computer network

Question

Consider a misbehaving TCP receiver. The receiver modifies its TCP such that upon receiving a data segment containing N bytes, the receiver divides the resulting acknowledgment into M, where M<=N, separate acknowledgments each covering one of distinct pieces of the received data segment.
For e.g. if it receives data acknowledging bytes 1 to 1000, then the receiver, for M=2 will send 2 ACKs for 501 and 1001.
Consider a normal TCP sender sending data to this misbehaving TCP receiver. The sender sends a 1500 byte
packet with sequence number 1. The receiver sends back M=3 ACKs.
What packets will the sender send next in response to the 3 ACKs?

(A) Response will be packets with byte sequence numbers 1501, 1505, 1509 and 1513.
(B) Response will be packets with byte sequence numbers 1501, 3003, 4506 and 6009.
(C) Response will be packets with byte sequence numbers 1501, 3002, 4503 and 6004.
(D) Response will be packets with byte sequence numbers 1501, 3001, 4501 and 6001.

Answer 1

2 TCP Security   https://www.cs.cmu.edu/~srini/15-441/F02/hw/hw2-sol.pdf

Answer 2

Option - D

The cwind will increase (Becomes 4 in 1 RTT) proportional to M..

 

Answer 3

Since in the slow start phase, the cwnd size is increased by 1 segment for every ack's received (irrespective of whether the Ack received is for size less than MSS). So the cwnd is increased by 3 after receiving 3 ACK's, i.e; cwnd= cwnd(initially 1) + 3 = 4 segments. So, the next responses would be packets with byte sequence numbers- 1501, 3001, 4501, 6001 [Since the segment size(MSS=1500B) is not changed].