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SET 4


  Question 1

Consider a 4-way set associative cache (initially empty) with total 16 cache blocks. The main memory consists of 256 blocks and the request for memory blocks is in the following order: 0, 255, 1, 4, 3, 8, 133, 159, 216, 129, 63, 8, 48, 32, 73, 92, 155. Which one of the following memory block will NOT be in cache if LRU replacement policy is used?

A : 3
B : 8
C : 129
D : 216


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     Correct answer is :D


  •   Question 2

    In the RSA public key cryptosystem, the private and public keys are (e, n) and (d, n) respectively, where n = p*q and p and q are large primes. Besides, n is public and p and q are private. Let M be an integer such that 0 < M < n and f(n) = (p- 1)(q-1). Now consider the following equations.
    I. M= Me mod n
    M = (M)d mod n
    II. ed = 1 mod n
    III. ed = 1 mod f(n)
    IV. M= Me mod f(n) M = (M)d mod f(n)
    Which of the above equations correctly represent RSA cryptosystem?


    A : I and II
    B : I and III
    C : II and IV
    D : III and IV


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     Correct answer is :C

     Solution :
      I is true because below is true in RSA-Cryptosystem.
    Encrypted-Text = (Plain-Text)e mod n
    Plain-Text = (Encrypted-Text)d mod n
    III is true because below is true
    d-1 = e mod φ(n)
    OR ed = 1 mod φ(n)

  •   Question 3

    While opening a TCP connection, the initial sequence number is to be derived using a time-of-day(ToD) clock that keeps running even when the host is down. The low order 32 bits of the counter of the ToD clock is to be used for the initial sequence numbers. The clock counter increments once per millisecond. The maximum packet lifetime is given to be 64s. Which one of the choices given below is closest to the minimum permissible rate at which sequence numbers used for packets of a connection can increase?

    A : 0.015/s
    B : 0.064/s
    C : 0.135/s
    D : 0.327/s


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     Correct answer is :A

     Solution :
      The maximum packet lifetime is given to be 64 seconds in the question. Thus, a sequence number increments after every 64 seconds. So, minimum permissible rate = 1 / 64 = 0.015 per second

  •   Question 4

    Let G(x) be the generator polynomial used for CRC checking. What is the condition that should be satisfied by G(x) to detect odd number of bits in error?

    A : G(x) contains more than two terms
    B : G(x) does not divide 1+x^k, for any k not exceeding the frame length
    C : 1+x is a factor of G(x)
    D : G(x) has an odd number of terms


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     Correct answer is :C

     Solution :
      Odd number of bit errors can be detected if G(x) contains (x+1) as a factor. See this for proof.

  •   Question 5

    Frames of 1000 bits are sent over a 10^6 bps duplex link between two hosts. The propagation time is 25ms. Frames are to be transmitted into this link to maximally pack them in transit (within the link). What is the minimum number of bits (i) that will be required to represent the sequence numbers distinctly? Assume that no time gap needs to be given between transmission of two frames

    A : i=2
    B : i=3
    C : i=4
    D : i=5


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     Correct answer is :D

     Solution :
      Transmission delay for 1 frame = 1000/(106) = 1 ms Propagation time = 25 ms The sender can atmost transfer 25 frames before the first frame reaches the destination. The number of bits needed for representing 25 different frames = 5

  •   Question 6

    Frames of 1000 bits are sent over a 10^6 bps duplex link between two hosts. The propagation time is 25ms. Frames are to be transmitted into this link to maximally pack them in transit (within the link). Suppose that the sliding window protocol is used with the sender window size of 2^i where is the number of bits identified in the previous question and acknowledgments are always piggybacked. After sending 2^i frames, what is the minimum time the sender will have to wait before starting transmission of the next frame? (Identify the closest choice ignoring the frame processing time.)

