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A TCP message consisting of 2800 bytes is passed to IP for delivery across two networks. The first network can carry a maximum payload of 1500 bytes per frame and the second network can carry a maximum payload of 500 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?

Doubt:- Answer is given as 120 Bytes but I am getting 140 Bytes
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@sarannya

Do it this way..may be it will help u to understand better ,

The concep behind fragmentation is dependent on MTU, and payload to be sent across that network's MTU.

So here we have MTU of 1500bytes ( 1st network)

Data to be sent 2800 bytes

Divide it multiples of 8, that to in such  a way that after adding 20 bytes of (header) to the fragment we get <= MTU,  

So here we have , 1480+20 and 1320+20,

As 1480+1320=2800

Now for second network divide above fragments(only data part not IP header one) into multiples of 8,

1.)    1480 can be divided  as

         480+20, 480+20(IP PART), 480+20,24(4 bits appended to make remaining 20 data bits multiple of 8)+20

As480+480+480+20=1480

Same For 1320

480+20,480+20,64(same as above 4 bits appended)+20

Now add all the IP headers for second network..u will get 140 bytes...

Hope it helps!
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First line 1500 exclude excluding network overhead

Second line 500 excluding network overhead

From first line -->1500+1300

Second line-->  500 +500+500 and 500+500+300

Total 6 frame

Total overhead=6*20  = 120 Byte

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