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GATE CS 2009 Q57 — Transport Layer: Flow Control, Congestion Control, UDP, TCP, Sockets

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Frames of 1000 bits\text{1000 bits} are sent over a 10610^6 bps\text{bps} duplex link between two hosts. The propagation time is 25 ms\text{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)(I) that will be required to represent the sequence numbers distinctly? Assume that no time gap needs to be given between transmission of two frames.

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Answer: D

To find the minimum number of bits (II) required to represent the sequence numbers distinctly, we first determine the maximum number of frames that can be present in transit (within the link) at any given instant.

1. Calculate Transmission Delay (TtT_t)

The transmission delay is the time taken to push all the bits of a frame onto the link:

Tt=Frame SizeBandwidth=1000 bits106 bps=10−3 seconds=1 msT_t = \frac{\text{Frame Size}}{\text{Bandwidth}} = \frac{1000 \text{ bits}}{10^6 \text{ bps}} = 10^{-3} \text{ seconds} = 1 \text{ ms}

2. Calculate the Number of Frames in Transit

The propagation delay (TpT_p) is given as 25 ms25 \text{ ms}.

The maximum number of frames that can be "in transit" (one-way on the link) at any time is determined by the ratio of propagation delay to transmission delay:

Frames in transit=TpTt=25 ms1 ms=25 frames\text{Frames in transit} = \frac{T_p}{T_t} = \frac{25 \text{ ms}}{1 \text{ ms}} = 25 \text{ frames}

Alternative approach using Bandwidth-Delay Product (BDP):

BDP=Bandwidth×Tp=106 bps×(25×10−3 s)=25,000 bits\text{BDP} = \text{Bandwidth} \times T_p = 10^6 \text{ bps} \times (25 \times 10^{-3} \text{ s}) = 25,000 \text{ bits} Number of frames=25,000 bits1000 bits/frame=25 frames\text{Number of frames} = \frac{25,000 \text{ bits}}{1000 \text{ bits/frame}} = 25 \text{ frames}

3. Determine the Number of Bits (II)

To uniquely identify each of the 2525 frames simultaneously present in transit, we need at least 2525 distinct sequence numbers.

The number of bits II required to represent at least 2525 distinct values must satisfy:

2I≥252^I \ge 25

Taking the ceiling logarithm:

I=⌈log⁡2(25)⌉=5 bitsI = \lceil \log_2(25) \rceil = 5 \text{ bits}

Thus, the minimum number of bits required is 55.

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