    A : 16ms
    B : 18ms
    C : 20ms
    D : 22ms


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     Correct answer is :B

     Solution :
      Size of sliding window = 2^5 = 32 Transmission time for a frame = 1ms Total time taken for 32 frames = 32ms The sender cannot receive acknoledgement before round trip time which is 50ms After sending 32 frames, the minimum time the sender will have to wait before starting transmission of the next frame = 50 32 = 18

  •   Question 7

    One of the header fields in an IP datagram is the Time to Live(TTL)field.Which of the following statements best explains the need for this field?

    A : It can be used to prioritize packets
    B : It can be used to reduce delays
    C : It can be used to optimize throughput
    D : It can be used to prevent packet looping


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     Correct answer is :D


  •   Question 8

    Which one of the following is not a client server application?

    A : Internet chat
    B : Web browsing
    C : E-mail
    D : ping


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     Correct answer is :D

     Solution :
      Ping is a computer network administration software utility used to test the reachability of a host on an Internet Protocol (IP) network. It measures the round-trip time for messages sent from the originating host to a destination computer that are echoed back to the source. It does not need any server for this.

  •   Question 9

    Suppose computers A and B have IP addresses 10.105.1.113 and 10.105.1.91 respectively and they both use the same netmask N. Which of the values of N given below should not be used if A and B should belong to the same network?

    A : 255.255.255.0
    B : 255.255.255.128
    C : 255.255.255.192
    D : 255.255.255.224


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     Correct answer is :D

     Solution :
      The first 3 blocks are the same and the Forth block has 113= 1110001 and 91=1101111. Only 2 bits should be same as in first 3 options

  •   Question 10

    A computer system has an L1 cache, an L2 cache, and a main memory unit connected as shown below. The block size in L1 cache is 4 words. The block size in L2 cache is 16 words. The memory access times are 2 nanoseconds. 20 nanoseconds and 200 nanoseconds for L1 cache, L2 cache and main memory unit respectively.
    When there is a miss in L1 cache and a hit in L2 cache, a block is transferred from L2 cache to L1 cache. What is the time taken for this transfer?




    A : 2 nanoseconds
    B : 20 nanoseconds
    C : 22 nanoseconds
    D : 88 nanoseconds


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     Correct answer is :D

     Solution :
      16/4 = 4 so we have to copy the block 4 times and place it in the L1 cache. So (20+2)*4 = 88

  •   Question 11

    A computer system has an L1 cache, an L2 cache, and a main memory unit connected as shown below. The block size in L1 cache is 4 words. The block size in L2 cache is 16 words. The memory access times are 2 nanoseconds. 20 nanoseconds and 200 nanoseconds for L1 cache, L2 cache and main memory unit respectively.
    When there is a miss in both L1 cache and L2 cache, first a block is transferred from main memory to L2 cache, and then a block is transferred from L2 cache to L1 cache. What is the total time taken for these transfers?




    A : 222 nanoseconds
    B : 888 nanoseconds
    C : 902 nanoseconds
    D : 968 nanoseconds


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     Correct answer is :D

     Solution :
      Time between memory and L2 cache = 200+20
    Time between L2 and L1 is 20+2
    now total there is a transfer of 4 words between each unit so 4*(220+22)

  •   Question 12

    Consider a network with 6 routers R1 to R6 connected with links having weights as shown in the following diagram
    All the routers use the distance vector based routing algorithm to update their routing tables. Each router starts with its routing table initialized to contain an entry for each neighbour with the weight of the respective connecting link. After all the routing tables stabilize, how many links in the network will never be used for carrying any data?




    A : 4
    B : 3
    C : 2
    D : 1


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     Correct answer is :C


  •   Question 13

    Consider a network with 6 routers R1 to R6 connected with links having weights as shown in the following diagram
    Suppose the weights of all unused links in the previous question are changed to 2 and the distance vector algorithm is used again until all routing tables stabilize. How many links will now remain unused?




    A : 0
    B : 1
    C : 2
    D : 3


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     Correct answer is :B


  • MY REPORT
    TOTAL = 13
    ANSWERED =
    CORRECT / TOTAL = /13
    POSITIVE SCORE =
    NEGATIVE SCORE =
    FINAL SCORE =