network layer

Question

station A in network 1 send a payload of 1600 B to station B in network 3 via network 2

if fragmentation is done the actual data size to be transmitted?

n/w 1: MTU=1500B

n/w 2: MTU=480B

n/w 3: MTU=1500B

Comments

total bytes recieved at dest B =total 1600 B data + 5*20(editted now....previous 5*100 was a mistake) ip header overhead=1700B

plz correct me if iam wrong..

and what should be the answer to total byte tranmitted...?

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total 1700B data is transmitted as 1600B payload and 100B header overhead.

Answer 1

As there are 3 networks, there will be 4 hosts.

Now, fragments for the 1st network will be as follows (as MTU is 1500 Byte):

1st fragment: 20B for IP + 1480B for payload
2nd fragment: 20B for IP + 120B for payload

Now, for 2nd network (as MTU is 480 Byte)

1st fragment: 20B IP header + 456B payload
2nd fragment: 20B IP header + 456B payload
3rd fragment: 20B IP header + 456B payload
4th fragment: 20B IP header + 112B payload

5th fragment: 20B IP header + 120B payload

Now, for the 3rd network:

Same as 2nd network because MTU for this greater than 2nd network.

So, total data received at host D = 100B for header + 1600B original payload = 1700B

ref: http://www.tcpipguide.com/free/t_IPMessageFragmentationProcess-2.htm

Comments

I didn't understand why the payload in the second network is 456 instead of 460 bytes?

and total data received at station D should be (40 bytes for header)+1600 byes(original payload)=1640

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@anky.  The payload must be some multiple of 8 bytes. Hence, 456.

Now, in the 3rd network, as 5 packets are passed to destination, 100B of header is received by network layer of destination.

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Dont you think that last two fragments of 112 and 120 Byte payload will be clubbed together in one fragment? So there will be total of 4 fragments adding additional (20*4=80B) of headers data. 1600+80 = 1680 B?

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bro they are two types  of fragmentation transparent and non transparent but in the question he did not mension any thing like that and they is no header length given and how can we take 20

Answer 2

Station 'A' (in Network 1 ) sends a payload of  $1600 B$  Via say Station 'AB'(in Network3) to Network B(in Network 2)

$\Rightarrow$ Station A  have $MTU=$1500 B$  so Maximum data Payload size $= $(1500-20)B=1480 B$

HEADER	PAYLOAD
20	1480

$\Rightarrow$ Station AB have $MTU=$1500 B$  so Maximum data Payload size $= $(1500-20)B=480 B$

HEADER	PAYLOAD
20	1480

$\Rightarrow$ Station B have $MTU=$480 B$ so Maximum data Payload size $= $(480-20)B=460 B$

HEADER	PAYLOAD
20	460

Now 1600 B has to be sent but maximum payload size=1480 so following fragments will be done by station A's Network Layer

Fragment1 :-

HEADER	PAYLOAD
20	1480

Fragment 2 :-

HEADER	PAYLOAD
20	120

1500 B and 140 B will reach Station AB and MTU of Station AB is 1500 B so following fragments will be there at network layer of station AB-:

Fragment 1:-

HEADER	PAYLOAD
20	1480

Fragment2:-

HEADER	PAYLOAD
20	20

Fragment 3:-

HEADER	PAYLOAD
20	140

all the 3 Fragments will reach Network layer of Station B,we will not bother about MTU of Station B as station B is destination ,it do not have to transfer data anyone.

P.S=> this answer is according to your data but i think you have swapped the value of MTU's of different Station. 

Comments

yes u r right threre is a little mistake in the que...i am correcting it

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where it has written that we need to take 20 bytes as header??

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for Network layer it is 20 byte and for transport layer it varies from 20 byte to 60 byte .so because of lack of sufficient information i took tcp header size as 20 B..what would be your choice ..???

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i thought question was incomplete. i didnt thought of any number as i didnt know about point mentioned by u.