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    Q1. Which one of the following is not a client server application?(A) Internet chat (B) Web browsing (C) E-mail (D)Ping C !"#"

    Ans$ DExplanation: Ping is %se& for 'nowing stat%s of a host b another host$

    PHYSICAL LAYER

    Q2. A serial transmission i %ses * information bits+ ! start bits+ # stop bit an& # pariforeach character$ A s nchrono%s transmission ! %ses , eight bit s nc characters follow,"eight bit information characters$ If the bit rate is #!"" bits secon& in both cases+ whathetransfer rates of i an& !?(A) #"" characters sec+ #., characters sec(B) *" characters sec+ #,/ characters sec

    (C) #"" characters sec+ #,/ characters sec(D)*"characters sec+#.,characters secI !""0

    Ans. CExplanation: in serial transmission * bit of act%al &ata is present in *1!1#1#(2#!) bits of eatransmission$ In one secon& #!"" bits are transferre&$ In #!"" bit onl #!""3(* #!) bitact%al &ata is present+ o act%al transfer rate is #!""3(* #!)2*"" bits 2#"" b tes2characters$

    imilarl in s nchrono%s transmission act%al &ata rate is2#!""3(," ,,) bps2#"4" bps2# b tes per secon&s$

    Q3. $ In the waveform (a) given below+ a bit stream is enco&e& b 5anchester scheme$ hsame bit stream is enco&e& in &ifferent co&ing scheme in waveform (b)$ he bit stream anco&ing scheme are6

    (a)

    (b)

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    (a) #""""#"### an& &ifferential 5anchester respectivel $(b) "####"#""" an& &ifferential 5anchester respectivel $(c) #""""#"### an& integral 5anchester respectivel $(&) "####"#""" an& integral 5anchester respectivel $

    Ans. a

    Explanation: in 5anchester represents # an& represents "$

    # " " " " # " # # # #

    o &ata is #""""#"#### an& for this &ifferential 5anchester enco&ing is

    # " " " " # " # # # #

    DATA LINK LAYER

    Q4. In a &ata lin' protocol+ the frame &elimiter flag is given b "###$ Ass%ming thst%ffing is emplo e&+ the transmitter sen&s the &ata se7%ence "###"##" as(A) "##"#"## (B) "##"#"##"(C) "###"##"" (D)"##"#"##"" I !""0

    Ans. DExplanation: three consec%tive ones are %se& for &elimiter so whenever in &ata

    consec%tive one comes st%ff a 8ero after them$Data is "###"##"After st%ffing "##"#"##""

    Q . Consi&er the following message 5 2 #"#"""##"#$ he c clic re&%n&anc chec' (Cfor this message %sing the &ivisor pol nomial :. 1 : 0 1 : ! 1 # is6(A) "###"(B) "#"##

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    (C) #"#"#(D)#"##" I !"".

    Ans. AExplanation: generator pol nomial is of &egree . so appen& . "s to the en& of &ata an&

    &ivi&e new &ata b generator pol nomial$ :.

    1 :0

    1 :!

    1 # 2##"#"## # " # " # ) # " # " " " # # " # " " " " " (

    # # " # " #

    " # # # " # #

    # # " # " #

    " " # # # " # "

    # # " # " #

    " " # # # # # "

    # # " # " #

    " " # " # # " "

    # # " # " #

    " # # " " # "

    # # " # " #

    " " " # # # "9emain&er is "###"

    Not!: if generator is divisible by x 3+1 then CRC can detect all odd number of errors.

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    Q". he message ##""#""# is to be transmitte& %sing the C9C pol nomial :, 1 # to protect itfrom errors$ he message that sho%l& be transmitte& is6(A) ##""#""#""" (B) ##""#""#"##

    (C) ##""#"#" (D) ##""#""#""## C !"";Ans. BExplanation: : , 1 # is generator pol nomial of &egree ,an& :, 1 # 2 #""#A&& three 8eros to the right of &ata+ new &ata is ##""#""#"""

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    Q#. What is the minim%m n%mber of bits (l) that will be re7%ire& to represent the se7n%mbers &istinctl ? Ass%me that no time gap nee&s to be given between transmissionsframes$(A) l2! (B) l2, (C) l20 (D) l2. C !""4

    Ans. DExplanation: :2#""" #"/ secon&s2# ms5a:im%m n%mber of frames that can be transmit to ma:imall pac' them is2( :1! p) :2 (!.1#) #2!/ which is win&ow si8e5inim%m se7%ence n%mbers re7%ire& 2 !/5inim%m n%mber of bits re7%ire& for se7%ence n%mber is .$

    Q1#. %ppose that the sli&ing win&ow protocol is %se& with the sen&er win&ow si8e of =+ where= is the n%mber of bits i&entifie& in the earlier part an& ac'nowle&gements are alwa s bac'e&$ After sen&ing != frames+ what is the minim%m time the sen&er will have to wait be

    starting transmission of the ne:t frame? (I&entif the closest choice ignoring the fr processing time$)(A) #/ms (B) #*ms (C) !"ms (D)!!ms C !""4

