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A TCP message consisting of 2800 bytes is passed to IP for delivery across two networks. The first network can carry a maximum payload of 1500 bytes per frame and the second network can carry a maximum payload of 500 bytes per frame, excluding network overhead. Assume that IP overhead per packet is 20 bytes. What is the total IP overhead in the second network for this transmission?

Doubt:- Answer is given as 120 Bytes but I am getting 140 Bytes

reopened | 379 views
+4
yes, it should be 140 bytes.

(496+20)*3 + (8+20) + (496+20)*2 + (312+20)
+1
Yes, it should be 140 Bytes.

Note that in the question MTU is given per frame, which is of form <20,20,1460> i.e.

+2
1500 bytes MTU is not including IP header
0
Thankyou!!

I too get the same answer.
0
pls explain this with solution
0

Can u please explain how u did that.

why is not  496*5 + 320*1 ?

0

@joshi_nitish.. i have a doubt that are the 2 networks one after the other or directly packet goes through the second network?

0
@gari,

that should have been mentioned in qsn to avoid ambiguity,

but according to intuition, it seems that the packet will cross n/w 1 before reaching n/w 2.

otherwise what the need of giving n/w 1.
0

https://gateoverflow.in/3731/gate2004-it-87

if we do by both ways this gate question gives same result =120

But in provided question

1)if we consider both networks we get 140

2)if only second network then 120

sometimes in the question some things are given to confuse .. so that why i was confused about it.

one more thing .. if we consider that  TCP message so definitely it will go by both networks..is this the correct catch or not?

0

Also see here they are following the strategy that the packet goes through second network https://www.geeksforgeeks.org/gate-gate-it-2004-question-87/

0

payload per frame means data portion of frame which includes ( IP header + TCP header + Application layer data ) so answer would be

(480+20) *3 , (40+20) , (480+20)*2 , (360+20)

​​​​

+1 vote
@sarannya

Do it this way..may be it will help u to understand better ,

The concep behind fragmentation is dependent on MTU, and payload to be sent across that network's MTU.

So here we have MTU of 1500bytes ( 1st network)

Data to be sent 2800 bytes

Divide it multiples of 8, that to in such  a way that after adding 20 bytes of (header) to the fragment we get <= MTU,

So here we have , 1480+20 and 1320+20,

As 1480+1320=2800

Now for second network divide above fragments(only data part not IP header one) into multiples of 8,

1.)    1480 can be divided  as

480+20, 480+20(IP PART), 480+20,24(4 bits appended to make remaining 20 data bits multiple of 8)+20

As480+480+480+20=1480

Same For 1320

480+20,480+20,64(same as above 4 bits appended)+20

Now add all the IP headers for second network..u will get 140 bytes...

Hope it helps!
by Junior (693 points)
0
yes, thanks.

But at 1st network its fragmented as (1496 + 20) and (1304 + 20) right?

(we should take largest number less than 1500 which is a multiple of 8. so its 1496, not 1480.)

At 2nd network it gets fragmented as: (496+20)*3 ,+ (8+20)3 and (496+20)*2  + (312+20)
+1
If you will take 1496+20 how can it pass the physical MTU LENGTH constraint !
0
MTU for 1st network is 1500 right? so 1496+20 satisfies that.

to go through 2nd network, it gets fragmented as  (496+20)*3 + (8+20)3 + (496+20)*2  + (312+20)
0
Yes...it's right... actually MTU payload is given...so..1496+20 is right...

If max MTU datagram limit would have been given...I think then...1480+20 would be right!
First line 1500 exclude excluding network overhead

Second line 500 excluding network overhead

From first line -->1500+1300

Second line-->  500 +500+500 and 500+500+300

Total 6 frame