Question

Frames of 1000 bits are sent over a 10⁶ bps duplex link between two hosts. The propagation time is 25 ms. Frames are to be transmitted into this link to maximally pack them in transit (within the link).

What is the minimum number of bits (i) that will be required to represent the sequence numbers distinctly? Assume that no time gap needs to be given between the transmission of two frames. 

1. i = 2
2. i = 3
3. i = 4
4. i = 5

Answer 1

Correct Answer - Option 4 : i = 5

Data:

Bandwidth = 106 bps

Propagation time = 25 ms

Calculation:

In 1 second, 106 bits can be sent

∴ 1 s → 106 b

1 ms →  103 b

25 ms → 25 × 103 b

Frame size = 1000 bits

Number of frames = ⌈(25 × 103) ÷ (1000)⌉

∴ Number of frames = 25 

Minimum number of bits = i = ⌈log2 25⌉ = 5