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
