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The round trip delay between X and Y is given as 60 ms and bottle neck bandwidth of link between X and Y is 512 KBps. What is the optimal window size (in packets) if the packet size is 64 bytes and channel is full duplex
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60B?
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No answer says $240$ but me and my friends agree on $480$
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480bits(60B) is correct.
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Does using full duplex affect it anyway?
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window size =(1+2a) where a =Tp/Tt
so it comes 481 bits
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@Rajesh,dont use formula

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it is 240 packets only. if the link is half duplex then in 1 rtt you will be able to send 1 window. but if link is full duplex you can send 2 windows in  1 rtt. so 2 windows size will become 60*512B. so number of packets per window will be 240
first find the bandwidth delay product = bandwidth * RTT = 512KBps * 60ms = 30720B.

If the size of the each packet is 64B.

then total no of packets = 30720B/64B = 480 packets.

IT is given wrong in made easy workbook

+1 vote
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