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).
- The propagation delay of the link is 100 milliseconds.
The minimum number of frames, the window size (rounded to the nearest integer) needed to achieve a link utilization of 50% is ___ .
GATE 2021 · Computer Networks · Network Protocols · medium
Answer: The minimum window size is 51 frames.
Calculate individual time components: Tt = 2000 / 10^6 = 2 ms
Tp = 100 ms (given)
Tack = 10 / 10^6 = 0.01 ms
Total round-trip cycle time = Tt + 2*Tp + Tack = 2 + 200 + 0.01 = 202.01 ms
Apply efficiency formula and solve for W: 0.50 <= W * 2 / 202.01
W >= 0.50 * 202.01 / 2
W >= 101.005 / 2
W >= 50.5025
Rounding to nearest integer: W = 51
Verify with W = 51: Utilization = 102 / 202.01 = 0.5049 = 50.49% >= 50%. Confirmed.
With W = 50: utilization = 100 / 202.01 = 49.5% < 50%. Not sufficient.