IP Packet question

Question

Consider and IP packet arrived at a router of size 1850 bytes. If the header length field in IPv4 header is 1010 and the packet is sent to a router whose Maximum transmission unit is 160 bytes then what is the number of fragments that the IP packet will be divided?

• 14

• 15

• 16

• 17

Answer 1

Header field = $(1010)_{2}$ = $10$,

So actual header length = $10*4$ = $40 B$

MTU = $160 B$ = header + data = $40 + x => x = 120$

Now $120$ is also divisible by $8$ so we can use it fully

data in the incoming packet = $1850 – 40 = 1810 B$

So number of fragments = $\left \lceil \frac{1810}{120} \right \rceil$= $\left \lceil 15.08 \right \rceil$= $16$

Option C is the correct answer

Similar question in GATE : https://gateoverflow.in/8255/gate-cse-2015-set-2-question-52