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A 3000 km long trunk operates at 1.536 mbps and is used to transmit 64 bytes frames and uses sliding window protocol. The propagation speed is 6 μ sec/km.
The minimum number of bits required in the sequence number field of the packet is
  • a)
    6 bits
  • b)
    7 bits
  • c)
    5 bits
  • d)
    4 bits
Correct answer is option 'B'. Can you explain this answer?
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A 3000 km long trunk operates at 1.536 mbps and is used to transmit 64...
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A 3000 km long trunk operates at 1.536 mbps and is used to transmit 64...
First we find the Transmission Delay
= Packet size / Bandwidth
= 64 bytes / 1.536 Mbps
= (64 × 8 bits) / (1.536 × 106 bits per sec)
= 333.33 μsec
Then we find the Propagation Delay
For 1 km, propagation delay = 6 μsec
For 3000 km, propagation delay = 3000 x 6 μsec = 18000 μsec
Value of a is calculated by  :
= Tp / Tt
= 18000 μsec / 333.33 μsec
= 54
Bits required in sequence number field
= ⌈log2(1+2 x a)⌉
= ⌈log2(1 + 2 x 54)⌉
= ⌈log2(109)⌉
= ⌈6.76⌉
= 7 bits
Hence, option 2) is correct.
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A 3000 km long trunk operates at 1.536 mbps and is used to transmit 64 bytes frames and uses sliding window protocol. The propagation speed is 6 μ sec/km.The minimum number of bits required in the sequence number field of the packet isa)6 bitsb)7 bitsc)5 bitsd)4 bitsCorrect answer is option 'B'. Can you explain this answer?
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