Computer Networks

Answer

True

Explanation

The Network layer is responsible primarily for logical addressing, routing, and forwarding packets from source to destination.

When a message is divided into multiple packets, the Network layer can treat those packets independently, particularly in a connectionless datagram network. Different packets may take different routes and may arrive out of order.

The Network layer does not provide the end-to-end relationship, sequencing, acknowledgements, or retransmissions required for reliable delivery. Those functions are normally provided by the Transport layer when a reliable transport protocol such as TCP is used.

Therefore, the statement that the Network layer does not establish a relationship among different packets of the same message is True.

Final answer: True.

Answer

(i) 245.248.136.0/21 and 245.248.128.0/22

Step-by-step solution

Given block: 245.248.128.0/20.

A /20 IPv4 block contains:
$2^{32-20}=2^{12}=4096$ addresses.

So the complete range is:
245.248.128.0 to 245.248.143.255.

1. Organization A needs half

Half of 4096 = 2048 addresses.
A 2048-address block corresponds to /21 because:
$2^{32-21}=2048$.

The valid /21 sub-blocks inside the /20 are:

• 245.248.128.0/21 → 245.248.128.0–245.248.135.255
• 245.248.136.0/21 → 245.248.136.0–245.248.143.255

2. Organization B needs one quarter

One quarter of 4096 = 1024 addresses.
A 1024-address block corresponds to /22 because:
$2^{32-22}=1024$.

If A receives 245.248.136.0/21, the remaining lower half is 245.248.128.0–245.248.135.255. B can receive 245.248.128.0/22, which covers 245.248.128.0–245.248.131.255.

The ISP then retains 245.248.132.0–245.248.135.255, which is another /22 block.

Thus the allocations do not overlap and exactly satisfy the required 1/2 and 1/4 sizes.

Checking the options

• (i) A = 245.248.136.0/21, B = 245.248.128.0/22 → valid, non-overlapping. ✅
• (ii) A = 245.248.128.0/21 and B = 245.248.128.0/22 → overlap. ❌
• (iii) A and B use overlapping address space and the block sizes/positions are invalid for the required allocation. ❌
• (iv) A = 245.248.136.0/22 is only 1024 addresses, not the required 2048 for half. ❌

Final answer: (i) 245.248.136.0/21 and 245.248.128.0/22.

Answer

(ii) 32 bits

Explanation

An IPv4 address is 32 bits long, normally written as four 8-bit octets in dotted-decimal notation, for example 192.168.1.10.

The IPv4 header contains two 32-bit address fields:

• Source address: 32 bits
• Destination address: 32 bits

Thus, each individual IPv4 address requires 32 bits.

Final answer: (ii) 32 bits.

Answer

(i) Error reporting

Explanation

IP is responsible for logical addressing, packet/datagram format, and packet handling/routing conventions. However, IP itself is a connectionless best-effort delivery protocol and does not provide error reporting as a core IP function.

Error reporting for IP-related delivery problems is provided by ICMP (Internet Control Message Protocol), which works alongside IP.

Therefore, among the given choices, error reporting is not directly an IP function.

Final answer: (i) Error reporting.

Answer

(i) Error reporting

Explanation

IP is responsible for logical addressing, datagram format, and packet forwarding/handling across interconnected networks. However, error reporting is provided by ICMP, which operates alongside IP rather than being a core function of IP itself.

Therefore, among the given options, error reporting is the one not directly applicable as an IP function.

Final answer: (i) Error reporting.

Answer

(ii) IP

Explanation

IP (Internet Protocol) is the Network-layer protocol responsible for logical addressing and forwarding datagrams between networks, making internetwork communication possible.

• TCP: Transport-layer protocol providing reliable, connection-oriented delivery.
• IP: Network-layer protocol providing logical addressing and datagram delivery across interconnected networks.
• ISP: Internet Service Provider, not a protocol.
• TCP/IP: A protocol suite, not the name of the individual internetworking protocol asked for here.

Final answer: (ii) IP.

Answer

(iv) Router

Explanation

A logical address is an IP address. Routers operate at the Network layer and use logical IP addresses to make forwarding decisions between different networks.