    Ans. CExplanation: given Win&ow si8e is !=2! . 2,!5a:im%m frame that sen&er can transmit in one 9 is !/+ when it is transmitting !;th frame itwill get ac' for #st frame + imilarl + when it is transmitting ,!th frame it will get ac' for /thframe$ After transmitting ,!th frame +sen&er have to wait %ntil ac'nowle&gement for !/th frameis arrive&$ Which is e7%als to (transmission time of !/ frame - transmission time of / framems

    Q$. Consi&er a parit chec' co&e with three &ata bits an& fo%r parit chec' bits$ hreeco&e wor&s are "#"#"##+ #""##"# an& ###"""#$ Which of the following are also co&e worI$ ""#"### II$ "##"##" III$ #"##"#" I>$ "###"#"(A) I an& III(B) I+ II an& III(C) II an& I>(D) I+ II+ III an& I> I !""0

    Ans. Dhis will be &isc%sse& in C or DI@I A=

    Q%.$ A broa&cast channel has #" no&es an& total capacit of #" 5bps$ It %ses polling for maccess$ nce a no&e finishes transmission+ there is a polling &ela of *" micro secon&s tone:t no&e$ Whenever a no&e is polle& + it is allowe& to transmit a ma:im%m of #""" bma:im%m thro%ghp%t of broa& cast channel is(a) # 5bps (b) #"" ## 5bps (c) #"5bp& (&) #"" 5bps I !"";

    Ans. b

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    Explanation: :2#""" b tes #"5bps 2*"" s$Dela beca%se of polling is 2 *" sEfficienc of channel + e 2transmission &ela (total &ela ) 2*"" (*""1*")2 #" ##5a:im%m thro%ghp%t is 2(#" ##) 3 #" 5bps2 #"" ## 5bps

    Q1&$Which one of the following statements is A= E?(A) Pac'et switching lea&s to better %tili8ation of ban&wi&th reso%rces than circ%it switc(B) Pac'et switching res%lts in less variation in &ela than circ%it switching$(C) Pac'et switching re7%ires more per pac'et processing than circ%it switching$(D) Pac'et switching can lea& to reor&ering %nli'e in circ%it switchiI !""0

    Ans. '

    Pa()!t s*i(+in,- !la/ 0a iation is o ! !(a s! o5 t+! sto ! an 5o *a !(+anis .

    6lo* Cont ol 7!(+anis in DLL

    Q11. tation A nee&s to sen& a message consisting of 4 pac'ets to tation B %sing a sliwin&ow (win&ow si8e ,) an& go-bac'-n error control strateg $ All pac'ets are rea&

    imme&iatel available for transmission$ If ever .th pac'et that A transmits gets lost (b%ac's from B ever get lost)+ then what is the n%mber of pac'ets that A will transmit for sen&the message to B?(A) #! (B) #0 (C) #/ (D) #* C !""/

    Ans. CExplanation: win&ow si8e is ,+so ma:im%m , pac'ets can be remaine& %nac'nowle&ge&go bac' n if ac'nowle&ge for a pac'et is not receive& then pac'ets after that pac'et is alretransmitte&$

    rame se7%ence for 4 frame is shown below$ rame with bol& se7%ence n%mber gets los1 2 3 4 " # " # $ % # $ % %

    Discar&e& pac'ets

    Q12. ost A is sen&ing &ata to host B over a f%ll &%ple: lin'$ A an& B+ are %sing thewin&ow protocol for flow control$ he sen& an& receive win&ow si8e are . pac'ets each pac'ets ( sent onl from A to B) are all #""" b tes long an& the transmission time for s%

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    pac'et is ." s$ Ac'nowle&gement pac'ets are ver small (sent onl from B to A) an& re7%ver negligible time$ he propagation &ela over the lin' is !"" s$ What is the ma:imachievable thro%ghp%t in this comm%nication?(A) ;$/4 : #"/ b tes per secon& (B) ##$## : #"/ b tes per secon&(C) #!$,, : #"/ b tes per secon& (D) #.$"" : #"/ b tes per secon& C !"",

    Ans $ BExplanation: ransmission rate + : is2 #""" b tes ." s2*"""bits ." s2#/" 5bpsEfficienc 2 .3 ." (."10"")2!." 0."2. 45a:im%m achievable thro%ghp%t2(. 4)3#/" 5bps2**$**5bps2##$## : #"/ b tes per secon&

    Q13. Consi&er a #"" 'm long cable r%nning at #$.0 5bps at propagation spee& !C , (i$e$ spee& of light2,3#"* meter secon&)$ At ma: how man bits can be fee& into the cable$

    Ans $ ;;" bitsExplanation 6 propagation &ela is 2 #"" 'm (!C ,)2#". meter (!3#"*meter secon&) 2.3#"-0 secon&s 2"$. ms5a:im%m n%mber of bits(it is also calle& capacit of a cable) that can be fee& into #"" mcable is2#$.0 5bps 3"$. ms 2 ;;" bits

    Q14. he ma:im%m win&ow si8e for &ata transmission %sing the selective reFect protoco bit frame se7%ence n%mbers is6

    (a) !n

    (b) !n-#

    (c) !n

    G # (&) !n-!

