MIT Practice Problems

Question

Ben Bitdiddle implements a reliable data transport protocol intended to provide "exactly once" semantics. Each packet has an 8-bit incrementing sequence number, starting at 0. As the connection progresses, the sender "wraps around" the sequence number once it reaches 255, going back to 0 and incrementing it for successive packets. Each packet size is S = 1000 bytes long (including all packet headers).

Suppose the link capacity between sender and receiver is C = 1 Mbyte per second and the round-trip time is R = 100 milliseconds.What is the highest throughput achievable if Ben's implementation is stop-and-wait?(in Kbytes/sec)

Answer 1

Throughput  = Efficiency * Link Capacity

So, first we will calculate the efficiency of stop and wait,

Efficiency (of stop and wait)= $tx/(tx+2tp)$

where tx = transmission time of packet.

and tp = one way latency

So, tx = L/B = $\frac{1000*10^3}{10^6}$ = 1msec

2*tp = 100msec (round trip time)

So, efficiency = $\frac{1}{1+100} = 0.0099 = 0.01(approx)$

Now throughput =efficiency * Link capacity = $\frac{0.01*10^6}{10^3} = 10KBps$

Comments

The answer is given as $10KBps$.

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sorry I had done mistake in units please check it now its right.