Solutions to Homework #5

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)

4 points for dropping packets
4 points for small window
4 points for no follow-up packets to transmit

Fast retransmit needs to receive three duplicate ACKS before it will retransmit a packet. If so many packets are lost (either the packets after the lost packet or their ACKS) that not even three duplicate ACKS make it back to the sender, then the mechanism cannot be activated, and a timeout will occur. This also can happen if the window is too small, or if three of more packets do not need to be transmitted after the lost packet.

2. 8.12 (max 12 points)

3 points for advantage of each
3 points for disadvantage of each

Marking a packet allows the endpoints to adjust to congestion more efficiently--they may be able to avoid losses (and timeouts) altogether by slowing their sending rates. However, transport protocols must be modified to understand and account for the congestion bit. Dropping packets leads to timeouts, and therefore may be less efficient, but current protocols (such as TCP) need not be modified to use RED. Also, dropping is a way to rein in an ill-behaved sender.

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.