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SET 3


  Question 1

A layer-4 firewall (a device that can look at all protocol headers up to the transport layer) CANNOT

A : block entire HTTP traffic during 9:00PM and 5:00AM
B : block all ICMP traffic
C : stop incoming traffic from a specific IP address but allow outgoing traffic to the same IP address
D : block TCP traffic from a specific user on a multi-user system during 9:00PM and 5:00AM


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     Correct answer is :A

     Solution :
      Since it is a layer 4 firewall it cannot block application layer protocol like HTTP.

  •   Question 2

    Consider different activities related to email. m1: Send an email from a mail client to a mail server m2: Download an email from mailbox server to a mail client m3: Checking email in a web browser Which is the application level protocol used in each activity?

    A : m1:HTTP m2:SMTP m3:POP
    B : m1:SMTP m2:FTP m3:HTTP
    C : m1: SMTP m2: POP m3: HTTP
    D : m1: POP m2: SMTP m3:IMAP


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     Correct answer is :C

     Solution :
      Sending an email will be done through user agent and message transfer agent by SMTP, downloading an email from mail box is done through POP, checking email in a web browser is done through HTTP

  •   Question 3

    Consider a network with five nodes, N1 to N5, as shown below
    The net work uses a Distance Vector Routing protocol. Once the routes have stabilized, the distance vectors at different nodes are as following
    N1: (0,1,7,8,4)
    N2 : (1,0,6,7,3)
    N3: (7,6,0,2,6)
    N4 : (8,7,2,0,4)
    N5 : (4,3,6, 4,0)
    Each distance vector is the distance of the best known path at that instance to nodes, N1 to N5, where the distance to itself is 0. Also, all links are symmetric and the cost is identical in both directions. In each round, all nodes exchange their distance vectors with their respective neighbors. Then all nodes update their distance vectors. In between two rounds, any change in cost of a link will cause the two incident nodes to change only that entry in their distance vectors
    The cost of link N2-N3 reduces to 2 in (both directions). After the next round of updates, what will be the new distance vector at node, N3?




    A : (3. 2, 0, 2, 5)
    B : (3, 2, 0, 2, 6)
    C : (7, 2, 0, 2, 5)
    D : (7, 2, 0, 2, 6)


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     Correct answer is :A



  •   Question 4

    Consider a network with five nodes, N1 to N5, as shown below
    The net work uses a Distance Vector Routing protocol. Once the routes have stabilized, the distance vectors at different nodes are as following
    N1: (0,1,7,8,4)
    N2 : (1,0,6,7,3)
    N3: (7,6,0,2,6)
    N4 : (8,7,2,0,4)
    N5 : (4,3,6, 4,0)
    Each distance vector is the distance of the best known path at that instance to nodes, N1 to N5, where the distance to itself is 0. Also, all links are symmetric and the cost is identical in both directions. In each round, all nodes exchange their distance vectors with their respective neighbors. Then all nodes update their distance vectors. In between two rounds, any change in cost of a link will cause the two incident nodes to change only that entry in their distance vectors
    After the update in the previous question, the link N1-N2 goes down. N2 will reflect this change immediately in its distance vector as cost, infinite . After the NEXT ROUND of update, what will be the cost to N1 in the distance vector of N3?




    A : 3
    B : 9
    C : 10
    D : infinite


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     Correct answer is :C

     Solution :
      N3 has neighbors N2 and N4
    N2 has made entry infinite
    N4 has the distance of 8 to N1
    N3 has the distance of 2 to N4
    So 2 + 8 = 10

  •   Question 5

    In the IPv4 addressing format, the number of networks allowed under Class C addresses is

    A : 2^14
    B : 2^7
    C : 2^21
    D : 2^24


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     Correct answer is :C

     Solution :
      For class C address, size of network field is 24 bits. But first 3 bits are fixed as 110; hence total number of networks possible is 2^21

  •   Question 6

    Which of the following transport layer protocols is used to support electronic mail?

