Chapter 3 HW Continued

P7. ACK packets do not need sequence numbers because the rdt3.0 can tell right away if duplicate ACKs have already gone through. The reason we don’t have sequence numbers is because we simply don’t need them. The duplicate ACK is ignored when sent through from the rdt3.0 sender because it does not need it.

P8. The rdt3.0 FSM receiver is the same as the rdt2.2 one.

P10. A timer would have to be added where the value of it is greater than the RTT propagation delay. A timeout event would also have to be added to wait for the ACK or NAK1 as well as wait for the ACK or NAK0. If the timeout event then occurs, the most recent transmitted packet is retransmitted. These modifications will work because if the timeout is caused by a lost packet or lost ACK, then the retransmission is sent to the receiver.

P16. Yes, this would cause the sender to send a lot of pipelined data into the channel. There are potential problems that can arise. For example, if data segments are lost, then the sender of the rdt3.0 won’t re-send them.

P24. A) True. If there is a sender that has a window size of 3, and sends packets 1, 2, and 3 at t0, at t1 the receiver ACKs 1, 2, and 3. At t2, the sender times out and resends 1, 2, and 3. Then at t3, the receiver will receive the duplicates and ACK 1, 2, and 3. At t4, the sender receives the ACKs that t1 sent and moves the window to 4, 5, 6. The finally at t5, the sender receives the ACKs 1, 2, and 3 sent at t2. These ACKs would then be outside the window of 4, 5, 6.

B) True, same reason as A).

C) True

D) True, a window size of 1 eliminates the possibility of out of order packets so the alternating bit protocol and the GBN would essentially be the same.

P26) A) The maximum size file that can be sent is the number of bytes representable by the total number of possible sequence numbers (2^32) which is about 4.19 GB.

B) 2^32/536 = 8,012,999

Next, the number of segments is 66 bytes of header that get added each segment making a total of 528,857,934 bytes of header. The total number of bytes if 2^32 + 528,857,934 = 4.824 * 10^9 bytes. Then considering the 155mbps link, the transmission would take 249 seconds.