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Which of the following statements are True ?

in Computer Networks by Loyal (6.6k points) | 200 views
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I am getting S1 and S2 as an answer. Please anybody Verify it
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I'm getting  S1 only, can you explain how s2 is valid
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I am getting S2 and S3. There will be total 8 fragments that will be sent with 184+20 B for 7 fragments and 50+20 B for the last one. So MF bit of 6th fragment will also be 1. Offset of 4th fragment comes out to be 69.

total data will be 1100+10(UDP)+20(IP) =1130. But 20B will be discarded by router and total data would now be 1110 which is to be sent.

Please confirm the answer.
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@Gupta731 sorry there was a calculation glitch, below is the solution I got 

1100 data +10 header ==> 1110 total packet from Transport layer.

MTU=206(DATA+IP header)==>Data = 186/8 ==> 23.25 it contains decimal so take 23.

So the Data should be 184.

Total length = 184+20 => 204

No of packets= 1110/184 ==>6.032608695652174.==> 7 packets(S3 is true and S1 is false)

length of data for $7^{th}$ packet =( 6.032608695652174 - 6 )* 184 =6

Offset of $4^{th}$ fragment = (3 * 184)/8  ==>552/8==>69  (S2 is true).

Conclusion:

S1 is false S2 is true S3 is true

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Yes exactly. But total packets will be 8 it seems.

As 184+20 =204 will be in each packet, but only 184 B of data will be present. It will take 8 packets to send the data completely along with the total overhead of IP headers.
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No of packets= 1110/184 ==>6.032608695652174.==> 7 packets(S3 is true and S1 is false

But I got 7 packets only 

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We could wait for @Na462 to verify with the answers provided.

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Answer is S2 and S3.
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There should be total of 6 fragmentation

See for network layer the payload is = 1100 + 10 = 1110 Bytes which will be fragmented = 1110 / (206-20) = 6

The fragments will be as = (186 + 20),(186+20),(186+20),(186+20),(186+20),(180,20)

I dont understand how 7 packets ???
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@Na462 that was the trap set.

You should not consider 186 as data length it should be 184 because it should be divisible by 8 as we scale down for offset.

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I didn't understand If I take 184 instead of 186 the utilization will decrease as overhead will increase. I don't understand your reason brother. Can u elaborate a little please. Or some reference
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See what will be the offset of 1st packet or 4th packet ?? You will get a decimal value which cannot be stored. so we we have to round value to nearest decimal but we should not take ceil because it crosses MTU so we take floor. Hope you got this
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I got your point brother but can't we take the floor of that decimal offset and store it ?
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Yes we have to take the floor not ceil

2 Answers

0 votes

Answer. S2 and S3.

From transport layer data is 1100Bytes. Transport layer will add header(size given 10Bytes) and will send to network layer.

Therefore total data send to network layer will be 1110 Bytes,

Now MTU = 206 including ip header therefore max. data that can be send is 206-20 = 186 Byte

since 186B is not divisible by 8, so fragment size will be 184B.

S1)#Fragements = Ceil(1110/8) = 7

So, Fragment 1 to fragment 6 size will 184B

and fragment 7 size will be 8B(although remaining is 6B, but to make it divisible by 8 we add padding to make it 8B)

S2) $184/8 = 23$

offset of 4$^{th}$ fragment is $3*23 = 69$

 

S3) 6$^{th}$ fragment is not the last fragment , therefore MF bit will 1. for 7$^{th}$ fragment mf bit will be 0.

by (469 points)
0 votes

S2,S3   are TRUE

S1 FALSE

by Active (3.9k points)

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