    A : SMTP
    B : IP
    C : TCP
    D : UDP


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     Correct answer is :C

     Solution :
      E-mail uses SMTP, application layer protocol which intern uses TCP transport layer protocol.

  •   Question 7

    The protocol data unit (PDU) for the application layer in the Internet stack is

    A : Segment
    B : Datagram
    C : Message
    D : Frame


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     Correct answer is :C

     Solution :
      The PDU for Datalink layer, Network layer , Transport layer and Application layer are frame, datagram, segment and message respectively.

  •   Question 8

    Consider an instance of TCPs Additive Increase Multiplicative decrease (AIMD) algorithm where the window size at the start of the slow start phase is 2 MSS and the threshold at the start of the first transmission is 8 MSS. Assume that a timeout occurs during the fifth transmission. Find the congestion window size at the end of the tenth transmission.

    A : 8 MSS
    B : 14 MSS
    C : 7 MSS
    D : 12 MSS


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     Correct answer is :C

     Solution :
      Given, initial threshold = 8
    Time = 1, during 1st transmission,Congestion window size = 2 (slow start phase)
    Time = 2, congestion window size = 4 (double the no. of acknowledgments)
    Time = 3, congestion window size = 8 (Threshold meet)
    Time = 4, congestion window size = 9, after threshold (increase by one Additive increase)
    Time = 5, transmits 10 MSS, but time out occurs congestion window size = 10
    Hence threshold = (Congestion window size)/2 = 10/5= 2 =
    Time = 6, transmits 2
    Time = 7, transmits 4
    Time = 8, transmits 5 (threshold is 5)
    Time = 9, transmits 6, after threshold (increase by one Additive increase)
    Time = 10, transmits 7
    During 10th transmission, it transmits 7 segments hence at the end of the tenth transmission the size of congestion window is 7 MSS

  •   Question 9

    Consider a source computer (S) transmitting a file of size 106 bits to a destination computer (D) over a network of two routers (R1 and R2) and three links (L1, L2, and L3). L1 connects S to R1; L2 connects R1 to R2; and L3 connects R2 to D. Let each link be of length 100km. Assume signals travel over each line at a speed of 108 meters per second. Assume that the link bandwidth on each link is 1Mbps. Let the file be broken down into 1000 packets each of size 1000 bits. Find the total sum of transmission and propagation delays in transmitting the file from S to D?

    A : 1005ms
    B : 1010ms
    C : 3000ms
    D : 3003ms


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     Correct answer is :A

     Solution :
      Transmission delay for 1 packet from each of S, R1 and R2 will take 1ms
    Propagation delay on each link L1, L2 and L3 for one packet is 1ms
    Therefore the sum of transmission delay and propagation delay on each link for one packet is 2ms.
    The first packet reaches the destination at 6th ms
    The second packet reaches the destination at 7th ms
    So inductively we can say that 1000th packet reaches the destination at 1005th ms

  •   Question 10

    An Internet Service Provider (ISP) has the following chunk of CIDR-based IP addresses available with it: 245.248.128.0/20. The ISP wants to give half of this chunk of addresses to Organization A, and a quarter to Organization B, while retaining the remaining with itself. Which of the following is a valid allocation of address to A and B?

    A : 245.248.136.0/21 and 245.248.128.0/22
    B : 245.248.128.0/21 and 245.248.128.0/22
    C : 245.248.132.0/22 and 245.248.132.0/21
    D : 245.248.136.0/24 and 245.248.132.0/21


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     Correct answer is :A

     Solution :
      Since half of 4096 host addresses must be given to organization A, we can set 12th bit to 1 and include that bit into network part of organization A, so the valid allocation of addresses to A is 245.248.136.0/21
    Now for organization B, 12th bit is set to 0 but since we need only half of 2048 addresses, 13th bit can be set to 0 and include that bit into network part of organization B so the valid allocation of addresses to B is 245.248.128.0/22

  •   Question 11

    Assume that source S and destination D are connected through two intermediate routers labeled R. Determine how many times each packet has to visit the network layer and the data link layer during a transmission from S to D.

