Solution

If the M bit is 1, it means that there is at least one more fragment. This fragment can be the first one or a middle one, but not the last one. We don’t know if it is the first one or a middle one; we need more information (the value of the fragmentation offset). However, we can say the original packet has been fragmented because the M bit value is 1.

MF= 0 Offset= 0 -> No fragment

MF= 0 Offset= !0 -> Last fragment

MF= 1 Offset= 0 -> First fragment

MF= 1 Offset= !0 -> Middle fragment

Solution

The HLEN value is 12 in decimal, which means the length of the header is 12 × 4 = 48 bytes. In which minimum header length is 20 bytes and remaining will be of options filed.

Therefore 48 - 20 = 28 bytes are the options field.

Solution

Prevent the packets from being involved in an infinite loop and also to prevent server crashes. That's why the TTL field is used in the IPv4 datagram field.

Solution

Whenever an IP packet is transmitted, the value in the Time to Live (TTL) field will be decremented on every single hop. Hence, TTL is changed on every hop.

Now, since TTL changes, hence the Checksum of the packet will also change.

For the Fragmentation offset and MF will change only when fragmentation happens. A packet will be fragmented if the packet has a size greater than the Maximum Transmission Unit (MTU) of the network.

Solution

Solve from given choices of Packet size

Solution

Payload means only data without the header.

The application layer can forward any size of the data to the transport layer.

The transport layer can hold the maximum size of the data (payload) is 65,495 bytes.

The Network layer can hold the maximum size of the data is 65,515 bytes.

The data link layer can hold the maximum size of the data is 1500 bytes.

Solution

Minimum header size is 20 byte in IP header. so, HLEN Field = 5 bytes

Total data bytes in the packet = 1800 - 20(Header) = 1780

Size of data in MTU = 510 - 20(header) = 490, But it should be multiple of 8 so, it’s 488

Possible no. of fragments = ceil(1780 / 488) = 4

First fragment packet = 488(Data) + 20(Header)

Second fragment packet= 488 (Data) + 20(Header)

Third fragment packet= 488 (Data) + 20(Header)

Remaining data bytes = (1780 - 488 x 3) = 336

Hence the size of the last packet = 316 + 20 = 336

Therefore the size of the last packet is 336 bytes, and the total no of the fragmented packet is 4.

Solution

All the statements are correct.

Solution

The acknowledgement field in a segment defines the sequence number of the byte which is to be received next i.e. sequence number of the byte that the sender should transmit next.

Solution

Current sender window size (W_S) = Minimum (congestion window(W_C), Receiver window(W_R))

Therefore, Sender's window size = Min(5 KB, 7 KB) = 5 KB = 5 x 2^10 bytes = 5120 bytes.

There is no need to think about bytes sent and acknowledged.

Solution

In 4 secs, the counter increases by 4,64,000

for every 1 sec, counter will increases by 4,64,000/ 4 = 1,16,000

The sequence no. is 32 bit long and it can hold only 2^32 – 1.

So it takes (2^32 – 1)/(1,16,000) = 37025.58 seconds

Solution

All options are correct.

Solution

1st segment : 1001 to 3000 (1001+1999=3000)

2nd segment: 3001 to 5000

3rd segment: 5001 to 7000

4th segment: 7001 to 9000

5th segment: 9001 to 11000

6th segment: 11001 to 13000

7th segment: 13001 to 15000

8th segment: 15001 to 17000

9th segment: 17001 to 18000(17001 + 999 = 18000)

10th segment: 18001 to 19000

Solution

Given, the congestion window size is 16KB and RTT = 50ms, Maximum Segment Size = 2 KB.

When Time Out occurs, Threshold = Size of congestion window / 2 = 16 KB / 2 = 8 KB

The initial phase is the slow start phase, and it grows exponentially, i.e.,

First RTT = 2KB

Second RTT = 4KB

Third RTT = 8 KB

(After Threshold reached, it starts congestion avoidance phase, and it grows linearly)

Fourth RTT =10KB

Fifth RTT = 12KB

Sixth RTT = 14KB

Seventh RTT = 16KB

So, the total number of RTT is 7 x 50 ms ==>> 350 msec.

Solution

SYN (synchronise) Used in conjunction with ACK to request a connection or accept a connection.

SYN= 1 and ACK = 0 indicate a connection request(Request packet).

SYN= 1 and ACK = 1 indicate a connection accepted(Reply packet )

SYN= 0 and ACK = 1 is an acknowledgement(Pure ACK packet) and Data packet

SYN= 0 and ACK = 0 not possible