Computer Networks - Computer Networks Section 3
- Option : A
- Explanation : Fourth octet of network address is 144 i.e.
10010000.
The first three bits of this octal is subnet bit (100), remaining bits can get maximum value as 11111.
So the maximum possible last octal IP address is 10011111 which is 159.
But the address with all 1s in host part is broadcast address and can't be assigned to a host. So the maximum possible host is 158.
- A
Query a NS for the root domain and then NS for the “dom” domains
- B
Directly query a NS for “dom” and then a NS for “mydomain.dom” domains
- C
Query a NS for in-addr.arpa and then a NS for 128.145.in-addr.arpa domains
- D
Directly query a NS for 145.in-addr.arpa and then a NS for 128.145.in-addr.arpa domains
- Option : C
- Explanation : We need to get NS for in-addr.arpa before doing query to 8.16.128.145.in-addr.arpa
33. Which of the following is NOT true with respect to a transparent bridge and a router?
- Option : B
- Explanation : Choice (b) is not true.
A bridge operates at layer 2 (Data Link Layer) so it uses MAC address while routes operates at layer 3 (Network Layer) so it uses IP addresses.
Consider three IP networks A, B and C. Host HA in network A sends messages each containing 180 bytes of application data to a host HC in network C. The TCP layer prefixes a 20 byte header to the message. This passes through an intermediate network B. The maximum packet size, including 20 byte IP header, in each network is
A : 1000 bytes
B : 100 bytes
C : 1000 bytes
The network A and B are connected through a 1 Mbps link, while B and C are connected by a 512 Kbps link (bps = bits per second).
- Option : D
- Explanation : Packet A sends an IP packet of 180 bytes of data
+20 bytes of TCP header +20 bytes of IP header
to B.
Ip layer of B now removes 20 bytes of IP header and has 200 bytes of data.
So, it makes 3 IP packets – (80 + 20, 80 + 20, 40 + 20) and sends to C as the Ip packet size of B is 100. So, C receives 260 bytes of data which includes 60 bytes of Ip headers and 20 bytes of TCP header.
Consider three IP networks A, B and C. Host HA in network A sends messages each containing 180 bytes of application data to a host HC in network C. The TCP layer prefixes a 20 byte header to the message. This passes through an intermediate network B. The maximum packet size, including 20 byte IP header, in each network is
A : 1000 bytes
B : 100 bytes
C : 1000 bytes
The network A and B are connected through a 1 Mbps link, while B and C are connected by a 512 Kbps link (bps = bits per second).
- Option : B
- Explanation : Ignoring errors, i.e. noise, we have
B = 2 H (Nyquist theorem)
= 354.5 kbps
