Question

Suppose that the stop-and-wait protocol is used on a link with a bit rate of 64 kilobits per second and 20 milliseconds propagation delay. Assume that the transmission time for the acknowledgement and the processing time at nodes are negligible. Then the minimum frame size in bytes to achieve a link utilization of at least 50% is ________.

Answer 1

Data:

 \(\eta ≥\frac{50}{100} ≥ \frac{1}{2}\)

T_p = 20 ms

BW = 64 Kb

Formula:

 

 \(\eta = \frac{1}{1 + 2a}\)

\(a = \frac{T_p}{T_t}\)

\(T_t = \frac{L}{BW}\)

Calculation:

\(\eta ≥ \frac{1}{2}\)

\(\frac{1}{1 + 2a} ≥ \frac{1}{2}\)
1 + 2a ≤ 2

2a ≤ 1

\(2\frac{T_P}{T_t} ≤ 1\)

2 × 20 × 10^-3   ≤ T_t

L ≥ 2 × 20 × 10-3  × 64 × 10³

L ≥ 40 × 64 bits

L ≥ 320 bytes

L_min = 320 bytes

he minimum frame size in bytes to achieve a link utilization of at least 50% is 320