Previous Gate Question Ans

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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) ."" me...

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