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GATE | GATE-IT-2004 | Question 87

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  • Difficulty Level : Medium
  • Last Updated : 15 Feb, 2018

A TCP message consisting of 2100 bytes is passed to IP for delivery across two networks. The first network can carry a maximum payload of 1200 bytes per frame and the second network can carry a maximum payload of 400 bytes per frame, excluding network overhead. Assume that IP overhead per packet is 20 bytes. What is the total IP overhead in the second network for this transmission?
(A) 40 bytes
(B) 80 bytes
(C) 120 bytes
(D) 160 bytes


Answer: (C)

Explanation:

Maximum Transmission Unit) is defined as the frame or packet of the largest size that an be sent over any packet or frame based network such as internet. The TCP protocol uses the MTU to determine the maximum size of every packet that is transmitted.

 

net_04_87

In the figure shown above MTU (Maximum Transmission Unit) contains IP header, TCP header and the payload or the TCP MSS (maximum segment size)

Now, according to the question maximum payload is given. Therefore during fragmentation we will consider only the payload size and not the header size.

For the first network the maximum allowed payload size =1200 bytes per frame and for the second network the maximum allowed payload size= 400 bytes per frame.

Per packet IP overhead is given as 20 bytes.

So first we will calculate the total number of packets formed.

For first network 2100 bytes will be divided into 2 packets of size 1200 and 900 bytes.

So IP overhead of 1st network =(2* 20=40 bytes)

For second network 2100 bytes will be divided into 6 packets 5 of 400 bytes and 1 of 100 bytes.

So, IP overhead of the 2nd network =( 6* 20=120 bytes)

Thus, the maximum IP overhead for the 2nd network = 120 bytes

This solution is contributed by Namita Singh.


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