Fragmentation

Question

Suppose a network:-

Suppose a TCP message contains 492 B of data and TCP header is of size 20B then deliver it to IP. IP header is 20 B.

Assume link A-R1 has the maximum frame size of 700 B including a frame size of 14 B, link R1-R2 can support a maximum frame size of 256 B including an 8 B of the frame header and link R2-B can support a maximum of 256 B including a 12 B frame header.

Then calculate the no of fragments at each link and show the format of fragments in each link.

Comments

Do R2 need to wait until all fragments  arrived or can it start sending the fregment as soon as it receives?

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You can choose any of the options you want but that doesn't matter here as we are not calculating time.

Answer 1

TCP data size = 492 B
TCP header size = 20 B
IP header size = 20 B
So packet size = 492 B + 20 B + 20 B = 532 B

20 B

512 B

Now our data will travel from A - R1

Maximum frame size allowed = 700 B (include frame header)
Frame Header Size = 14 B
So maximum packet size allowed = 700 B - 14 B = 686 B
Since 532 B < 686 B, so no need of fragmentation.

Now our data will travel from R1 - R2

Maximum frame size allowed = 256 B (include frame header)
Frame Header Size = 8 B
So maximum packet size allowed  = 256 B - 8 B = 248 B
Since 532 B > 248 B, so here we will do fragmentation

Maximum fragment data allowed = 248 B - 20 B = 228 B
But we know that fragment must carry data in multiple of 8.
So, First fragment will carry a data of 224 B and IP header of 20 B i.e 244 B
second second fragment will carry a data of 224 B and IP header of 20 B i.e 244 B
Third fragment will carry a data of 64 B and IP header of 20 B i.e 84 B

Now our data will travel from R2 - B

Maximum frame size allowed = 256 B (include frame header)
Frame Header Size = 12 B
So maximum packet size allowed = 256 B - 12 B = 244 B
Since 244 B = 244 B, so here we don't do any fragmentation

Comments

Maximum fragment data allowed = 248 B - 20 B = 228 B 

why are you subtracting 20 B ?? 256-8 =248 is already a frame size then what is mean by 20 B here??

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Given is 256 B as FRAME size, not the PACKET size. And it include 8 B of FRAME header. So subtracting 8 gives 248 B which is IP PACKET size including IP HEADER, so subtracting 20 B from this which gives 228 B.

Mind the PDU's of the layer.

DLL -> FRAME

NETWORK LAYER -> PACKET.

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Okk got it thank you

Answer 2

At $A$ only 1 segment $<frame \space header \space byte, data \space bytes>$ will be generated of the form $<14,532>$

At $R1$ $3$ segment $<frame \space  header \space byte, data \space bytes>$ will be generated of the form $<8,248>, <8,248>, <8,36>$ After padding by $R1$ - $<8,40>$

At $R2$ $5$ segment $<frame \space header \space byte, data \space bytes>$ will be generated of the form $<12,240>, <12,240>, <12,8>, <12,8>, <12, 36>$ After padding by $R2$ - $<12,40>$

Comments

which layer computer is?

It is in Physical layer.

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@Bikram Sir, by "router is layer 3 device" does it mean to say data will pass from physical layer to network layer and then return to physical layer?

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data will pass from physical layer to network layer and then return to physical layer?

yes, we can think like this..