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A channel has bit rate of 1 Mbps and propagation delay of 270 msec. Frame size is of 125 bytes. Acknowledgment is always piggybacked onto data frames. Four bit sequence number is used, ignore header size.
What is the maximum achievable channel utilization for Go-Back-N?
  • a)
    1.48%
  • b)
    0.18%
  • c)
    2.95%
  • d)
    2.78%
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
A channel has bit rate of 1 Mbps and propagation delay of 270 msec. Fr...
Data rate = 1 Mbps
Tp = 270 msec
F = 125 bytes
= 125 x 8 bits
= 1000 bits
Four bit sequence number
For Go-back-N, W= 16-1 = 15
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Most Upvoted Answer
A channel has bit rate of 1 Mbps and propagation delay of 270 msec. Fr...
To calculate the maximum achievable channel utilization for Go-Back-N, we need to consider the given parameters such as bit rate, propagation delay, frame size, and the fact that acknowledgments are piggybacked onto data frames.

1. Calculating the Transmission Time:
The transmission time can be calculated using the formula:
Transmission Time = Frame Size / Bit Rate
Given that the frame size is 125 bytes and the bit rate is 1 Mbps, we can calculate the transmission time as follows:
Transmission Time = 125 bytes / (1 Mbps / 8) = 1000 µs

2. Calculating the Round-trip Time (RTT):
Round-trip time is the time taken for a data frame to be sent from the sender to the receiver and the acknowledgment to be received back. It is calculated as the sum of the propagation delay and twice the transmission time. Since the acknowledgment is piggybacked, we only need to consider the transmission time for the acknowledgment.
RTT = 2 * Transmission Time + Propagation Delay
Given that the propagation delay is 270 msec (270,000 µs), we can calculate the RTT as follows:
RTT = 2 * 1000 µs + 270,000 µs = 271,000 µs

3. Calculating the Efficiency:
The efficiency of the channel can be calculated using the formula:
Efficiency = (Frame Size / RTT) * Bit Rate
Given that the frame size is 125 bytes and the bit rate is 1 Mbps, we can calculate the efficiency as follows:
Efficiency = (125 bytes / 271,000 µs) * 1 Mbps

Converting bytes to bits:
Efficiency = (125 bytes * 8 bits/byte) / 271,000 µs * 1 Mbps
Efficiency = 1000 bits / 271,000 µs * 1 Mbps
Efficiency = 1000 / 271 * 1 Mbps
Efficiency ≈ 3.69 × 10^(-3) Mbps

4. Converting Efficiency to Percentage:
To express the efficiency as a percentage, we multiply it by 100:
Efficiency (%) = Efficiency * 100
Efficiency (%) ≈ 3.69 × 10^(-3) Mbps * 100
Efficiency (%) ≈ 0.369%

Therefore, the maximum achievable channel utilization for Go-Back-N is approximately 0.369%, which rounds off to 0.37% (option B).
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