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Consider the sliding window flow-control protocol operating between a sender and a receiver over a full-duplex error-free link. Assume the following:

  • The time taken for processing the data frame by the receiver is negligible.
  • The time taken for processing the acknowledgement frame by the sender is negligible.
  • The sender has infinite number of frames available for transmission.
  • The size of the data frame is $2,000$ bits and the size of the acknowledgement frame is $10$ bits.
  • The link data rate in each direction is $1$ Mbps ($=10^6$ bits per second).
  • One way propagation delay of the link is $100$ milliseconds.

The minimum value of the sender's window size in terms of the number of frames, (rounded to the nearest  integer)  needed  to achieve a link utilization of $50\%$ is_____________.

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Let the sender window size be $N$.

$\text{One way propagation delay} =100\,ms=0.1\,s$

$\text{Transmission delay}_{packet}=\frac{\text{Size of data frame}}{\text{Link bit rate}}=\frac{2000}{10^{6}}=0.002\,s$

$\text{Transmission delay}_{ACK}=\frac{\text{Size of ACK frame}}{\text{Link bit rate}}=\frac{10}{10^{6}}=0.00001\,s$

$\text{Link Utilization}(\eta )=\frac{\text{Useful Data Transfer time}}{\text{Total time}}$

$\implies \eta = \frac{N.(T_{f})}{T_{f}+2(T_{p})+T_{ACK}}$

$\implies N = \left \lceil \frac{\eta.\left(T_{f}+2(T_{p})+T_{ACK}\right)}{T_{f}}\right \rceil = \left \lceil \frac{0.5(0.002+0.2+0.00001)}{0.002}\right \rceil =\lceil 50.5025 \rceil  =51$

Correct Answer: $51$
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let window size =N

a=Tp/Tt

a=100*(10^-3)*(10^6/2000)

a=100/2=50

so,to get 50% utilization

 

N/(1+2a)>=(1/2)

N>=(1+2a)(1/2)

N>=(1+100)(1/2)

N>=51

The minimum size of window size is 51
Answer:

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