    C !"".Ans. bIf the 7%estion is for sli&ing win&ow+ then option aIf 7%estion is for @B

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    Q1". he &istance between two stations 5 an& < is = 'ilometers$ All frames are H bits long$ propagation &ela per 'ilometer is t secon&s$ =et 9 bits secon& be the channel caAss%ming that processing &ela is negligible+ the minim%m n%mber of bits for the sn%mber fiel& in a frame for ma:im%m %tili8ation+ when the sli&ing win&ow protocol is %(A) =og!((!=t91!') ') (B) =og!(!=t9 ')

    (C) =og!((!=t91') ') (D) =og!((!=t91') !') C !"";Ans. AExplanation: 5a:im%m win&ow si8e(Wma:)2 ( :1! p) :$

    ere p 2 =t (loo' at the %nits of propagation &ela ):2 H 9

    Wma:2( :1!3 p) : 2 (H 9 1 !=t) (H 9)2(!=t91') ' =et minim%m n n%mber of bits re7%ire& then !n 2 Wma:2(!=t91!') ' n2log!((!=t91!') ')

    Q1$. In a sli&ing win&ow A9 scheme+ the transmitter s win&ow si8e is < an&receiver swin&ow si8e is 5$ he minim%m n%mber of &istinct se7%ence n%mbers reens%re correct operation of the A9 scheme is6(A) min(5+

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    (C) #/" b tes(D) #/" bits I !"".

    Ans. DExplanation:

    Efficienc of stop an& wait 2 # (#1!a)If # (#1!a) 2 "$. 2M !3 p 2 :$ 2M = 2 !3B3 p 2#/" bits

    Q21. n a wireless lin'+ the probabilit of pac'et error is "$!$ a stop an& wait protocol is %stransfer &ata across the lin'$ he channel con&ition is ass%me& to be in&epen&etransmission to transmission$ What is the average n%mber of transmission attempts re7%transfer #"" pac'ets?(a) #"" (b) #!. (c) #." (&)!""I !""/

    Ans. bExplanation: error rate "$! +In stop an& wait protocol sen&er will transmits#""3(#1"$!1"$!! 1"$!, 1"$!0 1NNNNN) pac'ets2 #"" 3(# (#-$"!))2#"" "$* 2 #!. (s%m of infinite @$P$ is a (a-r))

    7AC S la/!

    Q22. Which of the following is < tr%e with respect to a transparent bri&ge an& a ro%ter(a) Both bri&ge an& ro%ter selectivel forwar& &ata pac'ets(b) A bri&ge %ses IP a&&resses while a ro%ter %ses 5AC a&&resses(c) A bri&ge b%il&s %p its ro%ting table b inspecting incoming pac'ets(&)A ro%ter can connect between a =A< an& a WA< C !""0

    Ans. BExplanation: A bri&ge %se 5AC a&&resses(D== la er) an& ro%ter %ses IP a&&resses (

    la er)$

    Q23. Which of the following statements is A= E regar&ing a bri&ge(A) Bri&ge is a la er ! &evice(B) Bri&ge re&%ces collision &omain(C) Bri&ge is %se& to connect two or more =A< segments

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    (D) Bri&ge re&%ces broa&cast &omaI !"".

    Ans. DBri&ge will not &iscar& D== broa&cast pac'ets+ so it cannot re&%ce BC &omain$

    Q24. In a D5 me&i%m access control b%s =A

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    whichthis protocol can s%pport+ if each host has to be provi&e& a minim%m thro%ghp%frames per time slot?(A) # (B) ! (C) , (D)0 I !""0

    Ans. B

    Explanation: each host have thro%ghp%t of $#/ frames per time slot$ If there are n host total thro%ghp%t will be "$#/n + which is probabilit at which a frame can transmit s%cceo b the form%la in previo%s 7%estion

    np(#-p)n-# 2"$#/nn 3 "$! 3 ("$*)n-# 2"$#/nn2! satisf above con&ition$

    Q2#. he minim%m frame si8e re7%ire& for C 5A CD base comp%ter networ' r%nning aton !"" m cable with a lin' spee& of ! : #"* m s is(a) #!. b tes (b) !." b tes (c) ."" b tes (&) none of these

    I !""*Ans. bExplanation: se :2! p

    Q.2$ A an& B are the onl two stations on an Ethernet$ Each has a stea& 7%e%e of frasen&$ Both A an& B attempt to transmit a frame+ colli&e+ an& A wins the first bac'off racen& of this s%ccessf%l transmission b A+ both A an& B attempt to transmit an& coll

    probabilit that A wins the secon& bac'off race is(a) "$. (b) "$/!. (c) "$;. (&)#$"C !""0

    Ans. bExplanation: Sol0! it in t+! (lass.

