Homework Answers
Solution
1)
The formula for finding minimum distribution time is
Dcs = max{NF/us , F/dmin}
N=6
F=2 Gbits
us=56 Mbps
N F/us
=6(2000)/56
=12000/56
=214.28
F/dmin
=2000/12
=166.66
214.28>166.66
=166.66 sec
So the minimum time needed to distribute the file using P2P download
166.66 secs
---
2)
N=6
F=2 Gbits
us=56 Mbps
DP2P = max{F/us , F/dmin , NF/(us + ∑ui)}
F/us
2000/56=35.71
F/dmin
2000/12=166.66
NF/(us + ∑ui)
(6*2000)/(56+(27+21+24+29+29+21))
=57.97
=max{35.71,166.66,57.97}
---
all the best
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