Tuesday, Sep 27, 2011
This is an actual question from the 2010 mid-term exam.
Consider a network with a single 5 Mbps (megabits per second) link and a 15 ms (millisecond) propagation delay. The sender transmits a 3000-byte packet to a receiver.
Give all answers in units of milliseconds. Show your work. Circle your final answer.
1) 4.8 ms
dtrans = L/R
L = 3000 B = 24,000 b
24,000 b * 1000 ms 24
---------- = --- ms = 4.8 ms
5,000,000 b 5
2) 15 ms (this is the definition of propagation delay, which is given in the problem statement)
3) 19.8 ms
de2e = dtrans + dprop de2e = 4.8 ms + 15 ms = 19.8 ms
4) 24.6 ms
de2e = dtrans + dprop
dtrans = L/R
L = 6000 B = 48,000 b
48,000 b * 1000 ms 48
---------- = --- ms = 9.6 ms
5,000,000 b 5
de2e = dtrans + dprop
de2e = 9.6 ms + 15 ms = 24.6 ms
Note: dtrans for the 6000-byte packet could also be calculated by doubling dtrans for the 3000-byte packet.