    Q2%. # 'm long #" 5bps C 5A CD =A< has propagation spee& !""m s+ &ata frame is ! bit long incl%&ing ,! bits of hea&er +chec's%m an& other$ he first big slot after s%transmission is reserve for a receiver to capt%re channel an& to sen& ,! bit of ac'nowle&ge

    $ What is effective &ata rate e:cl%&ing overhea& an& ass%ming there is no collision$Ans6

    olve it if o% can+ else forget it$ Hee&a hai$

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    (A) #"""" bits(B) #"""" b tes(C) .""" bits(D).""" b tes I !"".

    Ans. AExplanation: p2(#'m) ( ! : #"*m s)2.: #"-/ econ&s=et is minim%m pac'et si8e5inim%m frame si8e can be fo%n& b form%la :2! p

    #@bps 2 !3.3#"-/2#"4 : #" -. 2#"0 bits

    Q31. A !'m long =A< has #";

    bps ban&wi&th an& %sers C 5A CD$ he signal travels althe wire at ! : #"* m s$ What is the minim%m pac'et si8e that can be %se& on this networ'?(A) ." b tes (B) #"" b tes(C) !"" b tes (D) none of the aboveC !"",

    Ans. DExplanation: minim%m pac'et si8e for a C 5A CD =A< is is the frame which cover who9 (ro%n& trip time)$i$e$ :2! p

    p2! : #" , (! : #"*) secon&s2 #"-. secon&s=et = bits be minim%m si8e of frame+ then :2 #"; secon&s

    :2! p#"; 2 ! : #" -. 2!"" bits 2(!"" *) b tes

    Q32. %ppose the ro%n& trip propagation &ela for a #" 5bps Ethernet having 0*-bit Famsignal is 0/$0 ms$ he minim%m frame si8e is6(a) 40 (b) 0#/ (c) 0/0 (&) .#! C !"".

    Ans. cExplanation: 9o%n& trip propagation &ela is !3 p5inim%m frame si8e of Ethernet can be fo%n& b %sing form%la :2!3 p$ =et = is minframe si8e then = #"5bps 2 0/$0 ms

    20/0 Hbits

    It has nothing to &o with Famming signal$$

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    Q33. Consi&er a 05bps to'en ring having to'en hol&ing time #"ms$ What is the si8e of longframe that can be sen& on ring$

    Ans. 0"HbitsExplanation: if to'en hol&ing time is #"ms then a %ser can transmit onl for #" ms+ ma:im%

    time till it can hol& to'en$ In # ms a %ser can transmit 05bps3 #"ms203#"/

    bits per secon&3#"3#"-, secon&s203#"0 bits20" Hbits$

    Q34. Consi&er a . 5bps to'en ring with propagation spee& !"" m s$ ow m%ch lengthcable is occ%pie& b # bit$

    Ans. 0" meter Explanation: convert bit to meters$$ &ivi&e b ban&wi&th an& m%ltipl with velocit

    Q3 . Consi&er a #" 5bps to'en ring =A< with a ring latenc of 0"" s$ A host that nee&stransmit sei8es the to'en$ hen it sen&s a frame of #""" b tes+ removes the frame after circ%late& all aro%n& the ring+ an& finall releases the to'en$ his process is repeate&frame$ Ass%ming that onl a single host wishes to transmit+ the effective &ata rate is6(A) # 5bps (B) ! 5bps (C) . 5bps (D)/ 5bps I !""0

    Ans. CExplanation: in this 7%estion &ela e& to'en reinsertion strateg is %se& $

    :2#""" b te #" 5bps 2*"" s

    9ing latenc 2 0"" s (=et %s ass%me p 2 9=+since no other info is given)herefore+ efficienc in &ela to'en reinsertion is # (#1 (

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    Q3#. In a to'en ring networ' the transmission spee& is #"; bps an& the propagation spee& is !""meters s$ he #-bit &ela in this networ' is e7%ivalent to6(A) ."" meters of cable$ (B) !"" meters of cable$(C) !" meters of cable$ (D) ." meters of cable$ C !"";

    Ans. CExplanation:We solve& one s%ch mo&el in the previo%s 7%estions$

    Q3$. Which of the following statements is 9 E abo%t C 5A CD?(A) IEEE *"!$## wireless =A< r%ns C 5A CD protocol(B) Ethernet is not base& on C 5A CD protocol(C) C 5A CD is not s%itable for a high propagation &ela networ' li'e satellite networ'$(D) here is no contention in a C 5A CD networ' I !"".

