Question

Host A is sending data to host B over a full duplex link. A and B are using the sliding window protocol for flow control. The send and receive window sizes are 5 packets each. Data packets (sent only from A to B) are all 1000 bytes long and the transmission time for such a packet is 50$\mu$s. Acknowledgment packets(sent only from B to A) are very small and require negligible transmission time. The propagation delay over the link is 200 $\mu$s. What is the maximum achievable throughput in this communication?

• A. $11.11\times {10}^6$ bps
• B. $7.69\times {10}^6$ bps
• C. $15.00\times {10}^6$ bps
• D. $12.33\times {10}^6$ bps

Answer 1

The correct option is A $11.11\times {10}^6$ bps

Given
Window size
n = 5 packets
Packet size = 1000 byte
Total packet size = 5 $\times$ 1000 = 5000 bytes
Total times = Transmission Time + Propagation Time
$=5\times 50+200\mu$s
$=250+200\mu$s
$=450\mu s$
$=450\times {10}^{-6}s$
Maximum Achievable throughput
$=\frac{TotalSize}{TotalTime}$
$=\frac{5000}{450\times {10}^{-6}}$
$=\frac{5000\times {10}^6}{450}$
$11.11\times {10}^6$bps