Maximum throughput for a sliding window data transmission

Question

I am trying to understand how to calculate the throughput for a sliding window data transmission, by solving some numerical examples. Below is the example followed by my analysis.

Example

Host A is sending data to Host B over a full duplex link. A and B are using sliding window protocol, with send and receive window sizes of 5 each. Data packets sent only from A to B, are 1000 bytes each in size, and transmission time for one such packet is 50 us. Propagation delay is 200 us. Assume Ack packets need negligible transmission time. What is the maximum achievable throughput in Mbps?
A. 7.69 B. 11.11 C. 12.33 D. 15.00

Analysis

• Round trip-time is 2 * 200 us = 400 us.       ... A
• Time required to fill the sending window = window size (5) * transmission time of 1 packet (50 us) = 250 us.       ... B
• Since B < A, sender has to wait for ack to 1st packet before sending the 6th packet. This ack appears at 450 us. (round-trip time is 400 us.)
• Between 250 us and 450 us, the sender is idle, that is no new data is being sent over the line.
• Assuming sender has an unlimited supply of data frames, the above cycle would repeat.
• Thus, in every 450 us, sender sends 5 packets = 5 * 1000 * 8 = 40000 bits of data.
• Hence, throughput = 40000 bits / 450 us = 84.7710 megabits per second. (84.7710 Mbps)

However, this is not one of the given options, not even close! Is there any mistake in my analysis above?

Answer 1

Throughput = Window /RTT

Here Window size = 5*1000 bytes = 5000 bytes.

RTT = 50us + 2*200 us=> 450us.

Therefore Throughput= 5000 bytes/450 us = 11.11Mpbs

Answer 2

As I stated in my comment, your analysis and calculation method is correct. However, I'd check my calculator if I were you because 40000 bits / 450 us = 88.88...Mbps, not 84.7710 Mbps.

I do not think it is mere coincidence that 88.88 is just 11.11*8, so it's a fair assumption that the question was actually asking to get megabytes per second instead of megabits per second.