    Ans. C

    Q3%. $ A ro%ter ha f%ll &%ple: Ethernet interfaces each operating at #"" 5bps$ Ethernet are at least *0 b tes long (incl%&ing the preamble an& inter pac'et gap)$ he ma:im%m processing time at the ro%ter for wirespee& forwar&ing to be possible is(in micro secon&s(a) "$"# (b) ,$,/ (c) /$;! (&) *I !""/

    Ans. cExplanation: here other information is not given so transmission time can be ta'en a processing time$ :2*0 b tes #""5bps 2(*03* #"") s 2 /$;! s$

    Stat! !nt 5o Lin)! Ans*! Q !stions 4& an 41 :Consi&er a to'en ring topolog with < stations (n%mbere& # to

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    $

    Q41. he ma:im%m %tili8ation of the to'en ring when t:2. ms+ t p 2 , ms+

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    Depth irst traversal is B#+B.+B!+B,+B0$

    Q43 $ Consi&er the correct spanning tree for the previo%s 7%estion$ =et host # sen& broa&cast ping pac'et$ Which of the following options represents the correct forwar&ing ta

    B,?

    (A) (B)P 9 P 9

    #+ !+ ,+ 0 , #+ ! 0.+ /+ 4+ # # ,+ 0 ,;+ *+ ##+ #! ! .+ / #

    ;+ *$ 4$ #"+ ##+ #! !

    (C) (D)P 9 P 9,+ 0 , #+ !+ ,+ 0 ,.+ /+ 4+ # # .+ ;+ 4+ # ##+ ! 0 ;+ * ##+ #! 0;+ *+ ##+ #! !

    C !""/

    Ans. CExplanation: se panning tree generate& in previo%s 7%estion$

    NET9 RK LAYER

    Q44. Which of the following assertion is A= E abo%t the Internal Protocol (IP) ?(A) It is possible for a comp%ter to have m%ltiple IP a&&ress$(B) IP pac'ets from the same so%rce to the same &estination ta'e &ifferent ro%tes innetwor'$(C) IP ens%res a pac'et is &iscar&e& if it is %nable to reach its &estination within an%mber of hopes$(D) pac'et so%rce cannot set ro%te of o%tgoing pac'etsR the ro%te is &etermine& onl btables in the ro%ters on the wa $ C !"",

    Ans. D

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    Explanation: strict source routing

    Q4 . he a&&ress resol%tion protocol (A9P) is %se& for6(a) in&ing the IP a&&ress from the Dia D(#!ms1

    entries inro%ting table

    of D)

    >ia @(/ms1 entries

    in ro%tingtable of @)

    5in

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    A * ,0 ,! !; *B 0* ,; !" ," !"C !! #; 0! !* #;D !. ," #! !. #!E !4 #" !/ !* #"

    #; !# #4 #/ "(&istance from to is ")

    @ ,! ,! ,0 / /

    o ro%ting table of is option a$

    Stat! !nt 5o Lin)! Ans*! Q !stions ## -#$ AND#%:

    ;;$Consi&er a simple graph with %nit e&ge costs$ Each no&e inthe graph represents a ro%ter$ Each no&e maintains a ro%tingtable in&icating the ne:t hop ro%ter to be %se& to rela a pac'etto its &estination an& the cost of the path to the &estinationthro%gh that ro%ter$ Initiall + the ro%ting table is empt $ hero%ting table is s nchrono%sl %p&ate& as follows$ In each%p&ation interval+ three tas's are performe&$(i) A no&e &etermines whether its neighbors in the graph areaccessible$ If so+ it sets thetentative cost to each accessible neighbor as #$ therwise+ the cost is set toinfinit $(ii) rom each accessible neighbor+ it gets the costs to rela to other no&es via thatneighb

    the ne:t hop)(iii) Each no&e %p&ates its ro%ting table base& on the information receive& in the previosteps b choosing the minim%m cost$

    Q#$. or the @raphs given above+ possible ro%ting tables for vario%s no&es after thestabili8e&+ are shown in the following options$ I&entif the correct table$

    (A) able for no&e A (B) able for no&e C (C) able for no&e B (D) able for no&e A - -B B #

    C C #D B ,E C ,

    C 0

    A A #B B #C - -

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    D D #E E #

    E ,

    A A #

    B - -C C #D D #E C !

    D !

    A B ,B B #C C !D - -E E #

    #

    I !"".

    Ans. CExplanation:(A) entr for D+ E an& are wrong$ D an& E are at &istance of ! hops from A an& &istance of , hops from A(B) Entr for is Wrong$ is at &istance of ! hops from C via D(D) entries for A an& C are wrong$

    Q#%. Contin%ing from the earlier problem+ s%ppose at some time t+ when the costsstabili8e&+ no&e A goes &own$ he cost rom no&e to no&e A at time(t 1 #"") is(A) M#"" b%t finite (B) infinite (C) , (D) M, an& U#""I !"".

