GATE 2009
March 14, 2025GATE 2009
| Question 57 |
Frames of 1000 bits are sent over a 106 bps duplex link between two hosts. The propagation time is 25ms. Frames are to be transmitted into this link to maximally pack them in transit (within the link).
What is the minimum number of bits (l) that will be required to represent the sequence numbers distinctly? Assume that no time gap needs to be given between transmission of two frames.
| I = 2 | |
| I = 3 | |
| I = 4 | |
| I = 5 |
Question 57 Explanation:
Transmission time (Tt) = 1000/106 seconds = 1 ms
Maximum number of frames that can be transmit to maximally pack them is = (Tt+2Tp)/Tx
= (25+1)/1 = 26 which is window size
Minimum sequence numbers required = 26
Minimum number of bits required for sequence number is 5.
Maximum number of frames that can be transmit to maximally pack them is = (Tt+2Tp)/Tx
= (25+1)/1 = 26 which is window size
Minimum sequence numbers required = 26
Minimum number of bits required for sequence number is 5.
Correct Answer: D
Question 57 Explanation:
Transmission time (Tt) = 1000/106 seconds = 1 ms
Maximum number of frames that can be transmit to maximally pack them is = (Tt+2Tp)/Tx
= (25+1)/1 = 26 which is window size
Minimum sequence numbers required = 26
Minimum number of bits required for sequence number is 5.
Maximum number of frames that can be transmit to maximally pack them is = (Tt+2Tp)/Tx
= (25+1)/1 = 26 which is window size
Minimum sequence numbers required = 26
Minimum number of bits required for sequence number is 5.
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