4 +15 gi tin ca A n B.

c. ng i c chia lm 5 phn, tr lam ta ca mi phn l 0.8.

tr t host A ti switch 1 l 15 = tr truyn. switch 1 s i 16 =15+0.8+0.2( tr tin trnh) cho

n khi n c gi tin chuyn ti switch 2. Tng t cho cc switch cn li. vy tng thi gian l: 16*4 +15 +

0.8 =79.8 micro giy.

P26: Trong chun Internet, ng i gi dng 96 ln bit gia 2 ln gi khung hnh lin tip. Thi gian t m

dng ny c gi l khong c ch gia c c khung hnh( inter-frame gap) v n c s dng cho php 1

thit b tip nhn hon thnh tin trnh nhn va chun b cho ln nhn frame k tip. T khi chun

Ethernet c quy nh, c 1 ci tin to ln trong cng ngh bao gm c tc ca vi x l, b nh v t

l Ethernet. Nu c c tiu chun c vit l i, s ci tin ny t c ng nh th no n khong c ch gia c c

khung hnh ?

Solution:



Nng cao tc ca vi x l m ch khong cch gia cc khung hnh c th c rt ngn, bt u t lc

n mt t thi gian hn hon thnh tin trnh nhn frame. Nhng, nng cao tc cp Ethernet m ch khong

cch gia cc khung hnh phi c tng ln.

P 27.

Cung cp a ch MAC v a ch IP cho interface t i Host A , tt c c c router v Host F . Gi s Host A

gi datagram n Host F .Hy ch ra a ch MAC ngun ( source ) v a ch MAC ch (destination )

trong frame ng gi IP datagram ny nh 1 frame c truy n .:

I , T A n left router .

II , T left router n right router .

III , T right router n F .

Ngoi ra , cn ch ra a ch IP ngun v a ch IP ch t i mi im .

Gii :

I , A to left router :

Source MAC : 00-00-00-00-00-00

Destination MAC : 22-22-22-22-22-22 .

Source IP : 111.111.111.001

Destination IP : 133.333.333.003

II , T left n right router :

Source MAC : 33-33-33-33-33-33

Destination MAC : 55-55-55-55-55-55 .

Source IP : 111.111.111.001




III , T right router n F :

Source MAC : 88-88-88-88-88-88

Destination MAC : 99-99-99-99-99-99

Source IP : 111.111.111.001


P28: Gi s by gi router tn cng bn tr i trong hnh 5.38 c thay th bng 1 switch. Host A,B,C v D v router bn phi l c c sao kt ni vo switch ny. Cung cp cho a ch MAC ngun v ch trong frame ng gi IP datagram nh 1 frame c truy n ( I ) t A n switch, ( II) t switch n router bn phi, ( III) t router bn phi n F. Ngoi ra cung cp a ch IP ngun v ch trong IP datagram c ng gi

trong frame t i nhng thi im ny.

Solution

I, T A n switch:

a ch Mac ngun: 00-00-00-00-00-00

a ch Mac ch: 55-55-55-55-55-55

IP ngun : 111.111.111.001

IP ch: 133.333.333.003

II, T switch n router bn phi

a ch Mac ngun: 00-00-00-00-00-00

a ch Mac ch: 55-55-55-55-55-55

a ch IP ngun: 111.111.111.001

a ch IP ch: 133.333.333.003

III, T router bn phi n F:

a ch Mac ngun: 88-88-88-88-88-88

a ch Mac ch: 99-99-99-99-99-99

a ch IP ngun: 111.111.111.001

a ch IP ch: 133.333.333.003

P29: Xem xt hnh 5.26. Gi s tt c c c c lin kt

u l 100 Mbps. Tng thng l ng ti a c th

t c trong 9 host v 2 server trong m ng ny

l bao nhiu ? B n c th gi nh bt k host hay

server no cng c th gi n bt k 1 host hay

server . T i sao?

Nu tt c 11=9+2 nt gi d liu tc ti a c th

l 100 Mbps, tng thng lng t c l 11*100 =

1100 Mbps



P 30 .Gi s 3 departmental Switch c thay th bi c c hub . Tt c link vn l 100 Mbps . Tng thng

l ng t c gia 9 host v 2 server trong m ng l bao nhiu ? Host v server c th gi n host v

server kh c c hay khng ? T i sao ?

Mi deparmental hub l 1 tn min duy nht c thng lng ti a l 100Mbps . Cc link kt ni vo web server

hay mail server cng c thng lng ti a l 100Mbps . Do , nu 3hub v 2 server ny gi data mc ti a l

100Mbps th thng lng tng cng l 500Mbps . Thng lng ti a ny c th t c gia 11 end system .

Cc host v server c th send n host v server khc .

P31: Gi s tt c c c switch trong Figure 5.26 c thay bi hub. Tt c c c link l 100 Mbps. Tr li c c

cu hi trong P29.

Tr li:

Tt c cc thit b u cui u nm trong cng 1 min xung t. Trong trng hp ny, tng ti a thng lng

100 Mbps ca tp hp ny l c th xy ra gia 11 thit b u cui. (dch xong k hiu g )

P32 : Xt ho t ng hc v switch (nghe ngu qu =)))) Figure 5.24. Gi s l (i)B gi 1 frame ti E, (ii) E

gi tr l i 1 frame ti B, (iii) A gi 1 frame ti B, (iv) B gi tr l i 1 frame ti A. Bng switch ban u trng

khng. Cho bit tr ng th i ca bng switch tr c v sau mi s kin. Vi mi s kin, cho bit ng i (s)

m mi mi frame s c forward i, v gii th ch ngn gn cu tr li.

Tr li :

Hnh

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