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.
Data Communication and Computer Networks MCQ - Functions of OSI and TCP/IP Layers
- A It is possible for a computer to have multiple IP addresses
- B IP packets from the same source to the same destination can take different routes in the network
- C IP ensures that a packet is forwarded if it is unable to reach its destination within a given number of hopes
- D 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
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.
TCP and UDP are Transport layer protocols.
Packets of same session may be routed through
different routes.
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
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
10
Match the following:
| Field | Length in bits |
| P. UDP Header’s Port Number | I . 48 |
| Q. Ethernet MAC Address | II . 8 |
| R. IPv6 Next Header | III . 32 |
| S. TCP Header’s Sequence Number | IV. 16 |
CODES :
| Â | P | Q | R | S |
| (A) | III | IV | II | I |
| (B) | II | I | IV | III |
| (C) | IV | I | II | III |
| (D) | IV | I | III | II |
UDP Header’s Port Number ⇒ 16 bit
Ethernet MAC Address ⇒ 48 bit
IPv6 Next Header ⇒ 8 bit
TCP Header’s Sequence Number ⇒ 32 bit