    Ans. AExplanation 6 Accor&ing to co%nt to infinit problem after t1#"" time cost will be M#""will be finite onl $

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    Q$1. two pop%lar ro%ting algorithm are &istance vector (D>) an& 'in' state (= ) ro%Which of the following are tr%e?( #) Co%nt to infinit is a problem onl with D> an& not = ro%ting$( !) In = + the shortest path algorithm is r%n onl at one no&e$( ,) In D>+ the shortest path algorithm is r%n onl at one no&e$

    ( 0) D> re7%ire lesser n%mber of networ' messages than = $(a) #+ ! an& 0 onl $(b) #+ , an& 0 onl $(c) ! an& , onl $(d) #an& 0onl $

    I !""*

    A< $ D

    Q$2. Co%nt to infinit is a problem associate& with(A) lin' state ro%ting protocol(B) &istance vector ro%ting protocol(C) D< while resolving host nameD) CPfor congestion control I !"".

    A< $ B

    Stat! !nt 5o Lin)! Ans*! Q !stions: Q $3 Q$4Consi&er a networ' with / ro%ters 9# to 9/ connecte&

    with lin's having weights as shown in the following&iagram

    Q$3. All the ro%ters %se the &istance vector base&ro%ting algorithm to %p&ate their ro%ting tables$ Eachro%ter starts with its ro%ting table initiali8e& to containan entr for each neighbor with the weight of therespective connecting lin'$ After all the ro%ting tablesstabili8e+ how man lin's in the networ' will never be%se& for carr ing an &ata?(A) 0 (B) , (C) ! (D) # C !"#"

    Ans. CExplanation: =in' 9#- 9! will not be %se& beca%se its cost is / an& lin' 9#-9,-9! costs .which is lesser than 9#-9! lin'$

    imilarl + lin' 90-9/ will not be %se&+ instea& this lin' we can %se 90-9.-9/ lin' which coonl . %nit$

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    Q$4. %ppose the weights of all %n%se& lin's in the previo%s 7%estion are change& to &istance vector algorithm is %se& again %ntil all ro%ting tables stabili8e$ ow man lnow remain %n%se&?(A) " (B) # (C) ! (D) ,C !"#"

    Ans. #Explanation:

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    .ach distance vector is the distance of the best known %ath at that instance to nodes, N1toN5,

    where the distance to itself is *& /lso, all links are s0 etric and also the cost is identical in

    both directions& n each round, all nodes e change their distance vectorswith their res%ective

    neighbours& !hen all nodes u%date their distancevectors& n between two rounds, an0 change

    in cost of a link will cause the two incident nodes to change onl0 that entr0 in their distance

    vectors&

    5&!he cost of link N2-N3 reduces to 2)in both directions & /fter the ne t round of

    u%dates, what will bethe new distance vector at node, N3

    a) )3,2,*,2,5

    b) )3,2,*,2,

    c) )+,2,*,2,5

    d) )+,2,*,2,

    ANS. /

    &/fter the u%date in the %revious 6uestion, the link N1-N2 goes down& N2 will reflect this

    change i ediatel0 in its distance vector as cost infinit0& /fter the N.7! $'8N" of

    u%date ,what will be the cost to N1 in the distance vector of N3

    a) 3

    b) 9

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    c) 1*

    d) nfinit0

    /N & ;ink n1-n2 goes down & N3 has neighbor N2 and N4 &N2 has ade entr0 infinit0 N4 has

    the distance of to N1, N3 has the distance of 2 to N4& o, 2< =1* )C

    TRANSP RT LAYER

    Q$#. Which of the following f%nctionalities m%st be implemente& b a transport protocoan& above the networ' protocol?(A) 9ecover from pac'et losses (B) Detection of &%plicate Pac'ets(C) Pac'et &eliver in the correct or&er (D) En& to en& Connectivit C !"",

    A< $ D

    Q$$. 5atch the following6(P) 5 P (#) Application la er ( ) B@P (!) ransport la er (9) CP (,) Data lin' la er ( ) PPP (0)

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    correctl b the receiver$ =et S be the amo%nt of &ata carrie& in the first segment (in ban& O be the ACH n%mber sent b the receiver$

    he val%e of S an& O(in that or&er) are(a) /" an& !4" (b)!," an& !4# (c)/" an& !,# (&) /" an& !,"I !"";

    Ans. &Explanation: In CP se7%ence n%mber is assign to each &ata b tes$

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    D%ring #"th transmission it transmit ;5 $ (C)

    Q%3. In the slow start phase of the CP congestion control algorithm+ the si8e of tcongestion win&ow(A) &oes not increase (B) increases linearl

    (C) increases7%a&raticall (D)increasese:ponentiallC !""*

    A< $ D

    Q%4. n a CP connection+ c%rrent congestion win&ow si8e is Congestion Win&ow 2 0he win&ow si8e a&vertise& b the receive& is A&vertise Win&ow 2 / HB$ he last b t

    the sen&er is =astB te ent 2 #"!0" an& the last b te ac'nowle&ge& b the receis=astB teAc'e& 2 *#4!$ he c%rrent win&ow si8e at the sen&er is6(A) !"0* b tes(B) 0"4/ b tes

    (C) /#00 b tes(D)*#4! b tes I !"".

