Chapter 6
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stations is again the time it takes for N/2 to send, which with the normalization above is N/2 time units. Power is thus 2/N . The graph is omitted. Ethernet throughput, delay, power (Exercise 3(a))
1 0.9
50
throughput
0.8
40
0.6
30
0.5 0.4
delay
throughput, power
0.7
20
delay
0.3 0.2
10
0.1
power
0
0 0
2
4
6
8
10
Load, N
Both power curves have their maximum at N = 1, the minimum N , which is somewhat artificial and is an artifact of the unnatural way we are measuring load. 4. Throughput here is min(x, 1), where x is the load. For x ≤ 1 the delay is 1 second, constantly. We cannot sustain x > 1 at all; the delay approaches infinity. The power curve thus looks like y = x for x ≤ 1 and is undefined beyond that. Another way to measure load might be in terms of the percentage of time the peak rate exceeds 1, assuming that the average rate remains less than 1.
5. Yes, particularly if the immediate first link is high-bandwidth, the first router has a large buffer capacity, and the delay in the connection is downstream. CongestionWindow can grow arbitrarily; the excess packets will simply pile up at the first router. 6. R1 cannot become congested because traffic arriving at one side is all sent out the other, and the bandwidths on each side are the same. We now show how to congest only the router R2 that is R1’s immediate left child; other R’s are similar.