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78 votes
78 votes

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 $50$ $μs$. Acknowledgment packets (sent only from $B$ to $A$) are very small and require negligible transmission time. The propagation delay over the link is $200$ $μs$. What is the maximum achievable throughput in this communication?

  1. $7.69 \times 10^6$ Bps
  2. $11.11 \times 10^6$ Bps
  3. $12.33 \times 10^6$ Bps
  4. $15.00 \times 10^6$ Bps
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7 Answers

Best answer
163 votes
163 votes
We need the maximum throughput, and for that we need to send as much data
as possible to fully utilize the bandwidth.

so, maximum packets that can be sent $=1 + 2a = 9$ (after calculation) for $100\%$ efficiency.

But we have a window size of 5 only, so we can send only $5$ packets at max.

Efficiency $=\dfrac{5}{9}$

Now, $\dfrac{A}{Q},$ Bandwidth of the channel $(BW) =\dfrac{L}{T_t}$
$=\dfrac{1000}{(50\times 10^{-6})}$

$= 20\times 10^{6} \text{bytes/sec}.$

So, max. throughput achievable $=\text{Efficiency$\times $BW}$

$=\dfrac{5}{9}\times 20\times 10^{6}= 11.11\times 10^{6} \text{bytes/sec}.$

Correct Answer: $B$
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56 votes
56 votes
I think options are given in bytes per sec instead of bits per sec.

Transmission time $=50 \text{ micro sec}$
Propagation time $=200\text{ micro sec}$

RTT$= 50+2\times 200=450 \text{ microsec}$

(Receiver can send an $\text{ACK}$ as soon as the first packet is received)

Total number of bits transmitted before first $\text{ACK}$ is received,
$=1000\times 5\times 8\text{ bits}=40000\text{ bits}$

After first $\text{ACK}$ is received, the same cycle of action repeats.

So, Throughput$=\left(\dfrac{40000}{450}\right)\times10^{6} \text{ bits}$

$= 88.88\times 10^{6} \text{bits per sec}$

$=11.11\times 10^{6}\text{ bytes per sec}$
edited by
6 votes
6 votes
5000 bytes are being sent in 200 + 50×5 us.so troughput =5000/450  ×10^6 =》11.11×10^6 option b.
2 votes
2 votes
In this question we need not to add acknowledgement time of last packet because here for throughput they are assuming that this tramsmission is continuous for indefinite time(It is a real system based question)

so Tt=250 micro sec

  Tp=200 microsecond

  N= 5000 bytes

so throughput= 5000/450 microsec = b option
Answer:

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