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Assume the configuration below:

A ---------------- B --------------- C

where A-B has an MTU of 1500 bytes and B-C has an MTU of 400 bytes.

Assume that link layer frames require 30 bytes of overhead, an IP header is 20 bytes, there is no loss in the network, and A always tries to send full frames to C.

For every 1000 bytes of IP data (not including the IP header) that A sends, how many bytes must B send to C (including link layer and IP header overhead)?

  1. 1050 bytes

  2. 1133 bytes

  3. 1135 bytes

  4. 1450 bytes

in Computer Networks by Active (1.1k points)
edited by | 213 views
0
1150 bytes ?
0

 (380 + 20(header) + 30( overhead)) + (380 + 20(header) + 30( overhead))+( 240 +20 + 30)

= 1150

+2

@Pawan Kumar 2

1150 bytes is correct answer.......but....segment would be like this....

1.368+20+30 

2.368+20+30

3.264+20+30

0

@hs_yadav sir   I have doubt if overhead is considered in MTU then 368+20+30 = 418 >400 

but if not considered can we take it as

(374 +20 + 30)

(374 +20 + 30)

(252 +20 + 30)

?

+1

1.MTU is maximum the payload length that DLL of any network could hold....means it includes...complete ip datagram ...

2. second point..

374 +20 + 30)

(374 +20 + 30)

(252 +20 + 30)   ....you can't do it like this becoz fragment offset(data) in every fragment (except last) must be divisible by 8....that is why...we would  take it 368...

0
sorry ( 376 +  20 +30 ) and ( 248 + 30+ 20) ?e arlier was typo
+1
Fragments will be

376+20+30

376+20+30

248+20+30
+1

thank u sir :)

0
But here the transmission is happening in the IP link (network layer) then why are we adding frame header in every IP datagrams? Don't y'all think 1000 Bytes includes 30 Bytes (frame header) and only the IP header will be added to every datagrams? (that is the answer should be 1060)
0
Given answer is option 3.
0
@Vijay Can you direct us to the website you have got this question from? Thanks

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