Data Communication and Computer Networks - Functions of OSI and TCP/IP Layers
- Option : A
- Explanation :
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.
- Option : C
- Explanation : 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
| 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_______.
- Option : A
- Explanation : 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.
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 |
