A link has a transmission speed of $10^6$ bits/sec. It uses data packets of size 1000 bytes each. Assume that the acknowledgment has negligible transmission delay, and that its propagation delay is the same as the data propagation delay. Also assume that the processing delays at nodes are negligible. The efficiency of the stop-and-wait protocol in this setup is exactly 25%. The value of the one-way propagation delay (in milliseconds) is ______.

Given parameters:

• Bandwidth, $B = 10^6\text{ bits/sec}$

• Packet size, $L = 1000\text{ bytes} = 8000\text{ bits}$

• Efficiency, $\eta = 25\% = 0.25$

1. Calculate transmission delay ($T_t$):

   $$T_t = \frac{L}{B} = \frac{8000\text{ bits}}{10^6\text{ bits/s}} = 8 \times 10^{-3}\text{ s} = 8\text{ ms}$$

2. Efficiency in Stop-and-Wait protocol is given by:

   $$\eta = \frac{T_t}{T_t + 2 T_p}$$

   $$0.25 = \frac{8}{8 + 2 T_p}$$

   $$8 + 2 T_p = \frac{8}{0.25} = 32$$

   $$2 T_p = 24 \implies T_p = 12\text{ ms}$$

Thus, the one-way propagation delay is 12 milliseconds.