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1 Answer

2 votes
2 votes
We have to send 800 + 20 = 820 bytes.

From  P to R1:

Maximum frame length supported is 1024 and 820 + 12 is within one frame- 832 bytes transferred.

From R1 to R2:

Maximum frame length is 256 bytes including 8 bytes header. We have 832 - 12 = 820 bytes (IP header added by P would be removed here). So, this requires ceil (820/(256 - 8)) = 4 frames - 3 * 256 + 820 - (248*3) + 8 = 852  bytes transferred.

From R2 to Q:

We need to send 248  bytes in a frame and maximum frame length is 512 bytes including 12 bytes header. So, four frames  frames are required - 3 * (248 + 12) + (76 + 12) = 868 bytes transferred.

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