Homework #5 Grade Distribution:
Points | Frequency |
---|---|
19 | 1 |
20 | 1 |
21 | 0 |
22 | 0 |
23 | 1 |
24 | 0 |
25 | 0 |
26 | 1 |
27 | 2 |
28 | 1 |
29 | 3 |
30 | 4 |
31 | 2 |
32 | 0 |
33 | 4 |
34 | 1 |
35 | 1 |
36 | 1 |
37 | 1 |
38 | 6 |
39 | 2 |
40 | 4 |
1. 8.8 (max 8 points)
2. 8.12 (max 12 points)
3. 8.20 (max 8 points)
An ATM network may be only one network (one type of network) in an internet. Making guarantees across an ATM network does not guarantee anything end-to-end if other networks without resource allocation must also be traversed. In other words, the problem of allocating resources and controlling congestion has to be solved end-to-end, and unless there is only an ATM network, it does little good to solve it in the ATM network.
The following answer was also accepted. ATM has limited knowledge about how it segmented another protocol's packets into cells. Therefore, in many cases ATM drops cells without regard to the effects on the other protocol's packets. It is more efficient to drop all the cells that are part of one packet than to drop cells from different packets. This is because bandwidth is not wasted on packets that cannot be reconstructed without retransmission.
4. 9.15 (max 12 points)
5 points for the rate
5 points for the bucket depth
1 point extra for each derivation
On average, 1.5 MB (.5*1 + 0*1.5 + 2*.5) are transmitted in 3 seconds. Hence, the rate is r =0.5 MB/s. Savings made during the 1.5 second interval, when we transmit at a rate less than 0.5 MB/s (in fact, at 0 MB/s here), are 1.5*.5 MB = 0.75 MB. We need to spend 1 MB during the last half second, and we can send 0.5*0.5 MB = 0.25 MB. Hence, the bucket depth needs to be 0.75 MB.