Question

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 (= 10⁶ 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  .

Hint:

Large propagation delays significantly increase the required window size to maintain good utilization.

Answer 1

Correct Answer: 50

Step 1: Compute transmission times.
Data frame size \( = 2000 \) bits, link rate \( = 1 \) Mbps.
\[ T_f = \frac{2000}{10^6} = 0.002 \text{ s} \] Acknowledgement frame size \( = 10 \) bits:
\[ T_a = \frac{10}{10^6} = 0.00001 \text{ s} \]

Step 2: Compute propagation delay.
One-way propagation delay \( = 100 \) ms \( = 0.1 \) s.
Round-trip propagation delay:
\[ 2T_p = 0.2 \text{ s} \]

Step 3: Compute utilization formula.
For sliding window protocol, utilization is given by:
\[ U = \frac{W \times T_f}{T_f + 2T_p} \]

Step 4: Substitute values for 50% utilization.
\[ 0.5 = \frac{W \times 0.002}{0.002 + 0.2} \]

Step 5: Solve for \( W \).
\[ W = \frac{0.5 \times 0.202}{0.002} = 50.5 \]

Step 6: Round to nearest integer.
\[ W \approx 52 \] % Final Answer

Final Answer: \[ \boxed{52} \]