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Frames of 20000 bytes are sent over 10 MBps full duplex link between 2 hosts. Propagation is 45 ms.Suppose that the sliding window protocol is used with the sender window size of 16 and acknowledgements are piggybacked.After sending 35 frames , what is the minimum time sender will have to wait before starting transmission of the next frame
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3.75 ms?
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How? I am getting 20..
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With given parameters max. Window size = 46, but here window size = 16 i.e. rest of the time is wasted. Assuming wasted time is equally divided between transmiitted frames.. I got 3.75 ms.

@anjali007 how you got 20?

1 Answer

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Bandwidth is given 10 MBps and the length of the packet is 20000 bytes ..

So Transmission time is L/B i.e 20000/10*10^6=2 ms

Now we are given the propagation delay that is one way trip time =45ms

So we are required to tell what is the minimum time we have to wait for the acknowledgement of the 1st packet and we have transmitted 35 packets.. we  will get the acknowledgement after 2*Tp=90ms and 2ms for Transmission of the ack

Time taken to send 35 packets is 35*2 =70 ms So we have to wait for 92-70=22ms

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why 35*2 and why not window_size*2 ?
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because it is clearly stated that we have transmitted 35 packets.. that is why I went with it
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But the window size is 16 so how can we have 35 frames which have not been acknowledged in the buffer?
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why you haven't consider this "acknowledgements are piggybacked"? so transmission time of each frame should be added.
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m not getting what is wrong here

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@Shubhgupta @jatin khachane 1 so how come we are transmitting 35 packets if the acknowledgements havent come for 1 st packet only .. the best I could infer was this only..

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as mentioned in question "ack are piggybacked" so frame size of ack will also be 20000 bytes? Isn't it? So we'll add transmission of that also?
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@Shubhgupta

Can you please point wht is wrong in above figure ?

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ya it should be 2ms for it also @Shubhgupta sorry I forgot to mention it.. 

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