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What is the bandwidth-delay product for a 50-Mbps channel on a geostationary satellite?
If the packets are all 1500 bytes (including overhead), how big should the window
be in packets?
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+1 vote

$T_{p}$=270ms=0.27s (Standard for geostationary satellites)

$\therefore$RTT=0.27*2=0.54s

Bandwidth(B)=50Mbps=$50\times10^{6}bps (\because 1Mbps=10^{6}bps)$

Bandwidth*Delay product=B*RTT=$50*10^{6}*0.54\ bits=27000000\ bits$  (This answers the first part)

Bandwidth-Delay product means maximum number of bits that can be present in the network at any given point in time.

Packet size(L) given=1500 B = $1500\times8\ bits(\because 1Byte =8bits)$

Now,if we manage to find maximum number of packets that can be present in the network,we also find the optimal window size.

N=$B*RTT\over L$=$27000000\over{1500*8}$=2250(Optimal window size)

by Junior (909 points)