• Hub → Physical layer; does not use logical addressing for forwarding.
• Switch → Data Link layer; normally forwards using MAC addresses.
• Bridge → Data Link layer; uses MAC addresses.
• Router → Network layer; uses IP/logical addresses.

Final answer: (iv) Router.

Answer

(iv) 60 bytes

Explanation

For IPv4, the base header is 20 bytes. The header length is specified by the IHL field in units of 32-bit words.

The IHL field is 4 bits, so its maximum value is 15 words.

Each word = 4 bytes.

Maximum header size:
$15 \times 4 = 60$ bytes.

Therefore, the maximum IPv4 header size is 60 bytes.

Final answer: (iv) 60 bytes.

Answer

(iv) None of the above

Explanation

All three listed functions can be associated with the Network layer:

1. Routing: Determines suitable paths for packets through an internetwork.
2. Internetworking: IP provides communication across interconnected networks.
3. Congestion control: Network-layer mechanisms and protocols can participate in congestion management, although congestion control is also handled at the Transport layer in systems such as TCP.

Therefore, none of options (i), (ii), or (iii) is the intended answer as a non-function.

Final answer: (iv) None of the above.

Answer

Flooding is a routing technique in which a router forwards an incoming packet on all outgoing links except the link on which it arrived, subject to rules that prevent indefinite circulation.

Key idea

Instead of calculating one best route, the packet is propagated throughout the network. A packet can reach the destination through multiple paths.

Advantages

• Very simple routing method.
• Does not require complete routing information.
• Can be highly robust because multiple copies may reach the destination.

Disadvantages

• Creates many duplicate packets and wastes bandwidth.
• Can cause congestion unless controlled.

Common controls include a hop count/TTL and packet sequence numbers to discard duplicates.

Exam-ready definition: Flooding forwards packets over all possible outgoing paths, with mechanisms such as hop limits used to control duplicates and loops.

Answer

(iii) DHCP

Explanation

The question asks which protocol is not used to resolve one form of address to another.

• DNS: Resolves domain names to IP addresses (and supports reverse/other mappings).
• ARP: Resolves an IPv4 address to a MAC address on a local network.
• RARP: Historically resolved a MAC address to an IP address.
• DHCP: Dynamically assigns network configuration such as an IP address, subnet mask, gateway, and DNS server information. It is not primarily an address-resolution protocol.

Therefore the correct answer is (iii) DHCP.

Final answer: (iii) DHCP.

Answer

(ii) 255.0.0.0

Explanation

In the traditional classful IPv4 addressing scheme, a Class A network uses:

• 8 bits for the network portion.
• 24 bits for the host portion.

Therefore its default prefix is /8.

The corresponding subnet mask is:

$255.0.0.0$

In binary:

11111111.00000000.00000000.00000000

Thus the correct answer is (ii) 255.0.0.0.

Note: 127.0.0.0/8 is reserved for loopback, and 255.255.255.0 is a /24 mask, not the default Class A mask.

Solution

The IP packet has:

• Header size = 20 bytes
• Payload = 4000 bytes
• MTU = 1500 bytes

The maximum payload that can fit in one fragment is:

$1500 - 20 = 1480$ bytes.

Since IPv4 fragment offsets are measured in units of 8 bytes, 1480 bytes is valid because $1480/8=185$.

Therefore the 4000-byte payload is divided as:

• Fragment 1: 1480 bytes payload
• Fragment 2: 1480 bytes payload
• Fragment 3: 1040 bytes payload

The fragment offsets are:

• Fragment 1: $0/8=0$
• Fragment 2: $1480/8=185$
• Fragment 3: $(1480+1480)/8=370$

Fragment	Payload	Total Length	Fragment Offset	MF bit
1	1480 bytes	1500 bytes	0	1
2	1480 bytes	1500 bytes	185	1
3	1040 bytes	1060 bytes	370	0

The MF (More Fragments) bit is 1 for the first two fragments and 0 for the final fragment.

Thus, the original packet is transmitted as three IPv4 fragments with offsets 0, 185, and 370.