1) In an IP packet, the value of HLEN is 5, and the value of the TOTAL LENGTH field is 1000 (in decimal). The number of data bytes in the packet will be __
Solution: Since HLEN = 5, the size of the IP header will be 5 x 4 = 20 bytes. The total size of the IP packet is given as 1000 bytes. Hence, the number of data bytes = 1000 – 20 = 980 bytes.
2) An IP packet arrives at the final destination with the M flag set as 1. What can you say about the packet?
a. The packet has not been fragmented.
b. The packet has been fragmented.
c. The packet has been fragmented, and this is not the last fragment.
d. The packet data was truncated on its way.
Answer: b, c
Solution: When the Mode (M) flag in a packet is 1, this indicates that the original packet has definitely been fragmented. Also, this is not the last fragment … there are more fragments after this. Hence, options (b) and (c) are true.
3) Which address class does the IP address 184.108.40.206 belong to?
a. Class A
b. Class B
c. Class C
d. Class D
Solution: Class A addresses start with “0”, class B addresses start with “10”, class C addresses start with “110”, and class D addresses start with “1110”. For the given IP address, the first byte 227 = 11100011 in binary, this starts with “1110”. Hence, this is a class D address.
4) What do the following IP address signify: 220.127.116.11 ?
a. It is a unicast address on the class B network 18.104.22.168
b. It is a broadcast address on the class B network 22.214.171.124
c. The subnet mask is 0.0.255.255
d. None of these
Solution: The first byte 144 = 10010000 in binary, which indicates that this is a class B address (starts with “10”). Therefore, the first two bytes (144.16) will indicate the network address. The last two bytes will indicate the host address, which is specified as 255.255 (all 1’s). The all 1’s pattern indicates a broadcast address. Thus the IP address signifies a broadcast address on the class B network 126.96.36.199.
5) An IP packet with 2500 bytes of data (plus header) passes through an IP network with MTU =
500 bytes. How many additional bytes will be delivered at the destination?
Solution: Each IP fragment will have a header of 20 bytes, and data part of maximum 500 – 20 = 480 bytes. The number of fragments for the 2500 bytes of data will be therefore [2500 / 480] = 6. The original data packet had a header of 20 bytes. So, number of additional bytes (for the fragment headers) = 6 x 20 – 20 = 100 bytes.
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