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

26 An Internet Service Provider (I SP) has the following chunk of CI DR-based I P 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 of Organization B, while retaining the remaining with itself. Which of the following is a valid allocation of addresses to A and B?

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 t o ‘0' and include that bit into network part of organization B so the valid allocation of addresses to B is 245.248.128.0/2.

27 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 if 300. The position of the datagram, the sequence numbers of the first and the last bytes of the payload, respectively are

Since M bit is 0, so there is no fragments after this fragment. Hence this fragment is the “last fragment” Tow, H LEN defines t he length of Header in datagram. Since H LEN is 10 so size of header is 10 * 4 = 40B.
Length of data = Total length – Leader length = 400 – 40 = 360 B.
Few, fragment offset of data in original datagram is measured in units of 8B. So to find first Byte of this fragment,
first Byte/B = fragment offset
first Byte = 300 * 8 = “2400” B
since length of data is 360B
so last Byte on this datagram will be “2759” B

28 An IP router implementing Classless Interdomain Routing (CIDR) receives a packet with address 131.23.151.76. The router’s routing table has the following entries:

Prefix	Output Interface Identifier
131.16.0.0 / 12	3
131.28.0.0 / 14	5
131.19.0.0 / 16	2
131.22.0.0 /15	1

The identifier of the output interface on which this packet will be forwarded is_______.

Given address 131.23. 151.76. coming to the first field of given routing table
⟹ 131.16. 0.0/12
131.0001 0111.151.76
131.0001 0000.0.0 ( given mask bits = 12)
⟹ 131.16.0.0 Matched
Coming to the 2nd field of given Routing table
⟹ 131.28.0.0/14
131.0001 0111. 151. 76
131.0001 0100.0.0 ( given mask bits = 14)
⟹ 131.20.0.0 not matched
Coming to the 3rd field of given Routing table
Error ! Not a valid link 131.19.0.0/16
131.0001 0111.151.76
131.0001 0111.0.0 ( given mask bits = 16)
⟹ 131.23.0.0 Not matched
Coming to the 4th field of given Routing table
⟹ 131.22.0.0/15
131.0001 0111.151.76
131.0001 0110.0.0 ( given mask bits = 15)
⟹ 131.22.0.0 Matched
We are getting 1st and 4th entries are matched so among them we have to picked up the longest mask bit, so output interface identifier is 1.

29 An IP router with a Maximum Transmission Unit (MTU) of 1500 bytes has received an IP packet of size 4404 bytes with an IP header of length 20 bytes. The values of the relevant fields in the header of the third IP fragment generated by the router for this packet are

30 Consider the following routing table at an IP

For each IP address in Group-I identify the correct choice of the next hop from Group-II using the entries from the routing table above.

List-I	List-II
A. 128.96.171.92	1. Interface 0
B. 128.96.167.151	2. Interface 1
C. 128.96.163.121	3. R2
D. 128.96.165.121	4. R3
	5. R4

Codes:

	A	B	C	D
(I)	1	3	5	4
(II)	1	4	2	5
(III)	2	3	4	5
(IV)	2	3	5	4

Do the BITWISE AND operation of group 1 with netmask given in the table.