Data Communication and Computer Networks - Functions of OSI and TCP/IP Layers

6. The address resolution protocol (ARP) is used for

  • Option : D
  • Explanation : The Address Resolution Protocol (ARP), allows a host to find the MAC (Physical) address of a target host on the same physical network, given only the targets IP address.
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7. Which of the following assertions is FALSE about the Internet Protocol (IP)?

  • Option : D
  • Explanation : The packet source cannot set the route of an outgoing packets; the route is determined only by the routing tables in the routers on the way.
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8. Packets of the same session may be routed through different paths in

  • Option : A
  • Explanation : TCP and UDP are Transport layer protocols. Packets of same session may be routed through different routes.
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9. Consider the store and forward packet switched network given below. Assume that the bandwidth of each link is 106 bytes/sec. A user on host A sends a file of size 103 bytes to host B through routers R1 and R2 in three different ways. In the first case a single packet containing the complete file is transmitted from A to B. In the second case, the file is split into 10 equal parts, and these packets are transmitted from A to B. In the third case, the file is split into 20 equal parts and these packets are sent from A to B. Each packet contains 100 bytes of header information along with the user data. Consider only transmission time and ignore processing, queuing and propagation delays. Also assume that there are no errors during transmission. Let Tl, T2 and T3 be the times taken to transmit the file in the first, second and third case respectively. Which one of the following is CORRECT?

functions of osi

  • Option : D
  • Explanation : The following data is given
    File Size = 1000 bytes
    Header Size = 100 bytes
    Transmission Speed of all links = 10^6 bytes/sec
    Case (1) :
    Transmission time for one link
    = packetsize/bandwidth
    = (1000 + 100)/10^6
    = 1100 microsecond.
    So, Total time(T1) is 3*1100
    = 3300 microsecond
    Case (2) :
    Transmission time for one link and one part
    = (100 + 100)/10^6
    = 200 microseccond
    Note : While a packet is being transmitted from R1 to R2, at the same time other packet is being transmitted from A to R1.
    So, Total time(T2) = 3 * 200 + 9 * 200
    = 2400 microsecond
    Case (3) :
    Transmission time for one link and one part
    = (50 + 100)/10^6
    = 150 microsecond
    Total time (T3) = 3 * 150 + 19 * 150
    = 3300 microsecond
    hence T1 = T3, T3 > T2
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10. Match the following:

FieldLength in bits
P. UDP Header’s Port NumberI . 48
Q. Ethernet MAC AddressII . 8
R. IPv6 Next HeaderIII . 32
S. TCP Header’s Sequence NumberIV. 16

CODES :

 PQRS
(A)IIIIVIII
(B)IIIIVIII
(C)IVIIIIII
(D)IVIIIIII

  • Option : C
  • Explanation : UDP Header’s Port Number ⇒ 16 bit Ethernet MAC Address ⇒ 48 bit IPv6 Next Header ⇒ 8 bit TCP Header’s Sequence Number ⇒ 32 bit
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