Test_series

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Is the Ans 180 ?

Answer 1

IP packet size=1500

HLEN=(10)₂=2, Actual Header size=2*4=8 bytes [Since the decimal value of HLEN signifies 4 bytes of header data]

However, 8 bytes of header is not practically implemented, min HLEN=(101)₂ i.e 5*4=20 bytes

Data in IP packet=1500-8(header)=1492

MTU=200, Data transfer limit=200-8=192 (since 192 is a multiple of 8, it's appropriate)

So, fragments will be like- 192(data) +8(header)

To transfer data of 1492 bytes of IP packet, we will need #fragments=ceil(1492/192)=8

1st fragment-192 +8, Frag. Offset=0

2nd fragment-192 +8, Frag. Offset=24

3rd fragment-192 +8, Frag. Offset=48

4th fragment-192 +8, Frag. Offset=72

5th fragment-192 +8, Frag. Offset=96

6th fragment-192 +8, Frag. Offset=120

7th fragment-192 +8, Frag. Offset=144

8th fragment-148 +8, Frag. Offset=168