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Station $A$ is sending data to station $B$ over a full duplex error free channel. A sliding window protocol is being used for flow control. The send and receive window size is being used for flow control. The send and receive window sizes are $6$ frames each. Each frame is $\text{1200 bytes}$ long and the transmission time for such a frame is $\text{70 micro sec}$. Acknowledgment frames sent by $B$ to $A$ are very small and require negligible transmission time. The propagation delay over the link is $\text{300 micro sec}$. What is the max achievable throughput in this communication?
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Throughput is the amount of useful data sent per unit time. Useful data means payload.

Since the window size is $6$, the sender can send a maximum of $6$ frames till the first frame is acknowledged.

Time is taken by the first frame to get acknowledged= transmission time of original frame + propagation delay from src to destination + transmission time of ack frame + propagation delay from receiver to src.

Since ack transmission is to be neglected, we get total time$= 70+300+300 = 670 micro sec$.

The data sent during this time $= 6 frames= 6*1200bytes = 7200 bytes$.

Hence, throughput achieved is $7200 bytes/670 micro sec = 85.97Mbps$.
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$\text{Sending Window Size} = \text{Receiver Window Size} = \text{6 frames}$

$\text{Each frame size (L) = 1200 bytes}$

$ \text{Transmission time} (T_t) \text{for one frame is } 70 \hspace{0.1cm} \mu s$

$\therefore Bandwidth $$= \dfrac{L}{T_t} \\ = \dfrac{1200}{70} \\ = 17.14 \hspace{0.1cm} bytes/\mu s$

$\text{Propagation Delay}(T_p) = 300  \hspace{0.1cm} \mu s$

Now, $Max^m \text{ achievable throughput will be } $ $= Efficiency \times Bandwidth \\ = \dfrac{6 \times 70}{70 + 2 \times 300} \times 17.14 \\ = \dfrac{420}{670} \times 17.14 \\ = 10.74 \hspace{0.1cm}bytes/\mu s \\ = 10.744 \times 8 \hspace{0.1cm} bits / \mu s \\ = 85.94 \hspace{0.1cm} bits / \mu s \\ = 85.94 \times 10^6  \hspace{0.1cm}  bits / sec \\ = 85.94 \hspace{0.1cm} Mb/s $
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Throughout =n*L/Tt+2*Tp

=85.97mbps
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As the channel is full duplex so sender has to wait for only PT time before transmission

efficiency =N/1+a ,a=PT/TT =>300 micro sec/70 microseconds =>30/7

window size of sender= window size of receiver =N =6

efficiency =6/(1+30/7) =42/37

TT=L/B =>B=L/TT =>1200*8/70*10^-6 =>960/7 Mbps

maximum throughput =efficiency*bandwidth =>(42/37)*(960/7) =>155.67 Mbps

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