    Ans. BExplanation: C%rrent Win&ow i8e 2min(congestion win&ow+ a&vertise& win&ow)

    here is no nee& to thin' abt b tes sent an& ac'nowle&ge&$

    Q% . %ppose that the ma:im%m transmit win&ow si8e for a CP connection is #!""" bEach pac'et consists of !""" b tes$ At some point of time+ the connection is slow-start ph

    with a c%rrent transmit win&ow of 0""" b tes$ %bse7%entl + the transmitter receivac'nowle&gements$ Ass%me that no pac'ets are lost an& there are no time-o%ts$ Whatma:im%m possible val%e of the c%rrent transmit win&ow?(A) 0""" b tes (B) *""" b tes(C) #"""" b tes (D) #!""" b tes I !""0

    Ans. B Explanation: in slow start phase if sen&er receive ac'nowle&gement for a win&ow thanenhances its c%rrent win&ow si8e b one win&ow$ o+ after two ac'nowle&gement win&ow si8e will be0"""1!"""1!"""2*""" b tes

    Q%". A client process P nee&s to ma'e a CP connection to a server process $ Consi&erfollowing sit%ation6 the server process e:ec%tes a soc'et ()+ a bin& () an& a listen () call in that or&er+ following which it is preempte&$ %bse7%entl + the client process P esoc'et () s stem call followe& b connect () s stem call to connect to the server process $server process has not e:ec%te& an accept () s stem call$ Which one of the following eco%l& ta'e place?

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    (A) connect () s stem call ret%rns s%ccessf%ll(B) connect () s stem call bloc's(C) connect () s stem call ret%rns an error (D) connect () s stem call res%lts in a core &%mp C !""*

    Ans. CExplanation: Connect() stem call is not bloc'ing s stem call b%t it bloc's %ntil connectiis establishe& or reFecte&$ If accept() is not e:ec%te& at server then connection will be ran& an error statement is ret%rne&$

    Q%#. Consi&er the following statements abo%t the timeo%t val%e %se& in CP$(#) he timeo%t val%e is set to the 9 (9o%n& rip ime) meas%re& &%ring CP coestablishment for the entire &%ration of the connection$

    (!) Appropriate 9 estimation algorithm is %se& to set the timeo%t val%e of CP connec (,) imeo%t val%e is set to twice the propagation &ela from the sen&er to receiver

    Which of the following choice hol&?(a) (#) is false+b%t (!) an& (,) are tr%e(b) (#) an& (,) are false+ b%t (!) is tr%e(c) (#) an& (!) are false+ b%t (,) is tr%e(&)(#) +(!) an& (,) are falseI !"";

    A< $ B+ we have seen Basic algo an& Vacobsons algo

    Q%$. While opening a CP connection+ the initial se7%ence n%mber is to be &erive& %sinof-&a ( oD) cloc' that 'eeps r%nning even when the host is &own$ he low or&er ,! bits co%nter of the oD cloc' is to be %se& for the initial se7%ence n%mbers$ he cloc' cincrements once per millisecon&$ he ma:im%m pac'et lifetime is given to be /0s$Which one of the choices given below is closest to the minim%m permissible rate at wse7%ence n%mbers %se& for pac'ets of a connection can increase?(A) "$"#. s (B) "$"/0 s (C) "$#,. s (D)"$,!; s C !""4

    Ans. B

    Q%%. In the CP IP protocol s%ite+ which one of the following is < part of the IP hea&er(A) ragment offset(B) o%rce IP a&&ress(C) Destination IP a&&ress(D) Destination port n%mber I !""0

    A< $ D

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    Q1&&. A CP message consisting of !#"" b tes is passe& to IP for &eliver across two netwhe first networ' can carr a ma:im%m pa loa& of #!"" b tes per frame an& the secon& ne

    can carr a ma:im%m pa loa& of 0"" b tes per frame+ e:cl%&ing networ' overhea&$ AssIP overhea& per pac'et is !" b tes$ What is the total IP overhea& in the secon& networ' fotransmission?

