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Official PYQ · 2016Set 1 · Q54Transport Layer: Flow Control, Congestion Control, UDP, TCP, SocketsNAT1 mark

GATE CS 2016 Set 1 Q54 — Transport Layer: Flow Control, Congestion Control, UDP, TCP, Sockets

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For a host machine that uses the token bucket algorithm for congestion control, the token bucket has a capacity of 11 megabyte\text{megabyte} and the maximum output rate is 2020 megabytes\text{megabytes} per second\text{second}. Tokens arrive at a rate to sustain output at a rate of 1010 megabytes\text{megabytes} per second\text{second}. The token bucket is currently full and the machine needs to send 1212 megabytes\text{megabytes} of data. The minimum time required to transmit the data is _____________ seconds\text{seconds}.
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Answer: 1.10:1.19

To find the minimum time required to transmit the data, we can divide the transmission process into two phases:

1. Burst Phase (Maximum Transmission Rate)

During this phase, the host transmits at the maximum output rate MM by utilizing both the tokens already present in the bucket and the new tokens arriving. Let tt be the duration of this burst phase.

  • Bucket Capacity (CC): 1 MB1\text{ MB}
  • Maximum Output Rate (MM): 20 MB/s20\text{ MB/s}
  • Token Arrival Rate (ρ\rho): 10 MB/s10\text{ MB/s}

The total data sent during the burst phase is given by:

Data Sent=M⋅t=C+ρ⋅t\text{Data Sent} = M \cdot t = C + \rho \cdot t

Solving for tt:

t=CM−ρ=1 MB20 MB/s−10 MB/s=0.1 secondst = \frac{C}{M - \rho} = \frac{1\text{ MB}}{20\text{ MB/s} - 10\text{ MB/s}} = 0.1\text{ seconds}

The amount of data transmitted during this burst phase of 0.1 seconds0.1\text{ seconds} is:

Dburst=M⋅t=20 MB/s×0.1 s=2 MBD_{\text{burst}} = M \cdot t = 20\text{ MB/s} \times 0.1\text{ s} = 2\text{ MB}

2. Sustained Phase (Token Arrival Rate)

After the bucket is emptied (at t=0.1 st = 0.1\text{ s}), the host can only transmit data as fast as new tokens arrive, which is at the rate of ρ=10 MB/s\rho = 10\text{ MB/s}.

  • Remaining Data to transmit: 12 MB−2 MB=10 MB12\text{ MB} - 2\text{ MB} = 10\text{ MB}
  • Transmission Rate: 10 MB/s10\text{ MB/s}

The time required to transmit the remaining data is:

tremaining=10 MB10 MB/s=1 secondt_{\text{remaining}} = \frac{10\text{ MB}}{10\text{ MB/s}} = 1\text{ second}

Total Time

The total minimum time required to transmit the entire 12 MB12\text{ MB} of data is:

Ttotal=t+tremaining=0.1 s+1 s=1.1 secondsT_{\text{total}} = t + t_{\text{remaining}} = 0.1\text{ s} + 1\text{ s} = 1.1\text{ seconds}

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