    A : Network layer 4 times and Data link layer-4 times
    B : Network layer 4 times and Data link layer-3 times
    C : Network layer 4 times and Data link layer-6 times
    D : Network layer 2 times and Data link layer-6 times


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     Correct answer is :C

     Solution :
      From the given diagram, its early visible that packet will visit network layer 4 times, once at each node [S, R, R, D] and packet will visit Data Link layer 6 times. One time at S and one time at D, then two times for each intermediate router R as data link layer is used for link to link communication.
    Once at packet reaches R and goes up from physical DL-Network and second time when packet coming out of router in order Network DL- Physical


  •   Question 12

    The transport layer protocols used for real time multimedia, file transfer, DNS and email, respectively are

    A : TCP, UDP, UDP and TCP
    B : UDP, TCP, TCP and UDP
    C : UDP, TCP, UDP and TCP
    D : TCP, UDP, TCP and UDP


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     Correct answer is :C

     Solution :
      Real time multimedia needs connectionless service, so under lying transport layer protocol used is UDP
    File transfer rums over TCP protocol with port no-21
    DNS runs over UDP protocol within port no-53
    Email needs SMTP protocol which runs over TCP protocol within port no 25

  •   Question 13

    Using public key cryptography, X adds a digital signature s to message M, encrypts , and sends it to Y, where it is decrypted. Which one of the following sequences of keys is used for the operations?

    A : Encryption: Xs private key followed by Ys private key; Decryption: Xs public key followed by Ys public key
    B : Encryption: Xs private key followed by Ys public key; Decryption: Xs public key followed by Ys private key
    C : Encryption: Xs public key followed by Ys private key; Decryption: Ys public key followed by Xs private key
    D : Encryption: Xs private key followed by Ys public key; Decryption: Ys private key followed by Xs public key


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     Correct answer is :D

     Solution :
      ENCRYPTION : Source has to encrypt with its private key for forming Digital signature for Authentication. source has to encrypt the , with Y 's public key to send it confidentially
    DECRYPTION : Destination Y has to decrypt first with its private key, then decrypt using source public key

  •   Question 14

    In an IPv4 datagram, the M bit is 0, the value of HLEN is 10, the value of total length is 400 and the fragment offset value is 300. The position of the datagram, the sequence numbers of the first and the last bytes of the payload, respectively are

    A : Last fragment, 2400 and 2789
    B : First fragment, 2400 and 2759
    C : Last fragment, 2400 and 2759
    D : Middle fragment, 300 and 689


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     Correct answer is :C

     Solution :
      M= 0 Means there is no fragment after this, i.e. Last fragment
    HLEN =10 - So header length is 410=40, as 4 is constant scale factor
    Total Length = 400(40 Byte Header + 360 Byte Payload)
    Fragment Offset = 300, that means 3008 Byte = 2400 bytes are before this last fragment
    So the position of datagram is last fragment
    Sequence number of First Byte of Payload = 2400 (as 0 to 2399 Sequence no are used)
    Sequence number of Last Byte of Payload = 2400+360-1=2759

  •   Question 15

    Determine the maximum length of cable (in km) for transmitting data at a rate of 500 Mbps in an Ethernet LAN with frames of size 10,000 bits. Assume the signal speed in the cable to be 2,00,000 km/s

    A : 1
    B : 2
    C : 2.5
    D : 5


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     Correct answer is :B

     Solution :
      500106 bits - - - - - - - - - 1 sec
    104bits - - - -2*105km
    1/5*104 sec - - - - -2*105/5*104 = 4km
    Maximum length of cable = 4/2 = 2km

  • MY REPORT
    TOTAL = 15
    ANSWERED =
    CORRECT / TOTAL = /15
    POSITIVE SCORE =
    NEGATIVE SCORE =
    FINAL SCORE =