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A channel has a bit rate of 4 kbps and one-way propagation delay of 20 ms. The channel uses stop and wait protocol. Thetransmission time of the acknowledgement frame is negligible. To get a channel efficiency of at least 50%, the minimumframe size should be
  • a)
    80 bytes
  • b)
    80 bits
  • c)
    160 bytes
  • d)
    160 bits
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
A channel has a bit rate of 4 kbps and one-way propagation delay of 20...
for 50% utilization
tt/(tt+2tp)>=1/2
2tt>=tt+2tp
tt>=2tp
L/B>=2*tp
L>=2*tp*B
so here L=2*20*10^-3*4*10^3
=160 bits
so ans is d
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Most Upvoted Answer
A channel has a bit rate of 4 kbps and one-way propagation delay of 20...
Explanation:

The channel bit rate is 4 kbps, which means that the maximum number of bits that can be transmitted in 1 second is 4,000 bits. The one-way propagation delay is 20 ms, which means that it takes 20 ms for a bit to travel from the sender to the receiver.

Stop and Wait Protocol

In stop and wait protocol, the sender sends a frame to the receiver and waits for an acknowledgement from the receiver. The receiver sends an acknowledgement back to the sender after receiving the frame. Only after receiving the acknowledgement, the sender can send the next frame.

Channel Efficiency

The channel efficiency is the ratio of the useful data transmitted to the total time taken to transmit it. To get a channel efficiency of at least 50%, the useful data transmitted should be at least half of the total time taken to transmit it.

Minimum Frame Size

The minimum frame size can be calculated using the following formula:

Minimum Frame Size = 2 * Propagation Delay * Channel Bit Rate

Substituting the given values, we get:

Minimum Frame Size = 2 * 20 ms * 4 kbps
Minimum Frame Size = 160 bits

Therefore, the minimum frame size should be 160 bits to get a channel efficiency of at least 50%.
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A channel has a bit rate of 4 kbps and one-way propagation delay of 20 ms. The channel uses stop and wait protocol. Thetransmission time of the acknowledgement frame is negligible. To get a channel efficiency of at least 50%, the minimumframe size should bea)80 bytesb)80 bitsc)160 bytesd)160 bitsCorrect answer is option 'D'. Can you explain this answer?
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