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Consider a TCP message that contains 1024 bytes of data and 20 bytes of TCP header is passed to IP for delivery across two networks interconnected by a router (i.e., it travels from the source host to a router to the destination host). The first network has an MTU of 1024 bytes and the second has an MTU of 576 bytes. If all packets are correctly delivered, the number of bytes, including headers, are delivered to the IP layer at the destination for TCP message, in the best
case is _________ bytes. (Assume all IP headers are 20 bytes)
asked in Computer Networks by (187 points) | 33 views
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1124 Bytes?
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How??
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I was considering segmentation of the data at L4 also... but since the question is asking about the best case, so we need not consider that. magma soln is correct with a edit that 1044B goes to IP of sender insead of 1024B.

2 Answers

+1 vote
Best answer

Sender A ------------------------------------------Router----------------------------------------Receiver B

At sender A to Router : 

TCP data={1024+20}=1044

IP= packet 1(1004+20) , packet 2(40+20)    {here MTU is 1024}

Router To Receiver : now router send data to receiver according to MTU

 Packet11 (556+20) ,  Packet12 (448+20) , Packet 2(40+20)    (here packet 1 coming from sender is divided into two packets {Packet11 and Packwt12} and packet 2 is send same because packet 2 has less then MTU)

so data recieved at reciever IP layer={556+20+448+20+40+20=1104 B}

answered ago by Active (2.3k points)
selected ago by
+1 vote

Ans : 1084  B

 

answered by Loyal (5.9k points)
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But the given ans is 1104 bytes......
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1024+20 is passed to IP for delivery. So adding that would give 1104B ans.
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How 1024+20 I could not get you .Please elaborate


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