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Stations $A$ and $B$ are connected through a line of bandwidth $64$ $\text{kbps}$. Station $A$ uses $16$ $\text{byte}$ packets to transmit messages to $B$ using a sliding window protocol. The round trip propagation delay between $A$ and $B$ is $50$ $\text{milliseconds}$. Determine the window size $A$ should use to maximize the line utilization. Assume that the ack frame is of negligible size and processing delay may be ignored. Justify your answer.
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Bandwidth between Stations A and B $=64\text{ kbps}=64000\text{ bps}$

Size of data packet $=16\text{ Byte}=128\text{ bits}$

So, Transmission time $(T_t)=\dfrac{\text{Size of data packet}}{\text{Bandwidth}}$

$=\dfrac{128}{64000}=.002\text{ sec}$

Round trip Propagation delay $(2\times T_p)= 50\text{ milisec}=0.05\text{ sec}$

For maximum Utilization in sliding window protocol window size of sender,

$=1+2\times \dfrac{T_p}{T_t}$

$=1+\dfrac{0.05}{0.002}=26$

Hence, window size of sender should be 26.

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Answer would be "25" for better understanding see this GATE question

https://gateoverflow.in/1820/gate2006-44

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Transmission time of frame $=\frac{ 16 * 8}{  64* 10^3} = 2ms $

Given $RTT=50 ms$

So number of frames transmitted in one $RTT = 50/2 = 25$ frames

To maximise the line utilisation $25$ frames must be sent ie, the window size of A should be is $25$.
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As soon as the first bit is uploaded in the link, it starts propogating, The acknowledgement for the frame is sent only once the last bit of the frame is received. So, the Round trip time is independent of the first frame we’re sending, 

Now, after the last bit of first frame is uploaded in the link, we stil have 50 msec left, We can still send 50/2 packets in the link, but the last frame sent will act as the first frame of the next window. Hence, effectively a total of 25 frames can be in one window including the first one.

For better clarification, consider the same parameters as above, but condier RTT = 4ms,

Now the diagram will look something like this

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