    (A) 0" b tes (B) *" b tes (C) #!" b tes (D) #/" b tes I !""0Ans. CExplanation: in 7%estion ma:im%m pa loa& can be transferre& is given not the ma:im%m si8e$ o &%ring fragmentation we have to consi&er onl pa loa& not hea&ers$At first networ' !#"" b te pa loa& will be &ivi&e& into #!"" b te an& 4"" b tes pa loasecon& networ' #!"" b tes pa loa& &ivi&e& into , pa loa&s of 0"" b tes an& similarl + 4 pa loa& will be &ivi&e& in ! pa loa& of 0"" b tes an& one pa loa& of #"" b te$

    o !#"" b tes of pa loa& in secon& networ' is &ivi&e& in si: pac'ets+ so total IP overhsecon& networ' is /3!"2#!" b tes

    Q1&1. A firewall is to be config%re& to allow hosts in a private networ' to freel open connections an& sen& pac'ets on open connections$ owever+ it will onl allow e:ternal hsen& pac'ets on e:isting open CP connections or connections that are being opene& (b inthost) b%t not allow them to open CP connections to hosts in the private networ'$ o achievminim%m capabilit of the firewall sho%l& be that of (a) A combinational circ%it(b) A finite a%tomation(c) A p%sh&own a%tomaton with one stac' (&)A p%sh&own a%tomaton with two stac'I !"";

    A< $ D(P%sh&own a%tomaton with two stac's can act as a comp%ter)

    Q1&2$ Which one of the following statements is A= E?(A) CP g%arantees a minim%m comm%nication rate(B) CP ens%res in-or&er &eliver(C) CP reacts to congestion b re&%cing sen&er win&ow si8e(D) CP emplo s retransmission to compensate for pac'et lossI !""0

    A< $ A( CP starts slow+ b%t &oesn t g%arantee minim%m comm%nication rate )

    >DPQ1&3. Pac'ets of the same session ma be ro%te& thro%gh &ifferent paths in6(a) CP+ b%t not DP

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    (b) CP an& DP(c) DP+ b%t not CP(&)

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    Ans. cExplanation: if n in&ivi&%al comm%nicate %sing secret 'e or s mmetric 'e techni7%%ni7%e 'e is re7%ire& for each connection$

    otal possible connections arenC!2n(n-#) !

    An& if n in&ivi&%al comm%nicate %sing p%blic 'e or As mmetric 'e techni7%e tin&ivi&%al sho%l& have p%blic an& private 'e pair$ o total n%mber of 'e 2 !n

    Q1&$.In the 9 A p%blic 'e cr ptos stem+ the private an& p%blic 'e s are (e+n) an&respectivel + where n2p37 an& p an& 7 are large primes$ Besi&es+ n isp%blic an& p private$ =et 5 be an integer s%ch that "X5Xn an&Y(n)2 (p Z #) (7 Z #)$ $5 25e mo& Y (n)5 2 (5 ) & mo& Y (n)

    Which of the above e7%ations correctl represent 9 A cr ptos stem?(A) I an& II (B) I an& III (C) II an& I> (D) III an& IC !""4

    Ans. C

    Q1&%. he minim%m positive integer p s%ch that , p mo&%lo #;2# is (a) . (b)* (c)#! (&) #/I !"";

    Ans. &

    Q11&. E:ponentiation is a heavil %se& operation in p%blic 'e cr ptograph $ Which ofollowing option is tightest %pper bo%n& on the n%mber of m%ltiplication re7%ire& to bn mo&%lo m+ "Ub+ nUm?(a) (log n) (b) (\n) (c) (n log n) (&) (n)I !"";

    Ans. a

    Q111. %ppose that two parties A an& B wish to set%p a common secret 'e (D- 'e betweenthemselves %sing the Diffie- ellman 'e e:change techni7%e$ he agree on ; amo&%l%san& , as the primitive root$ Part A chooses ! an& part B chooses . asrespectivesecrets$ heir D- 'e is6

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    (A) , (B) 0 (C) . (D)/I !"".

    Ans. BExplanation: D- 'e is gAB mo& n2 ,!3. mo& ;2 ,#" mo& ;2 0

    Q112. A sen&er is emplo ing p%blic 'e cr ptograph to sen& a secret message to a receiveWhich one of the following statements is 9 E?(A) en&er encr pts %sing receiver s p%blic 'e(B) en&er encr pts %sing his own p%blic 'e(C) 9eceiver &ecr pts %sing sen&er s p%blic 'e(D) 9eceiver &ecr pts %sing his own p%blic 'e

    I !""0

    Ans. a

    APPLICATI N LAYER

    Q113 E== an& P 9 +respectivel + are comman&s from the protocols(a) P an& P(b) E=

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    d) ock !C? traffic fro a s%ecific user on a ulti-user s0ste during 9:** % and 5:**a

    Cs 2*11

    /N & /

    115&Consider different activities related to e ail

    1: end an e ail fro a ail client to a ail server

    2: "ownload an e ail fro ailbo server to a ail client

    3: Checking e ail in a web browser

    a) & 1: >!!? 2: @!? 3:?'?

    b) & 1: @!? 2:A!? 3:>!!?

    c) & 1: @!? 2:?'? 3:>!!?

    d) & 1:?'? 2: @!? 3: @/?

    /N & C

    $11%# he protocol data unit3pdu)for the application la6er in the internet stack isa)segmentb)datagramc)messaged)frame

    !"S# C

    $117#8hich of the following transport la6er protocols is used to support electronic maila)smtpb)ipc)tcpd)udp

    !"S# C

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