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Question

If a datagram of size 4000 bytes from transport layer arrives at network layer and it has to be forwarded through a link with maximum capacity of 800 bytes. Then calculate the number of fragments needed if the header size is 20 bytes. Also calculate the data size of the last fragment.

1. 5 fragments ,100 bytes

2. 5 fragments ,104 bytes

3. 6 fragments ,100 bytes

4. 6 fragments ,104 bytes

Comments

I didn't get 1 thing here.

4000 bytes is the network layer payload or it is (NL data+20 bytes of header)?

and it is not the case that always the last fragment data size should be divisible by 8 all fragments preceding the last fragment must contain data size that is divisible by 8

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it is not the case that always the last fragment data size should be divisible by 8 all fragments preceding the last fragment must contain data size that is divisible by 8

yeah but what I read that we didn't check for the last fragment that is it divisible by 8 or not 

but whenever I solve  this type of question  I observe that  that last element is also divsible by 8 but

thanks I rectify my mistake from here  

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Now examine it that

5 x 780 B =  3900 B

4000 - 3900 =  100 B is as last fragment Ok ???

Now 780 is not divisible by  8

therefore each fragment should be  size of  776 B

5 x 4 = 20

120 B ---> size of the last Fragment

Answer 1

 

Byte Address	776 +(20 header)	776 +(20)	776 +(20)	776 +(20)	776 +(20)	100 +(20)
Fragment Offset	0	97	194	291	388	485
M-bit	1	1	1	1	1	0

 

 

IP-Payload= 3980bytes ; MTU= 780bytes

Fragment Offset: ByteAddress/8 ; 

Hence No. of fragments required to accomodate 4000bytes: 6

Answer 2

The number of fragments needed would be 6, and the data size of the last fragment would be 100 bytes. This is because, the total size of the datagram is 4000 bytes and header size is 20 bytes so, the total data size is 4000 - 20 = 3980 bytes. And the maximum capacity of the link is 800 bytes, so the number of fragments needed is 3980/800 = 4.975 = 6 (rounding up). The last fragment will have the remaining data, which is 3980 - (6-1)*800 = 100 bytes.