Suggested Solution Keys for HW#5
CMSC506, Fall 05
(a) The worst case is: all stations want to send and s is the lowest
numbered station. Total wait time is N+(N-1)d (bit time), where the
first term represents the contention time, and the second term
Homework #1 solutions
1. The dog can carry 21 gigabytes, or 168 gigabits. A speed of 18 km/hour equals 0.005
km/sec. The time to travel distance x km is s/0.005 = 200x sec, yielding a data rate of
168/200x Gbps or 840/x Mbps. For x < 5.6 km, the dog has a
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Chpt1-P25. Suppose two hosts, A and B, are separated by 20,000
kilometers and are connected by a direct link of R = 2 Mbps.
Suppose the propagation speed over the link is 2.5 108
a. Calculate the bandwidth-delay product, R_ d .