Hello experts Im trying to figure out this problem below Ple

Hello experts! I\'m trying to figure out this problem (below). Please explain where are the 776 and 404 coming from and how to solve this problem using the method in the solution. I\'m lost with the procedure for this.

3. An IPv4 datagram of length 1200 bytes (including a 20-byte header) needs to be sent from node A to node D in the network shown below. The MTUs for links A-B, B-C and C-D are 800, 700 and 600 bytes, respectively. Compute the length and offset fields for each of the fragments of the datagram received at D. (6 pts) 8 9620-Up B 404+24245--> 424 424B FI 5 0 72 F3

Solution

Hi,

Here several calucalation has been done.
We will go through it very easily.

Firstly,
we have frame of 1200 B = 1180 B + 20 B header

I called it \"OUR frame\". so not to confuse with MTU\'s frame size.

Now this frame will try to pass through first MTU with frame size of 800 B, so won\'t be able to pass through.
So that it will be divided into 2.

Division of Frame : (Carefully look into it)
Our frame = 1180 B
MTU\'s frame = 800 B

So first dividing MTU\'s frame size into 2 = 800/2 = 400 B
Now, subtract this from our frame which is 1180-400 = 780 B
remaining our frame is 1180-780 = 400 B

BUT here we have to add CRC bits for error checking sequences so that we have to subtract 4 B from 780 B and add it to 400 B frame

So, now we have 780-4 = 776 B and 400 + 4 = 404 B frame

I hope you got this point very clearly, no where it is written so clear if go and find.

Heading to the last :
In this way all the frame has been calculated, so that result is
F1 = 576 + 20 WITH OFFSET = 0 (becasuse here no frame is above it)
F2 = 104 + 20 WITH OFFSET = 576/8 = 72
F3 = 96 + 20 WITH OFFSET = (576+104)/8 = 85
F4 = 404 + 20 WITH OFFSET = (576+104+96)/8 = 97.

Hope got all of this sum now.
Cheers !!!